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CS61C L19 CPU Design: The Single-Cycle II & Control (1) Pearce, Summer 2010 © UCB

inst.eecs.berkeley.edu/~cs61c CS61C : Machine Structures

Lecture 19CPU Design: The Single-Cycle II & Control

2010-07-22

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Instructor Paul Pearce

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How to Design a Processor: step-by-step

1. Analyze instruction set architecture (ISA) => datapath requirements

• meaning of each instruction is given by the register transfers

• datapath must include storage element for ISA registers• datapath must support each register transfer

2. Select set of datapath components and establish clocking methodology

3. Assemble datapath meeting requirements4. Analyze implementation of each instruction to

determine setting of control points that effects the register transfer.

5. Assemble the control logic


Clocking Methodology

• Storage elements clocked by same edge• Being physical devices, flip-flops (FF) and

combinational logic have some delays • Gates: delay from input change to output change • Signals at FF D input must be stable before active clock

edge to allow signal to travel within the FF (set-up time), and we have the usual clock-to-Q delay

• “Critical path” (longest path through logic) determines length of clock period

Clk

.

.

.

.

.

.

.

.

.

.

.

.


Register-Register Timing: One complete cycleClk

PC

Rs, Rt, Rd, etc

ALUctr

Instruction Memory Access Time

Old Value New Value

RegWr Old Value New Value

Delay through Control Logic

busA, BRegister File Access Time

Old Value New Value

busWALU Delay

Old Value New Value

Old Value New Value

New ValueOld Value

Register WriteOccurs Here

32

ALUctr

clk

busW

RegWr

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs Rt

AL

U

5Rd


3c: Logical Operations with Immediate• R[rt] = R[rs] op ZeroExt[imm16]

op rs rt immediate

016212631

6 bits 16 bits5 bits5 bits

immediate

016 1531

16 bits16 bits

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

32

ALUctr

clk

busW

RegWr

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs RtA

LU

5Rd

But we’re writing to Rt register??


3c: Logical Operations with Immediate• R[rt] = R[rs] op ZeroExt[imm16] ]

op rs rt immediate

016212631


immediate

016 1531

16 bits16 bits

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

• Already defined 32-bit MUX; Zero Ext?

What about Rt register read??

32

ALUctr

clk

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

Rd

ZeroE

xt 3216imm16

ALUSrc

01

0

1

AL

U

5

RegDst


3d: Load Operations• R[rt] = Mem[R[rs] + SignExt[imm16]]Example: lw rt,rs,imm16

op rs rt immediate

016212631


32

ALUctr

clk

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

Rd

ZeroE

xt 3216imm16

ALUSrc

01

0

1

AL

U

5

RegDst


3d: Load Operations• R[rt] = Mem[R[rs] + SignExt[imm16]]Example: lw rt,rs,imm16

op rs rt immediate

016212631


32

ALUctr

clk

busW

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst

Exten

der 3216imm16

ALUSrcExtOp

MemtoReg

clk

Data In

32

MemWr01

0

1

AL

U 0

1

WrEn Adr

DataMemory

5

?


3e: Store Operations• Mem[ R[rs] + SignExt[imm16] ] = R[rt]

Ex.: sw rt, rs, imm16

op rs rt immediate016212631


32

ALUctr

clk

busW

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst

Exten

der 3216imm16

ALUSrcExtOp

MemtoReg

clk

Data In

32

MemWr01

0

1

AL

U 0

1

WrEn Adr

DataMemory

5


3e: Store Operations• Mem[ R[rs] + SignExt[imm16] ] = R[rt]

Ex.: sw rt, rs, imm16



32

ALUctr

clk

busW

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst

Exten

der 3216imm16

ALUSrcExtOp

MemtoReg

clk

Data In

32

MemWr01

0

1

AL

U 0

1

WrEn Adr

DataMemory

5


3f: The Branch Instruction

beq rs, rt, imm16• mem[PC] Fetch the instruction from memory

• Zero = R[rs] - R[rt] Calculate branch condition

• if (Zero) Calculate the next instruction’s address PC = PC + 4 + ( SignExt(imm16) x 4 )

else PC = PC + 4



CS61C L17 Combinational Logic Blocks and Intro to CPU Design (12) Pearce, Summer 2010 © UCB

Datapath for Branch Operations• beq rs, rt, imm16Datapath generates condition (zero)



Already have mux, adder, need special sign extender for PC, need equal compare (sub?)imm16

clk

PC

00

4nPC_sel

PC

Ext

Ad

derA

dder

Mu

x

Inst Address

32

ALUctr

clk

busW

RegWr

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs Rt

AL

U

5

z

Zero

CS61C L17 Combinational Logic Blocks and Intro to CPU Design (13) Pearce, Summer 2010 © UCB

