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CS141-L4-1Tarun Soni, Summer’03 Single Cycle CPU  Previously: built and ALU.  Today: Actually build a CPU Questions on CS140 ? Computer Arithmetic ?

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Presentation on theme: "CS141-L4-1Tarun Soni, Summer’03 Single Cycle CPU  Previously: built and ALU.  Today: Actually build a CPU Questions on CS140 ? Computer Arithmetic ?"— Presentation transcript:

1 CS141-L4-1Tarun Soni, Summer’03 Single Cycle CPU  Previously: built and ALU.  Today: Actually build a CPU Questions on CS140 ? Computer Arithmetic ? Attend office hours with TAs or me. Do the exercises in the text.

2 CS141-L4-2Tarun Soni, Summer’03  Instruction Set Architectures  Performance issues  2s complement, Addition, Subtraction  Multiplication, Division, Floating Point numbers The Story so far: Basically ISA & ALU stuff

3 CS141-L4-3Tarun Soni, Summer’03 CPU: Building blocks Adder MUX ALU 32 A B Sum Carry 32 A B Result OP 32 A B Y Select Adder MUX ALU CarryIn

4 CS141-L4-4Tarun Soni, Summer’03 CPU: Building blocks OP 32 A B Y Select MUX 32 A[31..0] B[31..0] 32 Sum[31..0] Carry Adder CarryIn 32 A[63..32] B[63..32] 32 Sum[63..32] Carry Adder CarryIn 32 Building a 64-bit adder from 2x32-bit adders

5 CS141-L4-5Tarun Soni, Summer’03 CPU: Building blocks 32 A B Sum[63..32] 32 Select MUX 32 A[31..0] B[31..0] 32 Sum[31..0] Carry Adder CarryIn 32 A[63..32] B[63..32] 32 S Cout Adder Cin=0 32 A[63..32] B[63..32] 32 S Cout Adder Cin=1 1 A B 1 Cout 1 Select MUX Silicon is cheap – sort-of

6 CS141-L4-6Tarun Soni, Summer’03 CPU Single Cycle CPU

7 CS141-L4-7Tarun Soni, Summer’03 CPU The Big Picture: Where are We Now? The Five Classic Components of a Computer Datapath Design, then Control Design Control Datapath Memory Processor Input Output

8 CS141-L4-8Tarun Soni, Summer’03 CPU: The big picture Instruction Fetch Instruction Decode Operand Fetch Execute Result Store Next Instruction ° Design hardware for each of these steps!!! Execute an entire instruction FetchDecode Fetch Execute Store Next

9 CS141-L4-9Tarun Soni, Summer’03 CPU: Clocking Clk Don’t Care SetupHold........................ SetupHold All storage elements are clocked by the same clock edge

10 CS141-L4-10Tarun Soni, Summer’03 CPU The Big Picture: The Performance Perspective Execution Time = Insts * CPI * Cycle Time Processor design (datapath and control) will determine: –Clock cycle time –Clock cycles per instruction Starting today: –Single cycle processor: Advantage: One clock cycle per instruction Disadvantage: long cycle time Execute an entire instruction

11 CS141-L4-11Tarun Soni, Summer’03 CPU We're ready to look at an implementation of the MIPS Simplified to contain only: –memory-reference instructions: lw, sw –arithmetic-logical instructions: add, sub, and, or, slt –control flow instructions: beq Generic Implementation: –use the program counter (PC) to supply instruction address –get the instruction from memory –read registers –use the instruction to decide exactly what to do All instructions use the ALU after reading the registers memory-reference? arithmetic? control flow? CPI Inst. CountCycle Time

12 CS141-L4-12Tarun Soni, Summer’03 CPU Review: The MIPS Instruction Formats optarget address 02631 6 bits26 bits oprsrtrdshamtfunct 061116212631 6 bits 5 bits oprsrt immediate 016212631 6 bits16 bits5 bits °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

13 CS141-L4-13Tarun Soni, Summer’03 CPU R-type –add rd, rs, rt –sub, and, or, slt LOAD and STORE –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 oprsrtdisplacement 016212631 6 bits16 bits5 bits

