cs61c l17 single cycle cpu datapath (1) garcia, fall 2005 © ucb lecturer psoe, new dad dan garcia...
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CS61C L17 Single Cycle CPU Datapath (1) Garcia, Fall 2005 © UCB
Lecturer PSOE, new dad Dan Garcia
www.cs.berkeley.edu/~ddgarcia
inst.eecs.berkeley.edu/~cs61c CS61C : Machine Structures
Lecture #17 Single Cycle CPU Datapath
2005-10-31There is one handout today at the front and
back of the room!
Halloween plans? Try the Castro, SF!
halloweeninthecastro.com
CPStoday!
Today 2005-10-31,from 7pm-mid($5 donation) go at least once…
Happy Halloween, everyone!
CS61C L17 Single Cycle CPU Datapath (2) Garcia, Fall 2005 © UCB
Wed’s talk : Jim Larus, µsoft (Cal Ph.D.)
“An Overview of the Singularity Project”• 306 Soda Hall, Wed 2005-11-02 @ 4-5pm
• Dr. Larus is the author of SPIM!• ftp://ftp.research.microsoft.com/pub/tr/TR-2005-135.pdf
“Singularity is a research project in Microsoft Research that started with the question: what would a software platform look like if it was designed from
scratch with the primary goal of dependability? Singularity is working to answer this question by building on advances in programming languages
and tools to develop a new system architecture and operating system (named Singularity), with the aim
of producing a more robust and dependable software platform. Singularity demonstrates the
practicality of new technologies and architectural decisions, which should lead to the construction of
more robust and dependable systems.”
CS61C L17 Single Cycle CPU Datapath (3) Garcia, Fall 2005 © UCB
Review
• Use muxes to select among input• S input bits selects 2S inputs• Each input can be n-bits wide, indep of S
• Implement muxes hierarchically
• ALU can be implemented using a mux• Coupled with basic block elements
• N-bit adder-subtractor done using N 1-bit adders with XOR gates on input
• XOR serves as conditional inverter
• Programmable Logic Arrays are often used to implement our CL
CS61C L17 Single Cycle CPU Datapath (4) Garcia, Fall 2005 © UCB
Anatomy: 5 components of any Computer
Personal Computer
Processor
Computer
Control(“brain”)
Datapath(“brawn”)
Memory
(where programs, data live whenrunning)
Devices
Input
Output
Keyboard, Mouse
Display, Printer
Disk (where programs, data live whennot running)
This weekand next
CS61C L17 Single Cycle CPU Datapath (5) Garcia, Fall 2005 © UCB
Outline of Today’s Lecture
• Design a processor: step-by-step
• Requirements of the Instruction Set
• Hardware components that match the instruction set requirements
CS61C L17 Single Cycle CPU Datapath (6) Garcia, Fall 2005 © UCB
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 requirements• 4. Analyze implementation of each instruction to determine setting of control points that effects the register transfer.
• 5. Assemble the control logic
CS61C L17 Single Cycle CPU Datapath (7) Garcia, Fall 2005 © UCB
Review: The MIPS Instruction Formats• All MIPS instructions are 32 bits long. 3 formats:
• R-type
• I-type
• J-type
• The different fields are:• op: operation (“opcode”) 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 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 address/immediate
016212631
6 bits 16 bits5 bits5 bits
CS61C L17 Single Cycle CPU Datapath (8) Garcia, Fall 2005 © UCB
Step 1a: The MIPS-lite Subset for today
• ADDU and SUBU•addu rd,rs,rt•subu rd,rs,rt
• OR Immediate:•ori rt,rs,imm16
• LOAD and STORE Word•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
op rs rt immediate
016212631
6 bits 16 bits5 bits5 bits
op rs rt immediate
016212631
6 bits 16 bits5 bits5 bits
CS61C L17 Single Cycle CPU Datapath (9) Garcia, Fall 2005 © UCB
Register Transfer Language• 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
CS61C L17 Single Cycle CPU Datapath (10) Garcia, Fall 2005 © UCB
Step 1: Requirements of the Instruction Set• Memory (MEM)
• instructions & data• Registers (R: 32 x 32)
• read RS• read RT• Write RT or RD
• PC• Extender (sign extend)• Add and Sub register or extended immediate
• Add 4 or extended immediate to PC
CS61C L17 Single Cycle CPU Datapath (11) Garcia, Fall 2005 © UCB
Step 2: Components of the Datapath•Combinational Elements
•Storage Elements• Clocking methodology
CS61C L17 Single Cycle CPU Datapath (12) Garcia, Fall 2005 © UCB
Combinational Logic Elements (Building Blocks)
•Adder
•MUX
•ALU
32
32
A
B32
Sum
CarryOut
32
32
A
B32
Result
OP
32A
B32
Y32
Select
Ad
der
MU
XA
LU
CarryIn
CS61C L17 Single Cycle CPU Datapath (13) Garcia, Fall 2005 © UCB
ALU Needs for MIPS-lite + Rest of MIPS• Addition, subtraction, logical OR, ==:
ADDU R[rd] = R[rs] + R[rt]; ...
