Chapter 3 Instruction Set Architectures Instruction Set Architecture • ISA Includes the information needed to interact with the microprocessor. • Does not include information as.

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Transcript Chapter 3 Instruction Set Architectures Instruction Set Architecture • ISA Includes the information needed to interact with the microprocessor. • Does not include information as.

Slide 1

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 2

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 3

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 4

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 5

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 6

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 7

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 8

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 9

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 10

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 11

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 12

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 13

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 14

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 15

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 16

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 17

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 18

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 19

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 20

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 21

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 22

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 23

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 24

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 25

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 26

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 27

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 28

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 29

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 30

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 31

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 32

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 33

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 34

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 35

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 36

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 37

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 38

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 39

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 40

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 41

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 42

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 43

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 44

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 45

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 46

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 47

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 48

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 49

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 50

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 51

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 52

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]


Slide 53

Chapter 3
Instruction Set Architectures

Instruction Set Architecture
• ISA Includes the information needed to interact
with the microprocessor.
• Does not include information as to how
microprocessor is designed or implemented
• Includes microprocessor instruction set, which
would be the set of all assembly languages
instructions.
• Also includes the complete set of accessible
registers.

3.1 Levels of Programming
Languages
• Programming languages are divided into
three categories.
• High level languages hide the details of the
computer and operating system.
• Are also referred to as platformindependent.
• Examples include C++, Java, and Fortran

3.1 Levels of Programming
Languages (contd…)
• Assembly language is an example of a
lower level language.
• Each microprocessor has its own assembly
language
• A program written in the assembly language
of one microprocessor cannot be run on a
different microprocessor

3.1 Levels of Programming
Languages (contd…)
• Backward compatibility used in order to
have old programs that ran on a old
microprocessor, can run on a newer model.
• Assembly language can manipulate the data
stored in a microprocessor.
• Assembly language are not platform
independent

3.1 Levels of Programming
Languages (contd…)
• Lowest level of languages are machine language.
• Contains binary values to cause microprocessor to
perform operations.
• Microprocessor understands the machine
language, and thus it is in this state that it executes
a instruction set.
• High level language and assembly language are
converted to machine language.

3.1.2 Compiling and Assembling
Programs
• High-level language
programs are compiled
• Assembly languages
are assembled.

3.1.2 Compilers
• Compiler checks statement in a program is valid.
• If every instruction is syntactically correct, then
the compiler generates a object code.
• Linker combines object code as an executable file.
• Executable file copied into memory, and
microprocessor then runs the machine code
contained in that file

3.1.2 Compilers (contd)
• A high-level language statement is usually
converted to a sequence of several machine
code instructions.
• Every high-level language statement might
have more then one valid conversion of a
statement.

3.1.2 Assemblers
• Every statement in assembly language
however corresponds to one unique
machine code instruction.
• The assembler converts source code to
object code, and then the linking, and the
loading of procedures occur.

Compiler vs Assembler
Assembly language
Program for processor X

Compiler for Pentium
Windows Pc

Other
Pentium
Object files

Assembler for
Processor X

Pentium
Object code

Processor X
Object code
Pentium Linker
Other processor X
Object files

Process X
linker

Pentium
Executable file
Processor X
Executable file
Windows
Pentium PC

Computer with
Processor X

3.2 A closer look at Assembly
Language
• Assembly language is
very important part of
a instruction set
architecture.
• Assembly instructions
can be grouped
together based on their
functions.

3.2.1.1 Data Transfer
Instructions
• Related with moving data from one place to
another.
• Not be mistaken as the idea that data is
literally “moved”, instead it is copied from
one location to another.
• This particular type of instruction set does
not modify data.

3.2.1.1 Data Transfer Instructions
(contd)
Instructions related with this category perform
the following transfers:
• Load data from memory into
microprocessor.
• Store data from the microprocessor into
memory.
• Move data within the microprocessor.
• Input data to the microprocessor.
• Output data from the microprocessor.

3.2.1.2 Data Operation
Instructions
• Data operation instructions modify their data
•
•
•
•

values.
They require one or two operands, and then they
store the result
Arithmetic instructions make up a large part of the
data operation instructions.
Logic instructions perform basic logical
operations on data.
Shift instructions shift bits of data values in a
register.

