CS 152 L07 Single Cycle 2 (1)UC Regents Fall 2004 © UCB 2004-09-21 John Lazzaro (lazzaro) Dave Patterson (patterson)

CS 152 L07 Single Cycle 2 (1) UC Regents Fall 2004 © UCB

2004-09-21

John Lazzaro(www.cs.berkeley.edu/~lazzaro)

Dave Patterson (www.cs.berkeley.edu/~patterson)

www-inst.eecs.berkeley.edu/~cs152/

CS152 – Computer Architecture andEngineering

Lecture 7 – (Design Notebook+)Single Cycle Control


Review° 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 (This Lecture)

° MIPS makes it easier• Instructions same size; Source registers, immediates always in same place

• Operations always on registers/immediates

° Single cycle datapath => CPI=1, CCT => long


Why should you keep a design notebook?

° Keep track of the design decisions and the reasons behind them

• Otherwise, it will be hard to debug and/or refine the design

• Write it down so that can remember in long project: 2 weeks ->2 yrs

• Others can review notebook to see what happened

° Record insights you have on certain aspect of the design as they come up

° Record of the different design & debug experiments• Memory can fail when very tired

° Industry practice: learn from others mistakes


Why do we keep it on-line?

° You need to force yourself to take notes!• Open a window and leave an editor running while you work

1) Acts as reminder to take notes

2) Makes it easy to take notes

• 1) + 2) => will actually do it

° Take advantage of the window system’s “cut and paste” features

° It is much easier to read your typing than your writing

° Also, paper log books have problems• Limited capacity => end up with many books

• May not have right book with you at time vs. networked screens

• Can use computer to search files/index files to find what looking for


How should you do it?° Keep it simple• DON’T make it so elaborate that you won’t use (fonts, layout, ...)

° Separate the entries by dates• type “date” command in another window and cut&paste

° Start day with problems going to work on today

° Record output of simulation into log with cut&paste; add date

• May help sort out which version of simulation did what

° Record key email with cut&paste

° Record of what works & doesn’t helps team decide what went wrong after you left

° Index: write a one-line summary of what you did at end of each day


On-line Notebook Example

°Refer to the handout “Example of On-Line Log Book” on CS 152 home page:

http://www-inst.eecs.berkeley.edu/~cs152/ handouts/online_notebook_example.html


Recap: Putting it All Together: 1 Cycle Datapathim

m16

32

ALUctr

busW

RegWr

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

Mu

x

MemtoReg

Clk

Data InWrEn32 Adr

DataMemory

MemWrA

LU

Zero

Instruction<31:0>

0

1

0

1

01

<21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRtRs

Ad

der

Ad

der

PC

Clk

00

Mu

x

4

PCSrc

PC

Ext

Adr

InstMemory

Clk


Recap: The MIPS-lite Subset

° ADD and subtract• add rd, rs, rt

• sub rd, rs, rt

° OR Imm:• ori rt, rs, imm16

° LOAD and STORE• lw rt, rs, imm16

• sw rt, rs, imm16

° BRANCH:• beq rs, rt, imm16

op rs rt rd shamt funct

061116212631

6 bits 6 bits5 bits5 bits5 bits5 bits

op rs rt immediate

016212631

6 bits 16 bits5 bits5 bits


Meaning of the Control Signals° ExtOp: “zero”, “sign”

° ALUsrc: 0 regB; 1 immed

° 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

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

Mu

x

MemtoReg

Clk

Data InWrEn32 Adr

DataMemory

MemWr

AL

U

Zero

0

1

0

1

01


Two equivalent ways to specify control

°Book does left version (Fig 5.18, p. 308)• Book combines all ALU instructions as “R-format” vs. separate instructions (add …)

• Good news: lecture different view than book

°We’ll do right by committee, 1 at a time

Controlline 0

Controlline 1

… Controlline n

AddU A B

SubU

ORI

LW

SW

BEQ X Y

AddU SubU ORI LW SW BEQ

Controlline 0

A X

Controlline 1…Controlline n

B Y

(Rotate about 45degree axis)


