Cycle Count Calculation in PIC -

you given subroutine written pic16f877 working @ 4mhz obtain approximate 30ms delay.

1 counter1 equ 0x20 2 counter2 equ 0x21 3 4 delay_loop: 5 6 movlw d'30' 7 movwf counter2 8 9 ; loop1 body takes 1ms 10 loop1: 11 movlw d'200' 12 movwf counter1 13 14 ; loop2 body takes 5us 15 loop2: 16 nop 17 nop 18 decfsz counter1, f 19 goto loop2 20 21 decfsz counter2, f 22 goto loop1 23 return  instruction cycles movlw          1 movwf          1  nop           1 decfsz         1* goto           1 return         2  

well, not end same values in lines 9 , 14. can you, please, show me detailed calculations?

firstly, instruction cycles not correct. goto 2 cycles , decfsz either 1 or 2 cycles depending on result.

so, loop2:

loop2:     nop                  ; 1 cycle     nop                  ; 1 cycle     decfsz counter1, f   ; 1 cycle except @ loop end, 2 cycles     goto loop2           ; 2 cycles 

so total 5 cycles, 4mhz clock having 1us cycle time¹ 5us. counter1 being 200, inner loop takes 200 * 5us = 1000us = 1ms.

now enclosing loop, loop1:

loop1:     movlw d'200'         ; 1 cycle     movwf counter1       ; 1 cycle      ; 1ms delay loop2 (removed clarity)      decfsz counter2, f   ; 1 cycle except @ loop end, 2 cycles     goto loop1           ; 2 cycles     return               ; 2 cycles 

so total of 5 cycles loop1 housekeeping (ignoring final return), added 1000us inner loop = 1005us. counter2 being 30, total delay 30 * 1005us = 30.15ms (give or take handful of initial housekeeping instructions , final return, idea).

¹remember these pics use 4 clock cycles per instruction, 4mhz clock executes @ 1mips, giving instruction time of 1us.