Transcript Basic Computer - California Lutheran University

Bus Architecture
Memory unit
111
4096x16
ALU
address
AR
001
PC
010
DR
011
AC
100
IR
101
TR
110
E
INPR
OUTR
clock
16-bit Bus
S2 Access
S1
S0 Select
Bus Architecture
• The three access select lines determine which register is
allowed to write to the bus at a given time (recall that
only one write at a time is allowed)
• Registers have load input signals (LD) that tell them to
read from the bus
• If registers are smaller than the bus (less bits) than
unused bits are set to 0
• Some registers have additional input signals
– Increment (INR) and Clear (CLR)
– See figure 5-4, page 130 of the textbook
Bus Architecture
• Memory has read/write input signals that tell it
when to take data from the bus and send data to
the bus
• Memory addresses (for both read and write
operations) are always specified via the Address
Register (AR)
– An alternative (used in many architectures) is a two
bus system
• One address bus
• One data bus
Bus Architecture
• Results of all ALU (arithmetic, logic, and shift
operations) are always sent to the Accumulator (AC)
register
– The ALU is the only way to set values into the accumulator
except for the clear (CLR) and increment (INR) control lines
• Inputs to the ALU come from
– The Accumulator (AC)
– The Data Register (DR)
– The Input Register (INPR)
• The E output from the ALU is the carry-out (Extended
AC) bit
– Many architectures pack this into a register with other status
bits such as overflow
Bus Architecture
• Some pairs of microoperations can be
performed in a single clock cycle
– The key is to make sure they don’t both try to
put data on the bus
– Consider the RTL statement
DR ← AC, AC ← DR
– This is allowed since the DR ← AC
microoperation uses the bus while the AC ←
DR microoperation does not
Instructions
• We said previously that there are two parts
to an instruction
– Opcode
– Operand
• Realistically the two parts should be called
– Opcode
– Everything else
Instructions
• Three basic types
– Those that reference memory operands
– Those that reference register operands
– Those that reference I/O devices
• Again, this is only for the fictitious
architecture in the textbook but you will
find similar categorizations in real
architectures
Memory Instructions
• There are 14 instructions in this class
– 7 direct memory address forms
– 7 indirect memory address forms
15
I
14
12
opcode
11
0
address
I = 0 means direct memory address
I = 1 means indirect memory address
Memory Instructions
Hex Code
Symbol
I=0
I=1
Description
AND
0xxx
8xxx
Mem AND AC
ADD
1xxx
9xxx
Mem + AC
LDA
2xxx
Axxx
Load AC from Mem
STA
3xxx
Bxxx
Store AC to Mem
BUN
4xxx
Cxxx
Unconditional Branch
BSA
5xxx
Dxxx
Branch to Subroutine
ISZ
6xxx
Exxx
Increment and Skip if Zero
Register Instructions
• There are 12 instructions in this class
– They can use the “operand field” to specify
the register and type of operation since no
memory address is required
15
14
0 1
12
1
1
11
0
Register operation
Register Instructions
Symbol
Hex Code
Description
CLA
7800
Clear AC
CLE
7400
Clear E bit
CMA
7200
Complement AC
CME
7100
Complement E bit
CIR
7080
Circulate right AC and E
CIL
7040
Circulate left AC and E
INC
7020
Increment AC
Register Instructions (cont.)
Symbol
Hex Code
Description
SPA
7010
Skip next instruction if AC is positive
SNA
7008
Skip next instruction if AC is negative
SZA
7004
Skip next instruction if AC is 0
SZE
7002
Skip next instruction if E is 0
HLT
7001
Halt
I/O Instructions
• There are 6 instructions in this class
– They can use the “operand field” to specify
the exact operation since no memory address
is required
15
14
1 1
12
1
1
11
0
I/O operation
I/O Instructions
Symbol
Hex Code
Description
INP
F800
Input character to AC
OUT
F400
Output character from AC
SKI
F200
Skip on input flag
SKO
F100
Skip on output flag
ION
F080
Interrupt on
IOF
F040
Interrupt off
Instruction Decoding
• The control unit evaluates bits 15 – 12 to
determine the instruction format
• At first glance it appears that there can be
only 8 unique instructions since the
opcode resides in 4 bits
• But, additional instructions are created
through the use of the I bit and unused bits
in the operand field
Instruction Set Design
• To be useful, an architecture’s instruction
set must contain enough instructions to
allow all possible computations
• Four categories are necessary
– Arithmetic, logical, shift operations
– Moving data to/from memory from/to registers
– Control such as branch and conditional
checks
– Input/output
Instruction Set Design
• The set in the book is complete in that all the
possible operations on binary numbers can be
performed through combinations of instructions
• But, the set is very inefficient in that highly used
operations require multiple instructions
• This is why the Pentium instruction set is so
large and complicated – it makes for efficient
programs
• So, how does the control unit know what
to tell the hardware what to do?