3.2.1.3 Program Control
Instructions
• For Assembly languages, the jump or branch
instruction is commonly used to go to another part
of the program.
• An assembly language instruction set may include
instructions to call and return from subroutines.
• Microprocessor can also be designed to accept
interrupts, which basically causes a
microprocessor to stop its current process, and
execute another set of instructions.

3.2.2 Data Types
Numeric data can be represented as integers:
• Unsigned integers of n-bit values can range
from 0 to 2^n-1.
• Signed n-bit integers can have values
between –2^n-1 to 2^n-1-1

3.2.2 Data Types (contd)
Other types include:
1. Float: Microprocessor may have special
registers only for floating point data, and its
corresponding instruction set.
2. Boolean: Instructions can perform logical
operations on these values.
3. Characters: Stored as binary values. Operations
include concatenation, replacing characters, or
character string manipulation.

3.2.3 Addressing Modes
• Microprocessor needs
memory address to
access data from the
memory.
• Assembly language
may use several
addressing modes to
accomplish this task.

3.2.3 Addressing Modes (contd)
3.2.3.1 Direct Mode
• Instruction includes memory access.
• CPU accesses that location in memory.
Example:
LDAC 5
Reads the data from memory location 5, and
stores the data in the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.2 Indirect Mode
• Address specified in instruction contains
address where the operand resides.
Example:
LDAC @5 or LDAC (5)
Retrieves contents of location 5, uses it to
access memory address

3.2.3. Addressing Modes (contd)
3.2.3.3 Register Direct and Register Indirect Modes
• Does not specify a memory address. Instead
specifies a register.
Example:
LDAC R
Where R is a register containing the value 5.
The instruction copies the value 5 from register and
into the CPU’s accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.4 Immediate Mode
• The operand specified in this mode is the
actual data it self.
Example:
LDAC #5
Moves the value 5 into the accumulator.

3.2.3 Addressing Modes (contd)
3.2.3.5 Implicit Mode
• Does not exactly specify an operand.
Instruction implicitly specifies the operand
because it always applies to a specific
register.
Example:
CLAC Clears the accumulator, and sets value
to zero. No operands needed.

3.2.3 Addressing Modes (contd)
3.2.3.6 Relative Mode
• Operand supplied is an offset, not the actual
address. Added to the contents of the CPU’s
program counter register to generate the required
address.
Example:
LDAC $5 is located at memory location 10, and it
takes up two blocks of memory. Thus the value
retrieved for this instruction will be 12 + 5, and
will be stored in the accumulator

3.2.3 Addressing Modes (contd)
3.2.3.7 Index Mode and Base Address Mode
• Address supplied by the instruction is added to the
contents of an index register.
• Base address mode is similar except, the index
register is replaced by a base address register.
Example:
LDAC 5(X) where X = 10
Reads data from location (5 + 10) = 15 and stores it
in the accumulator.

3.2.3 Addressing Modes (contd)
Summary
a)

0: LDAC 5 (instruction gets data from location 5)

b)

5: 10
stores value in CPU
0: LDAC @5 (instruction gets address from location 5)

h) 0: LDAC 5(X)
instruction gets value from index register
X: 10

then adds contents of X(10) to offset (5) to get
5: 10 (then gets data from location 10)

to get address (15)
15: 30

c)

d)

e)
f)

g)

10: 20
stores value in CPU
0: LDAC R
instruction gets address from register R
R: 5
stores value in CPU
0: LDAC R
instruction gets address from register R
: 5
then gets data from location 5
5: 10
stores value in CPU
0: LDAC # 5
stores value from instruction in CPU
0: LDAC (implicit)
instruction gets value from stack
stack
stores value in CPU
0: LDAC $5
1:
instruction adds address of next instruction (1) to
5:
offset (5) to get address (6)
6:
12
stores value in CPU

stores value in CPU

3.2.4 Instruction Formats
• Assembly language in machine language is
represented as binary value called
instruction code.
• Binary value for each instruction is in
different format.
• Representation of operation to be performed
is called opcode.

