I have build the above circuit on a breadboard and it is almost working but not quite. I've spent a number of hours on the problem to no avail. My background is not electrical but I've been building projects and circuit boards using arduinos, esp8266 and esp32 for a number of years. This is my first ever post....

The problem I'm trying to solve is negotiating two possible power supplies for the PCB, input 1 from linear regulator and input 2 from the USB bus. I do not want the USB bus providing power to the regulator as it currently does which back powers the 12V supply. I realise I could use diodes to achieve this, but I would like to get this to work!

I'm using LEDs at input 1 and 2 to confirm which power source is active.

If I only connect power to input 1 then load lights up and the LED at input 2 does not : intended behaviour.

If I only connect power to input 2 then load lights up and LED at input 1 does not: intended behaviour.

In reality power is likely to be supplied at input 1, then power supplied by input 2 and I want input 1 to be disconnected. This appears to happen. However then, when I disconnect power from input 2 leaving power on input 1, load does not light up: not intended behaviour.

I would like input 1 to be passed through to the load when input 2 is disconnected. If I measure the voltage at A it reads 0.8V which is enough to keep the logic level AO3402 on. If I then connect input 2 directly to ground, forcing the gates of Q7 and Q13 to 0V, the load lights up.

I am uncertain as to why the gates are kept at 0.8V. I have tried a lower value pull down at R36 but this does not work.

Any ideas appreciated, or different approaches also welcome. I do realise the two diodes would achieve the same result, but there may be significant power draw so I would like efficiency to be good.


Interesting problem!

As you have already determined, your circuit relies on the Q13 and Q7 turning completely off. However, you are stable in a state where there is a low current flowing through Q8, Q10, Q12, and Q14 sufficient to provide 0.8 volts on R36.

One way to solve this problem would be to make the connection at input 2 in such a way that it disconnects the connection between Q14 and R20 whenever you disconnect input 2. Then, point A will be pulled to ground and all will be well.

Good luck!


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you. I wonder if another n-channel mosfet with a pull down at the gate to ground and gate connected to input 2 would make this work? \$\endgroup\$ – SticilFace Apr 26 at 21:06
  • \$\begingroup\$ Your approach means that the circuit turn completely off whenever you transition between states, so that if power is present on both inputs, it must turn off when the one providing power is removed before turning on the other. You could try incorporating something with a threshold like the little power management chips used in microprocessor circuits that detect under-voltage conditions and generate processor reset signals. You could also incorporate some type of analog comparator. \$\endgroup\$ – John Birckhead Apr 27 at 18:26

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