1. CS 61C L16 Datapath (1) A Carle, Summer 2004 © UCB inst.eecs.berkeley.edu/~cs61c/su05 CS61C : Machine Structures Lecture #16 – Datapath 2005-07-18 Andy Carle

2. CS 61C L16 Datapath (2) A Carle, Summer 2004 © UCB Anatomy: 5 components of any Computer Personal Computer Processor Computer Control (“brain”) Datapath (“brawn”) Memory (where programs, data live when running) Devices Input Output Keyboard, Mouse Display, Printer Disk (where programs, data live when not running) This week

3. CS 61C L16 Datapath (3) A Carle, Summer 2004 © UCB Outline Design a processor: step-by-step Requirements of the Instruction Set Hardware components that match the instruction set requirements

4. CS 61C L16 Datapath (4) A Carle, Summer 2004 © 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

5. CS 61C L16 Datapath (5) A Carle, Summer 2004 © UCB Step 1: 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 optarget address 02631 6 bits26 bits oprsrtrdshamtfunct 061116212631 6 bits 5 bits oprsrt address/immediate 016212631 6 bits16 bits5 bits

6. CS 61C L16 Datapath (6) A Carle, Summer 2004 © UCB Step 1: The MIPS-lite Subset for today ADD and SUB 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 oprsrtrdshamtfunct 061116212631 6 bits 5 bits oprsrtimmediate 016212631 6 bits16 bits5 bits oprsrtimmediate 016212631 6 bits16 bits5 bits oprsrtimmediate 016212631 6 bits16 bits5 bits

7. CS 61C L16 Datapath (7) A Carle, Summer 2004 © UCB Step 1: 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 ADDUR[rd] = R[rs] + R[rt];PC = PC + 4 SUBUR[rd] = R[rs] – R[rt];PC = PC + 4 ORIR[rt] = R[rs] | zero_ext(Imm16); PC = PC + 4 LOADR[rt] = MEM[ R[rs] + sign_ext(Imm16)];PC = PC + 4 STOREMEM[ R[rs] + sign_ext(Imm16) ] = R[rt];PC = PC + 4 BEQ if ( R[rs] == R[rt] ) then PC = PC + 4 + sign_ext(Imm16)] << 2 else PC = PC + 4

8. CS 61C L16 Datapath (8) A Carle, Summer 2004 © 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

9. CS 61C L16 Datapath (9) A Carle, Summer 2004 © UCB Step 1: Abstract Implementation Data Out Clk 5 RwRaRb 32 32-bit Registers Rd ALU Clk Data In Data Address Ideal Data Memory Instruction Address Ideal Instruction Memory Clk PC 5 Rs 5 Rt 32 A B Next Address Control Datapath Control Signals Conditions

10. CS 61C L16 Datapath (10) A Carle, Summer 2004 © 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 (hard part!)

11. CS 61C L16 Datapath (11) A Carle, Summer 2004 © UCB Step 2a: Components of the Datapath Combinational Elements Storage Elements Clocking methodology

12. CS 61C L16 Datapath (12) A Carle, Summer 2004 © UCB Combinational Logic: More Elements Adder MUX ALU 32 A B Sum Carry 32 A B Result OP 32 A B Y Select Adder MUX ALU CarryIn

13. CS 61C L16 Datapath (13) A Carle, Summer 2004 © 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];... ORIR[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

14. CS 61C L16 Datapath (14) A Carle, Summer 2004 © 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 DataOut Address

15. CS 61C L16 Datapath (15) A Carle, Summer 2004 © UCB Clk Data In Write Enable NN 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

16. CS 61C L16 Datapath (16) A Carle, Summer 2004 © 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 32 busA 32 busB 555 RWRARB 32 32-bit Registers

17. CS 61C L16 Datapath (17) A Carle, Summer 2004 © UCB Administrivia Turn in your HW 45 in class! Project 2 due Sunday No extensions, No Excuses Hope you’ve already started Midterm Re-Grades Review the key that will be posted after lecture -I decided to hold up to let you have maximum fun with the HW question If you feel you have a legit complaint, write up a description of the error and staple it to the front of your test -Hand this in to Andy or a TA

