On Fri, 14 Nov 1997 11:43:27 +0100 =?ISO-8859-1?Q?Thomas_M=F8rch?=
<oz2afo@post8.tele.dk> writes:
...
>Is
>there
>any logical explanation of this behaviour?
...
>The
>program i'm using is as follows:
...
>start   bsf     STATUS,RP0
>        clrf    PORTB

This won't work.  RP0 must be clear to access PORTB, as it is in page 0.

>
>        movlw   00H
>        movwf   TRISB   ; Set port b as outputs
>        bsf     TRISB,7 ; set pin rb7 as input
>        bcf     OPTION_REG,7 ; make internal pull-up on portb;
>
You don't clear RP0 here.  So what your bsf/bcf below is doing, is
changing the LED pins from input to output, ouputting the random data in
the PORTB register (which your program never touches) leftover from
power-on.

>l001    call    Delay
>        bsf     PORTB,3 ; set portb pins high
>        bcf     PORTB,2

Even if you do clear RP0, doing bsf/bcf on PORTB is not very reliable.
If you have heavy DC loads (LEDs with no resistors?) or capacitance on
the output pins, doing a bsf/bcf on a port may corrupt the other bits in
the port.  The LED on port pin 3 may not stay as you expect it, due to
the second bit operation reading the entire port.  Do this instead:

        movlw   b'00001000'     ;B3 high, rest low.
        movwf   PORTB

This assumes the other pins of PORTB are inputs, or can stay at 0 all the
time.  If the other pins are outputting something that needs to be
independent of turning the LEDs on and off, keep a copy of the state for
the other pins in RAM, and change the LED bits:

        bsf     shadowB,3       ;LED 3 on
        bcf     shadowB,2       ;LED 3 off
        movfw   shadowB
        movwf   PORTB           ;Update port B.

Everytime you want to send something via PORTB, update the shadowB RAM
first, then copy it out.  The result will be correct even if the pins are
overloaded.