Well I was the OP. I did find out that the i/o pin
should not be set a float but setting the output to
only one state, preferably low. Trisx - output and
Portx as low. This applies to all the unused pins. It
is recommended that the unused pins are driven ground
through a high value resistor. Why? I believe this
happens when the output is set as high but want the
minimum draw in current from the i/o pins. Anyway as
you said, it is NOT economical. So during your PIC
startup just set the unused i/o pin to low and it will
be fine. Just imagine this. If the 'low' of the pins
are 'not really low' there will still be a potential
difference between that pin and ground causing a
current drain. Think about it. Use a meter.
You may want to read up on amplifier (MOSFET ,BJT) and
logic circuits to get a better understanding on what
goes on in PIC.

John


--- Christopher Head <chead@telus.net> wrote:

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> Hash: SHA1
>  
> Jan-Erik Soderholm wrote:
> > kravnus wolf wrote :
> >
> >> My experience with PIC sleep has been quite
> >> interesting. When I executed the sleep
> instruction,
> >> the system was still draining about 0.40mA. At
> times
> >> it would take up more......
> >> After reading the article on the PICLIST, I found
> that
> >> the output pins MUST not float or it would drain
> >> power.
> >
> > Are you talking about pins configured as *inputs*
> ?
> > If so, they should never float, sleeping or not.
> [snip]
> 
> Do you think you could just answer something that's
> been bugging me
> for a while? I've heard before that "inputs should
> never be floating,
> either make them outputs or put pull-resistors on
> them". Here's the
> problem: all the standard port pins are
> bidirectional, which means
> even those used as outputs "normally" will be inputs
> for a fraction of
> a second before the code has a chance to fix them.
> Does this mean that
> every output pin must carry a pull-resistor as well?
> That seems a
> little crazy, but it also seems like the only
> legitimate solution!
> 
> Thanks,
> Chris
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