14 CS141-L4-14Tarun Soni, Summer’03 CPU Memory –instruction & data Registers (32 x 32) –read RS –read RT –Write RT or RD PC Extender Add and Sub register or extended immediate Add 4 or extended immediate to PC Requirements to implement the ISA

15 CS141-L4-15Tarun Soni, Summer’03 CPU Combinational Elements Storage Elements –Clocking methodology State Element clk A B C = f(A,B,state) {State[n] = f(A,B,state[n-1])} Combinational Logic A B C = f(A,B)

16 CS141-L4-16Tarun Soni, Summer’03 CPU: Storage unit The set-reset latch –output depends on present inputs and also on past inputs

17 CS141-L4-17Tarun Soni, Summer’03 CPU: D-flip flop Two inputs: –the data value to be stored (D) –the clock signal (C) indicating when to read & store D Two outputs: –the value of the internal state (Q) and it's complement Output changes only on the clock edge

18 CS141-L4-18Tarun Soni, Summer’03 CPU: Clocking Methodology An edge triggered methodology Typical execution: –read contents of some state elements, –send values through some combinational logic –write results to one or more state elements

19 CS141-L4-19Tarun Soni, Summer’03 CPU: Storage block Register –Similar to the D Flip Flop except N-bit input and output Write Enable input –Write Enable: 0: Data Out will not change 1: Data Out will become Data In (on the clock edge) Clk Data In Write Enable NN Data Out

20 CS141-L4-20Tarun Soni, Summer’03 CPU: Register Files Register File consists of (32) registers: –Two 32-bit output buses: –One 32-bit input bus: busW Register is selected by: –RA selects the register to put on busA –RB selects the register to put on busB –RW selects the register to be written via busW when Write Enable is 1 Clock input (CLK) Factor only during write-enable=1; Otherwise, this unit acts just like combinational logic. Clk busW Write Enable 32 busA 32 busB 555 RWRARB 32 32-bit Registers

21 CS141-L4-21Tarun Soni, Summer’03 CPU: Register Files Built using D-flip flops Still use the real clock (not shown here) to do the actual write

22 CS141-L4-22Tarun Soni, Summer’03 CPU: 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

23 CS141-L4-23Tarun Soni, Summer’03 CPU: RTL is a mechanism for describing the movement and manipulation of data between storage elements: R[3] <- R[5] + R[7] PC <- PC + 4 + R[5] R[rd] <- R[rs] + R[rt] R[rt] <- Mem[R[rs] + immed] Register Transfer Language (RTL)

24 CS141-L4-24Tarun Soni, Summer’03 CPU: More building blocks

25 CS141-L4-25Tarun Soni, Summer’03 CPU: The big picture Instruction Fetch Instruction Decode Operand Fetch Execute Result Store Next Instruction ° Design hardware for each of these steps!!! Execute an entire instruction FetchDecode Fetch Execute Store Next

26 CS141-L4-26Tarun Soni, Summer’03 CPU: Instruction Fetch RTL version of the instruction fetch step: 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 PC Clk Next Address Logic

27 CS141-L4-27Tarun Soni, Summer’03 CPU: Binary arithmetic for PC In theory, the PC is a 32-bit byte address into the instruction memory: –Sequential operation: PC = PC + 4 –Branch operation: PC = PC + 4 + SignExt[Imm16] * 4 The magic number “4” always comes up because: –The 32-bit PC is a byte address –And all our instructions are 4 bytes (32 bits) long In other words: –The 2 LSBs of the 32-bit PC are always zeros –There is no reason to have hardware to keep the 2 LSBs In practice, we can simplify the hardware by using a 30-bit PC : –Sequential operation: PC = PC + 1 –Branch operation: PC = PC + 1 + SignExt[Imm16] –In either case: Instruction Memory Address = PC concat “00”

28 CS141-L4-28Tarun Soni, Summer’03 CPU: Instruction Fetch unit The common RTL operations –Fetch the Instruction: inst <- mem[PC] –Update the program counter: Sequential Code: PC <- PC + 4 Branch and Jump PC <- “something else”

29 CS141-L4-29Tarun Soni, Summer’03 CPU: Register-Register Operations (Add, Subtract etc.) 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 after decoding the instruction 3232 Result ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers RsRtRd ALU oprsrtrdshamtfunct 061116212631 6 bits 5 bits ° Worry about instruction decode to generate ALUctr and RegWr later.

