inst.eecs.berkeley.edu/~cs61c
CS61C : Machine Structures
Lecture 28 –
Single Cycle CPU Control I
2004-11-02
Lecturer PSOE Dan Garcia
www.cs.berkeley.edu/~ddgarcia
Déjà vu all over again! Who Won?
As of 2am 2004-11-03,
it looks like Bush was ahead but hadn’t
yet clinched it. We may have to wait for a
recount and an Ohio tabulation of
provisional ballots. A Country Divided!
cnn.com
CS 61C L28 Single Cycle CPU Control I (1)
Garcia, Fall 2004 © UCB
Review: How to Design a Processor: step-by-step
• 1. Analyze instruction set architecture (ISA)
=> datapath requirements
• meaning of each instruction is given by the
register transfers
• datapath must include storage element for ISA
registers
• datapath must support each register transfer
• 2. Select set of datapath components and
establish clocking methodology
• 3. Assemble datapath meeting requirements
• 4. Analyze implementation of each
instruction to determine setting of control
points that effects the register transfer.
•
5. Assemble the control logic
CS 61C L28 Single Cycle CPU Control I (2)
Garcia, Fall 2004 © UCB
Why do we have two dirty bits?
if (wEnb)
if (writeReg!=4'h0)
begin
array[writeReg] = writeD;
dirty1=1'b1;
dirty2=1'b1;
end
always @ (readReg1 or dirty1)
begin
readD1 = array[readReg1];
dirty1=0;
end
always @ (readReg2 or dirty2)
begin
readD2 = array[readReg2];
dirty2=0;
end
CS 61C L28 Single Cycle CPU Control I (3)
Garcia, Fall 2004 © UCB
Review : 3e: Store Operations
• Mem[ R[rs] + SignExt[imm16] ] = R[rt]
Ex.: sw rt, rs, imm16
31
26
21
op
rs
6 bits 5 bits
Rd Rt
RegDst
Mux
RegWr5 5
rt
5 bits
W_Src
32
ExtOp
CS 61C L28 Single Cycle CPU Control I (4)
32
Data In32
Clk
WrEn Adr
32
Data
Memory
Mux
Extender
16
immediate
16 bits
ALUctr MemWr
ALU
busA
Rw Ra Rb
32
32 32-bit
Registers busB
32
imm16
0
Rs Rt
5
Mux
busW
32
Clk
16
ALUSrc
Garcia, Fall 2004 © UCB
3f: The Branch Instruction
31
26
op
6 bits
21
rs
5 bits
16
rt
5 bits
0
immediate
16 bits
• 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
CS 61C L28 Single Cycle CPU Control I (5)
Garcia, Fall 2004 © UCB
Datapath for Branch Operations
• beq rs, rt, imm16
Datapath generates condition (equal)
26
op
6 bits
21
00
Adder
32
PC
Mux
Adder
PC Ext
imm16
0
rs
rt
immediate
5 bits 5 bits
16 bits
Inst Address
nPC_sel
4
16
Rs Rt
5
busA
Rw Ra Rb
32
32 32-bit
Registers
busB
32
Cond
RegWr 5 5
busW
Clk
Equal?
31
Clk
• Already MUX, adder, sign extend, zero
CS 61C L28 Single Cycle CPU Control I (6)
Garcia, Fall 2004 © UCB
Putting it All Together:A Single Cycle Datapath
Instruction<31:0>
<0:15>
<11:15>
Rs
<16:20>
<21:25>
Inst
Memory
Adr
Rt Rd Imm16
RegDst
ALUctr MemWr MemtoReg
Equal
Rt
Rd
1 0
Rs Rt
RegWr 5 5 5
busA
Rw
Ra
Rb
=
busW
32
32 32-bit
0
32
32
Registers busB
0
32
Clk
32
WrEn Adr 1
1 Data In
Data
imm16
32
Clk
16
Clk Memory
nPC_sel
imm16
Mux
ALU
Extender
PC Ext
Adder
Mux
PC
Mux
Adder
00
4
ExtOp ALUSrc
CS 61C L28 Single Cycle CPU Control I (7)
Garcia, Fall 2004 © UCB
An Abstract View of the Critical Path
Critical Path (Load Operation) =
• This affects
Delay clock through PC (FFs) +
how much you
Instruction Memory’s Access Time +
can overclock
Register File’s Access Time +
your PC!