Setting PC Source Control Signal° PCSrc: 0 PC <= PC + 4

1 PC <= PC + 4 + {SignExt(Im16), 2’b00}

° Later in lecture: higher-level connection between mux and branch cond

Adr

InstMemory

Ad

der

Ad

der

PC

Clk

00

Mu

x

4

PCSrc

PC

Extim

m16

Answer? AddU SubU ORI LW SW BEQ

0 0 0 0 0 0 01 0 0 0 0 0 12 0 0 0 0 1 13 0 0 0 1 1 14 0 0 1 1 1 15 0 1 1 1 1 16 1 1 1 1 1 17 1 1 1 1 1 X8 X X X X X 19 None of the above


Meaning of the Control Signals° ExtOp: 0 “zero” ; 1 “sign”

° ALUsrc: 0 regB; 1 immed

° ALUctr: “add”, “sub”, “or”

° MemWr: 1 write memory

° MemtoReg: 0 ALU; 1 Mem

° RegDst: 0 “rt”; 1 “rd”


32

ALUctr

Clk

busW

RegWr

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

Mu

x

MemtoReg

Clk

Data InWrEn32 Adr

DataMemory

MemWr

AL

U

Zero

0

1

0

1

01


Specify ALU source mux Control

° ALUsrc: 0 reg as ALU B input; 1 immediate as ALU B input

Clk

busW

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

Rd

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

Data In32

AL

U

0

1

01 Answer? AddU SubU ORI LW SW BEQ

0 0 0 0 0 0 01 0 0 0 0 0 12 0 0 0 0 1 13 0 0 1 1 1 04 0 0 0 1 1 15 0 0 0 1 1 X6 1 1 1 1 1 17 1 1 1 1 1 X8 X X X X X 19 None of the above


Administrivia°COD Reading for next lecture: Sections 5.5 “Multicycle”, 5.6 “Microprogramming” (on CD), “Fallacies and Pitfalls” 5.10

°Start Homework #2

°Lab 2 Verilog simulation Friday

°Find bugs in COD 3rd Edition?• $1 reward to first person to report a bug

• Send email to [email protected] Include Page number, line number on page,

BEFORE with bug, AFTER fix, why it’s a bug


Specify Immediate Extender Op Control

° ExtOp: 0 “zero extend immediate” ; 1 “sign extend imm.”

Clk

busW

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

Rd

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

Data In32

AL

U

0

1

01 Answer? AddU SubU ORI LW SW BEQ

0 0 0 0 0 0 11 0 0 0 0 1 12 0 0 0 1 1 13 0 0 1 1 1 14 0 1 1 1 1 15 1 1 1 1 1 16 X 0 1 1 1 17 X X 0 1 1 18 X X 1 0 0 09 None of the above


Specify Register Write Control


32

Clk

busW

RegWr

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

AL

U

0

1

01


0 0 0 0 0 0 0 1 0 0 0 0 0 1 2 0 0 0 0 1 1 3 0 0 0 1 1 1 4 0 0 1 1 1 1 5 0 1 1 1 1 1 6 1 1 1 1 1 1 7 1 1 1 1 0 0 8 1 1 1 1 X X 9 None of the above


Specify Register Destination Control° RegDst: 0 “rt”; 1 “rd”

Clk

busW

RegWr

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

AL

U

0

1

01


0 0 0 0 0 0 0 1 0 0 0 0 0 1 2 0 0 0 0 1 1 3 0 0 0 1 1 0 4 0 0 1 1 0 0 5 0 1 1 0 0 0 6 1 1 0 0 0 0 7 1 1 0 0 X X 8 1 1 0 0 0 X 9 None of the above

op rs rt immediate016212631


061116212631



0 0 0 0 0 0 0 1 0 0 0 0 0 1 2 0 0 0 0 1 0 3 0 0 0 1 0 0 4 0 0 1 0 0 0 5 0 1 0 0 0 0 6 1 0 0 0 0 0 7 0 0 0 0 1 X 8 X X X 0 1 X 9 None of the above

