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Timing Analysis

Spring 2019EE3954: Microprocessors and Microcontrollers

Avinash Karanth

Professor, Electrical Engineering and Computer Science

Ohio University

E-mail: karanth@ohio.edu

Acknowledgment: Harsha Chenji, Jim Goble, Maarten Uijt de Haag

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Course Administration

• Lab 1 this week!• Submit prelab, in-lab and post-lab reports!

• Quiz 1• On Blackboard, multiple choice questions (10), timed

quiz, randomized solutions, cannot back trace, once started, have to finish within 25 minutes, single attempt, so make sure you have reliable connection.

• Availability: Friday Feb 8 between 9:00 AM – 9:00 PM.

• Topics include everything including Timing Analysis (this slides)

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Decision

Process A Process B

instructions toset up conditionbtfss …goto ProcessAgoto ProcessB

Example: execute code at LOC_1 if a “0” occurs at PORTB,0execute code at LOC_2 if a “1” occurs at PORTB,0

btfss PORTB,0goto LOC_1goto LOC_2

ORbtfsc PORTB,0goto LOC_2goto LOC_1

Branching – if … then …

‘for’ loop and ‘while/repeat’ loop

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Counter0?

Set Counter

DecrementCounter

Process A

N

Y

Perform Process A“Counter” times

Condition False

Set up Condition

Process A

N

Y

Perform Process Awhile condition is true

Condition True

Set up Condition

Y

Process A

ContinueMain Program

ContinueMain Program

ContinueMain Program

N

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DECFSZ Command

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A useful command for doing a task a set number of times.

Loops – ‘for’ loop

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• Use ‘DECFSZ’

COUNT equ 0x20

; Set up the counter valuemovlw d’3’ ; SET COUNTER To “Three”movwf COUNT ; Store it in a GPR register.

NXT: …… ; PROCESS A…decfsz COUNT,F ; Decrement and check if zerogoto NXT…… ; Continue the main program…

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BTFSC Command

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Loops – ‘while/repeat until’ loop

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Example: Execute ProcessA subroutine/function until ‘w’ contains the value decimal 99

NXT: sublw d ’99’btfsc STATUS,Zgoto DONEcall ProcessAgoto NXT

DONE ………

NXT: call ProcessAsublw d ’99’btfss STATUS,Zgoto NXT

DONE ………

Always executes Process A at least once

Does not necessarilyexecute Process A

Recall: Sublw = literal – (W) -> W

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ADDFSR Command

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SUBWF Command

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Loops – ‘repeat until’ example

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movlw 0x20 ; set first address of the listmovwf FSR ; by setting FSR

NXT1: clrf INDF ; element of the listincf FSR,F ; increment the list pointermovlw 0x70 ; check if the end of the listsubwf FSR,W ; reached – here: 0x70btfss STATUS,Z ; skip next inst. If f-(w) = 0goto NXT1 ; goto next element in the listreturn

Question = will location 0x70 get cleared?

Timing Analysis

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Remember:

1 Instruction Cycle takes 4 Oscillator Cycles

osccy TT 4=

Most instructions take 1 instruction cycle to execute

Some instructions take 2 instruction cycles to execute: GOTO, CALL, RETURN, RETLW, RETFIE

Some instructions take either 1 or 2 instruction cycles to execute: DECFSZ, INCFSZ, BTFSC, BTFSS

See instruction summary tables

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Timing Analysis

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Execute 1

Fetch 2 Execute 2

Fetch 3

Fetch 4

Oops, the microcontroller just fetched the wrong instruction

Execute 4

Fetch 5

2 instruction cycles

DECFSZ, INCFSZ, BTFSC, BTFSS

if a SKIP occurs:if a SKIP does NOT occur:

Execute 1

Fetch 2 Execute 2

Fetch 3 Execute 3

Fetch 4

1 instruction cycle

Example 1 – Timing Analysis

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org 0x000

NXT:movlw 0x28 ; 1 instruction cyclemovwf PORTB ; 1 instruction cyclenop ; 1 instruction cyclemovf PORTC,W ; 1 instruction cyclesublw 0xAA ; 1 instruction cyclebtfss STATUS,Z ; 2 if taken; 1 is not takengoto NXT ; 2 instruction cycles

A ……

Disregarding first fetch instruction cycle time:If the ‘skip’ is taken the first time the time it takes to get to point A is:delay = 1 + 1 + 1 + 1 + 1 + 2 = 7 instruction cycles

If the ‘skip’ is taken the second time the time it takes to get to point A is:delay = (1 + 1 + 1 + 1 + 1 + 1 + 2) + (1 + 1 + 1 + 1 + 1 + 2) = 15 instruction cycles

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Timing Analysis

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With an external oscillator of 4MHz:

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10.25 10 seconds 0.25 sec

4 10osc

T −= = =

Oscillator period

4 1.0 seccy osc

T T = =

instruction cycle

7 instruction cycles = 7 * 1.0 μsec = 7 μsec15 instruction cycles = 15 * 1.0 μsec = 15 μsec