ALU to Perform a 32-bit Add +
Data Memory Access Time +
Ideal
Instruction
InstructionStable Time for Register File Write
Memory
Rd Rs Rt
5 5
5
PC
Clk
A
32
Rw Ra Rb
32 32-bit 32
Registers B
Clk
CS 61C L28 Single Cycle CPU Control I (8)
ALU
Next Address
Instruction
Address
Imm
16
32
Data
32 Address
Data
In
Ideal
Data
Memory
Clk
Garcia, Fall 2004 © UCB
An Abstract View of the Implementation
PC
Clk
Next Address
ALU
Control
Ideal
Instruction
Instruction Control Signals Conditions
Memory Rd Rs Rt
5 5
5
Instruction
Address
A
Data
Data
32 Address
Rw
Ra
Rb
32
Ideal
Out
32 32-bit 32
Data
Data
Registers B
Memory
In
Clk
32
Clk
Datapath
CS 61C L28 Single Cycle CPU Control I (9)
Garcia, Fall 2004 © UCB
Summary: A Single Cycle Datapath
• Rs, Rt, Rd, Imed16 connected to datapath
• We have everything except control signals
Instruction<31:0>
MemWr
Clk
MemtoReg
0
32
Data In32
ALUSrc
Rs Rd Imm16
WrEn Adr
Data
Memory
32
Mux
32
1
<0:15>
Extender
16
ALU
busA
Rw Ra Rb
32
32 32-bit
Registers busB
0
32
imm16
Rt
Zero
ALUctr
Mux
busW
32
Clk
Clk
<11:15>
Rt
RegDst
1 Mux0
Rs Rt
RegWr 5 5 5
<16:20>
Rd
Instruction
Fetch Unit
<21:25>
nPC_sel
1
ExtOp
CS 61C L28 Single Cycle CPU Control I (10)
Garcia, Fall 2004 © UCB
Anatomy Review: 5 components of any Computer
Personal Computer
Computer
Processor
Today
Control
(“brain”)
Yesterday Datapath
(“brawn”)
(& finish up)
Memory
(where
programs,
data
live when
running)
Devices
Input
Output
Keyboard,
Mouse
Disk
(where
programs,
data
live when
not running)
Display,
Printer
CS 61C L28 Single Cycle CPU Control I (11)
Garcia, Fall 2004 © UCB
Recap: Meaning of the Control Signals
0  PC <– PC + 4
1  PC <– PC + 4 +
“n”=next
{SignExt(Im16) , 00 }
• Later in lecture: higher-level connection
between mux and branch cond
• nPC_MUX_sel:
nPC_MUX_sel
Inst
Adr Memory
Adder
imm16
PC
Mux
Adder
PC Ext
CS 61C L28 Single Cycle CPU Control I (12)
00
4
Clk
Garcia, Fall 2004 © UCB
Recap: Meaning of the Control Signals
° MemWr: 1  write memory
• ExtOp: “zero”, “sign”
° MemtoReg: 0  ALU; 1  Mem
• ALUsrc: 0  regB;
1  immed
° RegDst: 0  “rt”; 1  “rd”
• ALUctr: “add”, “sub”, “or” ° RegWr: 1  write register
RegDst
ALUctr MemWr MemtoReg
=
32
WrEn Adr
Data In
Data
Clk Memory
ExtOp ALUSrc
CS 61C L28 Single Cycle CPU Control I (13)
32
0
Mux
ALU
Mux
Extender
Equal
Rd Rt
1 0
Rs Rt
RegWr 5 5 5
busA
Rw
Ra
Rb
busW
32
32 32-bit
32
Registers busB
0
32
Clk
1
imm16
32
16
1
Garcia, Fall 2004 © UCB
Great talk today – Don’t miss
306 Soda Hall @ 4pm
• Dr. David Anderson
• Space Sciences Laboratory,
U.C. Berkeley. SETI Director
“Public Resource Computing”
The majority of the world's computing power is no
longer concentrated in supercomputer centers and
machine rooms. Instead it is distributed around the
world in hundreds of millions of personal
computers and game consoles, many connected to
the Internet. A new computing paradigm, “publicresource computing”, uses these PCs to do
scientific supercomputing. This paradigm enables
new research in a number of areas and has social
implications as well: it catalyzes global
communities centered around common interests
and goals, it encourages public awareness of
current scientific research, and it may give the
public a measure of control over the directions of
science progress.
