Understanding an 'ideal' diode made from a p-channel MOSFET and PNP transistors
The idea of the transistors is that:


If the Left is low and the right is high R2 (and the left transistor a little) will negative-bias the base of the right transistor's base, allowing it to push the gate to the right voltage; closing the FET's channel and the body diode will block as well.
If the right is low and the left is high, the left transistor's b-e junction will work as a diode and pull the base of the right transistor high enough to close off, allowing R3 to pull the gate low, opening the transistor. Initially the right side will start to be powered by the body diode, but quite quickly the channel's low on resistance will take over causing a very low drop.


So the left transistor acts as a matched diode for the right transistor. The exact component values may hinge a bit on the chosen MOSFET and PNP matched pair. Similar tricks are available in other ways, but this is the most well known one.



If you tie the MOSFET's gate directly to ground, like this:







simulate this circuit &ndash; Schematic created using CircuitLab

You are effectively creating an always-on-link, with possibly some adjusted start-up behaviour. Usually this start-up behaviour is enhanced using capacitors and/or resistors on the gate-path.

Because if the left is high, and the right isn't, the right will get lifted up by the body diode, then the source becomes higher than the gate, causing the FET to turn on. If the right goes high, the source goes up relative to the gate right away and again the FET turns on. Not much for diode action.



In either case usually you'd seek a FET that has a very low On-Resistance at least 10 to 20 percent below the minimum operating voltage. So if you're using it on 3.3V, you'd want a FET that's fully on at 2.5V or so, which would probably mean 1.2V or less threshold, but that's down to datasheets.