Using picloops to calculate a timed delay

[ryan_ryan]

• 
  

  Delay        
  
                  movlw	D'6'
  	movwf	CounterC
  	movlw	D'24'
  	movwf	CounterB
  	movlw	D'168'
  	movwf	CounterA
  
  loop	
  
                  decfsz	CounterA,1
  	goto	loop
  	decfsz	CounterB,1
  	goto	loop
  	decfsz	CounterC,1
  	goto	loop
  	retlw         0x00

Hi, i was recommended to picloops to write delay subroutines for timed delays. For this particular 1 sec delay, it was calculated to have a total of 1000002 machine cycles and hence a 1.0000020sec delay.

My main problem now is that how do we actually calculate the total number of machine cycles, i have run the program on the simulator and have a rough idea but i am unable to work out the maths for the whole number of machine cycles.

Pls help.Thanks.

 

[checkmate]

You still havent go it eh?
Alright. Give you 3 simple scenarios.

   movlw   D'1' 
   movwf   A

loop
   decfsz A
   goto exit

exit
   nop

This will take (1+1)+(2)+(1) clock cycles

   movlw   D'2' 
   movwf   A

loop
   decfsz A
   goto exit

exit
   nop

This will take (1+1)+(1+2)+(1) clock cycles.

Get it?

   movlw   D'2' 
   movwf   A

loop
   decfsz A
   goto loop

exit
   nop

Now, you get (1+1)+(1+2)+(2)+(1) clock cycles. I hope this is clear enough.

Lastly, the "call" and "retlw" instructions each take 2 cycles.

 

[csaba911]

Just read thru this page **broken link removed**

STEVE