3.2.4 Instruction Formats (contd)
Examples
ADD = 1010
MOVE = 1000
LOAD = 0000
STORE = 0001
PUSH = 0100
POP = 1100

A = 00
B = 01
C = 10
D = 11

3.2.4 Instruction Formats
Examples (contd)
4 bits
opcode

2 bits
2 bits
2 bits
Operand #1 Operand #2 Operand #3

4 bits
opcode

2 bits
2 bits
Operand #1 Operand #2

4 bits
opcode

4 bits
Operand

2 bits
Operand

A)

B)

ADD A,B,C (A=B+C)
1010 00 01 10

MOVE A,B (A = B) 1000 00 01
ADD A,C (A = A + C) 1010 00 10
C) LOAD B (Acc = B) 0000 01
ADD C (Acc = Acc + C) 1010 10
STORE A (A = Acc)
0001 00
D) PUSH B (Stack = B) 0101
PUSH C (Stack = C,B) 0110
ADD (Stack = B + C) 1010
POP A (A = stack) 1100
Fewer bits requires less hardware, but more
instructions. Fewer bits allows faster
execution.

3.3 Instruction Set Architecture
Design
To design a optimal microprocessor, the following questions and issues have to be
addressed in order to come up with an optimized instruction set architecture
for the CPU:
1.
Completeness; does the instruction set have all of the instructions a program
needs to perform its required task.
2.
Issue of orthogonally, the concept of two instructions not overlapping, and
thus not performing the same function.
3.
The amount of registers to be added. More registers enables a CPU to run
faster, since it can access and store data on registers, instead of the memory,
which in turn enables a CPU to run faster. Having too many registers adds
unnecessary hardware.
4.
Does this processor have to be backward compatible with other
microprocessors.
5.
What types and sizes of data will the microprocessor deal with?
6.
Are interrupts needed?
7.
Are conditional instructions needed?

3.4 Creating a simple Instruction
Set
Designing a simple microprocessor fit for
maybe a microwave will involve integrating
the following models:
1. Memory model
2. Register model
3. Instruction set

3.4.1 Memory Model
• Microprocessor can access 64 K or 2^16
byes of memory
• Each byte has 8 bits or 64K x 8 of memory.
• I/O is treated as memory access, thus
requires same instruction to access I/O as it
does to access memory

3.4.2 Registers
•
•

•

•

Three registers in this microprocessor
First register is 8-bit accumulator where the
result is stored. Also provides one of the
operands for instructions requiring two
operands.
Second register R is a 8-bit register that provides
the second operands, and also stores in result so
that the accumulator can gain access to it.
Third register is a Z register which is 1 bit. It is
either 0 or 1. If a result of a instruction is 0 then
the register is set to 1 otherwise it is set to 0.

3.4.3 Instruction Set
16 instructions, 8-bit each:
Instruction

Instruction Code

Operation

NOP

0000 0000

No Operation

LDAC

0000 0001 

AC = M[]

STAC

0000 0010 

M[] = AC

MVAC

0000 0011

R = AC

MOVR

0000 0100

AC = R

JUMP

0000 0101 

GOTO 

JMPZ

0000 0110 

IF (Z = 1) THEN GOTO 

JPNZ

0000 0111 

IF (Z = 0) THEN GOTO 

ADD

0000 1000

AC=AC+R,If (AC+R=0) Then Z=1 Else Z = 0

SUB

0000 1001

AC-AC-R,If(AC-R=0) Then Z=1 Else Z = 0

INAC

0000 1010

AC=AC+1,If(AC+1=0) Then Z=1 Else Z = 0

CLAC

0000 1011

AC = 0, Z =1

AND

0000 1100

AC=ACR, If(AC R=0) Then Z=1 Else Z=0

OR

0000 1101

AC=ACR, If(ACR=0) Then Z=1 Else Z=0

XOR

0000 1110

AC=ACR,If(ACR=0) Then Z=1 Else Z=0

NOT

0000 1111

AC=AC’,If(AC’=0) Then Z=1 Else Z=0

3.4.3 Instruction Set (contd)
Note: LDAC uses direct addressing mode.
MOVR uses the implicit addressing mode.
JUMP uses immediate addressing mode.

