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Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering

ELEC 5200-001/6200-001 Computer Architecture and Design Spring 2007 Datapath and Control (Chapter 5). Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University, Auburn, AL 36849 http://www.eng.auburn.edu/~vagrawal

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Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering

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  1. ELEC 5200-001/6200-001Computer Architecture and DesignSpring 2007 Datapath and Control(Chapter 5) Vishwani D. Agrawal James J. Danaher Professor Department of Electrical and Computer Engineering Auburn University, Auburn, AL 36849 http://www.eng.auburn.edu/~vagrawal vagrawal@eng.auburn.edu ELEC 5200-001/6200-001 Lecture 8

  2. A Puzzle for the Day “I got the idea for the _______ while attending a talk at a computer conference. The speaker was so boring that I started daydreaming and hit upon the idea.” Doug Engelbart inventor of computer mouse ELEC 5200-001/6200-001 Lecture 8

  3. Von Neumann Kitchen Start ALU Control My choice PC Registers Processor Program Data Memory Output ELEC 5200-001/6200-001 Lecture 8

  4. Where Does It All Begin? • In a register called program counter (PC). • PC contains the memory address of the next instruction to be executed. • In the beginning, PC contains the address of the memory location where the program begins. ELEC 5200-001/6200-001 Lecture 8

  5. Where is the Program? Processor Memory Program counter (register) Machine code of program Start address ELEC 5200-001/6200-001 Lecture 8

  6. How Does It Run? Start Fetch instruction word from memory address in PC and increment PC ← PC + 4 Decode and execute instruction Program complete? No STOP Yes ELEC 5200-001/6200-001 Lecture 8

  7. Datapath and Control • Datapath: Memory, registers, adders, ALU, and communication buses. Each step (fetch, decode, execute) requires communication (data transfer) paths between memory, registers and ALU. • Control: Datapath for each step is set up by control signals that set up dataflow directions on communication bus and select ALU function. Control signals are generated by a control unit consisting of one or more finite-state machines. ELEC 5200-001/6200-001 Lecture 8

  8. Add Datapath for Instruction Fetch 4 Instruction Memory PC Instruction word to control unit and registers Address ELEC 5200-001/6200-001 Lecture 8

  9. RegWrite from control Register File: A Datapath Component 5 32 Registers (reg. file) reg 1 Read register numbers 32 reg 1 data 5 reg 2 Write register number 5 32 reg 2 data Write data 32 ELEC 5200-001/6200-001 Lecture 8

  10. Multi-Operation ALU Operation select from control Operation select ALU function 000 AND 001 OR 010 Add 110 Subtract 111 Set on less than 3 zero ALU result overflow zero = 1, when all bits of result are 0 ELEC 5200-001/6200-001 Lecture 8

  11. R-Type Instructions • Also known as arithmetic-logical instructions • add, sub, slt • Example: add $t0, $s1, $s2 • Machine instruction word 000000 10001 10010 01000 00000 100000 opcode $s1 $s2 $t0 function • Read two registers • Write one register • Opcode and function code go to control unit that generates RegWrite and ALU operation code. ELEC 5200-001/6200-001 Lecture 8

  12. Datapath for R-Type Instruction • 000000 10001 10010 01000 00000 100000opcode $s1 $s2 $t0 function (add) Operation select from control (add) 5 32 Registers (reg. file) 10001 $s1 Read register numbers 3 32 zero 5 $s2 10010 ALU result 32 overflow Write reg. number 5 $t0 01000 Write data 32 RegWrite from control activated ELEC 5200-001/6200-001 Lecture 8

  13. Load and Store Instructions • I-type instructions • lw $t0, 1200 ($t1) # incr. in bytes 100011 01001 01000 0000 0100 1011 0000 opcode $t1 $t0 1200 • sw $t0, 1200 ($t1) # incr. in bytes 101011 01001 01000 0000 0100 1011 0000 opcode $t1 $t0 1200 ELEC 5200-001/6200-001 Lecture 8

