An Example MicroArchitecture - CS Department

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CS 3A: Introduction to
Computer Organization
Example MicroArchitecture
Tannenbaum 4.1.1, 4.1.2
Department of Computer Science
Saddleback College
Shannon Alfaro
MicroArchitecture Level
• A brief look at our
levels again…
– We’ve looked at the
lowest level : Digital
Logic Level
– Now we’ll look at the
MicroArchitecture
Level
Big picture
Design of MicroArchitecture
• Depends on ISA being implemented
• ISA: RISC: 1 instruction/cycle; SPARC
CISC: 1 instruction/(>1cycle); Pentium
• Want to:
explain general principles of microarchitecture design but
 THERE ARE NONE!! EACH IS UNIQUE
Example MicroArchitecture
• Chosen a subset JVM for integer
operations: IJVM : Integer Java Virtual
Machine
• Our microarchitecture will contain a
microprogram (in ROM) whose job is to:
– Fetch
– Decode
– Execute _________ instructions
1 blank
Example MicroArchitecture
• Imagine the design of MicroArchitecture as the
following code:
• The microprogram has a set of variables
• Values represent the ________ of the computer
1 blank
Example MicroArchitecture
• IJVM Instructions: short; usually 1-2 fields
Opcode
Operation to
Perform
Operand
register locations/
variables
• Every instruction has an ________
• Many instructions have _________
• Model of Execution:
fetch-execute cycle
2 blanks
The Language to be
High Level
Implemented
Instructions
Compiler/
Assembler
MicroProgram
in CPU
MicroInstructions
IJVM: ISA Instructions
Memories…..
• Remember When???
CPU = Controller + Datapath
• Datapath: contains
– ALU
– Registers
– inputs & outputs (not shown)
IJVM Datapath
• Datapath: contains
– 32-bit registers
• registers are accessible only by
micro-program at the
microarchitecture level.
– B-BUS: Contents of Registers
– C-BUS: Output of ALU & can
write to multiple registers at
once
– ALU
IJVM Datapath
• Datapath: contains an ALU:
– just like what we constructed in chapter 3
(remember our decoder +
Adder/Subtractor & Logic gates?)
– Needs 2 inputs:
A(left)  register H (1 source)
B(right)  Bus B (9 sources)
Stack Based Machine
• Stack Based Machine: Operands are
placed on a stack & result is stored on the
stack
• SP: top of stack pointer
• LV: pointer to local variables
IJVM Instruction Set
Compiling Java to IJVM
IJVM Arithmetic Logic Unit
• ALU Operation:
– To load H:
• choose an ALU function
that
1. Passes the value at the B
input through the ALU
2. Writes value back into H
i.e. identity of B
ALU Truth Table
IJVM Arithmetic Logic Unit
• ALU Operation:
– Read & Write in Same Cycle
• Can happen with magic & timing
1st half of cycle you read register
2nd half you write
ALU Truth Table
Datapath Timing
•
Propagation Delay
– Just as in our homework problem, there is a delay
before the output of our gates is stable
•
•
•
•
Δx before values is stable; then ALU & shifter
can begin computation
Δy: ALU & Shifter outputs are stable
Δz: results propagated along C bus to
registers
Rising Edge: Registers Latch values into
•Falling Edge:
memory cells
•signals set up for output
•Δw time passes before valid
Datapath Timing
•
Propagation Delay
– Just as in our homework problem, there is a delay
before the output of our gates is stable
•
•
•
Δy: ALU & Shifter outputs are stable
Δz: results propagated along C bus to
registers
Rising Edge: Registers Latch values into
memory cells
•Δx before values is stable
•then ALU & shifter can begin computation
Datapath Timing
•
Propagation Delay
– Just as in our homework problem, there is a delay
before the output of our gates is stable
•
•
Δz: results propagated along C bus to
registers
Rising Edge: Registers Latch values into
memory cells
•Δy: ALU & Shifter outputs are stable
Datapath Timing
•
Propagation Delay
– Just as in our homework problem, there is a delay
before the output of our gates is stable
•
•
Δz: results propagated along C bus to
registers
Rising Edge: Registers Latch values into
memory cells
•Δz:
results propagated along C bus to
registers
Datapath Timing
•
Propagation Delay
– Just as in our homework problem, there is a delay
before the output of our gates is stable
•Rising Edge: Registers Latch values
into memory cells
Datapath Timing
•
To Implement this requires:
•
•
•
•
2 blanks
Rigid timing
______clock cycle
_________________ propagation delay
Fast load of registers from C Bus
Datapath Timing
•
Things to Note:
–
–
–
•
Falling Edge Signals Start of Bus Cycle
Rising Edge Signals End of Bus Cycle
____ units are operating ____ the time. The values
are garbage until the known delay has passed.