3.4.4 Implementation
1 + 2 + … + n, or
Total = 0
For I = 1 TO N do (Total = Total + I);
Break Down:
1: Total = 0, I = 0
2: I = I + 1
3: Total = Total + I
4: If I  n THEN GOTO 2

3.4.4 Implementation (contd)
Loop:

CLAC
STAC total
STAC i
LDAC i
INAC
STAC i
MVAC
LDAC total
ADD
STAC total
LDAC n
SUB
JPNZ Loop

Clear Accumulator
Store value 0 to address total
Store value 0 to address i
Load contents of address i into accumulator
Add 1 to the accumulator
Store result from accumulator back to address i
Move result from accumulator into Register R
Load Total into accumulator
Add contents of Register R and accumulator
and store it in accumulator
Store Total back to address total
Load n into accumulator
Subtract R (R = i) from AC (AC = n)
If result is not zero then jump back to loop:

3.4.4 Implementation (contd)
1st Loop

Instruction

2nd Loop

3rd Loop

4th Loop

5th Loop

CLAC

AC = 0

STAC total

Total = 0

STAC I

I=0

LDAC I

AC = 0

AC = 1

AC = 2

AC = 3

AC = 4

INAC

AC = 1

AC = 2

AC = 3

AC = 4

AC = 5

STAC I

I=1

I=2

I=3

I=4

I=5

MVAC

R=1

R=2

R=3

R=4

R=5

LDAC total

AC = 0

AC = 1

AC = 3

AC = 6

AC = 10

ADD

AC = 1

AC = 3

AC = 6

AC = 10

AC = 15

STAC total

Total = 1

Total = 3

Total = 6

Total = 10

Total = 15

LDAC n

AC = 5

AC = 5

AC = 5

AC = 5

AC = 5

SUB

AC = 4, Z = 0

AC = 3, Z = 0

AC = 2, Z = 0

AC = 1, Z = 0

AC = 0, Z = 1

JPNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

3.4.5 Analysis of Instruction Set,
and Implementation
• Cannot have value greater then 255, therefore n
has to be less then or equal to 22
• Is it complete? For simple hardware, maybe. Not
enough to be implemented in a PC.
• Fairly orthogonal; however by eliminating OR and
implementing by AND and NOT, we can reduce
the amount of hardware used.
• Not enough registers.

3.5 8085 Microprocessor
Instruction Set Architecture
• Processor has practical applications. Examples
•
•
•

•

include the Sojourner robot.
Contains several registers including the
accumulator register, A.
Other registers include B,C,D,E,H,L.
Some are accessed as pairs. Pairs are not
arbitrary. B and C, D and E, H and L.
SP is a 16 bit stack pointer register pointing to the
top of the stack.

3.5 8085 Microprocessor
Instruction Set Architecture
Contains five flags known as flag registers:
• Sign flag, S indicates sign of a value
• Zero flag, Z, tells if a arithmetic or logical
instruction produced 0 for a result.
• Parity flag, P, is set to 1 if result contains
even number of 1’s
• Carry flag, CY, is set when an arithmetic
operation generates a carry out.

3.5 8085 Microprocessor
Instruction Set Architecture
• Auxiliary carry flag, generates a carry out
from a lower half of a result to a upper half.
Example:
0000 1111 + 0000 1000 = 0001 0111
• IM register used for enable and disable
interrupts, and to check pending interrupts.

3.5.2 8085 Microprocessor Instruction Set
Contains a total of 74 instructions.

R, R1, R2 8 bit registers representing A, B, C, D, E, H or L
M

Indicates memory location

RP

Indicates register pair such as BC, DE, HL, SP



16 bit address representing address or data value.

n

8-bit address or data value stored in memory immediately after the
opcode

Cond

Condition for conditional instructions. NZ (Z = 0), Z (Z =
1),P (S = 0), N (S = 1), PO (P = 0), PE (P = 1), NC (CY = 0),
C (CY=1)

3.5.2 Data movemement instruction for the 80855 microprocessor
Operation

Instruction
NOP

No operation

MOV r1, r2

r1 = r2

MOV r, M

r1 = M[HL]

MOV M, r

M[HL] = r

MVI r, n

r=n

MVI M, n

M[HL] = n

LXI rp, 

rp = 

LDA 

A = M[]