30 CS141-L4-30Tarun Soni, Summer’03 CPU: Register - Register Timing 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 Clk-to-Q ALUct r 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

31 CS141-L4-31Tarun Soni, Summer’03 CPU: Logical Immediate Op. R[rt] <- R[rs] op ZeroExt[imm16] ] 32 Result ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs RtRd RegDst ZeroExt 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 Handle Rt as destination Handle Immediate as operand

32 CS141-L4-32Tarun Soni, Summer’03 CPU: Load Operations R[rt] <- Mem[R[rs] + SignExt[imm16]]Example: lw rt, rs, imm16 11 oprsrtimmediate 016212631 6 bits16 bits5 bits rd 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 Need data Memory! Reg-Write could be from result or data memory

33 CS141-L4-33Tarun Soni, Summer’03 CPU: Store Operations Mem[ R[rs] + SignExt[imm16] <- R[rt] ] Example: sw rt, rs, 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 Clk Data In WrEn 32 Adr Data Memory MemWr ALU oprsrtimmediate 016212631 6 bits16 bits5 bits 32 Mu x W_Src Reg can write to Data Memory

34 CS141-L4-34Tarun Soni, Summer’03 CPU: Branching beqrs, rt, imm16 –mem[PC]Fetch the instruction from memory –Equal <- R[rs] == R[rt]Calculate the branch condition –if (COND eq 0)Calculate the next instruction’s address PC <- PC + 4 + ( SignExt(imm16) x 4 ) –else PC <- PC + 4 oprsrtimmediate 016212631 6 bits16 bits5 bits

35 CS141-L4-35Tarun Soni, Summer’03 CPU: Datapath for Branching beq rs, rt, imm16Datapath 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 Calculate (PC+4) as well as (imm16+PC+4) and choose one Calculate the “condition” part of the branch op.

36 CS141-L4-36Tarun Soni, Summer’03 CPU: The Aggregate 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 Still need to worry about Instruction Decode

37 CS141-L4-37Tarun Soni, Summer’03 CPU: Datapath: High-level view 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 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 16 Imm 32 A B Next Address

38 CS141-L4-38Tarun Soni, Summer’03 CPU: Control Signals ALUctr RegDst ALUSrc ExtOp MemtoRegMemWr Equal Instruction Imm16RdRsRt nPC_sel Adr Inst Memory DATA PATH Control Op Fun RegWr

39 CS141-L4-39Tarun Soni, Summer’03 CPU: Control Signals: Meaning Adr Inst Memory Rs, Rt, Rd and Imed16 hardwired into datapath nPC_sel: 0 => PC PC <– PC + 4 + SignExt(Im16) || 00 Adder PC Clk 00 Mux 4 nPC_sel PC Ext imm16

40 CS141-L4-40Tarun Soni, Summer’03 CPU: Control Signals: Meaning ExtOp:“zero”, “sign” ALUsrc:0 => regB; 1 => immed ALUctr:“add”, “sub”, “or” 32 ALUct r 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 0 1 0 1 0 1 °MemWr:write memory °MemtoReg:1 => Mem °RegDst:0 => “rt”; 1 => “rd” °RegWr:write dest register =

