Jinx wrote:

>>> Input leakage current (pin at high impedance) is +/- 1uA when
>>> Vss =< Vpin =< Vdd

> GPIO0 (output) - normally '0'
> GPIO1 (output) - normally '0'
> GPIO2 (input) - 390k pulldown to 0V
> GPIO3 (input) - 100k pullup to Vbatt (Mclr)
> GPIO4 (output) - normally '0'
> GPIO5 (output) - normally '0'
> 
> It uses less than 15nA when SLEEPing, so although the d/s
> says "1uA", the true figure is or could be way less than that

Definitely. My question was more about what to do if you want guaranteed
lowest current and have no other requirements for the pin (that's the way I
understood the OP). I guess making it an (unloaded, of course) output is
probably best. 

Gerhard

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