STA 

M[] = A

LHLD 

HL = M[], M[ + 1]

SHLD 

M[], M[ + 1] = HL

LDAX rp

A = M[rp] (rp = BC, DE)

STAX rp

M[rp] = A (rp = BC, DE)

XCHG

DE  HL

PUSH rp

Stack = rp (rp  SP)

PUSH PSW

Stack = A, flag register

POP rp

rp = Stack (rp  SP)

POP PSW

A, flag register = Stack

XTHL

HL  Stack

SPHL

SP = HL

IN n

A = input port n

OUT n

Output port n = A

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ADD r

A=A+ r

All

ADD M

A = A + M[HL]

All

ADI n

A=A+ n

All

ADC r

A = A + r + CY

All

ADC M

A = A + M[HL] + CY

All

ACI n

A = A + n + CY

All

SUB r

A=A- r

All

SUB M

A = A - M[HL]

All

SUI n

A=A- n

All

SBB r

A = A - r - CY

All

SBB M

A = A - M[HL] - CY

All

SBI n

A = A - n - CY

All

INR r

r=r+1

Not CY

INR M

M[HL] = M[HL] + 1

Not CY

DCR r

r=r-1

Not CY

DCR M

M[HL] = M[HL] - 1

Not CY

INX rp

rp = rp + 1

None

DCX rp

rp = rp - 1

None

DAD rp

HL = HL + rp

CY

DAA

Decimal adjust

All

ANA r

A=A r

All

ANA M

A = A  M[HL]

All

3.5.2 Data operation instruction for the 80855 microprocessor
Instruction

Operation

Flags

ANI n

A=A n

All

ORA r

A=A r

All

ORA M

A = A  M[HL]

All

ORI n

A=A n

All

XRA r

A=A r

All

XRA M

A = A  M[HL]

All

XRI n

A=A n

All

CMP r

Compare A and r

All

CMP M

Compare A and M[HL]

All

CPI n

Compare A and n

All

RLC

CY = A7, A = A(6-0), A7

CY

RRC

CY = A0, A = A0, A(7-1)

CY

RAL

CY, A = A, CY

CY

RAR

A, CY = CY, A

CY

CMA

A = A’

None

CMC

CY = CY’

CY

STC

CY = 1

CY

3.5.2 Program control instruction
Instruction

Operation

JUMP 

GOTO 

J cond 

If condition is true then GOTO 

PCHL

GOTO address HL

CALL 

Call subroutine at 

C cond 

If condition is true then call subroutine at 

RET

Return from subroutine

R cond

If condition is true then return from
subroutine

RST n

Call subroutine at 8*n (n = 5.5, 6.5, 7.5)

RIM

A = IM

SIM

IM = A

DI

Disable interrupts

EI

Enable interrupts

HLT

Halt the CPU

3.5.3. A Simple 8085 Program
1: i = n, sum = 0
2: sum = sum + i, i = i - 1
3: IF i  0 then GOTO 2
4: total = sum

3.5.3 A Simple 8085 Program (contd)

Loop:

LDA n
MOV B, A
XRA A
ADD B
DCR B
JNZ Loop
STA total

i=n
sum = A  A = 0
sum = sum + i
i=i-1
IF i  0 THEN GOTO Loop
total = sum

3.5.3 Execution trace
Instruction

1st Loop

2nd Loop

3rd Loop

4th Loop

5th Loop

LDA n
MOV B, A

B=5

XRA A

A=0

ADD B

A=5

A=9

A = 12

A = 14

A = 15

DCR B

B = 4,
Z=0

B = 3,
Z=0

B = 2,
Z=0

B = 1,
Z=0

B = 0,
Z=1

JNZ Loop

JUMP

JUMP

JUMP

JUMP

NO JUMP

STA total

total = 15

3.5.4

Analyzing the 8085 ISA

• Instruction set more complete then the
simple CPU, however not sufficient enough
for a PC.
• Able to use subroutines, and interrupts
• It is fairly orthogonal.
• Has sufficient number of registers

Designed & Developed By

(Harman Jit Singh)

KANWER ACADEMY
98, New Model Town, Ludhiana
Contact No:- 98880-03277
E-Mail:- [email protected]