Hi Jon (Jon Hylands), in <354667ad.11930787@mail.huv.com> on Apr 28 you wrote:

> The question is, how do I isolate Vdd? I was thinking along the lines of
> running a trace from the Vdd pin directly to the power header pin, and to
> the rest of the circuit through a diode. The diode would be mounted to
> allow current to pass into the chip from the circuit power supply, but not
> from the chip to the rest of the circuit when powered from the programmer.
> Presumably, since the voltage drop would be about 0.6 volts, the power
> running into the PIC from the 5 volt circuit supply would end up being 4.4
> volts,  which is well within the limits.
> 
> Will this work? Is there a better way?

I don't think this is a good idea.  You wrote you have several PICs
interfacing to lots of peripherials.  When these peripherialy are powered
at 5.0V and output strong highs to the PIC, the PIC will be far beyond spec
(max input high voltage = VDD, see DC specs).  Probably it'll work, as the
protection diodes are 0.7V, too.  But it's definately not a good design.

Better:  Don't connect the 5V supply to the ISP header at all.  Require the
_circuit_ to supply power to the PICs also during programming.

Connect MCLR and RB6 and RB7 and GND to the ISP header.  Have all VDDs at
5V as normal in your circuit.  Tie each MCLR to its own PICs' VDD with a
5K1 resistor (as long as you don't have a power supply monitor chip in your
circuit).