Putting it All Together:A Single Cycle Datapath

imm16

32

ALUctr

clk

busW

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst

Exten

der

3216imm16

ALUSrcExtOp

MemtoReg

clk

Data In32

MemWrZero

Instruction<31:0><21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRtRs

clk

PC

00

4

nPC_sel

PC

Ext

Adr

InstMemory

Ad

derA

dder

Mu

x

01

0

1

z

AL

U 0

1

WrEn Adr

DataMemory

5


An Abstract View of the Implementation

DataOut

clk

5

Rw Ra Rb

RegisterFile

Rd

Data In

DataAddr Ideal

DataMemory

Instruction

InstructionAddress

IdealInstruction

Memory

PC

5Rs

5Rt

32

323232

A

B

Nex

t A

dd

ress

Control

Datapath

Control Signals Conditions

clk clk

AL

U


An Abstract View of the Critical Path Critical Path (Load Instruction) =

Delay clock through PC (FFs) + Instruction Memory’s Access Time + Register File’s Access Time, + ALU to Perform a 32-bit Add + Data Memory Access Time + Stable Time for Register File Write

clk

5

Rw Ra Rb

RegisterFile

Rd

Data In

DataAddr Ideal

DataMemory

Instruction

InstructionAddress

IdealInstruction

Memory

PC

5Rs

5Rt

32

323232

A

B

Nex

t A

dd

ress

clk clk

AL

U(Assumes a fast controller)


Administrivia• HW 7 due Saturday

• Specification update now online. Be sure to check it out. The changes are designed to make integration with project 2 easier. They are very minor.

• Project 2 online now!

• Reminder: Midterm regrades due Monday.

• Paul will be gone on Monday. Noah will be giving the lecture.

• Paul will have OH on Monday (barring flight delays)

• Be prepared for a kinetic learning activity


Review: A Single Cycle Datapath

32

ALUctr

clk

busW

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst

Exten

der

3216imm16

ALUSrcExtOp

MemtoReg

clk

Data In32

MemWrzero

01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs

nPC_sel instrfetchunitclk

• We have everything but the values of control signals


Recap: Meaning of the Control Signals• nPC_sel: “+4” 0 PC <– PC + 4

“br” 1 PC <– PC + 4 + {SignExt(Im16) , 00 }

• Later in lecture: higher-level connection between mux and branch condition

“n”=next

imm16

clk

PC

00

4nPC_sel

PC

Ext

Ad

derA

dder

Mu

x

Inst Address

0

1


Recap: Meaning of the Control Signals• ExtOp: zero, sign

• ALUsrc: 0 reg; 1 imm

• ALUctr: ADD, SUB, OR

° MemWr: 1 write memory° MemtoReg: 0 alu; 1 mem° RegDst: 0 rt; 1 rd° RegWr: 1 write register

32

ALUctr

clk

busW

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst

Exten

der 3216imm16

ALUSrcExtOp

MemtoReg

clk

Data In

32

MemWr01

0

1

AL

U 0

1

WrEn Adr

DataMemory

5


RTL: The Add Instruction

add rd, rs, rt•MEM[PC] Fetch the instruction

from memory

•R[rd] = R[rs] + R[rt] The actual operation

•PC = PC + 4 Calculate the next instruction’s address

op rs rt rd shamt funct

061116212631

6 bits 6 bits5 bits5 bits5 bits5 bits


Instruction Fetch Unit at the Beginning of Add• Fetch the instruction from Instruction memory: Instruction = MEM[PC]

• same for all instructions

imm16

clk

PC

00

4nPC_sel

PC

Ext

Ad

derA

dder

Mu

x

Inst Address

InstMemory Instruction<31:0>


The Single Cycle Datapath during Add

R[rd] = R[rs] + R[rt]op rs rt rd shamt funct

061116212631

32

ALUctr=ADD

clk

busW

RegWr=1

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=rd

Exten

der

3216imm16

ALUSrc=regExtOp=x

MemtoReg=ALU

clk

Data In32

MemWr=0

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs

nPC_sel=+4 instrfetchunitclk


Instruction Fetch Unit at the End of Add• PC = PC + 4

• This is the same for all instructions except: Branch and Jump

imm16

clk

PC

00

4nPC_sel=+4

PC

Ext

Ad

derA

dder

Mu

x

Inst Address

InstMemory


Single Cycle Datapath during Or Immediate?

op rs rt immediate

016212631

• R[rt] = R[rs] OR ZeroExt[Imm16]

32

ALUctr=

clk

busW

RegWr=

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=

Exten

der

3216imm16

ALUSrc=ExtOp=

MemtoReg=

clk

Data In32

MemWr=

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs

nPC_sel= instrfetchunitclk


• R[rt] = R[rs] OR ZeroExt[Imm16]op rs rt immediate

016212631

Single Cycle Datapath during Or Immediate?