Timing Analysis - Subroutines

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…

call SUB_1 ; 2

…

…

SUB_1: andlw 0x0F ; 1

movwf TEMP ; 1

movlw 0x12 ; 1

subwf TEMP,F ; 1

return ; 2

A) Subroutine itself takes: 6 instruction cycles

B) Calling the subroutine costs:2 instruction cycles

C) Totally, it costs: 6+2 = 8 instruction cycles

to call and execute the subroutine

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Timing Analysis – Building Delays

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1: A: movlw d’3’ ; 1 instruction cycle2: movwf COUNT ; 1 instruction cycle3: NXT:nop ; 1 instruction cycle4: decfsz COUNT,F ; 1 or 2 instruction cycles5: goto NXT ; 2 instruction cycles

B: …

Line: Instruction duration: Times Executed: Execution time:

1 1 1 1

2 1 1 1

3 1 COUNT 1*COUNT

4 1 COUNT-1 1*COUNT-1

4 2 1 2

5 2 COUNT-1 2*(COUNT-1)

Timing Analysis – Building Delays

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Time to go from A to B is:

(1 + 1) + COUNT + COUNT – 1 + 2 + 2•(COUNT – 1) =1+1+3+3-1+2+2(2) =

13 instruction cycles

So with an oscillator of 4MHz: delay takes 13 * 1 μsec = 13 μsec

1: A: movlw d’3’ ; 1 instruction cycle2: movwf COUNT ; 1 instruction cycle3: NXT:nop ; 1 instruction cycle4: decfsz COUNT,F ; 1 or 2 instruction cycles5: goto NXT ; 2 instruction cycles

B: …

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Timing Loops

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What is the general expression for this delay given any value for COUNT ?

COUNT*41nsInstructio# +=

So, what is the maximum delay we can do with this construction ?

COUNT is a byte so its maximum value is 255 !!Maximum value for delay is therefore: 1+4*255 = 1021 cycles, or 1021*1.0 μsec = 1021 μsec(0 gives 1+4*256) = 1025 cycles – because decfsz 0->255).

Timing Loops – Even Shorter

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COUNT equ 0x20 ; assign memory for countorg 0x100

A: movlw d’10’ ; 1movwf COUNT ; 1

NXT:decfsz COUNT,F ; 1 (2)goto NXT ; 2 instruction cycles

B: …

Time to go from A to B is:

(1 + 1) + 9 + 2 + 2*(10-1)= 31 instruction cycles

COUNT*31nsInstructio# +=

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Timing Loops – Even Shorter

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Counter0?

Set Counter

DecrementCounter

N

YContinueMain Program

Nested Loops

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OuterCounter

0?

Set Outer Counter

DecrementOuter Counter

N

YContinueMain Program

Set Inner Counter

InnerCounter

0?

DecrementInner Counter

Y

N

InnerLoop

OuterLoop

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Nested Timing Loops

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COUT equ 0x20 ; outer counter storageCIN equ 0x21 ; inner counter storage

org 0x0100 ; program location - start…

OUTER: movlw d’10’ ; 1movwf COUT ; 1

INNER: movlw d’255’ ; 1movwf CIN ; 1

NXT: decfsz CIN,F ; 1 (2)goto NXT ; 2decfsz COUT,F ; 1 (2)goto INNER ; 2

B: …

Nested Timing Loops - Computations

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Inner loop: CINInner •+= 31

Outer loop:( )

( )

COUTCINCOUT

COUTCIN

COUT

COUTCOUT

Inner

InnerOuter

••+•+=

••++=

•++=

+•+=

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For example, on the previous slide COUT = 10 and CIN = 255:

76625531Inner =•+=

(@ 4MHz -> 7.691 msec delay)

691,710)7663(1Outer =•++=

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Timing Loop Fine-tuning

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• By inserting instructions (most of the time nopinstructions) the inner loop delay can be changed and as a result the outer loop. Similarly, instructions can be added to the outer loop.

Fine-tuning Example

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COUT equ 0x20CIN equ 0x21

org 0x00…

OUTER: movlw 0x0A ; 1movwf COUT ; 1

INNER: movlw 0xFF ; 1movwf CIN ; 1nop ; 1

NXT: decfsz CIN,F ; 1 (2)goto NXT ; 2decfsz COUT,F ; 1 (2)goto INNER ; 2

B: …

2 instead of 1Since nop is insertedoutside the loop

( ) COUTInnerOuter •++= 31

CINInner •+= 32

Increment thisif nop is insertedinside the loop

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Delay Subroutines

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DELAY : movlw d’10’ ; 1movwf COUT ; 1

INNER: movlw d’255’ ; 1movwf CIN ; 1

NXT: decfsz CIN,F ; 1 (2)goto NXT ; 2decfsz COUT,F ; 1 (2)goto INNER ; 2return ; 2

Add 2 instruction cycles to the delay for the ‘call’ statement, and 2 instruction cycles for the ‘return’ statement

call DELAY ;2

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