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I would like to use a PFET as a diode to only let current from the 3.3V to 4.2V area (when 4.2V is not present).

When 4.2V is present, it should not charge the battery.

Is it possible to do this with mosfets. I would like minimal voltage drop and up to 600mA can be drawn. Circuit 2 is what I am trying to achieve. Circuit 1 is my attempt, but it isn't working.

Btw, this is a mod on a PCB, so I am limited in the part count.

enter image description here

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  • \$\begingroup\$ Your two FETs are arranged symmetrically. On what principle do you expect them to block current from the 4.2 V node from flowing back to the 3.3 V node? What stops you from just using a diode as a diode instead of trying to use something else as a diode? \$\endgroup\$
    – The Photon
    May 9 '20 at 14:24
  • \$\begingroup\$ I tried using a single PFET as a diode (just Q1), but it didn't work. The voltage drop of the diode is significant. I cannot afford 0.1 voltage drop \$\endgroup\$
    – Hamoudy
    May 9 '20 at 14:28
  • \$\begingroup\$ What you're looking for is an "ideal diode" circuit. There are several designs out there you can google. Or ICs specifically designed for this function, if you need absolute lowest forward voltage. \$\endgroup\$
    – The Photon
    May 9 '20 at 14:42
  • \$\begingroup\$ Ok thanks for pointing me in the right direction \$\endgroup\$
    – Hamoudy
    May 9 '20 at 15:09
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As the user The Photon already pointed out in the comments, there are some ics which can do what you want. For example, this evaluation board from Texas instrument uses two ideal diodes LM66100 to connect a load to two different supply voltages. If the second supply voltage is larger than the first one, the ideal diode of the latter is disabled and the load is supplied by the former.

If for some reason, you still want to implement something similar using discrete components, you can try the following. I must say upfront what this circuit was not optimized and may have some flaws, which I might have overseen.

How it works:

The upper mosfets act as ideal diodes operating in a complement fashion. The lower long tailed pair compares both supply voltages and turns on, the one which is higher. Some positive feedback might be needed somewhere in order to add some hysteresis during the transition. circuit

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  • \$\begingroup\$ Thank you very much, LM66100D is a suitable solution \$\endgroup\$
    – Hamoudy
    May 9 '20 at 21:45

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