Specify the Memory Write Control Signal° MemWr: 1 write memory

busW

32

ALUctr

Clk

RegWr

32

32

busA

32busB

55 5

Rw Ra Rb32 32-bitRegisters

Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

Mu

x

MemtoReg

Clk

Data InWrEn32 Adr

DataMemory

MemWr

AL

U

Zero

0

1

0

1

01


Specify Memory To Register File Mux Control° MemtoReg:0 ALU; 1 Mem

busW

32

ALUctr

Clk

RegWr

32

32

busA

32busB

55 5


Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

Mu

x

MemtoReg

Clk

Data InWrEn32 Adr

DataMemory

MemWr

AL

U

Zero

0

1

0

1

01


0 0 0 0 0 0 0 1 0 0 0 0 0 1 2 0 0 0 0 1 0 3 0 0 0 1 0 X 4 0 0 0 1 X X 5 0 1 0 0 0 0 6 1 0 0 0 0 0 7 0 0 0 0 1 X 8 0 0 0 1 0 X 9 None of the above


Specify the ALU Control Signals° ALUctr: 0 “add”, 1 “sub”, 2 “or”


0 0 0 0 0 0 01 0 0 0 0 0 12 0 1 0 0 0 13 0 1 2 0 0 14 0 1 2 0 0 X5 0 1 2 0 X X6 0 1 2 X X X7 X 1 2 X X X8 X X 2 0 0 19 None of the above

busW

32

ALUctr

Clk

RegWr

32

32

busA

32busB

55 5


Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

3216imm16

ALUSrcExtOp

Mu

x

MemtoReg

Clk

Data InWrEn32 Adr

DataMemory

MemWr

AL

U

Zero

0

1

0

1

01


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


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]• This is the same for all instructions

PC

Ext

Adr

InstMemory

Ad

der

Ad

der

PC

Clk

00

Mu

x4

PCSrc

imm

16

Instruction<31:0>


Instruction Fetch Unit at the End of Branchif (Zero == 1) PC = PC + 4 + {SignExt[imm16], 2’b00} ;

else PC = PC + 4

op rs rt immediate

016212631

° What is encoding of PCSrc?• Direct MUX select?

• Branch / not branch

° Let’s choose second option

PCSrc zero? MUX0 X 01 0 01 1 1

Adr

InstMemory

Ad

der

Ad

der

PC

Clk

00

Mu

x

4

PCSrc

imm

16

Instruction<31:0>

0

1

Zero

PCSrc


The Single Cycle Datapath during Load

32

ALUctr <= Add

Clk

busW

RegWr <= 1

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst = 0

Exten

der

Mu

x

Mux

3216imm16

ALUSrc = 1

ExtOp <= 1

Mu

x

MemtoReg <= 1

Clk

Data InWrEn

32

Adr

DataMemory

32

MemWr = 0A

LU

InstructionFetch Unit

Clk

Zero

Instruction<31:0>

0

1

0

1

01<

21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRsRt

° R[rt] <= Data Memory [R[rs] + SignExt[imm16]]

op rs rt immediate

016212631

PCSrc<= +4


The Single Cycle Datapath during Store

° Data Memory [R[rs] + SignExt[imm16]] <= R[rt]

op rs rt immediate

016212631

32

ALUctr <=

Clk

busW

RegWr <=

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst <=

Exten

der

Mu

x

Mux

3216imm16

ALUSrc <=

ExtOp <=

Mu

x

MemtoReg <=

Clk

Data InWrEn

32

Adr

DataMemory

32

MemWr <= A

LU


Clk

Zero

Instruction<31:0>

0

1

0

1

01<

21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRsRt

PCSrc <=


The Single Cycle Datapath during Store

32

ALUctr <= Add

Clk

busW

RegWr <= 0

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst <= x

Exten

der

Mu

x

Mux

3216imm16

ALUSrc <= 1

ExtOp <= 1

Mu

x

MemtoReg <= x

Clk

Data InWrEn

32Adr

DataMemory

32

MemWr <= 1A

LU


Clk

Zero

Instruction<31:0>

0

1

0

1

01<

21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRsRt

° Data Memory [R[rs] + SignExt[imm16]] <= R[rt]

