CS152 / Kubiatowicz Lec8.1 2/22/99©UCB Spring 1999 CS152 Computer Architecture and Engineering Lecture 8 Designing Single Cycle Control Feb 22, 1999 John Kubiatowicz (http.cs.berkeley.edu/~kubitron) lecture slides:

CS152 / Kubiatowicz Lec8.2 2/22/99©UCB Spring 1999 Recap: Summary from last time °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

CS152 / Kubiatowicz Lec8.3 2/22/99©UCB Spring 1999 Recap: The MIPS Instruction Formats °All MIPS instructions are 32 bits long. 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 registers specifier 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 optarget address bits26 bits oprsrtrdshamtfunct bits 5 bits oprsrt immediate bits16 bits5 bits

CS152 / Kubiatowicz Lec8.4 2/22/99©UCB Spring 1999 Recap: The MIPS 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 oprsrtrdshamtfunct bits 5 bits oprsrtimmediate bits16 bits5 bits

CS152 / Kubiatowicz Lec8.5 2/22/99©UCB Spring 1999 Recap: A Single Cycle Datapath °We have everything except control signals (underline) Today’s lecture will show you how to generate the control signals 32 ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb bit Registers Rs Rt Rd RegDst Extender Mux imm16 ALUSrc ExtOp Mux MemtoReg Clk Data In WrEn 32 Adr Data Memory 32 MemWr ALU Instruction Fetch Unit Clk Zero Instruction Imm16RdRsRt nPC_sel

CS152 / Kubiatowicz Lec8.6 2/22/99©UCB Spring 1999 The Big Picture: Where are We Now? °The Five Classic Components of a Computer °Today’s Topic: Designing the Control for the Single Cycle Datapath Control Datapath Memory Processor Input Output

CS152 / Kubiatowicz Lec8.7 2/22/99©UCB Spring 1999 Outline of Today’s Lecture °Recap and Introduction (10 minutes) °Control for Register-Register & Or Immediate instructions (10 minutes) °Questions and Administrative Matters (5 minutes) °Control signals for Load, Store, Branch, & Jump (15 minutes) °Building a local controller: ALU Control (10 minutes) °Break (5 minutes) °The main controller (20 minutes) °Summary (5 minutes)

CS152 / Kubiatowicz Lec8.8 2/22/99©UCB Spring 1999 RTL: The 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 bits 5 bits

CS152 / Kubiatowicz Lec8.9 2/22/99©UCB Spring 1999 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 Inst Memory Adder PC Clk 00 Mux 4 nPC_sel imm16 Instruction

CS152 / Kubiatowicz Lec8.10 2/22/99©UCB Spring 1999 The Single Cycle Datapath during Add 32 ALUctr = Add Clk busW RegWr = 1 32 busA 32 busB 555 RwRaRb bit Registers Rs Rt Rd RegDst = 1 Extender Mux imm16 ALUSrc = 0 ExtOp = x 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] Imm16RdRsRt oprsrtrdshamtfunct nPC_sel= +4

CS152 / Kubiatowicz Lec8.11 2/22/99©UCB Spring 1999 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 0 1

CS152 / Kubiatowicz Lec8.12 2/22/99©UCB Spring 1999 The Single Cycle Datapath during Or Immediate °R[rt] <- R[rs] or ZeroExt[Imm16] oprsrtimmediate ALUctr = Clk busW RegWr = 32 busA 32 busB 555 RwRaRb bit Registers Rs Rt Rd RegDst = Extender Mux imm16 ALUSrc = ExtOp = Mux MemtoReg = Clk Data In WrEn 32 Adr Data Memory 32 MemWr = ALU Instruction Fetch Unit Clk Zero Instruction Imm16RdRsRt nPC_sel =

CS152 / Kubiatowicz Lec8.13 2/22/99©UCB Spring 1999 The Single Cycle Datapath during Or Immediate 32 ALUctr = Or Clk busW RegWr = 1 32 busA 32 busB 555 RwRaRb bit Registers Rs Rt Rd RegDst = 0 Extender Mux 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] Imm16RdRsRt oprsrtimmediate nPC_sel= +4

CS152 / Kubiatowicz Lec8.14 2/22/99©UCB Spring 1999 The Single Cycle Datapath during Load 32 ALUctr = Add Clk busW RegWr = 1 32 busA 32 busB 555 RwRaRb bit Registers Rs Rt Rd RegDst = 0 Extender Mux 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 Imm16RdRsRt °R[rt] <- Data Memory {R[rs] + SignExt[imm16]} oprsrtimmediate nPC_sel= +4

CS152 / Kubiatowicz Lec8.15 2/22/99©UCB Spring 1999 Questions and Administrative Matters °Tomorrow: select groups for labs Unbalanced sections. Volunteers to come to afternoon? If you don’t come to section tomorrow, you may end up in random group. °Midterm next Wednesday 3/3: 5:30pm to 8:30pm, 277 Cory Hall Make-up quiz on Tuesday No class on that day °Midterm reminders: Pencil, calculator, two 8.5” x 11” pages of handwritten notes Sit in every other chair, every other row (odd row & odd seat) °Meet at LaVal’s pizza after the midterm -Need a headcount. How many are definitely coming?

