> >ALTHOUGH now that I think about it, in the case of no battery,
> >there would be NO conduction, because there's nowhere for the
> >current to go.  Which makes me an ass, and this thread moot,
> >since the programming would occur with no battery present.
> >
> >It would still be nice to be certain that, if someone were to
> >try to program the device with a battery in place, no smoke escapes.
> 
> Umm, hang on a minute mate. I know my maths is bad, but ...
> 
> If the battery is 9V, and the ICSP is at 5V, how is the diode going to
> conduct ??? It seems to me that you may have another danger, which you may
> like to verify first, and that is the risk of blowing up the regulator
> because you are applying a voltage greater than the design output voltage.
> You may like to try this on a sacrificial lamb first.

You're completely correct; the question of protection against
attempting ICSP with a battery in place is a separate issue.

The issue in that case is two competing sources, one trying to
hold the rail at ~5V and the other trying to hold it at 3.5V.  In
the absence of protection, the stronger one wins, which would
almost certainly NOT be the regulator I'm trying to protect.

Imagine a black box connecting the output of my regulator to
the rest of the circuit.  Inside that black box is some circuit
which isolates the regulator against voltages greater than the
current output voltage of the regulator (0V when no battery is
in place, 3.5V when the battery IS in place).  That's what I'm
after.

HOWEVER (bold letters again!), further consideration of all
elements of the design, including programming conditions, is
leading me to believe that it is highly unlikely that high voltage
ICSP will be a common event, which decreases the odds of
an accidental battery-connected program, and that the regulator
will survive without the battery connected, no problem.

Mike H.

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