CS 61C L28 Single Cycle CPU Control I (14)
Garcia, Fall 2004 © UCB
Administrivia
• Dan will be away at a conference on
Thursday and Friday, Andrew will
cover lecture.
• We regraded all the midterms and your
TAs have them to return to you.
CS 61C L28 Single Cycle CPU Control I (15)
Garcia, Fall 2004 © UCB
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
CS 61C L28 Single Cycle CPU Control I (16)
Garcia, Fall 2004 © UCB
Instruction Fetch Unit at the Beginning of Add
• Fetch the instruction from Instruction
memory: Instruction = MEM[PC]
• same for
all instructions
Inst
Memory
Adr
Instruction<31:0>
nPC_MUX_sel
imm16
Adder
PC Ext
CS 61C L28 Single Cycle CPU Control I (17)
PC
Mux
Adder
00
4
Clk
Garcia, Fall 2004 © UCB
The Single Cycle Datapath during Add
31
26
21
op
rs
16
11
rt
6
rd
shamt
• R[rd] = R[rs] + R[rt]
5
Zero
ALU
16
Extender
imm16
1
32
Rd
Clk
Imm16
MemtoReg = 0
MemWr = 0
0
32
Data In 32
ALUSrc = 0
Rs
WrEn Adr
32
Mux
busA
Rw Ra Rb
32
32 32-bit
Registers
busB
0
32
Rt
<0:15>
5
ALUctr = Add
Rt
<11:15>
5
Rs
Mux
32
Clk
Clk
1 Mux 0
RegWr = 1
busW
Rt
Instruction
Fetch Unit
<16:20>
RegDst = 1
Rd
funct
Instruction<31:0>
<21:25>
nPC_sel= +4
0
1
Data
Memory
ExtOp = x
CS 61C L28 Single Cycle CPU Control I (18)
Garcia, Fall 2004 © UCB
Instruction Fetch Unit at the End of Add
• PC = PC + 4
• This is the same for all instructions except:
Branch and Jump
Inst
Memory
Adr
Instruction<31:0>
nPC_MUX_sel
CS 61C L28 Single Cycle CPU Control I (19)
PC
Mux
Adder
imm16
Adder
0
00
4
1
Clk
Garcia, Fall 2004 © UCB
Peer Instruction
1:
Suppose we’re writing a MIPS interpreter in
2:
Verilog. Which sequence below is best
3:
organization for the interpreter?
4:
A. repeat loop that fetches instructions
5:
B. while loop that fetches instructions
6:
C. Decodes instructions using case statement 7:
D. Decodes instr. using chained if statements 8:
9:
E. Executes each instruction
0:
F. Increments PC by 4
CS 61C L28 Single Cycle CPU Control I (20)
ACEF
ADEF
AECF
AEDF
BCEF
BDEF
BECF
BEDF
EF
FAE
Garcia, Fall 2004 © UCB
Summary: Single cycle datapath
°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.
Processor
Input
• 5. Assemble the control logic
Control
°Control is the hard part
°MIPS makes that easier
Memory
Datapath
• Instructions same size
• Source registers always in same place
• Immediates same size, location
• Operations always on registers/immediates
CS 61C L28 Single Cycle CPU Control I (21)
Output
Garcia, Fall 2004 © UCB