CS152 / Kubiatowicz Lec8.16 2/22/99©UCB Spring 1999 The Single Cycle Datapath during Store °Data Memory {R[rs] + SignExt[imm16]} <- R[rt] oprsrtimmediate

CS152 / Kubiatowicz Lec8.17 2/22/99©UCB Spring 1999 The Single Cycle Datapath during Store 32 ALUctr = Add Clk busW RegWr = 0 32 busA 32 busB 555 RwRaRb bit Registers Rs Rt Rd RegDst = x Extender Mux 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 Imm16RdRsRt °Data Memory {R[rs] + SignExt[imm16]} <- R[rt] oprsrtimmediate nPC_sel= +4

CS152 / Kubiatowicz Lec8.18 2/22/99©UCB Spring 1999 The Single Cycle Datapath during Branch 32 ALUctr = Subtract Clk busW RegWr = 0 32 busA 32 busB 555 RwRaRb bit Registers Rs Rt Rd RegDst = x Extender Mux 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 Imm16RdRsRt °if (R[rs] - R[rt] == 0) then Zero <- 1 ; else Zero <- 0 oprsrtimmediate nPC_sel= “Br”

CS152 / Kubiatowicz Lec8.19 2/22/99©UCB Spring 1999 Instruction Fetch Unit at the End of Branch °if (Zero == 1) then PC = PC SignExt[imm16]*4 ; else PC = PC + 4 oprsrtimmediate °What is encoding of nPC_sel? Direct MUX select? Branch / not branch °Let’s choose second option Adr Inst Memory Adder PC Clk 00 Mux 4 nPC_sel imm16 Instruction 0 1 Zero

CS152 / Kubiatowicz Lec8.20 2/22/99©UCB Spring 1999 Step 4: Given Datapath: RTL -> Control ALUctr RegDst ALUSrc ExtOp MemtoRegMemWr Equal Instruction Imm16RdRsRt nPC_sel Adr Inst Memory DATA PATH Control Op Fun RegWr

CS152 / Kubiatowicz Lec8.21 2/22/99©UCB Spring 1999 A Summary of 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”

CS152 / Kubiatowicz Lec8.22 2/22/99©UCB Spring 1999 A Summary of the Control Signals addsuborilwswbeqjump RegDst ALUSrc MemtoReg RegWrite MemWrite nPCsel Jump ExtOp ALUctr x Add x Subtract Or Add x 1 x x 0 x x Subtract x x x x xxx optarget address oprsrtrdshamtfunct oprsrt immediate R-type I-type J-type add, sub ori, lw, sw, beq jump func op Appendix A See We Don’t Care :-)

CS152 / Kubiatowicz Lec8.23 2/22/99©UCB Spring 1999 The Concept of Local Decoding Main Control op 6 ALU Control (Local) func N 6 ALUop ALUctr 3 ALU

CS152 / Kubiatowicz Lec8.24 2/22/99©UCB Spring 1999 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 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) func N 6 ALUop ALUctr 3 R-typeorilwswbeqjump ALUop (Symbolic)“R-type”OrAdd Subtract xxx ALUop xxx

CS152 / Kubiatowicz Lec8.25 2/22/99©UCB Spring 1999 The Decoding of the “func” Field R-typeorilwswbeqjump ALUop (Symbolic)“R-type”OrAdd Subtract xxx ALUop xxx Main Control op 6 ALU Control (Local) func N 6 ALUop ALUctr 3 oprsrtrdshamtfunct R-type funct Instruction Operation add subtract and or set-on-less-than ALUctr ALU Operation Add Subtract And Or Set-on-less-than P. 286 text: ALUctr ALU

CS152 / Kubiatowicz Lec8.26 2/22/99©UCB Spring 1999 The Truth Table for ALUctr R-typeorilwswbeq ALUop (Symbolic) “R-type”OrAdd Subtract ALUop funct Instruction Op and subtract add or set-on-less-than

CS152 / Kubiatowicz Lec8.27 2/22/99©UCB Spring 1999 The Logic Equation for ALUctr ALUopfunc bit ALUctr 0x1xxxx1 1xx xx10101 °ALUctr = !ALUop & ALUop + ALUop & !func & func & !func This makes func a don’t care

