Transcript Assembler language – 8051

Assembler language – 8051
Mairtin O Conghaile
8051 Microcontroller
 Complete, highly-integrated microcomputer
 CPU, RAM, ROM, IO
 Port 0
 8-bit bidirectional I/O port OR
 multiplexed low-order address and data bus bytes
 Port 1
 8-bit bidirectional I/O port
 Port 2
 8-bit bidirectional I/O port OR
 high-order address byte
 Port 3
 8-bit bidirectional I/O port OR
 various special-function signals
2
8051 Microcontroller Interface
Signals (1)
PO/AD[7-0]
P2 / A[15-8]
Port 0 or Lo Addr/Data
Port 2 or Hi Addr
Addr Latch Ena/Program
Program Store Enable
Port 3/Special Functions:
Write
Read
Interrupt 0
Interrupt 1
Timer 0
Timer 1
Serial Input Port
Serial Output Port
Port 1
8
8
ALE /PROG*
PSEN
[Bus Timing]
P3.6 / WR*
P3.7 / RD*
[Int]
P3.2 /INT0*
P3.3 /INT1*
[I/O]
P3.4
P3.5
P3.0
P3.1
/ T0
/ T1
/RXD
/TXD
P1
8
3
8051 Microcontroller Interface
Signals (2)
External Access Enable/
Programming Voltage
Crystal/oscillator
Crystal
Reset
EA*/VPP
XTAL2*
(External)
XTAL1*
Clock / Reset
Generator
RST
4
Introduction to the 8051
microcontroller
 Type of memory





Code memory (64k max)
Internal RAM (128 bytes)
External RAM (64k max)
Special Function Registers (SFR)
Bit memory
Internal RAM
Basic Registers

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


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Accumulator
“R” registers
B register
Data Pointer (DPTR) – 16-bit register
Program Counter (PC) – 16-bit register
Stack Pointer (SP)
MOV instruction
 Simply stated, the MOV instruction copies
data from one location to another. It has
the following format
MOV destination, source
 This instruction tells the CPU to move (copy) the source
operand to the destination operand, without changing
the content of the source operand.
 Examples:




MOV
MOV
MOV
MOV
A,#55h
R0,A
R3, #95h
A,R3
; load 55h into register A
; copy contents of A into R0
; load value 95h into R3
;copy content of R3 into A
Some useful pointers
 Values can be loaded directly into any of the registers
A,B or R0-R7. However, to indicate that it is an
immediate value it must be preceeded with a # sign.
MOV A,#23H ;load 23H into A
MOV R6,#12 ;load 12d into R6
MOV R5,#0f9h
 If value is small, rest of bits are assumed to be all
zeros. E.g. mov a 4-bit value into an 8-bit register
 Moving a value that is too large into a register will
cause an error.
 To load a value into a register it must be preceeded
with a # sign. Otherwise it means to load from a
memory location.
MOV A,#17h ≠ MOV A,17h
ADD instruction
 The ADD instruction has the following
format:
ADD A, source ;Add the source operand to A
 This tells the CPU to add the source byte
to reg A and put the result in reg A
Example
 Calculate the content of the
accumulator after the program is
executed on the 8051.





MOV
MOV
MOV
ADD
ADD
R5,#25h
R7,#34H
A,#0
A,R5
A,R7
Addressing Modes
 An "addressing mode" refers to how you
are addressing a given memory location.
 In summary, the addressing modes are as
follows, with an example of each:
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
Immediate Addressing
Direct Addressing
Indirect Addressing
External Direct
Code Indirect
MOV A,#20h
MOV A,30h
MOV A,@R0
MOVX A,@DPTR
MOVC A,@A+DPTR
 Each of these addressing modes provides
important flexibility.
Immediate Addressing
 Immediate addressing is so-named because the value
to be stored in memory immediately follows the
operation code in memory. That is to say, the
instruction itself dictates what value will be stored in
memory.
MOV A,#20h
 This instruction uses Immediate Addressing because
the Accumulator will be loaded with the value that
immediately follows; in this case 20 (hexidecimal).
 Immediate addressing is very fast since the value to
be loaded is included in the instruction. However,
since the value to be loaded is fixed at compile-time it
is not very flexible.
Direct Addressing