41 CS141-L4-41Tarun Soni, Summer’03 CPU: Control Signals for various operations inst Register Transfer ADDR[rd] <– R[rs] + R[rt];PC <– PC + 4 ALUsrc = RegB, ALUctr = “add”, RegDst = rd, RegWr, nPC_sel = “+4” SUBR[rd] <– R[rs] – R[rt];PC <– PC + 4 ALUsrc = RegB, ALUctr = “sub”, RegDst = rd, RegWr, nPC_sel = “+4” ORiR[rt] <– R[rs] + zero_ext(Imm16); PC <– PC + 4 ALUsrc = Im, Extop = “Z”, ALUctr = “or”, RegDst = rt, RegWr, nPC_sel = “+4” LOADR[rt] <– MEM[ R[rs] + sign_ext(Imm16)];PC <– PC + 4 ALUsrc = Im, Extop = “Sn”, ALUctr = “add”, MemtoReg, RegDst = rt, RegWr, nPC_sel = “+4” STOREMEM[ R[rs] + sign_ext(Imm16)] <– R[rs];PC <– PC + 4 ALUsrc = Im, Extop = “Sn”, ALUctr = “add”, MemWr, nPC_sel = “+4” BEQif ( R[rs] == R[rt] ) then PC <– PC + sign_ext(Imm16)] || 00 else PC <– PC + 4 nPC_sel = EQUAL, ALUctr = “sub”

42 CS141-L4-42Tarun Soni, Summer’03 CPU: Control Signals: Logic Design nPC_sel <= if (OP == BEQ) then EQUAL else 0 ALUsrc <=if (OP == “000000”) then “regB” else “immed” ALUctr<= if (OP == “000000”) then funct elseif (OP == ORi) then “OR” elseif (OP == BEQ) then “sub” else “add” ExtOp <= _____________ MemWr<= _____________ MemtoReg<= _____________ RegWr:<=_____________ RegDst:<= _____________

43 CS141-L4-43Tarun Soni, Summer’03 CPU: Control Signals: Logic Design nPC_sel <= if (OP == BEQ) then EQUAL else 0 ALUsrc <=if (OP == “000000”) then “regB” else “immed” ALUctr<= if (OP == “000000”) then funct elseif (OP == ORi) then “OR” elseif (OP == BEQ) then “sub” else “add” ExtOp <= if (OP == ORi) then “zero” else “sign” MemWr<= (OP == Store) MemtoReg<= (OP == Load) RegWr:<= if ((OP == Store) || (OP == BEQ)) then 0 else 1 RegDst:<= if ((OP == Load) || (OP == ORi)) then 0 else 1

44 CS141-L4-44Tarun Soni, Summer’03 CPU: Example: Load R[rt] <- Mem[R[rs] + SignExt[imm16]] Viz., lw rt, rs, imm16

45 CS141-L4-45Tarun Soni, Summer’03 CPU: The abstract version Logical vs. Physical Structure 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

46 CS141-L4-46Tarun Soni, Summer’03 CPU: The real thing

47 CS141-L4-47Tarun Soni, Summer’03 CPU: 5 steps to design 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

48 CS141-L4-48Tarun Soni, Summer’03 CPU: Control Section The Five Classic Components of a Computer Control Datapath Memory Processor Input Output

49 CS141-L4-49Tarun Soni, Summer’03 CPU: Add Instruction addrd, rs, rt –mem[PC]Fetch the instruction from memory –R[rd] <- R[rs] + R[rt]The actual operation –PC <- PC + 4Calculate the next instruction’s address oprsrtrdshamtfunct 061116212631 6 bits 5 bits

50 CS141-L4-50Tarun Soni, Summer’03 CPU: The Add Instruction Instruction Fetch Unit at the Beginning of Add PC Ext Fetch the instruction from Instruction memory: Instruction <- mem[PC] –This is the same for all instructions Adr Inst Memory Adder PC Clk 00 Mux 4 nPC_sel imm16 Instruction

51 CS141-L4-51Tarun Soni, Summer’03 CPU: The Add Instruction The Single Cycle Datapath during Add 32 ALUctr = Add Clk busW RegWr = 1 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs Rt Rd RegDst = 1 Extender Mux 32 16 imm16 ALUSrc = 0 Mux MemtoReg = 0 Clk Data In WrEn 32 Adr Data Memory 32 MemWr = 0 ALU Instruction Fetch Unit Clk Zero Instruction R[rd] <- R[rs] + R[rt] 0 1 0 1 01 Imm16RdRsRt oprsrtrdshamtfunct 061116212631 nPC_sel= +4

52 CS141-L4-52Tarun Soni, Summer’03 CPU: The Add Instruction Instruction Fetch Unit at the End of Add PC <- PC + 4 –This is the same for all instructions except: Branch and Jump Adr Inst Memory Adder PC Clk 00 Mux 4 nPC_sel imm16 Instruction

