1 Copyright 1998 Morgan Kaufmann Publishers, Inc. All rights reserved. The single cycle CPU

1Copyright 1998 Morgan Kaufmann Publishers, Inc. All rights reserved.

The single cycle CPU

Shiftleft 2

PC

Instructionmemory

Readaddress

Instruction[31– 0]

Datamemory

Readdata

Writedata

RegistersWriteregister

Writedata

Readdata 1

Readdata 2

Readregister 1

Readregister 2

Instruction [15– 11]



Add

ALUresult

Zero


MemtoReg

ALUOp

MemWrite

RegWrite

MemRead

Branch

JumpRegDst

ALUSrc


4

Mux

Instruction [25– 0] Jump address [31– 0]

PC+4 [31– 28]

Signextend

16 32Instruction [15– 0]

1

Mux

1

0

Mux

0

1

Mux

0

1

ALUcontrol

Control

Add ALUresult

Mux

0

1 0

ALU

Shiftleft 2

26 28

Address


Performance of Single-Cycle Machines

• Memory Unit 2 ns• ALU and Adders 2 ns• Register file (Read or Write) 1 ns

Class Fetch Decode ALU Memory Write Back Total

R-format 2 1 2 0 1 6

LW 2 1 2 2 1 8

SW 2 1 2 2 7ns

Branch 2 1 2 5ns

Jump 2 2ns


What if we had a variable CK cycle?

Let’s check the following scenario:

• Rtype: 44%, LW: 24%, SW: 12% • BRANCH: 18%, JUMP: 2%

• I- number of instructions in program

• T- time of the CK cycle

• CPI - number of CK cycle per instruction (=1)

Execution=I*T*CPI= 8*24%+7*12%+6*44%+5*18%+2*2%=6.3 ns


The result:

EXE Single cycle T single clock * I T single clock 8

EXE Variable T variable clock * I T variable clock 6.3

We get a ratio of 1.27. The ratio is higher when more complicated instructions, e.g., floating point instructions are also implemented.

Since building a variable CK circuit is too complicated, we instead want instructions to take as many shorter CKs as required


Multicycle Approach

The idea of Multi-cycle approach:

•We’ll save time since each instruction takes only the necessary number of CK cycles (which are about 5 times shorter than the original CK cycle)

• We also save in components since we can use the same component in different phases of the same instruction


Building a Multi-Cycle CPU:

Split the instruction to steps (phases)

Make sure that the steps are balanced (same time required)

Reduce the job done at each step. In each step only one chore is done.

At the end of each CK cycle:

Store the result of the current step to be used by the next step. So, add more internal registers for storing the intermediate results.


A single cycle CPU capable of R-type & lw/sw instructions (data & control)

5[25:21=]Rs

5[20:16=]Rt

Reg File

InstructionMemoryPC ALU

Adder4

ck

ck

6[31:26]

RegWrite

16[15:0]

5

add

Sext16>-32

DataMemory

5[25:21=]Rs

6[5:0=]funct ALUcontrol

Rd

Address

D.In

D. Out

MemWrite


A single cycle CPU capable of R-type & lw/sw instructions - Data Path only

5[25:21=]Rs

5[20:16=]Rt

Reg File


Adder4

ck

ck16[15:0]

5

Sext16>-32

DataMemory

5[25:21=]Rs

Rd

Address

D.In

D. Out

lw

sw


PC 0x400000

Rs, Rt ALU inputs

ALU output)address(

Memory output

fetchWrite

backdecode execute

Memdata

memory

0x400004

new ALU inputs

New ALU output)new address(

new Memory output

fetchWrite

backdecode execute

Memdata

memory

D. Memdata

D.Mem adrs

I.Mem data

Timing of a single cycle CPU


PC

D. Memdata

D.Mem adrs

0x400000

Rs, Rt ALU inputs

ALU output)address(

Memory output

fetch Write backdecode execute

Memdata

memory

I.Mem data

PC

IR

A,B

ALUout

Mem data

MDR

fetch

Write back

decode

execute

memory

Timing of a lw instruction in a single cycle CPU

Timing of a lw instruction in a multi-cycle CPU

2ns

We want to replace a long single CK cyclewith 5 short ones:

