eem 486: computer architecture lecture 3 designing a single cycle datapath

EEM 486: Computer Architecture

Lecture 3

Designing a Single Cycle Datapath

Lec 3.2

The Big Picture: Where are We Now?

The Five Classic Components of a Computer

Today’s Topic: Design a Single Cycle Processor

Control

Datapath

Memory

ProcessorInput

Output

Lec 3.3

The Big Picture: The Performance Perspective

Performance of a machine is determined by:

• Instruction count

• Clock cycle time

• Clock cycles per instruction

Processor design (datapath and control) will determine:

• Clock cycle time

• Clock cycles per instruction

CPI

Inst. Count Cycle Time

Lec 3.4

Single-cycle datapath

All instructions execute in a single cycle of the clock (positive edge to positive edge)

Advantage: a great way to learn CPU

Unrealistic hardware assumptions, slow clock period

Lec 3.5

Single cycle data paths: Assumptions

Processor uses synchronous logicdesign (a “clock”)

f T

1 MHz 1 μs

10 MHz 100 ns

100 MHz 10 ns

1 GHz 1 ns

• All state elements act like positive edge-triggered flip flops

• Clocks arrive at all flip flops simultaneously.

D Q

clk

Lec 3.6

Review: Edge-Triggered D Flip Flops

D Q Value of D is sampled on positive clock edge

Q outputs sampled value for rest of cycle.

CLK

D

Q

Lec 3.7

How to Design a Processor: Step-by-Step

1. Analyze instruction set => datapath requirements• Meaning of each instruction is given by the register transfers

• Datapath must include storage element for ISA registers

- possibly more

• Datapath must support each register transfer

2. Select set of datapath components and establish clocking 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

Lec 3.8

The MIPS Instruction Formats

The three instruction formats:

• R-type

• I-type

• J-type

The different fields are:• op: operation of the instruction• rs, rt, rd: the source and destination register specifiers• shamt: shift amount• funct: selects the variant of the operation in the “op” field• address / immediate: address offset or immediate value• target address: target address of the jump instruction

op target address

02631

6 bits 26 bits

op rs rt rd shamt funct

061116212631

6 bits 6 bits5 bits5 bits5 bits5 bits

op rs rt immediate016212631

6 bits 16 bits5 bits5 bits

Lec 3.9

Step 1a: The MIPS-lite Subset for Today

ADD/SUB

• addU rd, rs, rt

• subU rd, rs, rt

OR Immediate:

• ori rt, rs, imm16

LOAD/STORE Word

• lw rt, rs, imm16

• sw rt, rs, imm16

BRANCH

• beq rs, rt, imm16


061116212631


op rs rt immediate

016212631


op rs rt immediate

016212631


op rs rt immediate

016212631


Lec 3.10

Step 1a: Executing MIPS Instructions

- Fetch next inst from memory - Get ready for the next instruction

opcode rs rt rd functshamt

- Decode fields to get a particular instruction

- Retrieve register values (rs, rt)

- Perform the operation (add, sub, or, lw, sw, beq)

- Place the result in a register (rt/rd) / memory, ormodify PC

InstructionFetch

InstructionDecode

OperandFetch

Execute

ResultStore

Lec 3.11

Step 1a: Logical Register Transfers RTL gives the meaning of the instructions

All start by fetching the instruction

op | rs | rt | rd | shamt | funct = MEM[ PC ]

op | rs | rt | Imm16 = MEM[ PC ]

inst Register Transfers

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

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

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

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

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

BEQ if ( R[rs] == R[rt] ) then PC <– PC + 4 + [sign_ext(Imm16) || 00] else PC <– PC + 4

Lec 3.12

Step 1: Requirements of the Instruction Set Memory

• instruction & data

Registers (32 x 32)

• read RS

• read RT

• Write RT or RD

PC

Extender

Add and Sub registers or register and extended immediate

Logical Or of a register and extended immediate

Add 4 or extended immediate to PC

Lec 3.13

Step 2: Components of the Datapath

Combinational Elements

Storage Elements

• Clocking methodology

Lec 3.14

Combinational Logic Elements (Basic Building Blocks)

Adder

MUX

ALU

32

32

A

B

32Sum

Carry

32

32

A

B

32Result

OP

32A

B32

Y32

Select

Adder

MU

XA

LU

CarryIn

Lec 3.15

Storage Element: Register (Basic Building Block)

Register• Similar to the D Flip Flop except

- N-bit input and output

- Write Enable input

• Write Enable:

- Negated (0): Data Out will not change

- Asserted (1): Data Out becomes Data InClk

Data In

Write Enable

N N

Data Out

Lec 3.16

Storage Element: Register File Register File consists of 32 registers:

• Two 32-bit output busses:

busA and busB

• One 32-bit input bus: busW

Register is selected by:• RA (number) selects the register to put on busA (data)

• RB (number) selects the register to put on busB (data)

• RW (number) selects the register to be writtenvia busW (data) when Write Enable is 1

Clock input (CLK) • The CLK input is a factor ONLY during write operation

• During read operation, behaves as a combinational logic block:

- RA or RB valid => busA or busB valid after “access time.”