  14. Datapath for lw Instruction 100011 01001 01000 0000 0100 1011 0000 opcode $t1 $t0 1200 MemWrite Operation select from control (add) 32 Registers (reg. file) 5 01001 $t1 Read register numbers 3 Read data 32 zero 5 result ALU Addr. Data memory overflow Write reg. number 5 $t0 01000 Write data Write data 32 32 RegWrite from control activated Sign extend MemRead activated 0000 0100 1011 0000 16 mem. data to $t0 ELEC 5200-001/6200-001 Lecture 8

  15. Datapath for sw Instruction 101011 01001 01000 0000 0100 1011 0000 opcode $t1 $t0 1200 MemWrite activated Operation select from control (add) 32 Registers (reg. file) 5 01001 $t1 Read register numbers 3 Read data 32 zero 5 $t0 result 01000 ALU Addr. Data memory overflow Write reg. number 5 32 Write data Write data $t0data to mem. 32 32 Sign extend RegWrite from control MemRead 0000 0100 1011 0000 16 ELEC 5200-001/6200-001 Lecture 8

  16. Branch Instruction (I-Type) • beq $s1, $s2, 25 # if $s1 = $s2, advance PC through 25 instructions 16-bits 000100 10001 10010 0000 0000 0001 1001 opcode $s1 $s2 25 Note: Can branch within ± 215 words from the current instruction address in PC. ELEC 5200-001/6200-001 Lecture 8

  17. 3 ALU Add Datapath for beq Instruction 16-bits 000100 10001 10010 0000 0000 0001 1001 opcode $s1 $s2 25 Operation select from control (subtract) 32 Registers (reg. file) 5 10001 $s1 Read register numbers 32 5 $s2 To branch control logic 10010 zero result 5 32 Write reg. number overflow Write data PC+4 Branch target 32 RegWrite from control From instruction fetch datapath 32 Sign extend Shift left 2 16 0000 0000 0001 1001 32 32 32 ELEC 5200-001/6200-001 Lecture 8

  18. J-Type Instruction • j 2500 # jump to instruction 2,500 26-bits 0000100000 0000 0000 0010 0111 0001 00 opcode2,500 32-bit jump address 0000 0000 0000 0000 0010 0111 00010000 bits 28-31 from PC+4 ELEC 5200-001/6200-001 Lecture 8

  19. 0 mux 1 Add Datapath for Jump Instruction Branch Jump Branch addr. 32 4 1 mux 0 32 32 PC+4 4 Shift left 2 28 Instruction Memory 32 PC 26 opcode (bits 26-31) to control 32 6 Address 32 Instruction word to control and registers ELEC 5200-001/6200-001 Lecture 8

  20. Jump 0-25 Shift left2 0 mux 1 4 Add 1 mux 0 ALU Branch opcode MemtoReg CONTROL RegDst 26-31 21-25 zero MemWrite MemRead ALU Instr. mem. PC Reg. File Data mem. 1 mux 0 16-20 0 mux 1 1 mux 0 11-15 Combined Datapaths ALU Cont. Sign ext. Shift left 2 0-15 0-5 ELEC 5200-001/6200-001 Lecture 8

  21. Control RegDst Jump Branch MemRead MemtoReg ALUOp MemWrite ALUSrc RegWrite Instruction bits 26-31 opcode Control Logic 2 Instruction bits 0-5 funct. ALUControl to ALU ELEC 5200-001/6200-001 Lecture 8

  22. Control Logic: Truth Table ELEC 5200-001/6200-001 Lecture 8

  23. How Long Does It Take? • Assume control logic is fast and does not affect the critical timing. Major time components are ALU, memory read/write, and register read/write. • Arithmetic-type (R-type) • Fetch (memory read) 2ns • Register read 1ns • ALU operation 2ns • Register write 1ns • Total 6ns ELEC 5200-001/6200-001 Lecture 8

  24. Time for lw and sw (I-Types) • ALU (R-type) 6ns • Load word (I-type) • Fetch (memory read) 2ns • Register read 1ns • ALU operation 2ns • Get data (mem. Read) 2ns • Register write 1ns • Total 8ns • Store word (no register write) 7ns ELEC 5200-001/6200-001 Lecture 8