SUBU R[rd] = R[rs] – R[rt]; ...
ORI R[rt] = R[rs] | zero_ext(Imm16)...
BEQ if ( R[rs] == R[rt] )...
• Test to see if output == 0 for any ALU operation gives == test. How?
• P&H also adds AND, Set Less Than (1 if A < B, 0 otherwise)
• ALU follows chap 5
CS61C L17 Single Cycle CPU Datapath (14) Garcia, Fall 2005 © UCB
Administrivia
• Project 2 graded• You have a week (2005-11-07) to regrade
• My wed OH this week moved to Fri @ 2p
• Final Exam location TBA (exam grp 14)•Sat, 2005-12-17, 12:30–3:30pm
•ALL students are required to complete ALL of the exam (even if you aced the midterm)
•Same format as the midterm- 3 Hours
- Closed book, except for 2 study sheets + green
- Leave your backpacks, books at home
CS61C L17 Single Cycle CPU Datapath (15) Garcia, Fall 2005 © UCB
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 memory
word 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
CS61C L17 Single Cycle CPU Datapath (16) Garcia, Fall 2005 © UCB
Clk
Data In
Write Enable
N N
Data Out
Storage Element: Register (Building Block)
• Similar to D Flip Flop except- N-bit input and output
- Write Enable input
• Write Enable:- negated (or deasserted) (0):
Data Out will not change
- asserted (1): Data Out will become Data In
CS61C L17 Single Cycle CPU Datapath (17) Garcia, Fall 2005 © UCB
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 written
via 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 5RWRA RB
32 32-bitRegisters
CS61C L17 Single Cycle CPU Datapath (18) Garcia, Fall 2005 © UCB
Step 3: Assemble DataPath meeting requirements
• Register Transfer Requirements Datapath Assembly
• Instruction Fetch
• Read Operands and Execute Operation
CS61C L17 Single Cycle CPU Datapath (19) Garcia, Fall 2005 © UCB
3a: Overview of the Instruction Fetch Unit
• The common RTL operations• Fetch the Instruction: mem[PC]• Update the program counter:
- Sequential Code: PC = PC + 4 - Branch and Jump: PC = “something else”
32
Instruction WordAddress
InstructionMemory
PCClk
Next AddressLogic
CS61C L17 Single Cycle CPU Datapath (20) Garcia, Fall 2005 © UCB
3b: Add & Subtract• R[rd] = R[rs] op R[rt] Ex.: addU rd,rs,rt
• Ra, Rb, and Rw come from instruction’s Rs, Rt, and Rd fields
• ALUctr and RegWr: control logic after decoding the instruction
32Result
ALUctr
Clk
busW
RegWr
3232
busA
32busB
5 5 5
Rw Ra Rb
32 32-bitRegisters
Rs RtRd
AL
U
op rs rt rd shamt funct061116212631
6 bits 6 bits5 bits5 bits5 bits5 bits
• Already defined register file, ALU
CS61C L17 Single Cycle CPU Datapath (21) Garcia, Fall 2005 © UCB
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, and we have the usual clock-to-Q delay
• “Critical path” (longest path through logic) determines length of clock period
Clk
.
.
.
.
.
.
.
.
.
.
.
.
CS61C L17 Single Cycle CPU Datapath (22) Garcia, Fall 2005 © UCB
Register-Register Timing: One complete cycle
32Result
ALUctr
Clk
busW
RegWr
3232
busA
32busB
5 5 5
Rw Ra Rb
32 32-bitRegisters
Rs RtRd
AL
U
Clk
PCRs, Rt, Rd,Op, Func
ALUctr
Instruction Memory Access Time
Old Value New Value
RegWr Old Value New Value
Delay through Control Logic
busA, BRegister File Access TimeOld Value New Value
busWALU Delay
Old Value New Value
Old Value New Value
New ValueOld Value
Register WriteOccurs Here
CS61C L17 Single Cycle CPU Datapath (23) Garcia, Fall 2005 © UCB
3c: Logical Operations with Immediate• R[rt] = R[rs] op ZeroExt[imm16] ]
32
Result
ALUctr
Clk
busW
RegWr
3232
busA
32busB
5 5 5
Rw Ra Rb32 32-bitRegisters
Rs
ZeroE
xt
Mu
x
RtRdRegDst
Mux
3216imm16
ALUSrc
AL
U
11
op rs rt immediate
016212631
6 bits 16 bits5 bits5 bits rd?
immediate
016 1531
16 bits16 bits
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Rt?