1ns 2ns 2ns 1ns

0x400000

Instruction in IR

ALU calculates something

0 1 3 4 5=(0)2


Therefore we should add registers to the single cycle CPU shown below:

5[25:21=]Rs

5[20:16=]Rt

Reg File


Adder4

ck

ck16]15:0[

5

Sext16>-32

DataMemory

Rd

Address

D.In

D. Out


Adding registers to “split” the instruction to 5 stages:

5[25:21=]Rs

5[20:16=]Rt

Reg File


Adder4

ck

ck16]15:0[

5

Sext16>-32

DataMemory

Rd

Address

D.In

D. Out

IR

ckck

ck

ck

ck

A

B

ALUout MDR

PCWrite

2

03

4

1

5


Here is the book’s version of the multi-cycle CPU:

Shiftleft 2

PC

Memory

MemData

Writedata

Mux

0

1


Writedata

Readdata 1

Readdata 2

Readregister 1

Readregister 2

Mux

0

1

Mux

0

1

4


Signextend

3216




Instructionregister

1 Mux

0

3

2

Mux

ALUresult

ALUZero

Memorydata

register


A

B

ALUOut

0

1

Address

Only PC and IR have write enable signalsAll other registers hold data for a single cycle


Here is our version of A mult--cycle CPU capable of R-type & lw/sw & branch instructions

5IR]20:16[=Rt

Reg FileInstruction & dataMemory

PC

ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck >> 2

>>2


Let us explain the multi-cycle CPU

• First we’ll look at a CPU capable of performing only R-type instructions

• Then, we’ll add the lw instruction

• And the sw instruction

• Then, the beq instruction

• And finally, the j instruction


Let us remind ourselves how works a single cycle CPU capable of performingR-type instructions.Here you see the data-path and the timing of an R-typeinstruction.

5[25:21=]Rs

5[20:16=]Rt

5[15:11=]Rd

Reg File


Adder4

ck

ck

6[31:26]

6[5:0=]funct

PC 0x400000 0x400004

Rs, Rt ALU inputs new ALU inputs

ALU output

Memory output New Memory output

New ALU output

fetchWrite

backdecode execute


A single cycle CPU demo: R-type instruction

5[25:21=]Rs

5[20:16=]Rt

5[15:11=]Rd

Reg FileInstruction

Memory

PCALU

ck

ck

4


A multi cycle CPU capable of performing R-type instructions

5IR]20:16[=Rt


PC

ALU

ck5

5IR]25:21[=Rs

Rd

IR

ck

ALUout

ck

A

ck

B

ck


A multi cycle CPU capable of R-type & instructions

fetch

5IR]20:16[=Rt

Reg File

Instruction & dataMemory

PC

ALU

ck5

5IR]25:21[=Rs

Rd

IR

ck

ALUout

ck

A

ck

B

ck

01



decode

5IR]20:16[=Rt


PC

ALU

ck5

5IR]25:21[=Rs

Rd

IR

ck

ALUout

ck

A

ck

B

ck

1

2



execute

5IR]20:16[=Rt


PC

ck5

5IR]25:21[=Rs

Rd

IR

ck

ALUout

ck

A

ck

B

ck

ALU

2

3



write back

5IR]20:16[=Rt


PC

ALU

ck5

5IR]25:21[=Rs

Rd

IR

ck

ALUout

ck

A

ck

B

ck

Rd

ck3

4


PC

GPR input

0x400000

Rs, Rt ALU inputs

ALU output(Data = result of cala.)