op rs rt immediate

016212631

PCSrc<= +4


The Single Cycle Datapath during Branch

32

ALUctr <=Sub

Clk

busW

RegWr <= 0

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst <= x

Exten

der

Mu

x

Mux

3216imm16

ALUSrc <= 0

ExtOp <= x

Mu

x

MemtoReg <= x

Clk

Data InWrEn

32

Adr

DataMemory

32

MemWr <= 0A

LU


Clk

Zero

Instruction<31:0>

0

1

0

1

01<

21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRsRt

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

016212631

PCSrc<= “Br”


Step 4: Given Datapath: RTL -> Control

ALUctrRegDst ALUSrcExtOp MemtoRegMemWr Zero

Instruction<31:0>

<21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRsRt

PCSrc

Adr

InstMemory

DATA PATH

Control

Op

<21:25>

Fun

RegWr


A Summary of Control Signals

inst Register Transfer

ADD R[rd] <= R[rs] + R[rt]; PC <= PC + 4

ALUsrc = RegB, ALUctr = “add”, RegDst = rd, RegWr, PCSrc = “+4”

SUB R[rd] <= R[rs] – R[rt]; PC <= PC + 4

ALUsrc = RegB, ALUctr = “sub”, RegDst = rd, RegWr, PCSrc = “+4”

ORi R[rt] <= R[rs] + zero_ext(Imm16); PC <= PC + 4

ALUsrc = Im, Extop = “Z”, ALUctr = “or”, RegDst = rt, RegWr, PCSrc = “+4”

LOAD R[rt] <= MEM[ R[rs] + sign_ext(Imm16)]; PC <= PC + 4

ALUsrc = Im, Extop = “Sn”, ALUctr = “add”, MemtoReg, RegDst = rt, RegWr, PCSrc = “+4”

STORE MEM[ R[rs] + sign_ext(Imm16)] <= R[rs]; PC <= PC + 4

ALUsrc = Im, Extop = “Sn”, ALUctr = “add”, MemWr, PCSrc = “+4”

BEQ if ( R[rs] == R[rt] ) then PC <= PC +4 + {sign_ext(Imm16)], 00’b2} else PC <= PC + 4

PCSrc = “Br”, ALUctr = “sub”


A Summary of the Control Signals

add sub ori lw sw beq

RegDst

ALUSrc

MemtoReg

RegWrite

MemWrite

PCSrcExtOp

ALUctr<2:0>

1

0

0

1

0

0x

Add

1

0

0

1

0

0x

Subtract

0

1

0

1

0

00

Or

0

1

1

1

0

01

Add

x

1

x

0

1

01

Add

x

0

x

0

0

1x

Subtract

op target address


061116212631

op rs rt immediate

R-type

I-type

J-type

add, sub

ori, lw, sw, beq

jump

func

op 00 0000 00 0000 00 1101 10 0011 10 1011 00 0100Appendix A10 0000See 10 0010 We Don’t Care :-)


How to implement control in Verilog?°Need to set control lines based on instruction

°Which statement in Verilog good for doing different operations depending on value in a field of a word?

case (selector) item {, item} : statement;item {, item} : statement;default : statement

endcase


The Single Cycle Datapath during Or Immediate

32

ALUctr <= Or

Clk

busW

RegWr <= 1

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst <= 0

Exten

der

Mu

x

Mux

3216imm16

ALUSrc <= 1

ExtOp <= 0

Mu

x

MemtoReg <= 0

Clk

Data InWrEn

32

Adr

DataMemory

32

MemWr <= 0A

LU


Clk

Zero

Instruction<31:0>

° R[rt] <= R[rs] or ZeroExt[Imm16]

