CS 61C: Great Ideas in Computer
Architecture (Machine Structures)
Lecture 30: Single-Cycle CPU
Datapath Control Part 2
Instructor: Miki Garcia
http://inst.eecs.berkeley.edu/~cs61c
FUNCTIONAL MRI REVEALS AMOUNT
PARALLEL PROCESSING IN BRAIN?
MIT Technology Review: Using independent
component analysis on fMRI data during visuomotor task, researchers from Greece conclude
that there are about 50 parallel processes
(“cores”) in the brain
http://science.slashdot.org/story/14/11/08/1913229/fmri-data-reveals-how-manyparallel-processes-run-in-the-brain
Design Steps – What we did
• MIPS light ISA:
+4
rd
rs
rt
Data
memory
registers
PC
instruction
memory
– ADDU, SUBU, ORI, LOAD, STORE, BEQ
ALU
imm
opcode, funct
==
zero
Controller
6/27/2016
Fall 2011 -- Lecture #29
3
Review: Processor Design 5 steps
Step 1: Analyze instruction set to determine datapath
requirements
– Meaning of each instruction is given by register transfers
– Datapath must include storage element for ISA registers
– Datapath must support each register transfer
Step 2: Select set of datapath components & establish
clock methodology
Step 3: Assemble datapath components that meet the
requirements
Step 4: Analyze implementation of each instruction to
determine setting of control points that realizes the
register transfer
Step 5: Assemble the control logic
Processor Design: 5 steps
Step 1: Analyze instruction set to determine datapath
requirements
– Meaning of each instruction is given by register transfers
– Datapath must include storage element for ISA registers
– Datapath must support each register transfer
Step 2: Select set of datapath components & establish
clock methodology
Step 3: Assemble datapath components that meet the
requirements
Step 4: Analyze implementation of each instruction to
determine setting of control points that realizes the
register transfer
Step 5: Assemble the control logic
Add & Subtract Ctrl + data timing
Clk
PC Old Value
Rs, Rt, Rd,
Op, Func
Old Value
ALUctr
Old Value
RegWr
Old Value
busA, B
Old Value
busW
Old Value
New Value
Instruction Memory Access Time
New Value
Delay through Control Logic
New Value
New Value
Register File Access Time
New Value
ALU Delay
New Value
ALUctr
RegWr Rd Rs Rt
5
Rw
busW
5
Ra Rb
busA
32
ALU
RegFile
clk
5
busB
32
32
Register Write
Occurs Here
3c: Logical Op (or) with Immediate
• R[rt] = R[rs] op ZeroExt[imm16]
31
26
21
op
rs
31 6 bits
RegDst
RegWr
Rw
Rs Rt
5 bits 16 15
16 bits
immediate
16 bits
5
ALUctr
5
Ra Rb
busA
busB
32
clk
16
ZeroExt
imm16
32
ALU
RegFile
32
5 bits
immediate
Writing to Rt register (not Rd)!!
0
5
0
rt
0000000000000000
16 bits
Rd Rt
1
16
0
1
32
ALUSrc
32
What about Rt
Read?
0
3d: Load Operations
• R[rt] = Mem[R[rs] + SignExt[imm16]]
Example: lw rt,rs,imm16
