simple processor control unit

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Simple Processor Control Unit. Instructor: Oluwayomi Adamo Digital Systems Design. Control Unit. Design a control unit for picking up instructions from memory address given by the program counter (PC). Interpret the instruction, Fetch the operands and feed them to the ALU, - PowerPoint PPT Presentation

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Instructor: Oluwayomi AdamoDigital Systems Design

Design a control unit for picking up instructions from memory address given by the program counter (PC).

Interpret the instruction, Fetch the operands and feed them to the

ALU, Store the result in destination registers Load the pc with destination address in case

of branch instruction, Contents of destination will be forwarded to

the LED or 7 segment display for display.

Op CC SRC DST

Register Instruction

01001011

Branch Instruction

11 CC ADDRESS

Halt and I/O Instruction

1100 L H DST

1100101111000110

11100011

CONT UNIT D MUX

MUX

MUX

PC

MUX A

MUX B

ALU

000G

DISPLAY

4 BITS

RCAOP

DMUXA

0 OR 1

SWICTH

MEMMUX R

EG

CC

DSTSRC

The control unit is like computer’s traffic cop. It coordinates and controls all operations occurring within

the processor. The control unit does not input, output, process, or store

data, it initiates and controls the sequence of these operations. Controls Data Movements in an Operational Circuit by

Switching Multiplexers and Enabling or Disabling Resources Follows Some ‘Program’ or Schedule Often Implemented as Finite State Machine or collection of

Finite State Machines

Any Circuit with Memory could be called a Finite State Machine◦ Even computers can be viewed as huge FSMs

Design of FSMs Involves◦ Defining states◦ Defining transitions between states◦ Optimization / minimization

Above Approach Is Practical for Small FSMs Only

pr_state nx_state

Combinational logic

Sequential logic

input

clock

rst

Output Is a Function of a Present State Only TYPE state IS (S0, S1, S2); SIGNAL Moore_state: state;

U_Moore: PROCESS (clock, reset) BEGIN IF(reset = ‘1’) THEN Moore_state <= S0; ELSIF (clock = ‘1’ AND clock’event) THEN CASE Moore_state IS WHEN S0 => IF input = ‘1’ THEN Moore_state <= S1; ELSE Moore_state <= S0;

END IF;

reset

WHEN S1 =>

IF input = ‘0’ THEN Moore_state <= S2; ELSE Moore_state <= S1; END IF; WHEN S2 => IF input = ‘0’ THEN Moore_state <= S0; ELSE Moore_state <= S1; END IF; END CASE; END IF; END PROCESS;

Output <= ‘1’ WHEN Moore_state = S2 ELSE ‘0’;

Output Is a Function of a Present State and Inputs

TYPE state IS (S0, S1); SIGNAL Mealy_state: state;

U_Mealy: PROCESS(clock, reset) BEGIN IF(reset = ‘1’) THEN Mealy_state <= S0; ELSIF (clock = ‘1’ AND clock’event) THEN CASE Mealy_state IS WHEN S0 => IF input = ‘1’ THEN Mealy_state <= S1; ELSE Mealy_state <= S0; END IF;

WHEN S1 => IF input = ‘0’ THEN Mealy_state <= S0; ELSE Mealy_state <= S1; END IF; END CASE; END IF; END PROCESS;

Output <= ‘1’ WHEN (Mealy_state = S1 AND input = ‘0’) ELSE ‘0’;

Fetch -> Decode -> Execute

Fetch Sequence t1: MAR <- (PC) t2: MBR <- (memory) PC <- (PC) +1 t3: IR <- (MBR) (tx = time unit/clock cycle)

Good Luck!!!

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