It is a 30 µs short. Stray inductance will not smooth this.

It is needed a turn on delay that does not delays turning off. For both mosfets.

 

It is a 30 µs short.

That won't hurt the MOSFETs or the power supplies.

But if you want to eliminate it, then you will need to use separate signals for the 5V and 12V enable, and separate their operation in time.

 

Alternately you could inhibit the output before switching the voltages.

 

I might be missing something, but can't this be solved with just two diodes?

Start with the schematic in post #3 (corrected from #2), and add a pullup resistor from the Q5 collector to the +5 V. Now add another Shottkey diode similar to D1, but going from the Q1 base to the Q5 collector. Now, whichever transistor (Q1 or Q5) comes on first, it clamps the other transistor's base to something below Vbe. I would increase R1, R2, R6, and R7 to decrease the available base current, which decreases the Vf of the clamping diodes, which reduces the possibility of a partial turn-on.

I also would delete R3, R5, and maybe R4.

ak

 

I proposed that in post #3 using only one diode giving the 12 volt side priority but could be configured to give the 5V side priority as well.

 

I meant your post #3. Typo, fixed.

ak