Delay Loop Question - Am I Missing Something?

Hello All.

I've a question regarding delay loops.

1) The book I am using gives this fragment of code as a means to produce a 42 microsecond delay with a PIC using a 4MHz crystal:

;=======================
; Procedure to delay
; 42 microseconds
;=======================
delay_125
movlw D'42' ; Repeat 42 machine cycles
movwf count1 ; Store value in counter
repeat
decfsz count1,f ; Decrement counter
goto repeat ; Continue if not 0
return ; End of delay


2) This makes sense to me: instructions run at 1/4 clock frequency, f, (here 4MHz / 4 = 1 MHz).

T (period) = 1 / f

So the period to execute each instruction is 1 microsecond.

Since the file count1 contains decimal 42, it should take 42 microseconds to count down to zero.

So far, so good (I think).



3) Now the confusion begins:

The same program then gives this fragment to produce a 5 millisecond delay. It relies on calling the 42 microsecond delay reproduced above 41 times:

;=======================
; Procedure to delay
; 5 milliseconds
;=======================
delay_5
movlw D'41' ; Counter = 41
movwf count2 ; Store in variable
delay
call delay_125 ; Delay
decfsz count2,f ; 40 times = 5 milliseconds
goto delay
return ; End of delay


4) It seems to me that calling a 42 microsecond delay 41 times produces this result:

(42x10-6 seconds) x 41 = 1.722x10-3 seconds. - That's not 5 milliseconds: it's too short.

-If I recall correctly, executing a "goto" instruction takes longer than one cycle. - Is that having an effect, or can anyone explain what I've missed, please?

Thanks.

Analogue.

The first function is not a 42uS delay. It loops 41 times giving a time of 3* 41 + 3 = 125 uS (The name of the function seems to indicate that even if the comment doesn't)

The first loop tales 3 cycles. 1 for the decfsz and 2 for the goto

Thanks for explaining how it works, jb99. - Makes a lot more sense now.

Analogue.