0

1

0

1

01<

21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRsRt

op rs rt immediate

016212631

PCSrc <= +4


Example for OR immediate casecase (Instruction<31:25>) 13 /* ORi */ :

beginRegDst = 0; ALUSrc = 1; MemtoReg = 0; RegWrite = 1; MemWrite = 0; PCSrc = 0; ExtOp = 0; ALUctr = 2’b10;

end…default : statement

endcase


Specify all control in one assignmentcase (Instruction<31:25>) 13 /* ORi */ :

{RegDst,ALUSrc,MemtoReg, RegWrite,MemWrite,PCSrc,Jump,

ExtOp,ALUctr} = {1’b0, 1’b1, 1’b0, 1’b1,

1’b0, 1’b0, 1’b0, 2’b01}; …

default : statement

endcase


Better way than specify as 0s and 1s°Hard to read and understand

• Easy to make mistakes; Alternative?

°Associate Names with control values!

parameter RegDstRt=0, RegDstRd=1,ALUSrcBReg = 0, ALUSrcBImm = 1, RegValALU = 0, RegValMem = 1, RegWr = 1, NoRegWr = 0,MemWr = 1, NoMemWr = 0,PCSrc4 = 0, PCSrcBr = 1,ZeroExt = 0, SignExt = 1,Add = 0, Sub = 1, Or = 2;


Specify all control symbolicallycase (Instruction<31:25>)

13 /* ORi */ :

{RegDst,ALUSrc,MemtoReg, RegWrite,MemWrite,PCSrc,Jump, ExtOp,ALUctr} = {RegDstRt, ALUSrcBImm, RegValALU, RegWr, NoMemWr, PCSrc4, ZeroExt, Or};

… default : statement

endcase


Local Decoding: R-type v. Add + Sub

R-type ori lw sw beq

RegDst

ALUSrc

MemtoReg

RegWrite

MemWrite

Branch

Jump

ExtOp

ALUop<N:0>

1

0

0

1

0

0

0

x

“R-type”

0

1

0

1

0

0

0

0

Or

0

1

1

1

0

0

0

1

Add

x

1

x

0

1

0

0

1

Add

x

0

x

0

0

1

0

x

Subtract

op 00 0000 00 1101 10 0011 10 1011 00 0100

MainControl

op

6

ALUControl(Local)

func

N

6ALUop

ALUctr

3

AL

U


The Encoding of ALUop

° In this exercise, ALUop has to be 2 bits wide to represent:

• (1) “R-type” instructions

• “I-type” instructions that require the ALU to perform:- (2) Or, (3) Add, and (4) Subtract

° To implement more of MIPS ISA, ALUop has to be bigger to represent more (4 bits in book to add NOR):

• (1) “R-type” instructions

• “I-type” instructions that require the ALU to perform:- (2) Or, (3) Add, (4) Subtract, and (5) Nor (Example: nor)

MainControl

op

6

ALUControl(Local)

func

N

6ALUop

ALUctr

3

R-type ori lw sw beq

ALUop (Symbolic) “R-type” Or Add Add Subtract

ALUop<1:0> 11 10 00 00 01


Drawback of this Single Cycle Processor°Long cycle time:• Cycle time must be long enough for the load instruction:

PC’s Clock -to-Q +

Instruction Memory Access Time +

Register File Access Time +

ALU Delay (address calculation) +

Data Memory Access Time +

Register File Setup Time +

Clock Skew

°Cycle time for load is much longer than needed for all other instructions


PreviewNext Time:

°MultiCycle Data Path•CPI 1, CycleTime much shorter (~1/5 of time)


° Single cycle datapath => CPI=1, CCT => long

° 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

° Control is the hard part

° MIPS makes control easier• Instructions same size

• Source registers always in same place

• Immediates same size, location

• Operations always on registers/immediates

Summary

Control

Datapath

Memory

ProcessorInput

Output


Where to get more information?