53 CS141-L4-53Tarun Soni, Summer’03 CPU: The Or Immediate Instruction R[rt] <- R[rs] or ZeroExt[Imm16] oprsrtimmediate 016212631 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 32 MemWr = ALU Instruction Fetch Unit Clk Zero Instruction 0 1 0 1 01 Imm16RdRsRt nPC_sel =

54 CS141-L4-54Tarun Soni, Summer’03 CPU: The Or Immediate Instruction The Single Cycle Datapath during Or Immediate 32 ALUctr = Or Clk busW RegWr = 1 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs Rt Rd RegDst = 0 Extender Mux 32 16 imm16 ALUSrc = 1 ExtOp = 0 Mux MemtoReg = 0 Clk Data In WrEn 32 Adr Data Memory 32 MemWr = 0 ALU Instruction Fetch Unit Clk Zero Instruction R[rt] <- R[rs] or ZeroExt[Imm16] 0 1 0 1 01 Imm16RdRsRt oprsrtimmediate 016212631 nPC_sel= +4

55 CS141-L4-55Tarun Soni, Summer’03 CPU: The Load Instruction The Single Cycle Datapath during Load 32 ALUctr = Add Clk busW RegWr = 1 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs Rt Rd RegDst = 0 Extender Mux 32 16 imm16 ALUSrc = 1 ExtOp = 1 Mux MemtoReg = 1 Clk Data In WrEn 32 Adr Data Memory 32 MemWr = 0 ALU Instruction Fetch Unit Clk Zero Instruction 0 1 0 1 01 Imm16RdRsRt R[rt] <- Data Memory {R[rs] + SignExt[imm16]} oprsrtimmediate 016212631 nPC_sel= +4

56 CS141-L4-56Tarun Soni, Summer’03 CPU: The Store Instruction The Single Cycle Datapath during Store Data Memory {R[rs] + SignExt[imm16]} <- R[rt] oprsrtimmediate 016212631 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 32 MemWr = ALU Instruction Fetch Unit Clk Zero Instruction 0 1 0 1 01 Imm16RdRsRt nPC_sel =

57 CS141-L4-57Tarun Soni, Summer’03 CPU: The Store Instruction The Single Cycle Datapath during Store 32 ALUctr = Add Clk busW RegWr = 0 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs Rt Rd RegDst = x Extender Mux 32 16 imm16 ALUSrc = 1 ExtOp = 1 Mux MemtoReg = x Clk Data In WrEn 32 Adr Data Memory 32 MemWr = 1 ALU Instruction Fetch Unit Clk Zero Instruction 0 1 0 1 01 Imm16RdRsRt Data Memory {R[rs] + SignExt[imm16]} <- R[rt] oprsrtimmediate 016212631 nPC_sel= +4

58 CS141-L4-58Tarun Soni, Summer’03 CPU: Datapath during branch 32 ALUctr = Subtract Clk busW RegWr = 0 32 busA 32 busB 555 RwRaRb 32 32-bit Registers Rs Rt Rd RegDst = x Extender Mux 32 16 imm16 ALUSrc = 0 ExtOp = x Mux MemtoReg = x Clk Data In WrEn 32 Adr Data Memory 32 MemWr = 0 ALU Instruction Fetch Unit Clk Zero Instruction 0 1 0 1 01 Imm16RdRsRt if (R[rs] - R[rt] == 0) then Zero <- 1 ; else Zero <- 0 oprsrtimmediate 016212631 nPC_sel= “Br”

59 CS141-L4-59Tarun Soni, Summer’03 CPU: Datapath during branch Instruction Fetch Unit at the End of Branch if (Zero == 1) then PC = PC + 4 + SignExt[imm16]*4 ; else PC = PC + 4 oprsrtimmediate 016212631 Adr Inst Memory Adder PC Clk 00 Mux 4 nPC_sel imm16 Instruction

60 CS141-L4-60Tarun Soni, Summer’03 CPU: Creating control from Datapath ALUctr RegDst ALUSrc ExtOp MemtoRegMemWr Equal Imm16RdRsRt nPC_sel Adr Inst Memory DATA PATH Control Op Fun RegWr