Memory output = the instruction

fetch decode execute Write Back

Inst. Mem data

Mem data

IR

A,B

ALUout

fetch

Write back

decode

execute

Timing of an R-type instruction in a single cycle CPU

Timing of an R-type instruction in a multi-cycle CPU

3 4 (=0)0 1 2

PC

Previous inst. Current instruction


Mem data

IR

A,B

ALUout

fetch

Write back

decode

execute

GPR outputs

ALUoutput

IR=M ( PC )

A= Rs, B= Rt

ALUuot= A op B

IRWrite

At the rising edge of CK:Rd=ALUoutR-Type instruction takes 4 CKs

PC

Previous inst.

Current instruction

Current instruction next inst.

IR=M(PC) A= Rs,B= Rt

ALUout = A op B Rd=ALUout

Rd = ALUout

The state diagram:


A multi-cycle CPU capable of R-type instructions (PC calc. )

5IR]20:16[=Rt


PC

ALU

4

ck5

5IR]25:21[=Rs

Rd

IR

ck

ALUout

ck

A

ck

B

ck


Mem data

IR

A,B

ALUout

fetch

Write back

decode

execute

GPR outputs

ALUoutput

ALUuot =A op B

At the rising edge of CK:Rd=ALUout

PC = PC+4

PC next PC = current PC+4current PC

next inst.Previous inst. current instruction

PCWrite



fetch

5IR]20:16[=Rt

Reg File

Instruction MemoryPC

ALU

ck5

5IR]25:21[=Rs

Rd

IR

ck

ALUout

ck

A

ck

B

ck

ALU

4


Fetch

WBR

ALU

Decode

1

6

0

7

R-type

The state diagram of a CPU capable of R-type instructions only

IR=M(PC)PC = PC+4

ALUout=A op B

A=RsB=Rt

Rd = ALUout


Fetch

WBR

Load

ALUAdrCmp

Decode

WB

1

26

0

74

3

lwR-type

lw

The state diagram of a CPU capable of R-type and lw

instructions

ALUout= A+sext(imm)

MDR = M(ALUout)

Rt = MDR


We added registers to “split” the instruction to 5 stages.Let’s discuss the lw instruction

5[25:21=]Rs

5[20:16=]Rt

Reg File


Adder4

ck

ck16]15:0[

5

Sext16>-32

DataMemory

Rd

Address

D.In

D. Out

IR

ckck

ck

ck

ck

A

B

ALUout MDR

PCWrite

2

03

4

1

5

In ths single-cycle we kept the “data flow” from left to right. Here we change that a little, since as we’ll see, we are some parts of the CPU more than once during the same instruction. So we prefer to move data the memory.

All parts related to lw only are blue


First we draw a multi-cycle CPU capable of R-type & lw instructions:

5IR]20:16[=Rt

Reg FileInstruction MemoryPC

ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck

ALUALU

We just moved the data memory All parts related to lw only are blue

Data Memory


A multi-cycle CPU capable of R-type & lw instructionsfetch

5IR]20:16[=Rt


ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck

ALUALU

Data Memory


A multi-cycle CPU capable of R-type & lw instructionsdecode

5IR]20:16[=Rt


ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck >> 2Data Memory


A multi-cycle CPU capable of R-type & lw instructionsAdrCmp

5IR]20:16[=Rt


ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck

ALU

Data Memory


A multi-cycle CPU capable of R-type & lw instructionsmemory

5IR]20:16[=Rt

Reg FileInstructionMemoryPC

ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

Branch Address

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck >> 2DataMemory


A multi-cycle CPU capable of R-type & lw instructionsWB

5IR]20:16[=Rt


ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck

DataMemory

ck

Rt


Can we unite the Instruction & Data memories? (They are not used simultaneously as in the single cycle CPU)

5IR]20:16[=Rt


ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck

DataMemory

ck


So here is a multi-cycle CPU capable of R-type & lw instructionsusing a single memory for instructions & data

