Microprocessors and Microcontrollers. Loops and Timing EE3954. by Maarten Uijt de Haag, Tim Bambeck. Loops & Timing.1

Transcription

1 Microprocessors and Microcontrollers Loops and Timing EE3954 by Maarten Uijt de Haag, Tim Bambeck Loops & Timing.1

2 Branching if then Decision Process A Process B instructions to set up condition btfss goto ProcessA goto ProcessB Example: execute code at LOC_1 if a 0 occurs at PORTB,0 execute code at LOC_2 if a 1 occurs at PORTB,0 btfss PORTB,0 goto LOC_1 goto LOC_2 OR btfsc PORTB,0 goto LOC_2 goto LOC_1 Loops & Timing.2

3 Loops for loop and while/repeat loop N Continue Main Program Set Counter Process A Decrement Counter Counter 0? Y Set up Condition Condition True Y Process A N Continue Main Program N Continue Main Program Process A Set up Condition Condition False Y Perform Process A Counter times Perform Process A while condition is true Loops & Timing.3

4 Loops for loop Use DECFSZ COUNT equ 0x20 ; Set up the counter value movlw d 3 ; SET COUNTER To Three movwf COUNT ; Store it in a GP register. NXT: ; PROCESS A decfsz COUNT,F ; Decrement and check if zero goto NXT ; Continue the main program Loops & Timing.4

5 Loops while/repeat until loop Example: Execute ProcessA subroutine until w contains the value decimal 99 Recall: Sublw = literal (W) -> W NXT: sublw d 99 btfsc STATUS,Z goto DONE call ProcessA goto NXT DONE Does not necessarily execute Process A NXT: call ProcessA sublw d 99 btfss STATUS,Z goto NXT DONE Always executes Process A at least once Loops & Timing.5

6 Loops Repeat Until Example CLR2070: movlw 0x20 ; set first address of the list movwf FSR ; by setting FSR NXT1: clrf INDF ; element of the list incf FSR,F ; increment the list pointer movlw 0x70 ; check if the end of the list subwf FSR,W ; reached here: 0x70 btfss STATUS,Z ; skip next inst. If f-(w) = 0 goto NXT1 ; goto next element in the list return Question = will location 0x70 get cleared? Loops & Timing.6

7 Timing Analysis Remember: 1 Instruction Cycle takes 4 Oscillator Cycles T = 4T cy osc 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 Loops & Timing.7

8 Timing Analysis DECFSZ, INCFSZ, BTFSC, BTFSS if a SKIP does NOT occur: if a SKIP occurs: 1 instruction cycle 2 instruction cycles Execute 1 Fetch 2 Execute 2 Fetch 3 Execute 3 Fetch 4 Execute 1 Fetch 2 Execute 2 Fetch 3 Fetch 4 Execute 4 Fetch 5 Oops, the microcontroller just fetched the wrong instruction Loops & Timing.8

9 Timing Analysis org 0x000 NXT: movlw 0x28 ; 1 instruction cycle movwf PORTB ; 1 instruction cycle nop ; 1 instruction cycle movf PORTC,W ; 1 instruction cycle sublw 0xAA ; 1 instruction cycle btfss STATUS,Z ; 2 if taken; 1 is not taken goto 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 = = 7 instruction cycles If the skip is taken the second time the time it takes to get to point A is: delay = ( ) + ( ) = 15 instruction cycles Loops & Timing.9

10 Timing Analysis With an external oscillator of 4MHz: 1 T osc = = = µ Oscillator period T = 4T = 1.0 µ sec cy osc seconds 0.25 sec instruction cycle 7 instruction cycles = 7 * 1.0 µsec = 7 µsec 15 instruction cycles = 15 * 1.0 µsec = 15 µsec Loops & Timing.10

11 Timing Analysis Subroutines 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 Loops & Timing.11

12 Timing Loops Building Delays 1: A: movlw d 3 ; 1 instruction cycle 2: movwf COUNT ; 1 instruction cycle 3: NXT: nop ; 1 instruction cycle 4: decfsz COUNT,F ; 1or2 instruction cycles 5: goto NXT ; 2 instruction cycles B: Line: Instruction duration: Times Executed: Execution time: COUNT 1*COUNT 4 1 COUNT-1 1*COUNT COUNT-1 2*(COUNT-1) Loops & Timing.12