61 CS141-L4-61Tarun Soni, Summer’03 CPU: Control Signals inst Register Transfer ADDR[rd] <– R[rs] + R[rt];PC <– PC + 4 ALUsrc = RegB, ALUctr = “add”, RegDst = rd, RegWr, nPC_sel = “+4” SUBR[rd] <– R[rs] – R[rt];PC <– PC + 4 ALUsrc = RegB, ALUctr = “sub”, RegDst = rd, RegWr, nPC_sel = “+4” ORiR[rt] <– R[rs] + zero_ext(Imm16); PC <– PC + 4 ALUsrc = Im, Extop = “Z”, ALUctr = “or”, RegDst = rt, RegWr, nPC_sel = “+4” LOADR[rt] <– MEM[ R[rs] + sign_ext(Imm16)];PC <– PC + 4 ALUsrc = Im, Extop = “Sn”, ALUctr = “add”, MemtoReg, RegDst = rt, RegWr, nPC_sel = “+4” STOREMEM[ R[rs] + sign_ext(Imm16)] <– R[rs];PC <– PC + 4 ALUsrc = Im, Extop = “Sn”, ALUctr = “add”, MemWr, nPC_sel = “+4” BEQif ( R[rs] == R[rt] ) then PC <– PC + sign_ext(Imm16)] || 00 else PC <– PC + 4 nPC_sel = “Br”, ALUctr = “sub”

62 CS141-L4-62Tarun Soni, Summer’03 CPU: Summary of Control Signals addsuborilwswbeqjump RegDst ALUSrc MemtoReg RegWrite MemWrite nPCsel Jump ExtOp ALUctr 1 0 0 1 0 0 0 x Add 1 0 0 1 0 0 0 x Subtract 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 x 0 x 0 0 1 0 x Subtract x x x 0 0 0 1 x xxx optarget address oprsrtrdshamtfunct 061116212631 oprsrt immediate R-type I-type J-type add, sub ori, lw, sw, beq jump func op00 0000 00 110110 001110 101100 010000 0010 Appendix A 10 0000See10 0010We Don’t Care :-)

63 CS141-L4-63Tarun Soni, Summer’03 CPU: Summary of Control Signals The Concept of Local Decoding Main Control op 6 ALU Control (Local) func N 6 ALUop ALUctr 3 ALU

64 CS141-L4-64Tarun Soni, Summer’03 CPU: 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 the full MIPS ISA, ALUop has to be 3 bits to represent: –(1) “R-type” instructions –“I-type” instructions that require the ALU to perform: (2) Or, (3) Add, (4) Subtract, and (5) And (Example: andi) Main Control op 6 ALU Control (Local) funcN 6 ALUop ALUctr 3 R-typeorilwswbeqjump ALUop (Symbolic)“R-type”OrAdd Subtract xxx ALUop 1 000 100 00 0 01 xxx

65 CS141-L4-65Tarun Soni, Summer’03 CPU: Decoding of the ‘func’ field R-typeorilwswbeqjump ALUop (Symbolic)“R-type”OrAdd Subtract xxx ALUop 1 000 100 00 0 01 xxx Main Control op 6 ALU Control (Local) func N 6 ALUop ALUctr 3 oprsrtrdshamtfunct 061116212631 R-type funct Instruction Operation 10 0000 10 0010 10 0100 10 0101 10 1010 add subtract and or set-on-less-than ALUctr ALU Operation 000 001 010 110 111 Add Subtract And Or Set-on-less-than Recall ALUctr ALU

66 CS141-L4-66Tarun Soni, Summer’03 CPU: Truth table for ALUctr R-typeorilwswbeq ALUop (Symbolic) “R-type”OrAdd Subtract ALUop 1 000 100 00 0 01 funct Instruction Op. 0000 0010 0100 0101 1010 add subtract and or set-on-less-than