• Already defined 32-bit MUX; Zero Ext?
What about Rt register read??
CS61C L17 Single Cycle CPU Datapath (24) Garcia, Fall 2005 © UCB
3d: Load Operations• R[rt] = Mem[R[rs] + SignExt[imm16]]Example: lw rt,rs,imm16
op rs rt immediate
016212631
6 bits 16 bits5 bits5 bits
32
ALUctr
Clk
busW
RegWr
32
32
busA
32
busB
5 5 5
Rw Ra Rb
32 32-bitRegisters
Rs
RtRd
RegDst
Exten
der
Mu
x
Mux
3216
imm16
ALUSrc
ExtOp
Clk
Data InWrEn
32
Adr
DataMemory
32
AL
U
MemWr Mu
x
W_Src
??
Rt
CS61C L17 Single Cycle CPU Datapath (25) Garcia, Fall 2005 © UCB
3e: Store Operations• Mem[ R[rs] + SignExt[imm16] ] = R[rt]
Ex.: sw rt, rs, imm16
op rs rt immediate016212631
6 bits 16 bits5 bits5 bits
32
ALUctr
Clk
busW
RegWr
3232
busA
32busB
55 5
Rw Ra Rb32 32-bitRegisters
Rs
Rt
Rt
RdRegDst
Exten
der
Mu
x
Mux
3216imm16
ALUSrcExtOp
Clk
Data InWrEn
32Adr
DataMemory
MemWr
AL
U
32
Mu
x
W_Src
CS61C L17 Single Cycle CPU Datapath (26) Garcia, Fall 2005 © UCB
3f: The Branch Instruction
•beq rs, rt, imm16• mem[PC] Fetch the instruction from memory
• Equal = R[rs] == R[rt] Calculate branch condition
• if (Equal) Calculate the next instruction’s address- PC = PC + 4 + ( SignExt(imm16) x 4 )
else- PC = PC + 4
op rs rt immediate016212631
6 bits 16 bits5 bits5 bits
CS61C L17 Single Cycle CPU Datapath (27) Garcia, Fall 2005 © UCB
Datapath for Branch Operations• beq rs, rt, imm16
Datapath generates condition (equal)
op rs rt immediate016212631
6 bits 16 bits5 bits5 bits
32
imm16
PC
Clk
00
Ad
der
Mu
x
Ad
der
4nPC_sel
Clk
busW
RegWr
32
busA
32busB
5 5 5
Rw Ra Rb32 32-bitRegisters
Rs Rt
Eq
ual
?