31
26
21
op
16
rs
6 bits
0
rt
5 bits
immediate
5 bits
16 bits
RegDst Rd Rt
1
RegWr
0
5
Rw
Rs Rt
5
ALUctr
5
Ra Rb
busA
busB
32
clk
16
ZeroExt
imm16
ALU
RegFile
32
32
0
1
32
ALUSrc
32
3d: Load Operations
• R[rt] = Mem[R[rs] + SignExt[imm16]]
Example: lw rt,rs,imm16
31
26
21
op
16
rs
6 bits
0
rt
5 bits
immediate
5 bits
16 bits
1
RegWr
0
Rs Rt
5
5
Rw
busW
5
Ra Rb
busA
busB
32
imm16
16
ExtOp
Extender
clk
32
ALU
RegFile
32
MemtoReg
ALUctr
RegDst Rd Rt
32
0
0
Adr
1
32
ALUSrc
clk
Data
Memory
1
3e: Store Operations
• Mem[ R[rs] + SignExt[imm16] ] = R[rt]
Ex.: sw rt, rs, imm16
31
26
21
op
6 bits
16
rs
5 bits
rt
5 bits
immediate
16 bits
RegDst Rd Rt
1
RegWr
ALUctr
0
5
5
5
busA
busB
32
imm16
16
ExtOp
Extender
clk
32
0
ALU
RegFile
32
MemtoReg
MemWr
Rs Rt
Rw Ra Rb
busW
0
32
0
32 WrEn Adr
1
32
ALUSrc
Data In
clk
Data
Memory
1
3e: Store Operations
• Mem[ R[rs] + SignExt[imm16] ] = R[rt]
Ex.: sw rt, rs, imm16
31
26
21
op
6 bits
16
rs
5 bits
rt
5 bits
immediate
16 bits
RegDst Rd Rt
1
RegWr
ALUctr
0
5
5
5
busA
busB
32
imm16
16
ExtOp
Extender
clk
32
0
ALU
RegFile
32
MemtoReg
MemWr
Rs Rt
Rw Ra Rb
busW
0
32
0
32 WrEn Adr
1
32
ALUSrc
Data In
clk
Data
Memory
1
3f: The Branch Instruction
31
26
op
6 bits
21
rs
5 bits
16
rt
5 bits
immediate
16 bits
0
beq rs, rt, imm16
– mem[PC] Fetch the instruction from memory
– Equal = R[rs] == R[rt] Calculate branch condition
– if (Equal) Calculate the next instruction’s address
• PC = PC + 4 + ( SignExt(imm16) x 4 )
else
• PC = PC + 4
Datapath for Branch Operations
beq rs, rt, imm16
31
26
op
6 bits
21
rs
5 bits
16
0
rt
5 bits
immediate
16 bits
Datapath generates condition (Equal)
Inst Address
00
RegWr
clk
busW
clk
5
Rs Rt
5
ALUctr
5
Rw Ra Rb
busA
RegFile
busB
32
=
ALU
PC
Mux
Adder
PC Ext
imm16
nPC_sel
Adder
4
Equal
32
32
Already have mux, adder, need special sign
extender for PC, need equal compare (sub?)
Instruction Fetch Unit including Branch
31
26
op
21
rs
16
rt
0
immediate
• if (Zero == 1) then PC = PC + 4 + SignExt[imm16]*4 ; else
PC = PC + 4
Inst
nPC_sel
Equal
MUX
ctrl
nPC_sel
• How to encode nPC_sel?
0
00
• Direct MUX select?
• Branch inst. / not branch inst.
Mux
PC
Adder
4
Adder
PC Ext
imm16
Instruction<31:0>
Memory
Adr
1
clk
• Let’s pick 2nd option
nPC_sel
0
1
1
zero?
x
0
1
MUX
0
0
1
Q: What logic
gate?
Putting it All Together:A Single Cycle Datapath
RegDst
32
0
5
5
5
Rw Ra Rb
RegFile
busA
busB
32
16
Extender
imm16
MemtoReg
MemWr
Rs Rt
clk
clk
ALUctr
Equal
32
=
ALU
busW
PC
PC Ext
Adder
Mux
00
RegWr
Adder
4
Rt Rd Imm16
Rd Rt
1
Instruction<31:0>
<0:15>
nPC_sel
Rs
<11:15>
Adr
<16:20>
<21:25>
Inst
Memory
0
32
1
32
Data In
clk
imm16
ExtOp
ALUSrc
32
0
WrEn Adr
Data
Memory
1
Datapath Control Signals
• ExtOp:
• ALUsrc:
• ALUctr:
•
•
•
•
“zero”, “sign”
0  regB;