Direct addressing is so-named because the value to be stored in
memory is obtained by directly retrieving it from another memory
location. For example:
MOV A,30h
This instruction will read the data out of Internal RAM address 30
(hexidecimal) and store it in the Accumulator.
Direct addressing is generally fast since, although the value to be
loaded isnt included in the instruction, it is quickly accessable since
it is stored in the 8051s Internal RAM.
It is also much more flexible than Immediate Addressing since the
value to be loaded is whatever is found at the given address-which may be variable.
Also, it is important to note that when using direct addressing any
instruction which refers to an address between 00h and 7Fh is
referring to Internal Memory. Any instruction which refers to an
address between 80h and FFh is referring to the SFR control
registers that control the 8051 microcontroller itself.
Indirect Addressing
 Indirect addressing is a very powerful addressing
mode which in many cases provides an exceptional
level of flexibility.
MOV A,@R0
 This instruction causes the 8051 to analyze the value
of the R0 register. The 8051 will then load the
accumulator with the value from Internal RAM which
is found at the address indicated by R0.
 For example, lets say R0 holds the value 40h and
Internal RAM address 40h holds the value 67h. When
the above instruction is executed the 8051 will check
the value of R0. Since R0 holds 40h the 8051 will get
the value out of Internal RAM address 40h (which
holds 67h) and store it in the Accumulator. Thus, the
Accumulator ends up holding 67h.
Indirect Addressing(cont’d)
 Indirect addressing always refers to Internal RAM; it
never refers to an SFR. Thus, in a prior example we
mentioned that SFR 99h can be used to write a value
to the serial port. Thus one may think that the
following would be a valid solution to write the value 1
to the serial port:
MOV R0,#99h ;Load the address of the serial port
MOV @R0,#01h ;Send 01 to the serial port WRONG!!
 On an 8051 these two instructions would produce an
undefined result since the 8051 only has 128 bytes of
Internal RAM.
External Direct Addressing
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
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External Memory is accessed using "External Direct"
addressing.
There are only two commands that use External Direct
addressing mode:
MOVX A,@DPTR
MOVX @DPTR,A
As you can see, both commands utilize DPTR.
In these instructions, DPTR must first be loaded with the
address of external memory that you wish to read or write.
Once DPTR holds the correct external memory address, the
first command will move the contents of that external
memory address into the Accumulator.
The second command will do the opposite: it will allow you
to write the value of the Accumulator to the external
memory address pointed to by DPTR.
External Indirect Addressing
 External memory can also be accessed using a form
of indirect addressing.
 This form of addressing is usually only used in
relatively small projects that have a very small
amount of external RAM. An example of this
addressing mode is:
MOVX @R0,A
 Once again, the value of R0 is first read and the value
of the Accumulator is written to that address in
External RAM. Since the value of @R0 can only be 00h
through FFh the project would effectively be limited to
256 bytes of External RAM.
Question 1
 Write a short program for the 8051
microcontroller that carries out the
following instructions:

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Loads the accumulator with the value 40h
Loads R7 with 12d
Copies the content of R7 to address 30h directly
Loads the register R0 with 30h
Indirectly moves the contents of 30h to reg B
 Indicate clearly the contents, at each stage,
of all memory locations and registers
involved.
Structure of Assembly language
 Consists of a series of assembly
language instructions.
 Instruction consists of four fields:
[label1:] mnemonic [operands] [;comment]
 Label field allows the program to refer to a line
by name.
 Comment field must begin with a semicolon
Example of Assembly Program
HERE:
ORG 0H
MOV R5,#25h
MOV R7,#34H
MOV A,#0
ADD A,R5
ADD A,R7
ADD A,#12h
SJMP HERE
END
;start at mem loc 0
;load 25h in r5
;load 34h into r7
;clear A
;A=A+R5
;A=A+R7
;A=A+12
;stay in this loop
;end of asm source
;file
8051 Data Types and Directives
 Only one type of data type – 8-bit
 It’s the job of the programmer to break
down data larger than 8-bits.
DB
 The DB directive is the most widely used data
directive in assembler.
 Used to define the 8-bit data.
 When DB is used, the numbers can be decimal,
binary, hex or ASCII formats.
 The only directive that can be used to define ASCII
strings larger than two characters.
DATA1:
DATA2:
DATA3:
ORG 500H
DB 39H
DB “2591” ;ASCII NUMBERS
ORG 518H
DB “Computer Engineering”
Assembler Directives
 ORG: Used to indicate the beginning
of the address.
 EQU: Used to define a constant
without occupying a memory
location.

e.g.
Count EQU
25
 END: Indicates the end of the source
file.
PSW (Program Status Word)
 Addresses D0h, Bit-Addressable
 The Program Status Word is used to store a number of
important bits that are set and cleared by 8051
instructions.
 The PSW SFR contains the carry flag, the auxiliary carry
flag, the overflow flag, and the parity flag.
 Additionally, the PSW register contains the register bank
select flags which are used to select which of the "R"
register banks are currently selected.
Example
 State the contents of the RAM
locations after the following program:
SETB PSW.4
MOV R0,#99H
MOV R1,#85H
MOV R2,#99H
MOV R7,#85H
MOV R5,#99H
Solution
 By default PSW.3 = 0 and PSW.4 =0;
therefore line 1 sets RS1=1 and RS0 = 0,
thereby selecting register bank 2.
 Register Bank 2 uses RAM locations 10H –
17H. After execution of this program we
have the following:





RAM
RAM
RAM
RAM
RAM
location
location
location
location
location
10H
11H
12H
17H
15H
has
has
has
has
has
value
value
value
value
value
99H
85H
3FH
63H
12H
Stack in the 8051
 Section of RAM used by the CPU to
store information temporarily.
 Information can be data or an
address.
 The CPU needs this storage area
since there are only a limited number
of registers.
How stacks are accessed in the
8051
 The register used to access the stack is called the SP
(stack pointer) and is 8-bits wide (00h-ffh).
 When the 8051 is powered up the SP contains the
value 07.
 This means that RAM location 08 is the first location
used for the stack.
 Final location is 1F (20h -> used for bit- addressable
memory)
 Storing of a CPU register in the stack is called a
PUSH.
 Loading the contents of the stack back into a CPU
register is called a POP.
Pushing onto the stack
 The SP points to the last used
location of the stack.
 As we push data onto the stack, the
stack pointer is incremented by one.
 When you pop a value off the stack,
the 8051 returns the value from the
memory location indicated by SP, and
then decrements the value of SP.
Example
 Show the stack and stack pointer for the following:
MOV R6,#25H
MOV R1,#12H
MOV R4,#0F3H
PUSH 6
;push onto stack from R6
PUSH 1
;push onto stack from R1
PUSH 4
;push onto stack from R4
POP 3
;pop stack into R3
POP 5
;pop stack into R5
POP 2
;pop stack into R2
Upper Limit
 Ram locations 08 – 1F used for the
stack.
 If more than 24bytes of stack required,
then the SP must be changed to point
to RAM locations 30h-7Fh using the
instruction
MOV SP,#xx
 Also may need to shift SP if a given
progam needs register bank1,2 or 3.
Example
 Show the stack and stack pointer for the
following instructions:
MOV SP,#5FH
MOV R2,#25H
MOV R1,#12H
MOV R4,#0F3H
PUSH 2
PUSH 1
PUSH 4
Questions

Write a simple program in which the value 55h is added five times.