CS152 / Kubiatowicz Lec8.28 2/22/99©UCB Spring 1999 The Logic Equation for ALUctr ALUopfunc bit 000xxxx1 ALUctr 0x1xxxx1 1xx xx xx10101 °ALUctr = !ALUop & !ALUop + ALUop & !func & !func

CS152 / Kubiatowicz Lec8.29 2/22/99©UCB Spring 1999 The Logic Equation for ALUctr ALUopfunc bit ALUctr 01xxxxx1 1xx xx10101 °ALUctr = !ALUop & ALUop + ALUop & !func & func & !func & func + ALUop & func & !func & func & !func

CS152 / Kubiatowicz Lec8.30 2/22/99©UCB Spring 1999 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

CS152 / Kubiatowicz Lec8.31 2/22/99©UCB Spring 1999 Step 5: Logic for each control signal °nPC_sel <= if (OP == BEQ) then “Br” else “+4” °ALUsrc <=if (OP == “Rtype”) then “regB” else “immed” °ALUctr<= if (OP == “Rtype”) then funct elseif (OP == ORi) then “OR” elseif (OP == BEQ) then “sub” else “add” °ExtOp <= _____________ °MemWr<= _____________ °MemtoReg<= _____________ °RegWr:<=_____________ °RegDst:<= _____________

CS152 / Kubiatowicz Lec8.32 2/22/99©UCB Spring 1999 Step 5: Logic for each control signal °nPC_sel <= if (OP == BEQ) then “Br” else “+4” °ALUsrc <=if (OP == “Rtype”) then “regB” else “immed” °ALUctr<= if (OP == “Rtype”) 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

CS152 / Kubiatowicz Lec8.33 2/22/99©UCB Spring 1999 The “Truth Table” for the Main Control R-typeorilwswbeqjump RegDst ALUSrc MemtoReg RegWrite MemWrite nPC_sel Jump ExtOp ALUop (Symbolic) x “R-type” Or Add x 1 x x 0 x x Subtract x x x x xxx op ALUop x x x Main Control op 6 ALU Control (Local) func 3 6 ALUop ALUctr 3RegDst ALUSrc :

CS152 / Kubiatowicz Lec8.34 2/22/99©UCB Spring 1999 The “Truth Table” for RegWrite R-typeorilwswbeqjump RegWrite op °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

CS152 / Kubiatowicz Lec8.35 2/22/99©UCB Spring 1999 PLA Implementation of the Main Control RegWrite ALUSrc MemtoReg MemWrite Branch Jump RegDst ExtOp ALUop

CS152 / Kubiatowicz Lec8.36 2/22/99©UCB Spring 1999 A Real MIPS Datapath (CNS T0)

CS152 / Kubiatowicz Lec8.37 2/22/99©UCB Spring 1999 Putting it All Together: A Single Cycle Processor 32 ALUctr Clk busW RegWr 32 busA 32 busB 555 RwRaRb bit Registers Rs Rt Rd RegDst Extender Mux imm16 ALUSrc ExtOp Mux MemtoReg Clk Data In WrEn 32 Adr Data Memory 32 MemWr ALU Instruction Fetch Unit Clk Zero Instruction Imm16RdRsRt Main Control op 6 ALU Control func 6 3 ALUop ALUctr 3 RegDst ALUSrc : Instr nPC_sel

CS152 / Kubiatowicz Lec8.38 2/22/99©UCB Spring 1999 Worst Case Timing (Load) Clk PC Rs, Rt, Rd, Op, Func Clk-to-Q ALUctr Instruction Memoey Access Time Old ValueNew Value RegWrOld 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 MemtoRegOld ValueNew Value AddressOld ValueNew Value busWOld ValueNew Delay through Extender & Mux Register Write Occurs Data Memory Access Time

CS152 / Kubiatowicz Lec8.39 2/22/99©UCB Spring 1999 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

CS152 / Kubiatowicz Lec8.40 2/22/99©UCB Spring 1999 °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 Processor Input Output

CS152 / Kubiatowicz Lec8.41 2/22/99©UCB Spring 1999 Where to get more information? °Chapter 5.1 to 5.3 of your text book: David Patterson and John Hennessy, “Computer Organization & Design: The Hardware / Software Interface,” Second Edition, Morgan Kaufman Publishers, San Mateo, California, °One of the best PhD thesis on processor design: Manolis Katevenis, “Reduced Instruction Set Computer Architecture for VLSI,” PhD Dissertation, EECS, U C Berkeley, °For a reference on the MIPS architecture: Gerry Kane, “MIPS RISC Architecture,” Prentice Hall.