Wow, thanks. I can't think of any more questions to ask :)
That's what I call a good answer!

Thx,
Steve

> > I'm looking for some clairity on what the Latch does in the PIC18
> > devices.
>
> The PORT register is the immediate state of the port pins, whereas the LAT
> register holds what you want the output pins to be.
>
> This means on reading, PORT will reflect the actual pin voltage in real
> time, and LAT simply acts as a RAM location.  In other words, LAT always
> reads back what you last set it to.
>
> However, LAT is most useful for writing.  A parallel 8 bit write to either
> register will have the same effect (MOVWF PORTB for example), but there is
> a
> significant difference on bit manipulation (BCF PORTB, 2  for example).
> All
> writes to PORT are actually read/modify/writes.  This matters when the
> final
> write doesn't update all the bits, or you are performing an arithmetic
> operation where the original bit values effect the result.  In those cases,
> the read will read the instantaneous value of the port pins, which is NOT
> necessarily what you last set them to.  The pin input and output circuits
> are independent.  The pin input circuits read the actual pin voltage.  If
> the external circuit is forcing the pin to a particular state, then this
> state will be read whether the drivers are trying to drive to that state or
> not.  Note that this can happen for short times even with a normal external
> circuit.  The output drivers can only source/sink a finite current, and
> there is always some capacitance from a port pin to ground.  This means
> that
> after being set to a different state, the port pin will take a finite
> amount
> of time to reach that new state.  If a read occurs before it reached the
> new
> state, then the read will return the old value, not the value just written.
> To make matters worse, the write to the port register is done near the end
> of the instruction cycle, and the read is done near the beginning.  If I
> remember right, there is only one Q cycle of time between the write from
> one
> instruction and the read from the next.  This amounts to 1/4 instruction
> time, not a whole instruction time as you might expect.
>
> For example, on a 20MHz 16F:
>
>      clrf   portb       ;set all port B pins to 0V
>      ...
> ;
> ;   A "long" time later:
> ;
>      bsf    portb, 1    ;set RB1 to 5V
>      bsf    portb, 2    ;and then RB2 also to 5V 200nS later
>
> The second BSF will quite likely set RB1 back to 0V.  Can you see why?
>
> However, if these BSF were done on LATB instead of PORTB, then they would
> have the intended effect.
>
>
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>
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