5IR]20:16[=Rt

Reg File

PC

ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck

Instruction & dataMemory


PC

D. Memdata

D.Mem adrs

0x400000

Rs, Rt ALU inputs

ALU output)address(

Memory output

fetch Write backdecode execute

Memdata

memory

I.Mem data

PC

IR

A,B

ALUout

Mem data

MDR

fetch

Write back

decode

execute

memory

Timing of a lw instruction in a single cycle CPU

Timing of a lw instruction in a multi-cycle CPU

PC+4

Previous inst. current instruction

Data address

Data to Rt


Mem data

IR

A,B

ALUout

Mem data

MDR

fetch

Write back

decode

execute

memory

GPR outputs

ALUoutput

IR=M ( PC )PC= PC+4

A= Rs, B= Rt

ALUuot=A+sext(imm)

MDR=M(ALUout)

At the rising edge of CK:Rt=MDR

PC

Previous inst. current instruction

Data address

Data address

Data to Rt

PCWrite, IRWrite


Fetch

WBR

Load

ALUAdrCmp

Decode

WB

1

26

0

74

3

lwR-type

The state diagram of a CPU capable of R-type and lw

instructions

ALUout= A+sext(imm)

MDR = M(ALUout)

Rt = MDR

IR=M(PC)PC = PC+4

ALUout=A op B

A=RsB=Rt

Rd = ALUout


A multi-cycle CPU capable of R-type & lw & sw instructions

5IR]20:16[=Rt


PC

ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

Branch Address

IR

ck

MDR

ck

ALUout

ck

A

ck

B

ck >> 2

lwsw


Fetch

WBR

Load

ALUAdrCmp

Store

Decode

WB

1

5

26

0

74

3

lw+swR-type

swlw

The state diagram of a CPU capable of R-type and lw and sw

instructions

M(ALUout)=B

IR=M(PC)PC = PC+4

ALUout=A op B

A=RsB=Rt

Rd = ALUout

ALUout= A+sext(imm)

MDR = M(ALUout)

Rt = MDR


A multi-cycle CPU capable of R-type & lw/sw & branch instructions

5IR]20:16[=Rt


PC

ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

IR

ck

IR

ck

ALUout

ck

A

ck

B

ck

>>2


Calc PC=PC+sext(imm)>>2

Adding the instruction beq to the state diagram:

Calc Rs -Rt (just to produce the zero signal)

Fetch

WBR

Load

BranchALUAdrCmp

Store

Decode

WB

1

5

2 86

0

74

3

lw+swR-type beq

zero

swlw

not zero


Adding the instruction beq to the state diagram, a more efficient way: Let’s use the decode state in which the ALU is doing nothing to compute the branch address.We’ll have to store it for 1 more CK cycle, until we know whether to branch or not! (We store it in the ALUout reg.)

Fetch

WBR

Load

BranchALUAdrCmp

Store

Decode

WB

1

5

2 86

0

74

3

lw+swR-type beq

swlw

Calc ALUout=PC+sext(imm)>>2

Calc Rs - Rt. If zero, load the PC with ALUout data, else do not load the PC


A multi-cycle CPU capable of R-type & lw/sw & branch instructions

5IR]20:16[=Rt


PC

ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

Branch Address

IR

ck

IR

ck

ALUout

ck

A

ck

B

ck

>>2

PC+4


Fetch

Jump

WBR

Load

BranchALUAdrCmp

Store

Decode

WB

1

5

2 86

9

0

74

3

lw+swR-type beq

j

swlw

Adding the instruction j to the state diagram:

PC = PC]31:28[ || IR]25:0[>>2


A multi-cycle CPU capable of R-type & lw/sw & branch & jump instructions

5IR]20:16[=Rt


PC

ALU

4

ck

16IR]15:0[

5

Sext16>-32

5IR]25:21[=Rs

Rd

Branch Address

IR

ck

IR

ck

ALUout

ck

A

ck

B

ck

>>2

PC+4= next address

Jump addressIR]25:0[

>>2+ PC]31:28[


The phases (steps) of all instructions

Step nameAction for R-type

instructionsAction for memory-reference

instructionsAction 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]]

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] IIcomputation, branch/ (IR[15-0]) PC = ALUOut (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]] = MDR5