67 CS141-L4-67Tarun Soni, Summer’03 CPU: Logic Equation ALUctr[2] The Logic Equation for ALUctr ALUopfunc bit ALUctr 0x1xxxx1 1xx00101 1xx10101 ALUctr = !ALUop & ALUop + ALUop & !func & func & !func This makes func a don’t care

68 CS141-L4-68Tarun Soni, Summer’03 CPU: Logic Equation ALUctr[1] The Logic Equation for ALUctr ALUopfunc bit 000xxxx1 ALUctr 0x1xxxx1 1xx00001 1xx00101 1xx10101 ALUctr = !ALUop & !ALUop + ALUop & !func & !func

69 CS141-L4-69Tarun Soni, Summer’03 CPU: Logic Equation ALUctr[0] The Logic Equation for ALUctr ALUopfunc bit ALUctr 01xxxxx1 1xx01011 1xx10101 ALUctr = !ALUop & ALUop + ALUop & !func & func & !func & func + ALUop & func & !func & func & !func

70 CS141-L4-70Tarun Soni, Summer’03 CPU: ALU Control block The ALU Control Block ALU Control (Local) func 3 6 ALUop ALUctr 3 ALUctr = !ALUop & ALUop + ALUop & !func & func & !func ALUctr = !ALUop & !ALUop + ALUop & !func & !func ALUctr = !ALUop & ALUop + ALUop & !func & func & !func & func + ALUop & func & !func & func & !func

71 CS141-L4-71Tarun Soni, Summer’03 CPU: Main Control R-typeorilwswbeqjump RegDst ALUSrc MemtoReg RegWrite MemWrite Branch 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 x 0 x 0 0 1 0 x Subtract x x x 0 0 0 1 x xxx op00 000000 110110 001110 101100 010000 0010 ALUop 1000 0 x 0100 0 x 0000 1 x Main Control op 6 ALU Control (Local) func 3 6 ALUop ALUctr 3 RegDst ALUSrc :

72 CS141-L4-72Tarun Soni, Summer’03 CPU: Main Control The “Truth Table” for RegWrite R-typeorilwswbeqjump RegWrite111000 op00 000000 110110 001110 101100 010000 0010 RegWrite = R-type + ori + lw = !op & !op & !op & !op & !op & !op (R-type) + !op & !op & op & op & !op & op (ori) + op & !op & !op & !op & op & op (lw) RegWrite

73 CS141-L4-73Tarun Soni, Summer’03 CPU: Main Control PLA Implementation of the Main Control RegWrite ALUSrc MemtoReg MemWrite Branch Jump RegDst ExtOp ALUop

74 CS141-L4-74Tarun Soni, Summer’03 CPU Putting it All Together: A Single Cycle Processor 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 32 MemWr ALU Instruction Fetch Unit Clk Zero Instruction 0 1 0 1 01 Imm16RdRsRt Main Control op 6 ALU Control func 6 3 ALUop ALUctr 3 RegDst ALUSrc : Instr nPC_sel

75 CS141-L4-75Tarun Soni, Summer’03 CPU Worst Case Timing (Load) Clk PCPC Rs, Rt, Rd, Op, Func Clk-to-Q ALUct r Instruction Memoey Access Time Old ValueNew Value RegW r Old ValueNew Value Delay through Control Logic busA Register File Access Time Old ValueNew Value busB ALU Delay Old ValueNew Value Old ValueNew Value Old Value ExtOpOld ValueNew Value ALUSrcOld ValueNew Value MemtoRe g Old ValueNew Value Addre ss Old ValueNew Value busWOld ValueNew Delay through Extender & Mux Register Write Occurs Data Memory Access Time

76 CS141-L4-76Tarun Soni, Summer’03 CPU: Single Cycle Solution 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

77 CS141-L4-77Tarun Soni, Summer’03 CPU: Single Cycle Solution °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 Control Datapath Memory Processor Input Output

78 CS141-L4-78Tarun Soni, Summer’03 CPU: Interrupts °Datapath for interrupts °Interrupt: basically hardware line requesting an immediate jump °PC = Int[I] if Int[I] = 1; °May or maynot save registers °May or maynot be maskable. °Useful for multitasking control & real-time processing °Signal Processing °Harder to implement in case of a multi-cycle/pipelines system !


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