>>
>
> But I think that since the gate will be more negative than the source,
> the (P-channel) mosfet must be on, not just the body diode conducting.

Yes, sorry that's what I meant, it is on this way - the source is tied to a 
load (not shown), then it is biased using the body diode(?), so the the D-S 
channel is on. Obviously if the source is tied to ground, it's off and only 
the body diode is conducting. Sorry to confuse things, I'm confusing myself 
now - not slept for too long.. :-) The main point I had is that I am not 
sure if much would be gained doing it this way as opposed to using the 
schottky, plus maybe using the zeners would be a bit inefficient too 
(depending on the rest of the circuit). 


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