° Chapter 5.1 to 5.4 of your text book:• David Patterson and John Hennessy, “Computer

Organization & Design: The Hardware / Software Interface,” Third Edition, Morgan Kaufman Publishers, San Mateo, California, 2003.

° One of the best PhD thesis on processor design:

• Manolis Katevenis, “Reduced Instruction Set Computer Architecture for VLSI,” PhD Dissertation, EECS, U C Berkeley, 1982.

° For a reference on the MIPS architecture:• Gerry Kane, Joe Heinrich “MIPS RISC

Architecture,” Prentice Hall, 2nd edition, 1992


Bonus Slides°The following slides show how to go from tables that describe inputs and control lines into gates

°Can be done by CAD tools instead of by hand for 2-level logic equations very efficiently


The Truth Table for ALUctr

R-type ori lw sw beqALUop(Symbolic) “R-type” Or Add Add Subtract

ALUop<2:0> 1 00 0 10 0 00 0 00 0 01

ALUop func

bit<2> bit<1> bit<0> bit<2> bit<1> bit<0>bit<3>

0 0 0 x x x x

ALUctrALUOperation

Add 0 1 0

bit<2> bit<1> bit<0>

0 x 1 x x x x Subtract 1 1 0

0 1 x x x x x Or 0 0 1

1 x x 0 0 0 0 Add 0 1 0

1 x x 0 0 1 0 Subtract 1 1 0

1 x x 0 1 0 0 And 0 0 0

1 x x 0 1 0 1 Or 0 0 1

1 x x 1 0 1 0 Set on < 1 1 1

funct<3:0> Instruction Op.

0000

0010

0100

0101

1010

add

subtract

and

or

set-on-less-than


The Logic Equation for ALUctr<2>ALUop func

bit<2> bit<1> bit<0> bit<2> bit<1> bit<0>bit<3> ALUctr<2>

0 x 1 x x x x 1

1 x x 0 0 1 0 1

1 x x 1 0 1 0 1

° (ALUctr<2> = ! ALUop<2>) & ALUop<0> + ALUop<2> & !func<2> & func<1> & !

func<0>

This makes func<3> a don’t care


The Logic Equation for ALUctr<1>

ALUop func

bit<2> bit<1> bit<0> bit<2> bit<1> bit<0>bit<3>

0 0 0 x x x x 1

ALUctr<1>

0 x 1 x x x x 1

1 x x 0 0 0 0 1

1 x x 0 0 1 0 1

1 x x 1 0 1 0 1

° (ALUctr<1> = !ALUop<2>) & !ALUop<1> + ALUop<2> & !func<2> & !func<0>


The Logic Equation for ALUctr<0>

ALUop func

bit<2> bit<1> bit<0> bit<2> bit<1> bit<0>bit<3> ALUctr<0>

0 1 x x x x x 1

1 x x 0 1 0 1 1

1 x x 1 0 1 0 1

° ALUctr<0> = !ALUop<2> & ALUop<1> + ALUop<2> & !func<3> & func<2>

& !func<1> & func<0>

+ ALUop<2> & func<3> & !func<2> & func<1> & !func<0>


The ALU Control Block

ALUControl(Local)

func

3

6ALUop

ALUctr

3

° ALUctr<2> = !ALUop<2> & ALUop<0> + ALUop<2> & !func<2> & func<1> & !func<0>

° ALUctr<1> = !ALUop<2> & !ALUop<1> + ALUop<2> & !func<2> & !func<0>

° ALUctr<0> = !ALUop<2> & ALUop<1> + ALUop<2> & !func<3> & func<2> & !func<1> &

func<0>

+ ALUop<2> & func<3> & !func<2> & func<1> & !func<0>


Step 5: Logic for each control signal

° PCSrc <= (OP == `BEQ) ? `Br : `plus4;

° ALUsrc <= (OP == `Rtype) ? `regB : ìmmed;

° ALUctr <= (OP == `Rtype`) ? funct :(OP == ÒRi) ? ÒRfunction :(OP == `BEQ) ? `SUBfunction :

