Back to my original simple circuit. (Switch, 1 Transistor, 1 diode, 1C, 3R).

If an  _open-drain_ PIC pin  is used then I don't think there is a power-off
sneak current issue.

Here is my thinking:

Internal to the PIC the pin is connected to two things:

1: the gate of a FET -- this is extremely high impedance

2: the cathode of the protection diode, the anode of which is connected to
the substrate. Since the substrate is connected to ground, this diode will
be reverse biased and also very high impedance.

Thus, there should not be any significant leakage into the PIC pin.

Am I right?

Bob Ammerman
RAm Systems

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