  25. Time for beq (I-Type) • ALU (R-type) 6ns • Load word (I-type) 8ns • Store word (I-type) 7ns • Branch on equal (I-type) • Fetch (memory read) 2ns • Register read 1ns • ALU operation 2ns • Total 5ns ELEC 5200-001/6200-001 Lecture 8

  26. Time for Jump (J-Type) • ALU (R-type) 6ns • Load word (I-type) 8ns • Store word (I-type) 7ns • Branch on equal (I-type) 5ns • Jump (J-type) • Fetch (memory read) 2ns • Total 2ns ELEC 5200-001/6200-001 Lecture 8

  27. How Fast Can the Clock Be? • If every instruction is executed in one clock cycle, then: • Clock period must be at least 8ns to perform the longest instruction. • This is a single cycle machine. • It is slower because many instructions take less than 8ns but are still allowed that much time. • Method of speeding up: Use multicycle datapath. ELEC 5200-001/6200-001 Lecture 8

  28. A Single Cycle Example Delay of 1-bit full adder = 1ns Clock period ≥ 32ns a31 . . . a2 a1 a0 c32 1-b full adder s31 . . . s2 s1 s0 1-b full adder b31 . . . b2 b1 b0 1-b full adder 1-b full adder Time of adding words ~ 32ns Time of adding bytes ~ 32ns 0 ELEC 5200-001/6200-001 Lecture 8

  29. A Multicycle Implementation a31 . . . a2 a1 a0 Delay of 1-bit full adder = 1ns Clock period ≥ 1ns Time of adding words ~ 32ns Time of adding bytes ~ 8ns Shift 1-b full adder b31 . . . b2 b1 b0 s31 . . . s2 s1 s0 c32 FF Shift Shift Initialize to 0 ELEC 5200-001/6200-001 Lecture 8

  30. Multicycle Datapath Instr. reg. (IR) A Reg. PC Addr. Memory ALU Register file ALUOut Reg. Data 4 Mem. Data (MDR) B Reg. One-cycle paths ELEC 5200-001/6200-001 Lecture 8

  31. Multicycle Datapath Requirements • Only one ALU, since it can be reused. • Single memory for instructions and data. • Five registers added: • Instruction register (IR) • Memory data register (MDR) • Three ALU registers, A and B for inputs and ALUOut for output ELEC 5200-001/6200-001 Lecture 8

  32. Multicycle Datapath PCSource PCWrite etc. Shift left 2 26-31 to Control FSM 0-25 RegWrite 21-25 28-31 16-20 Instr. reg. (IR) A Reg. PC Addr. 11-15 Memory ALU Register file ALUSrcA ALUSrcB ALUOut Reg. IorD Data Mem. Data (MDR) B Reg. out RegDst MUX control IRWrite 4 in1 in2 MemRead Sign extend Shift left 2 MemtoReg 0-15 MemWrite ALU control ALUOp 0-5 ELEC 5200-001/6200-001 Lecture 8

  33. 3 to 5 Cycles for an Instruction ELEC 5200-001/6200-001 Lecture 8

  34. Cycle 1 of 5: Instruction Fetch (IF) • Read instruction into IR, M[PC] → IR • Control signals used: • IorD = 0 select PC • MemRead = 1 read memory • IRWrite = 1 write IR • Increment PC, PC + 4 → PC • Control signals used: • ALUSrcA = 0 select PC into ALU • ALUSrcB = 01 select constant 4 • ALUOp = 00 ALU adds • PCSource = 00 select ALU output • PCWrite = 1 write PC ELEC 5200-001/6200-001 Lecture 8

  35. Cycle 2 of 5: Instruction Decode (ID) 31-26 25-21 20-16 15-11 10-6 5-0 R I J opcode | reg 1 | reg 2 | reg 3 | shamt | fncode • Control unit decodes instruction • Datapath prepares for execution • R and I types, reg 1→ A reg, reg 2 → B reg • No control signals needed • Branch type, compute branch address in ALUOut • ALUSrcA = 0 select PC into ALU • ALUSrcB = 11 Instr. Bits 0-15 shift 2 into ALU • ALUOp = 00 ALU adds opcode | reg 1 | reg 2 | word address increment opcode | word address jump ELEC 5200-001/6200-001 Lecture 8