32

ALUctr=OR

clk

busW

RegWr=1

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=rt

Exten

der

3216imm16

ALUSrc=immExtOp=zero

MemtoReg=alu

clk

Data In32

MemWr=0

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs



The Single Cycle Datapath during Load?

• R[rt] = Data Memory {R[rs] + SignExt[imm16]}op rs rt immediate

016212631

32

ALUctr=

clk

busW

RegWr=

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=

Exten

der3216

imm16

ALUSrc=ExtOp=

MemtoReg=

clk

Data In32

MemWr=

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs



The Single Cycle Datapath during Load

• R[rt] = Data Memory {R[rs] + SignExt[imm16]}op rs rt immediate

016212631

32

ALUctr=ADD

clk

busW

RegWr=1

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=rt

Exten

der3216

imm16

ALUSrc=immExtOp=sign

MemtoReg=mem

clk

Data In32

MemWr=0

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs



The Single Cycle Datapath during Store?

op rs rt immediate

016212631

• Data Memory {R[rs] + SignExt[imm16]} = R[rt]

32

ALUctr=

clk

busW

RegWr=

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=

Exten

der3216

imm16

ALUSrc=ExtOp=

MemtoReg=

clk

Data In32

MemWr=

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs



The Single Cycle Datapath during Store

• Data Memory {R[rs] + SignExt[imm16]} = R[rt]op rs rt immediate

016212631

32

ALUctr=ADD

clk

busW

RegWr=0

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=x

Exten

der3216

imm16

ALUSrc=immExtOp=sign

MemtoReg=x

clk

Data In32

MemWr=1

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs



The Single Cycle Datapath during Branch?

• if (R[rs] - R[rt] == 0) then Zero = 1 ; else Zero = 0op rs rt immediate

016212631

32

ALUctr=

clk

busW

RegWr=

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=

Exten

der3216

imm16

ALUSrc=ExtOp=

MemtoReg=

clk

Data In32

MemWr=

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs



The Single Cycle Datapath during Branch

• if (R[rs] - R[rt] == 0) then Zero = 1 ; else Zero = 0op rs rt immediate

016212631

32

ALUctr=SUB

clk

busW

RegWr=0

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst=x

Exten

der3216

imm16

ALUSrc=regExtOp=x

MemtoReg=x

clk

Data In32

MemWr=0

zero01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs

nPC_sel=br instrfetchunitclk


Instruction Fetch Unit at the End of Branch• if (Zero == 1) then PC = PC + 4 + SignExt[imm16]*4 ;

else PC = PC + 4

op rs rt immediate

016212631

• What is encoding of nPC_sel?

• Direct MUX select?

• Branch inst. / not branch

• Let’s pick 2nd option

• nPC_sel = +4 0

• nPC_sel = br 1nPC_sel zero? MUX

0 x 01 0 01 1 1

Adr

InstMemory

nPC_selInstruction<31:0>

Zero

nPC_sel

Q: What logic gate?

imm16 clk

PC

00

4

PC

Ext

Ad

derA

dder

Mu

x

0

1

MUX ctrl


Peer Instruction

12a) FFb) FTc) TFd) TT


Peer Instruction Answer

12a) FFb) FTc) TFd) TT

1) Memory has by far a higher access time than combinational logic. TRUE

2) Let’s look back at the datapath. Take a sw instructions. Can we also write to a register? If so, which reg? What value? TRUE


Summary: A Single Cycle Datapath

32

ALUctr

clk

busW

RegWr

32

32busA

32

busB

5 5

Rw Ra Rb

RegFile

Rs

Rt

Rt

RdRegDst

Exten

der

3216imm16

ALUSrcExtOp

MemtoReg

clk

Data In32

MemWrzero

01

0

1

Z

AL

U 0

1

WrEn Adr

DataMemory

5


<16:20>

<11:15>

<0:15>

Imm16RdRtRs

nPC_sel instrfetchunitclk

• We have everything! Now we just need to know how to BUILD

CONTROL


°5 steps to design a processor• 1. Analyze instruction set datapath requirements• 2. Select set of datapath components & establish clock

methodology• 3. Assemble datapath meeting the requirements• 4. Analyze implementation of each instruction to

determine setting of control points that effects the register transfer.

• 5. Assemble the control logic• Formulate Logic Equations• Design Circuits

Summary: Single-cycle Processor

Control

Datapath

Memory

ProcessorInput

Output

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