18. CS 61C L16 Datapath (18) A Carle, Summer 2004 © UCB Step 3: Assemble DataPath meeting requirements Register Transfer Requirements  Datapath Assembly Instruction Fetch Read Operands and Execute Operation

19. CS 61C L16 Datapath (19) A Carle, Summer 2004 © 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 Word Address Instruction Memory PCClk Next Address Logic

20. CS 61C L16 Datapath (20) A Carle, Summer 2004 © 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 32 Result ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers RsRtRd ALU oprsrtrdshamtfunct 061116212631 6 bits 5 bits Already defined register file, ALU

21. CS 61C L16 Datapath (21) A Carle, Summer 2004 © 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........................

22. CS 61C L16 Datapath (22) A Carle, Summer 2004 © UCB Register-Register Timing: One complete cycle 32 Result ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers RsRtRd ALU Clk PC Rs, Rt, Rd, Op, Func ALUctr Instruction Memory Access Time Old ValueNew Value RegWrOld ValueNew Value Delay through Control Logic busA, B Register File Access Time Old ValueNew Value busW ALU Delay Old ValueNew Value Old ValueNew Value Old Value Register Write Occurs Here

23. CS 61C L16 Datapath (23) A Carle, Summer 2004 © UCB 3c: Logical Operations with Immediate R[rt] = R[rs] op ZeroExt[imm16] ] 32 Result ALUct r Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs ZeroExt Mux RtRd RegDst Mux 32 16 imm16 ALUSrc ALU 11 oprsrtimmediate 016212631 6 bits16 bits5 bits rd? immediate 0161531 16 bits 0 0 0 0 0 0 0 0 Rt? Already defined 32-bit MUX; Zero Ext? What about Rt register read??

24. CS 61C L16 Datapath (24) A Carle, Summer 2004 © UCB 3d: Load Operations R[rt] = Mem[R[rs] + SignExt[imm16]] Example: lw rt,rs,imm16 oprsrtimmediate 016212631 6 bits16 bits5 bits 32 ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs RtRd RegDst Extender Mux 32 16 imm16 ALUSrc ExtOp Clk Data In WrEn 32 Adr Data Memory 32 ALU MemWr Mu x W_Src ?? Rt

25. CS 61C L16 Datapath (25) A Carle, Summer 2004 © UCB 3e: Store Operations Mem[ R[rs] + SignExt[imm16] ] = R[rt] Ex.: sw rt, rs, imm16 oprsrtimmediate 016212631 6 bits16 bits5 bits 32 ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs Rt Rd RegDst Extender Mux 32 16 imm16 ALUSrc ExtOp Clk Data In WrEn 32 Adr Data Memory MemWr ALU 32 Mux W_Src

26. CS 61C L16 Datapath (26) A Carle, Summer 2004 © 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 oprsrtimmediate 016212631 6 bits16 bits5 bits

27. CS 61C L16 Datapath (27) A Carle, Summer 2004 © UCB Datapath for Branch Operations beq rs, rt, imm16 Datapath generates condition (equal) oprsrtimmediate 016212631 6 bits16 bits5 bits 32 imm16 PC Clk 00 Adder Mux Adder 4 nPC_sel Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs Rt Equal? Cond PC Ext Inst Address Already MUX, adder, sign extend, zero

28. CS 61C L16 Datapath (28) A Carle, Summer 2004 © UCB Putting it All Together:A Single Cycle Datapath imm16 32 ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs Rt Rd RegDst Extender Mux 32 16 imm16 ALUSrc ExtOp Mux MemtoReg Clk Data In WrEn 32 Adr Data Memory MemWr ALU Equal Instruction 0 1 0 1 0 1 Imm16RdRtRs = Adder PC Clk 00 Mux 4 nPC_sel PC Ext Adr Inst Memory

29. CS 61C L16 Datapath (29) A Carle, Summer 2004 © 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.

30. CS 61C L16 Datapath (30) A Carle, Summer 2004 © 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 Processor Input Output