Vasile Surducan <PICLIST@mitvma.mit.edu> wrote:

> I need a finest resolution so every time the minimum period will be 10
> seconds and no more, for 10 minutes and 10% I need to do 60 times 1
> second on and 9 secons off.

    Ah, ok... NOW I understand.  I, too, was puzzled when I saw that
    question coming from you.

    Something like this should work; it is -- as so many things are --
 a
    variation on the Bresenham line-drawing algorithm:

    ; Enter with duty-cycle (0-100) in DUTY.  This routine uses a
    ; register called SWITCH which must not change between calls
    ; to the routine.  Note that I'm assuming RADIX=DEC.

    TIMER_INTERRUPT:

        TSTF    DUTY
        BNZ     TESTSW

        MOVLW   -1
        MOVWF   SWITCH

    TESTSW:

        BTFSS   SWITCH,7
        GOTO    TURNON

        BCF     OUTPUT_PIN
        GOTO    FINISH

    TURNON:

        BSF     OUTPUT_PIN

        MOVLW   100
        SUBWF   SWITCH

    FINISH:

        MOVF    DUTY,W
        ADDWF   SWITCH

    This will ALWAYS produce the minimum PWM period necessary for a
    particular duty-cycle.  In other words, DUTY=50 will result in

        ON-OFF
        ON-OFF
        etc.

    and DUTY=10 will result in

        ON-OFF-OFF-OFF-OFF-OFF-OFF-OFF-OFF-OFF
        ON-OFF-OFF-OFF-OFF-OFF-OFF-OFF-OFF-OFF
        etc.

    -Andy

=== Andrew Warren - aiw@cypress.com
===
=== Principal Design Engineer
=== Cypress Semiconductor Corporation

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