2 8 96

1

0

7 4

3


MultiCycle implementation with Control

Shiftleft 2

PCMux

0

1


Writedata

Readdata 1

Readdata 2

Readregister 1

Readregister 2


Mux

0

1

Mux

0

1

4


Signextend

3216




Instructionregister

ALUcontrol

ALUresult

ALUZero

Memorydata

register

A

B

IorD

MemRead

MemWrite

MemtoReg

PCWriteCond

PCWrite

IRWrite

ALUOp

ALUSrcB

ALUSrcA

RegDst

PCSource

RegWrite

Control

Outputs

Op[5– 0]

Instruction[31-26]


Mux

0

2

Jumpaddress [31-0]Instruction [25– 0] 26 28

Shiftleft 2

PC [31-28]

1

1 Mux

0

3

2

Mux

0

1ALUOut

Memory

MemData

Writedata

Address

Final State Machine

PCWritePCSource = 10

ALUSrcA = 1ALUSrcB = 00ALUOp = 01PCWriteCond

PCSource = 01

ALUSrcA =1ALUSrcB = 00ALUOp= 10

RegDst = 1RegWrite

MemtoReg = 0

MemWriteIorD = 1

MemReadIorD = 1

ALUSrcA = 1ALUSrcB = 10ALUOp = 00

RegDst = 0RegWrite

MemtoReg =1

ALUSrcA = 0ALUSrcB = 11ALUOp = 00

MemReadALUSrcA = 0

IorD = 0IRWrite

ALUSrcB = 01ALUOp = 00

PCWritePCSource = 00

Instruction fetchInstruction decode/

register fetch

Jumpcompletion

BranchcompletionExecution

Memory addresscomputation

Memoryaccess

Memoryaccess R-type completion

Write-back step

(Op = 'LW') or (Op = 'SW') (Op = R-type)

(Op

= 'B

EQ')

(Op

= 'J

')

(Op = 'SW

')

(Op

= 'L

W')

4

01

9862

753

Start


Fetch

Jump

WBR

Load

BranchALUAdrCmp

Store

Decode

WB

1

5

2 86

9

0

74

3

lw+swR-type beq

j

swlw

The final state diagram:


Relevant control signalsRTL descriptionState name

IorD=0, MemRead=1, IRWrite=1

ALUSrcA=0, ALUSrcB=01,ALUop=00 (add), PCSrc=00,PCWrite=1

IR= M(PC)

PC=PC+4

Fetch

0

(no signals are needed)

ALUSrcA=0, ALUSrcB=11,ALUop=00 (add) (for branch)

A = RsB = Rt

ALUOut = PC+( sext(imm)>>2 )

Decode

1

ALUSrcA=1, ALUSrcB=10,ALUop=00 (add) (for lw & sw)

(for sw)

ALUOut = A+( sext(imm) )

(B = Rt)

AdrCmp

2

ALUSrcA=1, ALUSrcB=00,ALUop=10 (funct bits determines op)

ALUOut = A op BALU6

ALUSrcA=1, ALUSrcB=00,ALUop=01(sub),PCSrc=01, PCWriteCond=1

if (A = = B) PC=ALUOutelse do nothing

Branch

8

PCSrc=10, PCWrite=1

PC= PC]31:28[||(IR]25:0[>>2)

Jump9

IorD=1, MemRead=1MDR = M(ALUOut)Load3

IorD=1, MemWrite=1M(ALUOut) = BStore5

RegDest=1, MemtoReg=0RegWrite=1

Rd = ALUOutWBR7

RegDest=0, MemtoReg=1RegWrite=1

Rt = MDRWB4


• Implementation:

Finite State Machine for Control (The book’s version)

PCWrite

PCWriteCond

IorD

MemtoReg

PCSource

ALUOp

ALUSrcB

ALUSrcA

RegWrite

RegDst

NS3NS2NS1NS0

Op5

Op4

Op3

Op2

Op1

Op0

S3

S2

S1

S0

State register

IRWrite

MemRead

MemWrite

Instruction registeropcode field

Outputs

Control logic

Inputs


Opcode= IR]31:26[zero, neg, etc.