Clock Length >= Δw + Δx + Δy + Δz
2 blanks
Memory Operations
•
2-ways to address memory:
– 32-bit word-addressable memory
–
–
Memory Address Register (MAR)
Memory Data Register (MDR)
– 8-bit byte-addressable memory port
–
Program Counter (PC) = MBR[7…0]; Read Only
Memory Operations
•
Register Combinations
–
–
•
MAR/MDR: used to read/write ISA-level data words
PC/MBR: used to read the executable ISA-level
program(consists of a byte stream)
NOTE: MBR: additional open arrow
–
determines whether the MBR value placed on Bus
is +ive or –ive
MicroInstructions
•
Recall… ALUs, Register, Buses
–
–
•
All have control signals
Actions determined by their truth
table.
IJVM Datapath
– 29 signals needed to control all of our
components.
– values of signals control which
portions of the circuit are contributing
to the final result of the bus cycle.
These signals together create our
Binary Micro-Instruction
MicroInstructions
•
Signals are divided up into 5 functional groups:
•
•
•
•
•
9 signals to control writing data from the C bus into
registers
9 signals to control enabling registers onto the B bus for
ALU input
8 signals to control the ALU and shifter functions
2 signals (not shown) to indicate memory read/write via
MAR/MDR
1 signal (not shown) to indicate memory fetch via PC/MBR
Memory Read Operation
•
1st Bus Cycle:
–
•
2nd Bus Cycle:
–
•
Memory Address is loaded into MAR
Data is fetched from Memory & stored in registers
3rd Bus Cycle:
– Data can now be used in an instruction
Memory Read Operation
•
NOTE:
–
–
We can start another instruction during the 2nd bus
cycle
not one that needs this information from the
previous read
MicroInstruction Format
•
•
36-signals to 1 IJVM instruction
Groups:
Addr: Contains the address of a potential next
microinstruction
JAM: Determines how next microinstruction is
selected
MicroInstruction Format
•
Groups:
JAM: Determines how next microinstruction is
selected
N: ALU result was negative
A-B= -ive  A < B
Z: ALU result was zero
A-B = 0  A = B
MicroInstruction Format
•
Groups:
ALU: controls the ALU & shifter functions
C: Selects which registers are written from
the C Bus
Mem: Memory Functions
B: Selects the B bus source
Finally….
• Here we are!!!
Our Controller
+ Datapath!!!

Sequencer
• Responsible for
stepping through
the sequence of
operations
necessary for
execution of a
single ISA
instruction.
• Produce 2 kinds
of information on
each cycle
– state of every
control signal in
the system
– address of the
microinstruction
that is to be
executed next
Control Store
• Holds
microprogram:
can be
implemented
as memory
(memory cells)
or logic gates
Control Store
• Accessed through
MBR & MDR
• Holds microinstructions… NOT
ISA instructions
• Properties:
– 512 words
– 1 word = 36-bit microinstruction
• Has it’s own address
register: MicroProgram
Counter (MPC)
• Has it’s own data
register:
MicroInstruction
Register (MIR)
Important Difference….
• Main Memory:
– Program Instructions; executed
sequentially(except for branches) :
a = a+b; a = a-b; a = a + b;
• Control Memory:
– Control words contain information on how
to control datapath for the operation.
– It is a Read Only Memory.
– Next Instruction is part of instruction is in
the Control Memory
Micro Program Responsibilities
• The microprogram needs to:
• __________________
• ______________________________________
2 blanks
Simulation of Architecture
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