1  immed
“ADD”, “SUB”, “OR”
MemWr:
MemtoReg:
RegDst:
RegWr:
ALUctr
MemtoReg
MemWr
RegDst Rd Rt
1
Inst Address
4
RegWr
0
Rs Rt
5
5
Rw
busW
Ra Rb
busA
RegFile
clk
imm16
16
Extender
PC Ext
Adder
1
32
0
0
32
clk
32
busB
PC
Mux
32
ExtOp
imm16
5
ALU
Adder
0
00
nPC_sel & Equal
1  write memory
0  ALU; 1  Mem
0  “rt”; 1  “rd”
1  write register
32 WrEn Adr
1
Data In
ALUSrc
clk
32
Data
Memory
1
Given Datapath: RTL  Control
Instruction<31:0>
Rd
<0:15>
Rs
<11:15>
Rt
<16:20>
Op Fun
<21:25>
<0:5>
<26:31>
Inst
Memory
Adr
Imm16
Control
nPC_sel RegWr RegDst ExtOp ALUSrc ALUctr
DATA PATH
MemWr MemtoReg
RTL: The Add Instruction
31
26
op
6 bits
21
rs
5 bits
16
rt
5 bits
11
6
0
rd
shamt
funct
5 bits
5 bits
6 bits
add rd, rs, rt
– MEM[PC]
Fetch the instruction from memory
– R[rd] = R[rs] + R[rt] The actual operation
– PC = PC + 4 Calculate the next instruction’s address
Instruction Fetch Unit at the Beginning of Add
• Fetch the instruction from Instruction
memory: Instruction = MEM[PC]
– same for
all instructions
Inst
Memory
nPC_sel
00
Adder
4
PC
Mux
Adder
PC Ext
imm16
clk
Inst Address
Instruction<31:0>
Single Cycle Datapath during Add
31
26
op
21
16
rs
11
rt
rd
6
shamt
R[rd] = R[rs] + R[rt]
5
5
Rw
busW
5
busA
Ra Rb
busB
32
imm16
16
ExtOp=x
Extender
clk
Rs Rt Rd Imm16
zero ALUctr=ADD
MemtoReg=0
MemWr=0
32
=
ALU
RegFile
32
<0:15>
Rs Rt
<11:15>
RegWr=1
0
<16:20>
1
<21:25>
Rd Rt
funct
Instruction<31:0>
instr
fetch
unit
nPC_sel=+4
RegDst=1
clk
0
32
0
0
32
1
Data In
32
ALUSrc=0
clk
WrEn Adr
Data
Memory
1
Instruction Fetch Unit at End of Add
• PC = PC + 4
– Same for all
instructions except:
Branch and Jump
nPC_sel=+4
00
Adder
4
Inst
Memory
PC
Mux
Adder
PC Ext
imm16
clk
Inst Address
P&H Figure 4.17
Summary of the Control Signals (1/2)
inst
Register Transfer
add
R[rd]  R[rs] + R[rt]; PC  PC + 4
ALUsrc=RegB, ALUctr=“ADD”, RegDst=rd, RegWr, nPC_sel=“+4”
sub
R[rd]  R[rs] – R[rt]; PC  PC + 4
ALUsrc=RegB, ALUctr=“SUB”, RegDst=rd, RegWr, nPC_sel=“+4”
ori
R[rt]  R[rs] + zero_ext(Imm16); PC  PC + 4
ALUsrc=Im, Extop=“Z”, ALUctr=“OR”, RegDst=rt,RegWr, nPC_sel=“+4”
lw
R[rt]  MEM[ R[rs] + sign_ext(Imm16)]; PC  PC + 4
ALUsrc=Im, Extop=“sn”, ALUctr=“ADD”, MemtoReg, RegDst=rt, RegWr,
nPC_sel = “+4”
sw
MEM[ R[rs] + sign_ext(Imm16)]  R[rs]; PC  PC + 4
ALUsrc=Im, Extop=“sn”, ALUctr = “ADD”, MemWr, nPC_sel = “+4”
beq
if (R[rs] == R[rt]) then PC  PC + sign_ext(Imm16)] || 00
else PC  PC + 4
nPC_sel = “br”,
ALUctr = “SUB”
Summary of the Control Signals (2/2)
See
Appendix A
func 10 0000 10 0010
We Don’t Care :-)
op 00 0000 00 0000 00 1101 10 0011 10 1011 00 0100 00 0010
add
sub
ori
lw
sw
beq
jump
RegDst
1
1
0
0
x
x
x
ALUSrc
0
0
1
1
1
0
x
MemtoReg
0
0
0
1
x
x
x
RegWrite
1
1
1
1
0
0
0
MemWrite
0
0
0
0
1
0
0
nPCsel
0
0
0
0
0
1
?