next state

current state

control signalsnext statecalculation

Outputs decoder

State reg

ck

The Control Finite State Machine:

For 10 states coded 0-9, we need 4 bits, i.e., ]S3,S2,S1,S0[


The control signals decoder

We just implement the table of slide 54: Let’s look at ALUSrcA: it is “0” in states 0 and 1 and it is “1” in states 2, 6 and 8. In all other states we don’t care. let’s look at PCWrite: it is “1” in states 0 and 9. In all other states it must be “0”.And so, we’ll fill the table below and build the decoder.

S3 S2 S1 S0 ALUSrcA

state Control signals

0 10 0 0 0

01 00 0 0

0

0 0

10

1 0

X

1 X 0

0

0 00 1 0 1

PCWrite

0

1 0

11

0 0

X

1 X 0

0

All other combinations X

0 01 1

0 11 0 1

1

X

1 00 1 1

PCWriteCond

0

0

0

0

0

0

0

0

0

0

1

0

0

0

fetch

decode

AdrCmp

load

WB

store

ALU

WBR

branch

jump


The state machine “next state calc.” logic

R-type=000000, lw=100011, sw=101011, beq=000100, bne=000101, lui=001111, j=0000010, jal=000011, addi=001000

Fetch

0

Jump

9

WBR

7

Load

3

Branch

8ALU 6

AdrCmp 2

Store

5

Decode

1

WB

4

lw+sw

R-typebeq j

swlw

IR31 IR30 IR29 IR28 IR27 IR26

opcode

S3 S2 S1 S0

current state

S3 S2 S1 S0

next state

X 0X X X X X 0 0 0 0 0 0 1

0 0 0 1 0 1 1 00 0 00 0 0

X

X 1

0X

X X X

X X X

X

0 0 1 0

0 0 1 0

0 0 1 1

0 1 0 1

1 0X X X X X 0 0 1 0 0 1 0

R-type

lw

sw

lw+sw


Opcode = IR]31:26[

next state

current state

control signalsnext statecalculation

Outputs decoder

State reg

ck

The Control Finite State Machine:

Meally machine

PCWrite

PCWriteCond

zero

Mooremachine

to PC


Microprogramming

PCWritePCWriteCondIorD

MemtoRegPCSourceALUOpALUSrcBALUSrcARegWrite

AddrCtl

Outputs

Microcode memory

IRWrite

MemReadMemWrite

RegDst

Control unit

Input

Microprogram counter

Address select logic

Op[

5–

0]

Adder

1

Datapath

Instruction registeropcode field

BWrite


Microinstruction

LabelALU

control SRC1 SRC2Register control Memory

PCWrite control Sequencing

Fetch Add PC 4 Read PC ALU SeqAdd PC Extshft Read Dispatch 1

Mem1 Add A Extend Dispatch 2LW2 Read ALU Seq

Write MDR FetchSW2 Write ALU FetchRformat1 Func code A B Seq

Write ALU FetchBEQ1 Subt A B ALUOut-cond FetchJUMP1 Jump address Fetch

Microinstruction formatField name Value Signals active Comment

Add ALUOp = 00 Cause the ALU to add.ALU control Subt ALUOp = 01 Cause the ALU to subtract; this implements the compare for

branches.Func code ALUOp = 10 Use the instruction's function code to determine ALU control.

SRC1 PC ALUSrcA = 0 Use the PC as the first ALU input.A ALUSrcA = 1 Register A is the first ALU input.B ALUSrcB = 00 Register B is the second ALU input.