13 Timing Loops Building Delays 1: A: movlw d 3 ; 1 instruction cycle 2: movwf COUNT ; 1 instruction cycle 3: NXT: nop ; 1 instruction cycle 4: decfsz COUNT,F ; 1or2 instruction cycles 5: goto NXT ; 2 instruction cycles B: Time to go from A to B is: (1 + 1) + COUNT + COUNT (COUNT 1) = (2) = 13 instruction cycles So with an oscillator of 4MHz: delay takes 13 * 1 µsec = 13 µsec Loops & Timing.13

14 Timing Loops What is the general expression for this delay given any value for COUNT? # Instructions = 1+ 4*COUNT 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). Loops & Timing.14

15 Timing Loops Even shorter COUNT equ 0x20 org 0x100 A: movlw d 10 ; 1 movwf COUNT ; 1 NXT: decfsz COUNT,F ; 1 (2) goto NXT ; 2 instruction cycles B: Time to go from A to B is: (1 + 1) *(10-1) = 31 instruction cycles # Instructions = 1+ 3*COUNT Loops & Timing.15

16 Timing Loops Even shorter Set Counter Decrement Counter N Counter 0? Continue Main Program Y Loops & Timing.16

17 Set Outer Counter Set Inner Counter Timing Loops Nested Loops Flowchart Decrement Inner Counter Inner Loop N Inner Counter 0? Y Decrement Outer Counter Outer Loop N Outer Counter 0? Continue Main Program Y Loops & Timing.17

18 Timing Loops Nested Loops - Assembly F equ d 1 ; destination-> file reg. COUT equ 0x20 ; outer counter storage CIN equ 0x21 ; inner counter storage org 0x0100 ; program location - start OUTER: movlw d 10 ; 1 movwf COUT ; 1 INNER: movlw d 255 ; 1 movwf CIN ; 1 NXT: decfsz CIN,F ; 1 (2) goto NXT ; 2 decfsz COUT,F ; 1 (2) goto INNER ; 2 B: Loops & Timing.18

19 Timing Loops Nested Loops - Computations Inner loop: Outer loop: Δ Outer Δ = 1+ 3 CIN Inner = 1+ 3 COUT = 1+ = 1+ ( 3 + Δ Inner ) ( CIN ) = 1+ 4 COUT + Δ Inner COUT COUT COUT + 3 CIN COUT For example, on the previous slide COUT = 10 and CIN = 255: ΔInner = = ΔOuter = 1+ (3+ 766) 10 = 766 7,691 (@ 4MHz -> msec delay) Loops & Timing.19

20 Timing Loops Fine-Tuning By inserting instructions (most of the time nop instructions) the inner loop delay can be changed and as a result the outer loop. Similarly, instructions can be added to the outer loop. Loops & Timing.20

21 Timing Loops Fine Tuning with nop s : COUT equ 0x20 CIN equ 0x21 org 0x00 OUTER: movlw 0x0A ; 1 movwf COUT ; 1 INNER: movlw 0xFF ; 1 movwf CIN ; 1 nop ; 1 NXT: decfsz CIN,F ; 1 (2) goto NXT ; 2 decfsz COUT,F ; 1 (2) goto INNER ; 2 B: Δ = 1+ 3 ( + Δ ) COUT Outer Inner ΔInner = CIN Increment this if nop is inserted inside the loop 2 instead of 1 Since nop is inserted outside the loop Loops & Timing.21

22 Timing Loops Design Subroutines call DELAY ;2 DELAY : movlw d 10 ; 1 movwf COUT ; 1 INNER: movlw d 255 ; 1 movwf CIN ; 1 NXT: decfsz CIN,F ; 1 (2) goto NXT ; 2 decfsz COUT,F ; 1 (2) goto INNER ; 2 return ; 2 Add 2 instruction cycles to the delay for the call statement, and 2 instruction cycles for the return statement Loops & Timing.22