Thanks for the vote of confidence
As far as I can see, the original problem was that the current could flow back into a PSU that was turned off, because current could flow through the zener backwards, with a minimal voltage drop (up to 1.3).
So by switching the drain and source, all I'm doing is switching the problem around: the PSU can now feed the power bus, with minimal voltage drop (up to 1.3) through the same diode, the same way.
plus... in this backwards configuration, when the MOSFET is actually activated, the "minimal" voltage drop actually goes away.
Which part of the above is wrong? What am I misunderstanding here?
Otherwise, I'm only aware that I may not get enough voltage to drive the MOSFET for the +12 line, on each PSU, but I've already mentionned that. Since Vgs needs to be 2.0 to 4.0, given a 12v supply, the gate voltage needs to be at 14 to 16 volts, and preferably higher, where the driver output appears to be clamped at 14. Yo_Duh and I discussed that, and he assures me that the output will be sufficient, based on the maximum rated voltage increase of the output over the supply, of 13v which, above a supply of 5, would output 18v: plenty to drive the MOSFET of the +12 line.
I already knew that.
As for the current, I have yet to confirm that the driver can actually supply that, yet Yo_Duh tells me that it could.
So what else am I missing?
My solution to the above, which I have yet to verify, is to swap the MOSFET for an IRL1404, instead of the IRF1404, only for the +12 line. The Vgs then drop to a range between 1.0 and 3.0, which is slightly less of a probleem, but still a problem.
Either that, or build my own driver from scratch.
I'm also considering leaving the circuit with the MOSFETs in the right (orthodox) placement, and forcing all the drivers to turn on all at the same time, once and only once all three PSUs are "ready", which would effectively mean hardwiring the PSUs together.
I really don't want to use the MOSFETs in an odd way, like I've suggested.:shrug: