Exact Delay Time calculations for Delay(10ms)

Hello.

I came across this code generated by the MikroC Pro for PIC and it delays 10 mS when using an oscillator of 4 Mhz.


According to my calculations when using a 4 Mhz oscillator the PIC will divide the frequency by 4 to obtain an internal clock of 1 Mhz. Is that right?

Each MOVLW and MOVWF operation takes 1 clock cycle.
The DECFSZ operation takes 2 if it skips, or 1 cycle if it does not. right?
The GOTO takes 2 cycles. right?

The delay loop using R13 and R12 will sum a total of

DECFSZ R13+0, 1
GOTO L_pause3 <--- this will catch here most of the time, except when R13 reaches 0. This is 250*3 + 2 = 752 clock cycles

Then the next part

; 752 cycles so far
DECFSZ R12+0, 1 ; 753 cycles if we don't skip
GOTO L_pause3 ; 755 cycles and we repeat this 12 times. 755*12 = 9060 clock cycles.

On the last pass we do another 752+2 cycles because we skip

and finally
NOP
NOP
adds another 2 cycles.

That yields 4+9060 + 754 + 2 = 9816 clock cycles.. or 9.816 mS

Just_Delayed_10ms:

mmm, seems i'm missing some clock cycles.

Please correct me. I feel I must be wrong somewhere.

The basic delay will delay for initial_value*3-1 instructions. So for your example, the first inner loop takes innerloop(251) = 251*3-1 cy, and every subsequent inner loop takes innerloop(256) = 256*3-1 cy. The outer loop takes 13*3-1 + innerloop(251) + 12*innerloop(256) = 13*3-1 + 251*3-1 + 12*(256*3-1). You then have the 4 instructions before the loops, and the 2 instructions after. The total is therefore 13*3-1 + 251*3-1 + 12*(256*3-1) + 4 + 2 = 10000 cy.

Hi,

Working out your own software delay times is a useful grey matter exercise, however you might find this link easier.
https://www.golovchenko.org/cgi-bin/delay

If you use MPLABs Sim you can check every single step of your delay loop and monitor both time and instructions.

dougy83 said:

The basic delay

Code:

cnt = initial_value
loop:
   decfsz cnt
   goto loop

will delay for initial_value*3-1 instructions. So for your example, the first inner loop takes innerloop(251) = 251*3-1 cy, and every subsequent inner loop takes innerloop(256) = 256*3-1 cy. The outer loop takes 13*3-1 + innerloop(251) + 12*innerloop(256) = 13*3-1 + 251*3-1 + 12*(256*3-1). You then have the 4 instructions before the loops, and the 2 instructions after. The total is therefore 13*3-1 + 251*3-1 + 12*(256*3-1) + 4 + 2 = 10000 cy.

Click to expand...

Hello dougy83, you are right. I still don't understand why. Thank you.
Thankyou WP, it takes a time to understand how the stopwatch works.

Both of you are right. My brain is dry at this point.

Finally! I got it.

The first loop counts from 251 to 0. The register R13 rolls to 255 and the following inner loops count from 0 to 0. I was very confused. Now it is clear. First loop is (251*3) - 1, the following are (256*3)-1 because it rolls from 0 to 255, 254,253, ...2,1,0.

Thanks again.