  36. Cycle 3 of 5: Execute (EX) • R type: execute function on reg A and reg B, result in ALUOut • Control signals used: • ALUSrcA = 1 A reg into ALU • ALUsrcB = 00 B reg into ALU • ALUOp = 10 instr. Bits 0-5 control ALU • I type, lw or sw: compute memory address in ALUOut ← A reg + sign extend IR[0-15] • Control signals used: • ALUSrcA = 1 A reg into ALU • ALUSrcB = 10 Instr. Bits 0-15 into ALU • ALUOp = 00 ALU adds ELEC 5200-001/6200-001 Lecture 8

  37. Cycle 3 of 5: Execute (EX) • I type, beq: subtract reg A and reg B, write ALUOut to PC • Control signals used: • ALUSrcA = 1 A reg into ALU • ALUsrcB = 00 B reg into ALU • ALUOp = 01 ALU subtracts • If zero = 1, PCSource = 01 ALUOut to PC • If zero = 1, PCwriteCond =1 write PC • Instruction complete, go to IF • J type: write jump address to PC ← IR[0-25] shift 2 and four leading bits of PC • Control signals used: • PCSource = 10 • PCWrite = 1 write PC • Instruction complete, go to IF ELEC 5200-001/6200-001 Lecture 8

  38. Cycle 4 of 5: Reg Write/Memory • R type, write destination register from ALUOut • Control signals used: • RegDst = 1 Instr. Bits 11-15 specify reg. • MemtoReg = 0 ALUOut into reg. • RegWrite = 1 write register • Instruction complete, go to IF • I type, lw: read M[ALUOut] into MDR • Control signals used: • IorD = 1 select ALUOut into mem adr. • MemRead = 1 read memory to MDR • I type, sw: write M[ALUOut] from B reg • Control signals used: • IorD = 1 select ALUOut into mem adr. • MemWrite = 1 write memory • Instruction complete, go to IF ELEC 5200-001/6200-001 Lecture 8

  39. Cycle 5 of 5: Reg Write • I type, lw: write MDR to reg[IR(16-20)] • Control signals used: • RegDst = 0 instr. Bits 16-20 are write reg • MemtoReg = 1 MDR to reg file write input • RegWrite = 1 read memory to MDR • Instruction complete, go to IF For an alternative method of designing datapath, see N. Tredennick, Microprocessor Logic Design, the Flowchart Method, Digital Press, 1987. ELEC 5200-001/6200-001 Lecture 8

  40. 1-bit Control Signals zero(ALU) PCWriteCond PCWrite etc. PCWrite ELEC 5200-001/6200-001 Lecture 8

  41. 2-bit Control Signals ELEC 5200-001/6200-001 Lecture 8

  42. Control: Finite State Machine Start State 0 Clock cycle 1 Instruction fetch State 1 Clock cycle 2 Instruction decode and register fetch FSM-M Memory access instr. FSM-R FSM-B FSM-J Clock cycles 3-5 R-type instr. Branch instr. Jump instr. ELEC 5200-001/6200-001 Lecture 8

  43. State 0: Instruction Fetch (CC1) PCSource=00 PCWrite etc.=1 Shift left 2 26-31 to Control FSM 0-25 RegWrite 21-25 28-31 16-20 Instr. reg. (IR) A Reg. PC Addr. Memory ALU Register file ALUSrcB=01 ALUOut Reg. ALUSrcA=0 IorD=0 Data Mem. Data (MDR) B Reg. out RegDst MUX control IRWrite =1 4 Add in1 in2 MemRead = 1 Sign extend Shift left 2 MemtoReg 0-15 MemWrite ALUOp =00 ALU control 0-5 ELEC 5200-001/6200-001 Lecture 8

  44. MemRead =1 ALUSrcA = 0 IorD = 0 IRWrite = 1 ALUSrcB = 01 ALUOp = 00 PCWrite = 1 PCSource = 00 State 0 Control FSM Outputs Start State 1 Instruction decode/ Register fetch/ Branch addr. State0 Instruction fetch Outputs? ELEC 5200-001/6200-001 Lecture 8

