On Sun, 8 Oct 2000 12:22:59 +1000 Roman Black <fastvid@EZY.NET.AU>
writes:
> Harold Hallikainen wrote:
>
> >         When interfacing to the outside world, I like to put a 10K
> pull-up, then
> > a 10K series resistor to a schmitt input on the PIC. I also then
> put an
> > appropriately sized capacitor between the PIC input and ground.
> This
> > keeps power consumption down while slowing down edges and uses a
> PIC
> > input that can deal with slow edges.
> >
> > Harold
>
> Harold, won't this give you a 50/50 voltage divider? So when the
> switch
> goes low the pin voltage only goes down to 2.5v? Or are you
> suggesting
> the
> switch goes at the point between the two resistors. This will work
> but
> could
> mean you need to run three wires to a externally mounted switch?
> Please explain? :o)
> -Roman
>
>

        Yes, the switch goes from the junction of the two resistors to ground.
Since the PIC input is high impedance, there is no DC drop across the
resistor between the switch and the PIC. The PIC gets either +5V or
ground, but with a slow edge (due to the series R and the C to ground),
but the schmitt input deals with slow edges well. You only need to run
the one wire from the junction out to the switch along with a ground
wire.

Harold


FCC Rules Online at http://www.hallikainen.com/FccRules/

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