multi cycle in microprocessor 8086 sy B-tech

1
(Based on text: David A. Patterson & John L. Hennessy, Computer Organization and Design:
The Hardware/Software Interface, 3rd
Ed., Morgan Kaufmann, 2007)
Processor: Multi-
Processor: Multi-
Cycle Datapath &
Cycle Datapath &
Control
Control

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COURSE CONTENTS
COURSE CONTENTS
 Introduction
 Instructions
 Computer Arithmetic
 Performance
 Processor: Datapath
Processor: Datapath
 Processor: Control
Processor: Control
 Pipelining Techniques
 Memory
 Input/Output Devices

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PROCESSOR:
PROCESSOR:
DATAPATH & CONTROL
DATAPATH & CONTROL
 Multi-Cycle Datapath
 Multi-Cycle Control
 Additional Registers and
Multiplexers

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 Break up an instruction into steps, each step takes a
cycle:
 balance the amount of work to be done
 restrict each cycle to use only one major functional unit
 Different instructions take different number of cycles to
complete
 At the end of a cycle:
 store values for use in later cycles (easiest thing to do)
 introduce additional “internal” registers for such
temporal storage
 Reusing functional units (reduces hardware cost):
 Use ALU to compute address/result and to increment PC
 Use memory for both instructions and data
Multicycle Approach
Multicycle Approach

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 Additional “internal registers”:
 Instruction register (IR) -- to hold current instruction
 Memory data register (MDR) -- to hold data read from memory
 A register (A) & B register (B) -- to hold register operand values from register files
 ALUOut register (ALUOut) -- to hold output of ALU, also serves as memory address
register (MAR)
 All registers except IR hold data only between a pair of adjacent cycles and thus do
not need write control signals; IR holds instructions till end of instruction, hence
needs a write control signal
Multi-Cycle Datapath:
Multi-Cycle Datapath:
Additional Registers
Additional Registers
Shift
left 2
PC
Memory
Data
Write
data
M
u
x
0
1
Registers
Write
register
Write
data
Read
data 1
Read
data 2
Read
register 1
Read
register 2
M
u
x
0
1
M
u
x
0
1
4
Instruction
[15– 0]
Sign
extend
32
16
Instruction
[25– 21]
Instruction
[20– 16]
Instruction
[15– 0]
Instruction
register
1 M
u
x
0
3
2
M
u
x
ALU
result
ALU
Zero
Memory
data
register
Instruction
[15– 11]
A
B
ALUOut
0
1
Address
Inst /
Note: we ignore jump inst here

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 Additional multiplexors:
 Mux for first ALU input -- to select A or PC (since we use ALU for both
address/result computation & PC increment)
 Bigger mux for second ALU input -- due to two additional inputs: 4 (for normal PC
increment) and the sign-extended & shifted offset field (in branch address
computation)
 Mux for memory address input -- to select instruction address or data address
Multicycle Datapath:
Multicycle Datapath:
Additional Multiplexors
Additional Multiplexors
Shift
left 2
PC
Memory
Data
Write
data
M
u
x
0
1
Registers
Write
register
Write
data
Read
data 1
Read
data 2
Read
register 1
Read
register 2
M
u
x
0
1
M
u
x
0
1
4
Instruction
[15– 0]
Sign
extend
32
16
Instruction
[25– 21]
Instruction
[20– 16]
Instruction
[15– 0]
Instruction
register
1 M
u
x
0
3
2
M
u
x
ALU
result
ALU
Zero
Memory
data
register
Instruction
[15– 11]
A
B
ALUOut
0
1
Address
Inst /
Note: we ignore jump inst here

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Multi-Cycle
Multi-Cycle
Datapath & Control
Datapath & Control
Note the reason for each control signal; also note that we have included the jump
instruction
2
2
2

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 Note:
 three possible sources for value to be written into PC (controlled
by PCSource): (1) regular increment of PC, (2) conditional
branch target from ALUOut, (3) unconditional jump (lower 26
bits of instruction in IR shifted left by 2 and concatenated with
upper 4 bits of the incremented PC)
 two PC write control signals: (1) PCWrite (for unconditional
jump), & (2) PCWriteCond (for “zero” signal to cause a PC write
if asserted during beq inst.)
 since memory is used for both inst. & data, need IorD to select
appropriate addresses
 IRWrite needed for IR so that instruction is written to IR (IRWrite
= 1) during the first cycle of the instruction and to ensure that
IR not be overwritten by another instruction during the later
cycles of the current instruction execution (by keeping IRWrite =
0)
 other control signals
Control Signals for
Control Signals for
Multi-Cycle Datapath
Multi-Cycle Datapath

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1. Instruction Fetch (All instructions)
2. Instruction Decode (All instructions), Register Fetch & Branch
Address Computation (in advance, just in case)
3. ALU (R-type) execution, Memory Address Computation, or
Branch Completion (Instruction dependent)
4. Memory Access or R-type Instruction Completion (Instruction
dependent)
5. Memory Read Completion (only for lw)
At end of every clock cycle, needed data must be stored into register(s)
or memory location(s).
Each step (can be several parallel operations) is 1 clock cycle -->
Instructions take 3 to 5 cycles!
Breaking the Instruction
Breaking the Instruction
into 3 - 5 Execution Steps
into 3 - 5 Execution Steps
Data ready
operation
Clock in result
Clock
Events during a cycle, e.g.:

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 Use PC to get instruction (from memory) and put it in the
Instruction Register
 Increment of the PC by 4 and put the result back in the PC
 Can be described succinctly using RTL "Register-Transfer
Language"
IR <= Memory[PC];
PC <= PC + 4;
 Which control signals need to be asserted?
 IorD = 0, MemRead = 1, IRWrite = 1
 ALUSrcA = 0, ALUSrcB = 01, ALUOp = 00, PCWrite = 1, PCSource = 00
 Why can instruction read & PC update be in the same step? Look at state element
timing
 What is the advantage of updating the PC now?
Step 1: Instruction
Step 1: Instruction
Fetch
Fetch

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 In this step, we decode the instruction in IR (the opcode enters
control unit in order to generate control signals). In parallel,
we can
 Read registers rs and rt, just in case we need them
 Compute the branch address, just in case the instruction is a
branch beq
 RTL:
A <= Reg[IR[25:21]];
B <= Reg[IR[20:16]];
ALUOut <= PC + (sign-extend(IR[15:0]) << 2);
 Control signals:
 ALUSrcA = 0, ALUSrcB = 11, ALUOp = 00 (add)
 Note: no explicit control signals needed to write A, B, & ALUOut.
They are written by clock transitions automatically at end of step
Step 2: Instruction Decode,
Step 2: Instruction Decode,
Reg. Fetch, & Branch Addr.
Reg. Fetch, & Branch Addr.
Comp.
Comp.

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 One of four functions, based on instruction type:
 Memory address computation (for lw, sw):
ALUOut <= A + sign-extend(IR[15:0]);
 Control signals: ALUSrcA = 1, ALUSrcB = 10, ALUOp = 00
 ALU (R-type):
ALUOut <= A op B;
 Control signals: ALUSrcA = 1, ALUSrcB = 00, ALUOp = 10
 Conditional branch:
if (A==B) PC <= ALUOut;
 Control signals: ALUSrcA = 1, ALUSrcB = 00, ALUOp = 01 (Sub),
PCSource = 01, PCWriteCond = 1 (to enable zero to write PC if 1)
What is the content of ALUOut during this step? Immediately after this step?
 Jump:
PC <= PC[31:28] || (IR[25:0]<<2);
 Control signals: PCSource = 10, PCWrite = 1
 Note: Conditional branch & jump instructions completed at this step!
Step 3: Instruction
Step 3: Instruction
Dependent Operation
Dependent Operation

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 For lw or sw instructions (access memory):
MDR <= Memory[ALUOut];
or
Memory[ALUOut] <= B;
 Control signals (for lw): IorD = 1 (to select ALUOut as address),
MemRead = 1, note that no write signal needed for writing to MDR, it is written
by clock transition automatically at end of step
 Control signals (for sw): IorD = 1 (to select ALUOut as address),
MemWrite = 1
 For ALU (R-type) instructions (write result to register):
Reg[IR[15:11]] <= ALUOut;
 Control signals: RegDst = 1 (to select register address), MemtoReg = 0,
RegWrite = 1
Step 4: Memory Access or
Step 4: Memory Access or
ALU
ALU
(R-type) Instruction
(R-type) Instruction
Completion
Completion

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 For lw instruction only (write data from MDR to register):
Reg[IR[20:16]]<= MDR;
 Control signals: RegDst = 0 (to select register address),
MemtoReg = 1, RegWrite = 1
 Note: lw instruction completed at this step!
Step 5: Memory Read
Step 5: Memory Read
Completion
Completion

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Summary of Execution
Summary of Execution
Steps
Steps
Step name
Action for R-type
instructions
Action for memory-reference
instructions
Action for
branches
Action for
jumps
Instruction fetch IR <= Memory[PC]
PC <= PC + 4
Instruction A <= Reg [IR[25:21]]
decode/register fetch B <= Reg [IR[20:16]]
/branch addr comp ALUOut <= PC + (sign-extend (IR[15:0]) << 2)
Execution, address ALUOut <= A op B ALUOut <= A + sign-extend if (A ==B) then PC <= PC [31:28]
computation, branch/ (IR[15:0]) PC <= ALUOut II(IR[25:0]<<2)
jump completion
Memory access or R-type Reg [IR[15:11]] <= Load: MDR <= Memory[ALUOut]
completion ALUOut or
Store: Memory[ALUOut] <= B
Memory read completion Load: Reg[IR[20:16]] <= MDR
Some instructions take shorter number of cycles, therefore next instructions can start earlier.
Hence, compare to single-cycle implementation where all instructions take same amount of time, multi-cycle
implementation is faster!
Multi-cycle implementation also reduces hardware cost (reduces adders & memory, increases number of

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 How many cycles will it take to execute this code?
lw $t2, 0($t3)
lw $t3, 4($t3)
beq $t2, $t3, Label #assume not
add $t5, $t2, $t3
sw $t5, 8($t3)
Label: ...
 What is going on during the 8th cycle of execution?
 In what cycle does the actual addition of $t2 and $t3 takes
place?
Simple Questions
Simple Questions

multi cycle in microprocessor 8086 sy B-tech

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