  45. State 1: Instr. Decode/Reg. Fetch/ Branch Address (CC2) PCSource PCWrite etc. Shift left 2 26-31 to Control FSM 0-25 RegWrite 21-25 28-31 16-20 Instr. reg. (IR) A Reg. PC Addr. Memory ALU Register file ALUSrcA=0 ALUSrcB=11 ALUOut Reg. IorD Data Mem. Data (MDR) B Reg. out RegDst MUX control Add IRWrite 4 in1 in2 MemRead Sign extend Shift left 2 MemtoReg 0-15 MemWrite ALUOp = 00 ALU control 0-5 ELEC 5200-001/6200-001 Lecture 8

  46. MemRead =1 ALUSrcA = 0 IorD = 0 IRWrite = 1 ALUSrcB = 01 ALUOp = 00 PCWrite = 1 PCSource = 00 FSM-M FSM-R FSM-B FSM-J State 1 Control FSM Outputs Start State 1 Instruction decode (ID) / Register fetch / Branch addr. State0 Instruction fetch (IF) ALUSrcA = 0 ALUSrcB = 11 ALUOp = 00 Opcode = lw, sw Opcode = R-type Opcode = BEQ Opcode = J-type ELEC 5200-001/6200-001 Lecture 8

  47. State 1 (Opcode=“lw”) → FSM-M (CC3-5) PCSource CC4 PCWrite etc. Shift left 2 26-31 to Control FSM 0-25 RegWrite=1 21-25 28-31 16-20 CC3 Instr. reg. (IR) A Reg. PC Addr. Memory ALU Register file ALUSrcA=1 ALUSrcB=10 ALUOut Reg. IorD=1 Data Mem. Data (MDR) B Reg. out CC5 MUX control Add IRWrite 4 RegDst=0 in1 in2 MemRead=1 Sign extend Shift left 2 MemtoReg=1 0-15 MemWrite ALUOp = 00 ALU control 0-5 ELEC 5200-001/6200-001 Lecture 8

  48. State 1 (Opcode=“sw”)→FSM-M (CC3-4) CC4 PCSource PCWrite etc. Shift left 2 26-31 to Control FSM 0-25 RegWrite 21-25 28-31 16-20 CC3 Instr. reg. (IR) A Reg. PC Addr. Memory ALU Register file ALUSrcA=1 ALUSrcB=10 ALUOut Reg. IorD=1 Data Mem. Data (MDR) B Reg. out RegDst=0 MUX control Add IRWrite 4 CC4 in1 in2 MemRead Sign extend Shift left 2 MemtoReg 0-15 MemWrite=1 ALUOp = 00 ALU control 0-5 ELEC 5200-001/6200-001 Lecture 8

  49. FSM-M (Memory Access) From state 1 Opcode = “lw” or “sw” Compute mem addrress ALUSrcA =1 ALUSrcB = 10 ALUOp = 00 Opcode = “lw” Opcode = “sw” Read Memory data Write memory MemRead = 1 IorD = 1 MemWrite = 1 IorD = 1 Write register To state 0 (Instr. Fetch) RegWrite = 1 MemtoReg = 1 RegDst = 0 ELEC 5200-001/6200-001 Lecture 8

  50. State 1(Opcode=R-type)→FSM-R (CC3-4) PCSource PCWrite etc. Shift left 2 26-31 to Control FSM 0-25 RegWrite 21-25 28-31 16-20 Instr. reg. (IR) A Reg. PC Addr. Memory ALU Register file ALUSrcA=1 CC3 ALUSrcB=00 ALUOut Reg. 11-15 IorD Data Mem. Data (MDR) B Reg. out RegDst=0 MUX control IRWrite 4 “funct. code” CC4 in1 in2 MemRead Sign extend Shift left 2 MemtoReg=0 0-15 MemWrite ALUOp = 10 ALU control 0-5 ELEC 5200-001/6200-001 Lecture 8

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