The Control Unit
Instruction Register (IR)
15 14 - 12
11 - 0
3x8
Decoder
Other Inputs
12
D7 – D0
I
T15 – T0
n
Control
Unit
4x16
Decoder
Sequence
Counter
Increment
Clear
Master Clock
Lots of combination logic
goes in here
CSC321
Control Timing/Sequence Counter
• Due to the nature of the Fetch-Execute instruction cycle,
instructions require more than one clock pulse to
complete
• But, not all instructions require the same number of clock
pulses
• Thus, we divide the master clock into unique time steps
– Each time step will provide conditional input to the logic within
the control unit (recall the RTL notation for conditional operation)
• Clock division is performed by the sequence counter
– Recall that you designed one of these on the exam
CSC321
Sequence Counter Timing
Master Clock
T0
T1
T2
T3
•
•
•
•
•
Note that only one timing signal is a logic 1 at any given time
The counter is modulo 16, T15 returns to T0
The Master Clock always increments the counter from Tn to Tn+1
A clear input resets the counter circuit back to T0
An increment moves the counter from Tn+1 to Tn+2
CSC321
Sequence Counter Timing
• With the set of timing signals we can now
implement RTL statements such as
T0: AR ← PC
• What does this RTL statement mean?
CSC321
Instruction Cycle Revisited
•
•
•
•
Fetch an instruction from memory
Decode the instruction
Read the operands from memory/registers
Execute the instruction
– The actual implementation of each phase is
dependent on the instruction, although the
fetch and decode phases are common to all
CSC321
Fetch and Decode
• Initially…
– Something (the operating system in the case
of computers that have them, hardware in the
case of computers that don’t have an OS)
places the address of the first program
statement into the Program Counter (PC)
– The Sequence Counter (SC) is cleared to 0
– Program execution begins
CSC321
Instruction Fetch
• Starting at time T0, fetch an instruction
from memory
– What are the RTL statements to perform this
step?
– Hint: the key words are “instruction” and “from
memory”
– Refer to the bus architecture: pg 130 of the
textbook
CSC321
Bus Architecture
Memory unit
111
4096x16
ALU
address
AR
001
PC
010
DR
011
AC
100
IR
101
TR
110
E
INPR
OUTR
clock
CSC321
16-bit
Bus
S2 Access
S1
S0 Select
Instruction Fetch
• RTL
T0: AR ← PC
T1: IR ← M[AR], PC ← PC + 1
• Note that after fetching an instruction we
always increment the program counter
CSC321
Instruction Decode
• Instruction fetch took two cycles (T0 and
T1) so instruction decode starts at T2
– What are the RTL statements to perform this
step?
– Hint: the key words are “instruction” and
“decode”
– Refer to the bus architecture and the control
unit: pgs 130 and 137 of the textbook
CSC321
Bus Architecture
Memory unit
111
4096x16
ALU
address
AR
001
PC
010
DR
011
AC
100
IR
101
TR
110
E
INPR
OUTR
clock
CSC321
16-bit
Bus
S2 Access
S1
S0 Select
The Control Unit
Instruction Register (IR)
15 14 - 12
11 - 0
3x8
Decoder
Other Inputs
12
D7 – D0
I
T15 – T0
4x16
Decoder
Sequence
Counter
Increment
Clear
Master Clock
CSC321
n
Control
Unit
Instruction Decode
• RTL
T2: D0, … D7 ← Decode IR(12-14), AR ← IR(0-11), I ← IR(15)
CSC321
Fetch and Decode
T0: AR ← PC
T1: IR ← M[AR], PC ← PC + 1
T2: D0, … D7 ← Decode IR(12-14), AR ← IR(0-11), I ← IR(15)
• Figure 5-8 on page 140 of the textbook
shows the schematic interactions
between the timing signals, the register
controls, and the bus
CSC321
Fetch and Decode
T1
Bus
T0
Memory
Address
111
Read
AR
001
LD
PC
010
INR
IR
LD
101
Clock
CSC321
• What happens at
time T0?
• What happens at
time T1?
• What do we need
to add to perform
at time T2?
Decoding the Instruction
• Once the instruction is fetched from memory (T0,
T1) and the opcode is passed to the decoder (T2)
the control unit must figure out what to do next
(T3)
– Based on the opcode and the I bit (b15) it must make
decisions regarding operation type (memory, register,
I/O) and operand address mode (direct, indirect)
– Figure 5-9, pg 142 shows the RTL in flow chart form
CSC321
Decoding the Instruction
Start
SC <- 0
T0
AR <- PC
Of course, all this stuff gets
implemented in
combinational logic
T1
IR <- M[AR], PC <- PC + 1
Three instruction
types
T2
Addressing modes
Decode opcode IR(14-12)
AR <- IR(11-0), I<- IR(15)
= 1, register or I/O
= 0, memory reference
D7
= 1, I/O
T3
= 0, register
I
Execute I/O
Instruction
SC <- 0
Execute Register
Instruction
SC <- 0
T3
= 1, indirect
= 0, direct
I
T3
AR <- M[AR]
Execute Memory
Instruction
SC <- 0
CSC321
T3
Nothing
Next Time
• We start looking at RTL for each individual
instruction
• You should be reading/studying the
material in chapter 5 (we’re up to page
143)
• Don’t leave studying this material until the
night before an exam (which should be
coming up soon)
CSC321
Homework
• 5-1, 5-2, 5-3, 5-4, 5-5, 5-6
• Due next lecture