Show the stack and the stack pointer for each line of the following:
Org 0
MOV
MOV
MOV
MOV
PUSH
PUSH
PUSH
CLR A
MOV
MOV
POP 7
POP 2
POP 5
SP,#70H
R5,#66H
R2,#7FH
R7,#5DH
5
2
7
R2,A
R7,A
Program Flow
 When an 8051 is first initialized, it resets the PC to
0000h.
 The 8051 then begins to execute instructions
sequentially in memory unless a program instruction
causes the PC to be otherwise altered.
 There are various instructions that can modify the
value of the PC; specifically, conditional branching
instructions, direct jumps and calls, and "returns"
from subroutines.
 Additionally, interrupts, when enabled, can cause the
program flow to deviate from its otherwise sequential
scheme.
Loop and Jump Instructions
 Repeating a sequence of instructions a
certain number of times is called a loop.
 The loop action is performed by the
instruction
DJNZ reg,label
 In this instruction, the register is
decremented; if it is not zero, it jumps to
the target address referred to by the label.
 Prior to the start of the loop the register is
loaded with the counter for the number of
repetitions.
Example
 Write a program to clear the Acc, then add
3 to the accumulator ten times.
AGAIN:
MOV
MOV
ADD
DJNZ
MOV
A,#0
R2,#10
A,#03
R2,AGAIN
R5,A
Question
 Write a program to load the
accumulator with the value 10h and
then complement the Acc 700times.
(Hint: Try two separate loops to
achieve the overall of 700)
Solution
NEXT:
AGAIN:
MOV
MOV
MOV
CPL
DJNZ
DJNZ
A,#10H
R3,#10
R2,#70
A
R2,AGAIN
R3,NEXT
Other Conditional Jumps
Instruction
Action
JZ
Jump if A = 0
JNZ
Jump if A ≠ 0
DJNZ
Decrement and jump if A ≠ 0
CJNE A,byte
Jump if A ≠ byte
CJNE reg,#data
Jump if byte ≠ #data
JC
Jump if CY = 1
JNC
Jump if CY = 0
JB
Jump if bit = 1
JNB
Jump if bit = 0
JBC
Jump if bit = 1 and clear bit
Example
 Find the sum of the values 79H, F5H
and E2H. Put the sum of the registers
in R0 (low byte) and R5 (high byte)
Solution
MOV
MOV
ADD
JNC
INC
N_1: ADD
JNC
INC
N_2: ADD
JNC
INC
OVER: MOV
A,#0
R5,A
A,#79H
N_1
R5
A,#0F5H
N_2
R5
A,#0E2H
OVER
R5
R0,A
;clear A
;clear R5
;A=A+79h
;if no carry, add next
;if CY=1, increment R5
;A=79H+F5H=6EH and CY1=1
;jump if CY=0
;if CY=1, increment R5
;A=6E+E2=50H and CY=1
;jump if CY=0
;if CY=1, increment R5
;now R0=50H and R5=02
Unconditional Jump Instructions
 All conditional jumps are short jumps,
meaning that the address of the
target must be within -128 and +127
bytes of the contents of the program
counter (PC).
 Unconditional jump instructions are:
 LJMP (Long jump) – 3 byte instruction
 SJMP (Short jump) – 2 byte instruction
CALL instructions
 CALL instruction is used to call a subroutine
 LCALL (long call) – 3 byte instruction
 ACALL (absolute call) – 2 byte instruction
 When a subroutine is called, control is
transferred to that subroutine.
 After finishing execution of the subroutine, the
instruction RET (return) transfers control back to
the caller.
Time Delay Generation and
Calculation
 For the CPU to execute an instruction takes
a certain number of clock cycles.
 In the 8051 family, these clock cycles are
referred to as machine cycles.
 We can calculate a time delay using the
available list of instructions and their
machine cycles.
 In the 8051, the length of the machine
cycle depends on the frequency of the
crystal oscillator connected to the 8051
system.
Time Delay Generation and
Calculation (cont’d)
 The frequency of the crystal
connected to the 8051 family can
vary from 4MHz to 30MHz.
 In the 8051, one machine cycle lasts
12 oscillator periods.
 Therefore, to calculate the machine
cycle, we take 1/12 of the crystal
frequency and then take the inverse.
Example
 The following shows crystal frequency
for three different 8051-based
systems. Find the period of the
machine cycle in each case.
(a)
11.0592MHz
(b)
16MHz
(c)
20MHz
Solution
 1/11.0592MHz = period per oscillation
Machine cycle = 12x
= 1.085μs
 1/16MHz = period per oscillation
Machine cycle = 12x
= 0.75μs
 1/20MHz = period per oscillation
Machine cycle = 12x
= 0.6μs
Question
 For an 8051 system of 11.0592MHz,
find how long it takes to execute each
of the following instructions:
(a)
MOV
R3,#55
(b)
DEC
R3
(c)
DJNZ
R2,target
(d)
NOP
(e)
MUL
AB
Solution
(a)
(b)
(c)
(d)
(e)
MOV
DEC
DJNZ
NOP
MUL
R3,#55
R3
R2,target
AB
1x1.085μs
1x1.085μs
2x1.085μs
1x1.085μs
4x1.085μs
Delay Calculation
 A delay subroutine consists of two parts:
(a) setting a counter
(b) a loop
 Most of the time delay is performed by the body of
the loop.
 Very often we calculate the time delay based on the
instructions inside the loop and ignore the clock cycles
associated with the instructions outside the loop.
 Largest value a register can hold is 255; therefore,
one way to increase the delay is to use the NOP
command.
 NOP, which stands for “No Operation” simply wastes
time.
Example
 Find the size of the delay in the following program, if
the crystal frequency is 12MHz.
MOV
A,#55H
AGAIN:
MOV
P1,A
ACALL DELAY
CPL
A
SJMP
AGAIN
DELAY:
MOV
R3,#200
HERE:
DJNZ
R3,HERE
RET
 What does the above program do?
Solution
DELAY:MOV
HERE:
R3,#200
DJNZ
R3,HERE
RET
Crystal Cycle
12
24
12
Therefore, we have a delay of [(200X24)+12+24]x0.083μs
= 402μs
Loop inside a loop delay
 Another way to get a large delay is to use a loop
inside a loop, which is also called a nested loop. E.g.
Crystal Cycle
DELAY:
MOV
R3,#250
12
HERE:
NOP
12
NOP
12
NOP
12
NOP
12
DJNZ
R3,HERE
24
RET
24
Time Delay=[250.(12+12+12+12+24)]x0.083μs +
(12+24)x0.083μs = μs
Question
 For a machine cycle of 1μs, find the time delay of the
following subroutine.
DELAY:
AGAIN:
HERE:
MOV
MOV
NOP
NOP
DJNZ
DJNZ
RET
R2,#200
R3,#250
R3,HERE
R2,AGAIN
Solution
 HERE Loop: (4x250)x1μs = 1000 μs
 AGAIN Loop: Repeats the HERE loop 200
times i.e. 200x1000μs = 200ms
 The instructions “MOV R3,#250” and
“DJNZ R2,AGAIN” at the beginning at end
of the AGAIN loop will add (3x200x1μs) =
600μs to the delay time.
 Total execution time ≈ 200.6ms (an
approximation since we have ignored the
first and last instructions in the subroutine)