ECE3055 Computer Architecture and Operating Systems Lecture 5 Datapath

1. ECE3055 Computer Architecture and Operating SystemsLecture 5 Datapath Prof. Hsien-Hsin Sean Lee School of Electrical and Computer Engineering Georgia Institute of Technology

2. The Processor: Datapath and Control • 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, j • 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 Why? memory-reference? arithmetic? control flow?

3. More Implementation Details • Abstract / Simplified View:Two types of functional units: • Elements that operate on data values (combinational) • Elements that contain state (sequential)

4. State Elements • Unclocked vs. Clocked • Clocks used in synchronous logic • when should an element that contains state be updated? falling edge cycle time rising edge

5. No Change Reset Set Undefined An unclocked state element • The set-reset latch • output depends on present inputs and also on past inputs R Q QN S

6. Latches and Flip-flops • Output is equal to the stored value inside the element (don't need to ask for permission to look at the value) • Change of state (value) is based on the clock • Latches: whenever the inputs change, and the clock is asserted • Flip-flop: state changes only on a clock edge (edge-triggered methodology) "logically true", — could mean electrically low A clocking methodology defines when signals can be read and written — wouldn't want to read a signal at the same time it was being written

7. D-latch • 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

8. Q 10T D Latch w/ Transmission Gates C C Q D En

9. D D Q 10T D Latch w/ Transmission Gates C=1 C Q D D En Writing Data

10. D D Q 10T D Latch w/ Transmission Gates C=0 C Q D D_new En Writing Data

11. En D Q Oscillating  Unstable Unstable Problem of Transparency D Q D Transparent D-Latch C 1

12. D flip-flop • Output changes only on the clock edge

13. Our Implementation • 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

14. Register File • Built using D flip-flops

15. Register File • Note: we still use the real clock to determine when to write

16. Simple Implementation • Include the functional units we need for each instruction A L U c o n t r o l 5 3 R e a d r e g i s t e r 1 R e a d d a t a 1 5 R e g i s t e r R e a d Z e r o r e g i s t e r 2 n u m b e r s R e g i s t e r s D a t a A L U A L U 5 W r i t e r e s u l t r e g i s t e r R e a d d a t a 2 W r i t e D a t a d a t a R e g W r i t e a . R e g i s t e r s b . A L U

17. Building the Datapath • Use multiplexers to stitch them together

18. P C S r c 1 M A d d u x A L U 0 4 A d d r e s u l t S h i f t R e g W r i t e l e f t 2 I n s t r u c t i o n [ 2 5 – 2 1 ] R e a d r e g i s t e r 1 R e a d M e m W r i t e R e a d P C d a t a 1 I n s t r u c t i o n [ 2 0 – 1 6 ] a d d r e s s R e a d M e m t o R e g A L U S r c r e g i s t e r 2 Z e r o I n s t r u c t i o n R e a d 1 A L U A L U [ 3 1 – 0 ] 1 R e a d W r i t e d a t a 2 1 A d d r e s s r e s u l t M r e g i s t e r M d a t a u M I n s t r u c t i o n u I n s t r u c t i o n [ 1 5 – 1 1 ] x W r i t e u x m e m o r y R e g i s t e r s x 0 d a t a 0 D a t a 0 W r i t e m e m o r y R e g D s t d a t a 1 6 3 2 S i g n I n s t r u c t i o n [ 1 5 – 0 ] e x t e n d A L U M e m R e a d c o n t r o l I n s t r u c t i o n [ 5 – 0 ] A L U O p R-Type Instructions (e.g. add $2, $3, $4; Not JR/JALR)

19. P C S r c 1 M A d d u x A L U 0 4 A d d r e s u l t S h i f t R e g W r i t e l e f t 2 I n s t r u c t i o n [ 2 5 – 2 1 ] R e a d r e g i s t e r 1 R e a d M e m W r i t e R e a d P C d a t a 1 I n s t r u c t i o n [ 2 0 – 1 6 ] a d d r e s s R e a d M e m t o R e g A L U S r c r e g i s t e r 2 Z e r o I n s t r u c t i o n R e a d 1 A L U A L U [ 3 1 – 0 ] 1 R e a d W r i t e d a t a 2 1 A d d r e s s r e s u l t M r e g i s t e r M d a t a u M I n s t r u c t i o n u I n s t r u c t i o n [ 1 5 – 1 1 ] x W r i t e u x m e m o r y R e g i s t e r s x 0 d a t a 0 D a t a 0 W r i t e m e m o r y R e g D s t d a t a 1 6 3 2 S i g n I n s t r u c t i o n [ 1 5 – 0 ] e x t e n d A L U M e m R e a d c o n t r o l I n s t r u c t i o n [ 5 – 0 ] A L U O p I-Type Instructions (e.g. lw $4, 1000($15))