Clk

busW

Write Enable

3232

busA

32busB

5 5 5RW RA RB

32 32-bitRegisters

Lec 3.17

Storage Element: Idealized Memory

Memory (idealized)• One input bus: Data In

• One output bus: Data Out

Memory word is selected by:• Address selects the word to put on Data Out

• Write Enable = 1: address selects the memoryword to be written via the Data In bus

Clock input (CLK) • The CLK input is a factor ONLY during write operation

• During read operation, behaves as a combinational logic block:

- Address valid => Data Out valid after “access time.”

Clk

Data In

Write Enable

32 32DataOut

Address

Lec 3.18

Clocking Methodology

All storage elements are clocked by the same clock 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, and we have the usual clock-to-Q delay

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

.

.

.

.

.

.

.

.

.

.

.

.

Clk

Don’t Care

Setup HoldSetup Hold

Lec 3.19

Step 3: Assemble Datapath Meeting Requirements

Register Transfer Requirements Datapath Assembly

Instruction Fetch

Read Operands and Execute Operation

Lec 3.20

3a: Overview of the Instruction Fetch Unit The common RTL operations

• Fetch the Instruction: mem[PC]

- PC == Program Counter, points to next instruction• Update the program counter:

- Sequential Code: PC <- PC + 4

- Branch and Jump: PC <- “something else”

32

Instruction WordAddress

InstructionMemory

PCClk

Next AddressLogic

Lec 3.21

Straight-line Instruction Fetch

32

Addr

Data

32

InstrMem

Why +4 and not +1?

CLK

Address

Data IMem[PC + 8]

IMem[PC + 4]IMem[PC]

PC + 8PC + 4PC

op target address02631

6 bits 26 bits

op rs rt rd shamt funct061116212631


op rs rt immediate016212631


Lec 3.22

3b: Add & Subtract R[rd] <- R[rs] op R[rt] Example: addU rd, rs, rt

• Ra, Rb, and Rw come from instruction’s rs, rt, and rd fields

• ALUctr and RegWr: control logic produces after decoding the instruction

32

Result

ALUctr

Clk

busW

RegWr

32

32

busA

32

busB

5 5 5

Rw Ra Rb

32 32-bitRegisters

Rs RtRd

ALU


061116212631


Lec 3.23

How data flows after posedge

32

RegFile

32

WE32

5Ra

5Rb

5Rw

32ALU

32

32

op

Control Logic

Addr Data

InstrMem

D

PC

Q

Adder

4

Lec 3.24

Register-Register Timing: One complete cycle

32

Result

ALUctr

Clk

busW

RegWr

3232

busA

32busB

5 5 5

Rw Ra Rb

32 32-bitRegisters

Rs RtRd

ALU

Clk

PC

Rs, Rt, Rd,Op, Func

Clk-to-Q

ALUctr

Instruction Memory Access Time

Old Value New Value

RegWr Old Value New Value

Delay through Control Logic

busA, B

Register File Access Time

Old Value New Value

busW

ALU Delay

Old Value New Value

Old Value New Value

New ValueOld Value

Register WriteOccurs Here

Lec 3.25

3c: Logical Operations with Immediate R[rt] <- R[rs] op ZeroExt[imm16]

op rs rt immediate

016212631


32

Result

ALUctr

Clk

busW

RegWr

32

32

busA

32

busB

5 5 5

Rw Ra Rb

32 32-bitRegisters

Rs

Zero

Ext

Mu

x

RtRdRegDst

Mux

3216imm16

ALUSrc

ALU

Rt?

Lec 3.26

3d: Load Operations R[rt] <- Mem[R[rs] + SignExt[imm16]] E.g.: lw rt, rs,

imm16

op rs rt immediate

016212631


32

ALUctr

Clk

busW

RegWr

3232

busA

32busB

5 5 5

Rw Ra Rb32 32-bitRegisters

Rs

RtRdRegDst

Exte

nd

er

Mu

x

Mux

3216

imm16

ALUSrc

ExtOp

Clk

Data InWrEn

32

Adr

DataMemory

32

ALU

MemWr Mu

x

W_Src

??