Cond
PC
Ext
Inst Address
• Already MUX, adder, sign extend, zero
CS61C L17 Single Cycle CPU Datapath (28) Garcia, Fall 2005 © UCB
Putting it All Together:A Single Cycle Datapath
imm
16
32
ALUctr
Clk
busW
RegWr
3232
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
MemWrA
LU
Equal
Instruction<31:0>
0
1
0
1
01
<21:25>
<16:20>
<11:15>
<0:15>
Imm16RdRtRs
=
Ad
der
Ad
der
PC
Clk
00Mu
x
4
nPC_sel
PC
Ext
Adr
InstMemory
CS61C L17 Single Cycle CPU Datapath (29) Garcia, Fall 2005 © UCB
An Abstract View of the Implementation
DataOut
Clk
5
Rw Ra Rb32 32-bitRegisters
Rd
AL
U
Clk
Data In
DataAddress Ideal
DataMemory
Instruction
InstructionAddress
IdealInstruction
Memory
Clk
PC
5Rs
5Rt
32
323232
A
B
Nex
t A
dd
ress
Control
Datapath
Control Signals Conditions
CS61C L17 Single Cycle CPU Datapath (30) Garcia, Fall 2005 © UCB
Peer Instruction
A. If the destination reg is the same as the source reg, we could compute the incorrect value!
B. We’re going to be able to read 2 registers and write a 3rd in 1 cycle
C. Datapath is hard, Control is easy
ABC1: FFF2: FFT3: FTF4: FTT5: TFF6: TFT7: TTF8: TTT
CS61C L17 Single Cycle CPU Datapath (31) Garcia, Fall 2005 © UCB
Peer Instruction
A. Our ALU is a synchronous device
B. We should use the main ALU to compute PC=PC+4
C. The ALU is inactive for memory reads or writes.
ABC1: FFF2: FFT3: FTF4: FTT5: TFF6: TFT7: TTF8: TTT
CS61C L17 Single Cycle CPU Datapath (32) Garcia, Fall 2005 © UCB
Peer Instruction
A. SW can peek at HW (past ISA abstraction boundary) for optimizations
B. SW can depend on particular HW implementation of ISA
C. Timing diagrams serve as a critical debugging tool in the EE toolkit
ABC1: FFF2: FFT3: FTF4: FTT5: TFF6: TFT7: TTF8: TTT
CS61C L17 Single Cycle CPU Datapath (33) Garcia, Fall 2005 © UCB
Peer Instruction
A. (a+b)• (a+b) = b
B. N-input gates can be thought of cascaded 2-input gates. I.e., (a ∆ b ∆ c ∆ d) = a ∆ (b ∆ (c ∆ d))where ∆ is one of AND, OR, XOR, NAND
C. You can use NOR(s) with clever wiring to simulate AND, OR, & NOT
ABC1: FFF2: FFT3: FTF4: FTT5: TFF6: TFT7: TTF8: TTT
CS61C L17 Single Cycle CPU Datapath (34) Garcia, Fall 2005 © UCB
A. (next slide)
B. (next slide)
C. You can use NOR(s) with clever wiring to simulate AND, OR, & NOT.
° NOR(a,a)= a+a = aa = a° Using this NOT, can we make a NOR an OR? An And?° TRUE
Peer Instruction Answer
A. (a+b)• (a+b) = b
B. N-input gates can be thought of cascaded 2-input gates. I.e., (a ∆ b ∆ c ∆ d) = a ∆ (b ∆ (c ∆ d))where ∆ is one of AND, OR, XOR, NAND
C. You can use NOR(s) with clever wiring to simulate AND, OR, & NOT
ABC1: FFF2: FFT3: FTF4: FTT5: TFF6: TFT7: TTF8: TTT
CS61C L17 Single Cycle CPU Datapath (35) Garcia, Fall 2005 © UCB
A. (a+b) • (a+b) =?= b
(a+b)•(a+b)
aa+ab+ba+bb distribution
0+b(a+a)+b complimentarity,commutativity,distribution,idempotent
b(1)+b identity, complimentarity
b+b identity
b idempotent
Peer Instruction Answer (A)
TRUE
CS61C L17 Single Cycle CPU Datapath (36) Garcia, Fall 2005 © UCB
A.
B. N-input gates can be thought of cascaded 2-input gates. I.e., (a ∆ b ∆ c ∆ d) = a ∆ (b ∆ (c ∆ d))where ∆ is one of AND, OR, XOR, NAND…FALSE
Let’s confirm!
CORRECT 3-inputXYZ|AND|OR|XOR|NAND000| 0 |0 | 0 | 1 001| 0 |1 | 1 | 1 010| 0 |1 | 1 | 1 011| 0 |1 | 0 | 1 100| 0 |1 | 1 | 1 101| 0 |1 | 0 | 1 110| 0 |1 | 0 | 1 111| 1 |1 | 1 | 0
CORRECT 2-inputYZ|AND|OR|XOR|NAND00| 0 |0 | 0 | 1 01| 0 |1 | 1 | 1 10| 0 |1 | 1 | 1 11| 1 |1 | 0 | 0
0 0 0 1 0 1 1 1 0 1 1 1 0 1 0 1 0 1 1 0 0 1 0 0 0 1 0 0 1 1 1 1
Peer Instruction Answer (B)
CS61C L17 Single Cycle CPU Datapath (37) Garcia, Fall 2005 © UCB
Peer Instruction
A. Truth table for mux with 4-bits of signals has 24 rows
B. We could cascade N 1-bit shifters to make 1 N-bit shifter for sll, srl
C. If 1-bit adder delay is T, the N-bit adder delay would also be T
ABC1: FFF2: FFT 3: FTF4: FTT5: TFF6: TFT7: TTF8: TTT
CS61C L17 Single Cycle CPU Datapath (38) Garcia, Fall 2005 © UCB
Peer Instruction Answer
A. Truth table for mux with 4-bits of signals is 24 rows long
B. We could cascade N 1-bit shifters to make 1 N-bit shifter for sll, srl
C. If 1-bit adder delay is T, the N-bit adder delay would also be T
ABC1: FFF2: FFT 3: FTF4: FTT5: TFF6: TFT7: TTF8: TTT
A. Truth table for mux with 4-bits of signals controls 16 inputs, for a total of 20 inputs, so truth table is 220 rows…FALSE
B. We could cascade N 1-bit shifters to make 1 N-bit shifter for sll, srl … TRUE
C. What about the cascading carry? FALSE
CS61C L17 Single Cycle CPU Datapath (39) Garcia, Fall 2005 © UCB
°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°Next time!
Summary: Single cycle datapath
Control
Datapath
Memory
ProcessorInput
Output