ÀDDfunction;

° ExtOp <= _____________

° MemWr <= _____________

° MemtoReg <= _____________

° RegWr: <=_____________

° RegDst: <= _____________


Step 5: Logic for each control signal

° PCSrc <= (OP == `BEQ) ? `Br : `plus4;

° ALUsrc <= (OP == `Rtype) ? `regB : ìmmed;

° ALUctr <= (OP == `Rtype`) ? funct :(OP == ÒRi) ? ÒRfunction :

(OP == `BEQ) ? `SUBfunction : ÀDDfunction;

° ExtOp <= (OP == ÒRi) : `ZEROextend : `SIGNextend;

° MemWr <= (OP == `Store) ? 1 : 0;

° MemtoReg <= (OP == `Load) ? 1 : 0;

° RegWr: <= ((OP == `Store) || (OP == `BEQ)) ? 0 : 1;

° RegDst: <= ((OP == `Load) || (OP == ÒRi)) ? 0 : 1;


The “Truth Table” for the Main Control

R-type ori lw sw beq jump

RegDst

ALUSrc

MemtoReg

RegWrite

MemWrite

PCSrc

Jump

ExtOp

ALUop (Symbolic)

1

0

0

1

0

0

0

x

“R-type”

0

1

0

1

0

0

0

0

Or

0

1

1

1

0

0

0

1

Add

x

1

x

0

1

0

0

1

Add

x

0

x

0

0

1

0

x

Subtract

x

x

x

0

0

0

1

x

xxx

op 00 0000 00 1101 10 0011 10 1011 00 0100 00 0010

ALUop <2> 1 0 0 0 0 x

ALUop <1> 0 1 0 0 0 x

ALUop <0> 0 0 0 0 1 x

MainControl

op

6

ALUControl(Local)

func

3

6

ALUop

ALUctr

3

RegDst

ALUSrc

:


The “Truth Table” for RegWrite


RegWrite 1 1 1 0 0 0

op 00 0000 00 1101 10 0011 10 1011 00 0100 00 0010

° RegWrite = R-type + ori + lw= !op<5> & !op<4> & !op<3> & !op<2> & !op<1> & !op<0> (R-type)

+ !op<5> & !op<4> & op<3> & op<2> & !op<1> & op<0> (ori)

+ op<5> & !op<4> & !op<3> & !op<2> & op<1> & op<0> (lw)

op<0>

op<5>. .op<5>. .<0>

op<5>. .<0>

op<5>. .<0>

op<5>. .<0>

op<5>. .<0>


RegWrite


PLA Implementation of the Main Control

op<0>

op<5>. .op<5>. .<0>

op<5>. .<0>

op<5>. .<0>

op<5>. .<0>

op<5>. .<0>

R-type ori lw sw beq jumpRegWrite

ALUSrc

MemtoReg

MemWrite

Branch

Jump

RegDst

ExtOp

ALUop<2>

ALUop<1>

ALUop<0>


A Real MIPS Datapath (CNS T0)


Putting it All Together: A Single Cycle Processor

32

ALUctr

Clk

busW

RegWr

32

32

busA

32

busB

55 5

Rw Ra Rb

32 32-bitRegisters

Rs

Rt

Rt

RdRegDst

Exten

der

Mu

x

Mux

3216imm16

ALUSrc

ExtOp

Mu

x

MemtoReg

Clk

Data InWrEn

32

Adr

DataMemory

32

MemWrA

LU


Clk

Zero

Instruction<31:0>

0

1

0

1

01<

21:25>

<16:20>

<11:15>

<0:15>

Imm16RdRsRt

MainControl

op

6

ALUControlfunc

6

3

ALUopALUctr

3RegDst

ALUSrc

:Instr<5:0>

Instr<31:26>

Instr<15:0>

PCSrc

Documents

CS 152 L07 Single Cycle 2 (1)UC Regents Fall 2004 © UCB 2004-09-21 John Lazzaro (lazzaro) Dave Patterson (patterson)