Single-cycle processor

this page covers chapter 4. but some of terms and concepts was already covered in Logic Circuits

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• A single cycle processor is a processor that carries out one instruction in a single clock cycle. (that is $\text{CPI}=1cc/ins$)

• These are the instructions we will implement:

  • R-type instructions: add, sub, and, or, slt
  • I-type instructions: lw, sw, beq
  • J-type instructions: j

1. Fetch
2. Decode
3. Execute
4. Memory
5. Write Back

Logic Design Conventions

• Datapath elements are classified into two types:

  • Combinational elements (e.g. ALU, AND gate)
  • State elements (e.g. Registers, Memory (data, instruction))

• clock methodology: edge-triggered (positive, rising edge)

• nearly all elements are 32-bit wide

• buses are indicated with thicker lines

Control

Main Control Unit

• Input: 6-bit opcode (instruction[31-26])
• Output:
  • RegDst: Whether write to register rt (0) or rd (1)
  • RegWrite: Whether Write register register is written with Write data value
  • ALUSrc:
    • 1 = the 2nd ALU operand is the sign-extended 16-bit immediate value,
    • 0 = the 2nd ALU operand comes from the 2nd register file output (Read data 2)
  • MemRead: 1 = read from memory
  • MemWrite: 1 = write to memory
  • MemtoReg:
    • 0 = the ALU result is written to the register file
    • 1 = the memory data is written to the register file
  • Branch: 1 = branch
  • ALUOp: 2-bit control signal for ALU operation
  • Jump: 1 = jump, 0 = don’t jump

Instr.	Opcode	RegDst	ALUOp1	ALUOp0	ALUSrc	Branch	MemRead	MemWrite	RegWrite	MemtoReg	Jump
R	000000 (0)	1	1	0	0	0	0	0	1	0	0
lw	100011 (35)	0	0	0	1	0	1	0	1	1	0
sw	101011 (43)	X	0	0	1	0	0	1	0	X	0
beq	000100 (4)	X	0	1	0	1	0	0	0	X	0
addi	001000 (8)	0	0	0	1	0	0	0	1	0	0
j	000010 (2)	X	X	X	X	X	0	0	0	X	1
jm	0x3f=111111	X	0	0	1	1	1	0	0	X	1

ALU Control Unit

Determine the 4-bit ALU control signal based on ALUOp signals and funct code.

• Input:
  • ${\mathrm{ALUOp}}_1$: 1-bit control signal (from the control unit)
  • ${\mathrm{ALUOp}}_0$: 1-bit control signal (from the control unit)
  • $\mathrm{Funct}=F_5{F}_4{F}_3{F}_2{F}_1{F}_0$: 6-bit function code (from instruction[5-0])
• Output:
  •  4-bit control signal for ALU operation (will be used to determine as the ALU operation in the ALU)
    • ${\mathrm{ALUControl}}_3=0$
    • ${\mathrm{ALUControl}}_2=(F_1\cdot {\mathrm{ALUOp}}_1)+{\mathrm{ALUOp}}_0$
    • ${\mathrm{ALUControl}}_1=(\overline{{\mathrm{ALUOp}}_1})+(\overline{F_2})$
    • ${\mathrm{ALUControl}}_0=(F_0+F_3)\cdot {\mathrm{ALUOp}}_1$

Input 1		Input 2		Output
${\mathrm{ALUOp}}_{1,0}$	opcode	Funct ($F_{5-0}$)	Instruction Operation	Operation
00	LW, SW	XXXXXX	load,store	0010	add
01	beq	XXXXXX	branch equal	0110	sub
10	R	100000	add	0010	add
10	R	100010	subtract	0110	sub
10	R	100100	AND	0000	and
10	R	100101	OR	0001	or
10	R	101010	slt	0111	slt

Datapath

Instruction Memory

Register File

Data Memory

ALU