SRC2 4 ALUSrcB = 01 Use 4 as the second ALU input.Extend ALUSrcB = 10 Use output of the sign extension unit as the second ALU input.Extshft ALUSrcB = 11 Use the output of the shift-by-two unit as the second ALU input.Read Read two registers using the rs and rt fields of the IR as the register

numbers and putting the data into registers A and B.Write ALU RegWrite, Write a register using the rd field of the IR as the register number and

Register RegDst = 1, the contents of the ALUOut as the data.control MemtoReg = 0

Write MDR RegWrite, Write a register using the rt field of the IR as the register number andRegDst = 0, the contents of the MDR as the data.MemtoReg = 1

Read PC MemRead, Read memory using the PC as address; write result into IR (and lorD = 0 the MDR).

Memory Read ALU MemRead, Read memory using the ALUOut as address; write result into MDR.lorD = 1

Write ALU MemWrite, Write memory using the ALUOut as address, contents of B as thelorD = 1 data.

ALU PCSource = 00 Write the output of the ALU into the PC.PCWrite

PC write control ALUOut-cond PCSource = 01, If the Zero output of the ALU is active, write the PC with the contentsPCWriteCond of the register ALUOut.

jump address PCSource = 10, Write the PC with the jump address from the instruction.PCWrite

Seq AddrCtl = 11 Choose the next microinstruction sequentially.Sequencing Fetch AddrCtl = 00 Go to the first microinstruction to begin a new instruction.

Dispatch 1 AddrCtl = 01 Dispatch using the ROM 1.Dispatch 2 AddrCtl = 10 Dispatch using the ROM 2.


Interrupt and exception

Type of event From Where ? MIPS terminology

Interrupt External I/O device request

------------------------------------------------------------------------------------

Invoke Operation system Internal Exception

From user program

-------------------------------------------------------------------------------------

Arithmetic Overflow Internal Exception

Using an undefined

Instruction Internal Exception--------------------------------------------------------------------------------------

Either Exception or interrupt Hardware malfunctions


Exceptions handling

Exception type Exception vector address (in hex)

Undefined instruction c0 00 00 00

Arithmetic Overflow c0 00 00 20

We have 2 ways to handle exceptions:

Cause register or Vectored interrupts

MIPS – Cause register


Handling exceptions

10


Handling exceptions

Shiftleft 2

Memory

MemData

Writedata

Mux

0

1


Mux

0

1

4


Signextend

3216




Instructionregister

ALUcontrol

ALUresult

ALUZero

Memorydata

register

A

B

IorD

MemRead

MemWrite

MemtoReg

PCWriteCond

PCWrite

IRWrite

Control

Outputs

Op[5– 0]

Instruction[31-26]


Mux

0

2

Jumpaddress [31-0]Instruction [25– 0] 26 28

Shiftleft 2

PC [31-28]

1

Address

EPC

CO 00 00 00 3

Cause

ALUOp

ALUSrcB

ALUSrcA

RegDst

PCSource

RegWrite

EPCWriteIntCauseCauseWrite

1

0

1 Mux

0

3

2

Mux

0

1

Mux

0

1

PC

Mux

0

1


Writedata

Readdata 1

Readdata 2

Readregister 1

Readregister 2

ALUOut


Fetch

Jump

WBR

Load

BranchALU

AdrCmp

Store

Decode

WB

1

5

2 86

9

0

74

3

lw+swR-type be

q

j

sw

lw

SavePC

10

IRET

1

JumpInt

11

Handling interrupts:

int

int iret


D Q

“1”

irq

int (to the state machine)

eint

clr_irq~

The interrupt source

Handling an interrupt: remembering it in a FF until it is serviced


Jumping to the interrupt routine

C0000000

Iret

Returning from interrupt

Interrupt


Jumping to the interrupt routine

C0000000

Iret

Returning from interrupt

Interrupt

irq eint

0 10 1


Fetch > decode >ex >wb




Fetch > Save_PC >JumpInt

C0000000

Iret Fetch > decode > Iret

The state machine in action during interrupt


End of multi-cycle implementation

Documents

1 Copyright 1998 Morgan Kaufmann Publishers, Inc. All rights reserved. The single cycle CPU