20. P C S r c 1 M A d d u x A L U 0 4 A d d r e s u l t S h i f t R e g W r i t e l e f t 2 I n s t r u c t i o n [ 2 5 – 2 1 ] R e a d r e g i s t e r 1 R e a d M e m W r i t e R e a d P C d a t a 1 I n s t r u c t i o n [ 2 0 – 1 6 ] a d d r e s s R e a d M e m t o R e g A L U S r c r e g i s t e r 2 Z e r o I n s t r u c t i o n R e a d 1 A L U A L U [ 3 1 – 0 ] 1 R e a d W r i t e d a t a 2 1 A d d r e s s r e s u l t M r e g i s t e r M d a t a u M I n s t r u c t i o n u I n s t r u c t i o n [ 1 5 – 1 1 ] x W r i t e u x m e m o r y R e g i s t e r s x 0 d a t a 0 D a t a 0 W r i t e m e m o r y R e g D s t d a t a 1 6 3 2 S i g n I n s t r u c t i o n [ 1 5 – 0 ] e x t e n d A L U M e m R e a d c o n t r o l I n s t r u c t i o n [ 5 – 0 ] A L U O p I-type Instruction for Branches(e.g. beq $4, $5, Label7)

21. Control • Selecting the operations to perform (ALU, read/write, etc.) • Controlling the flow of data (multiplexer inputs) • Information comes from the 32 bits of the instruction • Example:add $8, $17, $18 Instruction Format:000000 10001 10010 01000 00000 100000 op rs rt rd shamt funct • ALU's operation based on instruction type and function code

22. A L U c o n t r o l 3 Z e r o A L U A L U r e s u l t Control • e.g., what should the ALU do with this instruction • Example: lw $1, 100($2) 35 2 1 100 op rs rt 16 bit offset • ALU control input000 AND 001 OR 010 add 110 subtract 111 set-on-less-than • Why is the code for subtract 110 and not 011? What do you need for slt instruction?

23. ALUOp computed from instruction type A L U c o n t r o l 3 ALUOp Funct field ALU Control ALUOp1 ALUOp0 F5 F4 F3 F2 F1 F0 Z e r o A L U A L U 0 0 X X X X X X 010 r e s u l t X 1 X X X X X X 110 1 X X X 0 0 0 0 010 1 X X X 0 0 1 0 110 1 X X X 0 1 0 0 000 1 X X X 0 1 0 1 001 1 X X X 1 0 1 0 111 Control the ALU • Must describe hardware to compute 3-bit ALU control input • given instruction type 00 = lw, sw 01 = beq, 11 = arithmetic (incl. slt) • function code for arithmetic • Describe it using a truth table (can turn into gates): ALUOp funct = inst[5:0] ALU control lw/sw add beq sub add sub arith and or slt Generated from Decoding inst[31:26] inst[5:0]

24. ALU Control • Simple combinational logic (truth tables)

25. Memto- Reg Mem Mem Instruction RegDst ALUSrc Reg Write Read Write Branch ALUOp1 ALUp0 R-format 1 0 0 1 0 0 0 1 0 lw 0 1 1 1 1 0 0 0 0 sw X 1 X 0 0 1 0 0 0 beq X 0 X 0 0 0 1 0 1 0 M u x A L U A d d 1 r e s u l t A d d S h i f t l e f t 2 R e g D s t 4 B r a n c h M e m R e a d M e m t o R e g I n s t r u c t i o n [ 3 1 – 2 6 ] C o n t r o l A L U O p M e m W r i t e A L U S r c R e g W r i t e I n s t r u c t i o n [ 2 5 – 2 1 ] R e a d R e a d r e g i s t e r 1 P C R e a d a d d r e s s d a t a 1 I n s t r u c t i o n [ 2 0 – 1 6 ] R e a d Z e r o r e g i s t e r 2 I n s t r u c t i o n 0 R e g i s t e r s A L U R e a d A L U [ 3 1 – 0 ] 0 R e a d W r i t e M d a t a 2 A d d r e s s r e s u l t 1 d a t a I n s t r u c t i o n r e g i s t e r M u M u m e m o r y x u I n s t r u c t i o n [ 1 5 – 1 1 ] W r i t e x 1 D a t a x d a t a 1 m e m o r y 0 W r i t e d a t a 1 6 3 2 I n s t r u c t i o n [ 1 5 – 0 ] S i g n e x t e n d A L U c o n t r o l I n s t r u c t i o n [ 5 – 0 ] Use rt not rd

26. Control Unit Signals Inst[31:26] To harness the datapath

27. Our Simple Control Structure • All of the logic is combinational • We wait for everything to settle down, and the right thing to be done • ALU might not produce “right answer” right away • we use write signals along with clock to determine when to write • Cycle time determined by length of the longest path We are ignoring some details like setup and holdtimes

28. Single Cycle Implementation • Calculate cycle time assuming negligible delays except: • memory (2ns), ALU and adders (2ns), register file access (1ns)

29. Where we are headed • Single Cycle Problems: • what if we had a more complicated instruction like floating point? • wasteful of area • One Solution: • use a “smaller” cycle time • have different instructions take different numbers of cycles • a “multicycle” datapath: