Ok, developing a circuit that will use in circuit programming (for a PIC micro). I want to isolate the Vdd pin so that when the programmer is programming the chip, it doesn't power the rest of the circuit (which would use too much power, causing the programmer to not work)

I could use a diode on Vdd, however this would result in a voltage drop durring normal opperation...

so I came up with this solution... would this work?

http://thomas.clubplus.net/isolate.jpg
My threory is durring normal opperations, the transistor will power it's own base allowing current to pass, but when the circuit is unpowered, and programming voltage is applied, the transistor will stop the power from going to the rest of the circuit. Somehow I have the nagging feeling I'm overlooking something, which is why I'm asking here

Thanks!

 

The VBE for a transistor is approx. 0.6V, which is needed for the transistor to conduct. In your configuration, then the VCE is dictated by the VBE, and not VCE(sat), i.e. more than ~0.6V, which is similar to a diode drop.

If you want lower drop, you could use a good schottky diode, probably 0.3 ~ 0.4V drop, which is better than regular silicon diode.