Rt?

Lec 3.27

3e: Store Operations

Mem[ R[rs] + SignExt[imm16] ] <- R[rt] E.g.: sw rt, rs, imm16

op rs rt immediate

016212631


32

ALUctr

Clk

busW

RegWr

3232

busA

32busB

55 5


Rs

Rt

Rt

RdRegDst

Exte

nd

er

Mu

x

Mux

3216imm16

ALUSrcExtOp

Clk

Data InWrEn

32Adr

DataMemory

MemWr

ALU

32

Mu

x

W_Src

Lec 3.28

3f: The Branch Instruction

beq rs, rt, imm16

• mem[PC] Fetch the instruction from memory

• Equal <- R[rs] == R[rt] Calculate the branch condition

• if (Equal) Calculate the next instruction’s address

- PC <- PC + 4 + ( SignExt(imm16) x 4 )

• else

- PC <- PC + 4

op rs rt immediate

016212631


Lec 3.29

Datapath for Branch Operations

beq rs, rt, imm16 Datapath generates condition (equal)

op rs rt immediate

016212631


32

imm16P

C

Clk

00

Ad

der

Mu

x

Ad

der

4nPC_sel

Clk

busW

RegWr

32

busA

32busB

5 5 5


Rs Rt

Eq

ual?

Cond

PC

Ext

Inst Address

Lec 3.30

Putting it All Together: A Single Cycle Datapath

imm

16

32

ALUctr

Clk

busW

RegWr

32

32

busA

32busB

55 5


Rs

Rt

Rt

RdRegDst

Exte

nd

er

Mu

x

3216imm16

ALUSrcExtOp

Mu

x

MemtoReg

Clk

Data InWrEn32 Adr

DataMemory

MemWrA

LU

Equal

Instruction<31:0>

0

1

0

1

01

<2

1:2

5>

<1

6:2

0>

<1

1:1

5>

<0

:15

>

Imm16RdRtRs

=

Ad

der

Ad

der

PC

Clk

00

Mu

x

4

nPC_sel

PC

Ext

Adr

InstMemory

Lec 3.31

A Single Cycle Datapath

PC

Instructionmemory

Readaddress

Instruction[31– 0]

Instruction [20– 16]


Add


MemtoReg

ALUOp

MemWrite

RegWrite

MemRead

BranchRegDst

ALUSrc


4

16 32Instruction [15– 0]

0

0Mux

0

1

Control

Add ALUresult

Mux

0

1

RegistersWriteregister

Writedata

Readdata 1

Readdata 2

Readregister 1

Readregister 2

Signextend

Shiftleft 2

Mux

1

ALUresult

Zero

Datamemory

Writedata

Readdata

Mux

1


ALUcontrol

ALUAddress

Lec 3.32

An Abstract View of the Critical Path Register file and ideal memory:

• The CLK input is a factor ONLY during write operation

• During read operation, behave as combinational logic:

- Address valid => Output valid after “access time.”

Critical Path (Load Operation) = PC’s Clk-to-Q + Instruction Memory’s Access Time + Register File’s Access Time + ALU to Perform a 32-bit Add + Data Memory Access Time + Setup Time for Register File Write + Clock Skew

Clk

5

Rw Ra Rb

32 32-bitRegisters

RdA

LU

Clk

Data In

DataAddress

IdealData

Memory

Instruction

InstructionAddress

IdealInstruction

Memory

Clk

PC

5Rs

5Rt

16Imm

32

323232

A

B

Next

Ad

dre

ss

Lec 3.33

An Abstract View of the Implementation

DataOut

Clk

5

Rw Ra Rb

32 32-bitRegisters

Rd

ALU

Clk

Data In

DataAddress

IdealData

Memory

Instruction

InstructionAddress

IdealInstruction

Memory

Clk

PC

5Rs

5Rt

32

323232

A

B

Next

Ad

dre

ss

Control

Datapath

Control SignalsConditions

Lec 3.34

Steps 4 & 5: Implement the control

Next time

Lec 3.35

Summary 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

MIPS makes it easier

• Instructions same size

• Source registers always in same place

• Immediates same size, location

• Operations always on registers/immediates

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

Next time: implementing control

eem 486: computer architecture lecture 3 designing a single cycle datapath

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