Won't work.  At least, not the way you think it will.

PICs use a read/modify/write sequence to set/clear output bits. The read=20
portion is a read of the actual voltage on the I/O pins, NOT what is in=20
the latch.  So, if PIN_A0 is set to output a 1 but is set as an input,=20
setting PIN_A3 will cause the PIC to read port A (including the 0 on=20
PIN_A0) set bit 3 and output that byte, thus setting the PIN_A0 latch to=20
0 and outputting a 0 when you changed the TRIS.
You would need to do one of the following:
1) Not use any other pin on that port as an output.
2) Use a byte of memory to hold the port value and only output that byte=20
to the port after modifying it.
3) Set PIN_A0 before changing the TRIS each time.

And you would not gain anything.  If it is 0.5mA with a 10K pull down,=20
it is 0mA without it.

Kerry

On 7/19/2014 5:32 AM, Richard R. Pope wrote:
> Hello all,
>       I just had a thought. Oh, what a dangerous thing. Oh well here it
> is. What if I pull the output pins low with a 10K resistor hooked up to
> ground and leave them in input mode. When I want to output a high I then
> change the pin from input mode to output mode. I have already setup the
> pin to output a high. So I am only changing the TRIS bit for the pin and
> since the pins are only high for a very short period of time this would
> minimize the current draw. This would only be 0.5mA of current. Am I
> thinking clearly on this? What about using a 100K resistor?
> Thanks,
> rich!
>

--=20
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