Jump
0
0
0
0
0
0
1
ExtOp
x
x
0
1
1
x
Add
Subtract
Or
Add
Add
x
Subtract
ALUctr<2:0>
31
26
21
16
R-type
op
rs
rt
I-type
op
rs
rt
J-type
op
11
rd
6
shamt
immediate
target address
x
0
funct
add, sub
ori, lw, sw, beq
jump
Boolean Expressions for Controller
RegDst
ALUSrc
MemtoReg
RegWrite
MemWrite
nPCsel
Jump
ExtOp
ALUctr[0]
ALUctr[1]
=
=
=
=
=
=
=
=
=
=
add + sub
ori + lw + sw
lw
add + sub + ori + lw
sw
beq
jump
lw + sw
sub + beq
(assume ALUctr is 00 ADD, 01 SUB, 10 OR)
or
Where:
rtype
ori
lw
sw
beq
jump
=
=
=
=
=
=
~op5
~op5
op5
op5
~op5
~op5






~op4
~op4
~op4
~op4
~op4
~op4






~op3
op3
~op3
op3
~op3
~op3






~op2
op2
~op2
~op2
op2
~op2






~op1
~op1
op1
op1
~op1
op1
 ~op0,
 op0
 op0
 op0
 ~op0
 ~op0
How do we
implement this in
gates?
add = rtype  func5  ~func4  ~func3  ~func2  ~func1  ~func0
sub = rtype  func5  ~func4  ~func3  ~func2  func1  ~func0
Controller Implementation
opcode
func
“AND” logic
add
sub
ori
lw
sw
beq
jump
“OR” logic
RegDst
ALUSrc
MemtoReg
RegWrite
MemWrite
nPCsel
Jump
ExtOp
ALUctr[0]
ALUctr[1]
Peer Instruction
1) We should use the main ALU to
compute PC=PC+4 in order to save
some gates
2) The ALU is inactive for memory reads
(loads) or writes (stores).
a)
b)
c)
d)
e)
12
FF
FT
TF
TT
Help!
Summary: Single-cycle Processor
• Five steps to design a processor:
Processor
1. Analyze instruction set 
Input
datapath requirements
Control
Memory
2. Select set of datapath
components & establish
Datapath
Output
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
• Formulate Logic Equations
• Design Circuits
Bonus Slides
• How to implement Jump
Single Cycle Datapath during Jump
31
26 25
J-type
0
op
target address
jump
• New PC = { PC[31..28], target address, 00 }
Instruction<31:0>
Jump=
Clk
ALUSrc =
ExtOp =
Rd Imm16 TA26
MemtoReg =
MemWr =
0
32
Data In 32
<0:25>
32
Rs
WrEn Adr
Data
Memory
32
Mux
1
<0:15>
16
Extender
imm16
Zero
Mux
32
Clk
busA
Rw Ra Rb
32
32 x 32-bit
Registers busB
0
32
ALU
busW
5
Rt
ALUctr =
Rs Rt
5 5
<11:15>
RegWr =
Clk
<16:20>
RegDst =
<21:25>
Rd
Rt
1 Mux 0
Instruction
Fetch Unit
nPC_sel=
1
Single Cycle Datapath during Jump
31
26 25
0
op
J-type
target address
jump
• New PC = { PC[31..28], target address, 00 }
Instruction<31:0>
Jump=1
ExtOp = x
<0:25>
Clk
ALUSrc = x
Rd Imm16 TA26
MemtoReg = x
MemWr = 0
0
WrEn Adr
Data
Memory
32
Mux
32
Rs
32
Data In 32
<0:15>
1
<11:15>
16
Extender
imm16
Zero
Mux
32
Clk
busA
Rw Ra Rb
32
32 x 32-bit
busB
Registers
0
32
ALU
busW
5
Rt
ALUctr =x
Rs Rt
5 5
<16:20>
RegWr = 0
Clk
<21:25>
RegDst = x
Rd
Rt
1 Mux 0
Instruction
Fetch Unit
nPC_sel=?
1
Instruction Fetch Unit at the End of Jump
31
26 25
J-type
0
op
target address
jump
• New PC = { PC[31..28], target address, 00 }
Jump
Inst
Memory
nPC_sel
Adr
Instruction<31:0>
Zero
nPC_MUX_sel
Adder
00
4
0
imm16
PC
Mux
Adder
1
Clk
How do we modify this
to account for jumps?
Instruction Fetch Unit at the End of Jump
31
26 25
0
op
J-type
target address
jump
• New PC = { PC[31..28], target address, 00 }
Inst
Memory
Jump
nPC_sel
Instruction<31:0>
Adr
Zero
imm16
00
TA
Mux
Adder
1
4 (MSBs)
1
PC
Adder
0
26
Mux
4
00
nPC_MUX_sel
0
Clk
Query
• Can Zero still
get asserted?
• Does nPC_sel
need to be 0?
• If not, what?