Mike Keitz <mkeitz@BEV.NET> wrote:

>burst
>        movlw   .2              ;2 byte cycle counter - initialize to 770 dec.
>        movwf   ccntl           ;Need 2 RAM variables.
>        movlw   .3
>        movwf   ccnth
>burstfl
>        nop                     ;Equalize loop time if only low byte dec.
>        nop
>burstsl
>        nop                     ;May be removed for slower clocks.
>        bsf     PORT_B,0        ;Set output high. (can use any port here)
>        goto    $+1             ;2- cycle NOPs.
>        goto    $+1
>        nop
>        bcf     PORT_B,0        ;Output was high for 6 cycles, now low.
>; From here, always 5 cycles back to burstsl.
>        decfsz  ccntl
>        goto    burstfl
>        decfsz  ccnth
>        goto    burstsl
>;Fall thru to here when burst is done.

    Mike:

    Won't work.  Instead of generating a 10-ms burst, your code will generate on
ly
    a 6.67-ms burst.  To make your DECFSZs work properly, you'll need to load
    ccnth with 4, not 3.

    If 9.974 ms is close enough, the following bit of code will do the job using
    only one register:

            CLRF    COUNT

        L1: MOVLW   00000001B   ;The "1" in W corresponds to the PORTB bit that
            GOTO    $+1         ;we'll be toggling.  This example is for PORTB,0
.

        L2: XORWF   PORTB
            BTFSC   PORTB,0
            GOTO    L1

        L3: GOTO    $+1
            NOP
            XORWF   PORTB
            BTFSC   PORTB,0
            GOTO    L3

        L4: GOTO    $+1
            NOP
            XORWF   PORTB
            BTFSC   PORTB,0
            GOTO    L4

            DECFSZ  COUNT
            GOTO    L2

    -Andy

--
Andrew Warren - fastfwd@ix.netcom.com
Fast Forward Engineering, Vista, California