I have a device powered by either USB or a single-cell NiMH, like in the diagram below. The boost converter is resistant to backfeeding, but I need to prevent backfeeding the buck converter when powered by battery alone. What would be the best way? I could add a diode on the 5 V line before the buck converter, but I assume it would still backfeed through GND? Perhaps a MOSFET on the output of the buck?

Update: I'm now considering two solutions:

A). Boost the battery to 3.6 V and then feeding that and USB through the buck converter to reduce both to 3.3 V. The boost converter isn't susceptible to backfeeding, so I should be okay with this. I was hoping to avoid this because of increased losses by boosting the battery higher than necessary, but the boost is 84% efficient and the buck is 92% efficient, so hopefully the losses would be acceptable. This approach has the advantage of leveraging the lower output ripple of the buck converter and ensuring that VCC is exactly the same with either power source.

B) Use the design in the diagram below and put a Schottky on the 5V input. When I asked about this on the TI designers' forum they said that this will prevent backfeeding. The advantage of this is that I get about 9% greater efficiency and battery life, and I can also use a more common 3.3V boost converter instead of a 3.6 V boost.

enter image description here

  • 1
    \$\begingroup\$ How much power does the 3.3V load require? A Schottky diode could work but you have to account for the forward voltage drop. e.g. you would need to set the output of the buck/boost to 3.3 plus the drop of the diode. \$\endgroup\$
    – JYelton
    Jun 22, 2023 at 23:30
  • \$\begingroup\$ How can backfeed happen through ground, if a diode is on 5V? Please draw a diagram. Also why can two regulator outputs be paralleled? If the battery boost converter outputs 3.31V and buck converter 3.29V, the circuit will be powered by the battery alone. \$\endgroup\$
    – Justme
    Jun 22, 2023 at 23:35
  • \$\begingroup\$ I'm not sure if backfeeding can occur through GND or other pins, but I wanted to be sure. Ideally I'd rather have the diode on the input rather than the output so I don't have to account for voltage drop, which can vary with temp, etc. I have the ability to shutdown (disconnect) the input to the boost converter when there is 5 V power, so I don't need to worry about powering the device with the battery if the boost output is slightly higher than the buck output. \$\endgroup\$ Jun 22, 2023 at 23:39
  • \$\begingroup\$ Alternatively I could boost the battery higher and run them both through the buck, but I'd rather not do that because of loss of efficiency. \$\endgroup\$ Jun 22, 2023 at 23:40
  • \$\begingroup\$ How about a high-side switch with reverse-current blocking on the output of the buck converter, like this one?: diodes.com/assets/Datasheets/AP2161A_71A.pdf \$\endgroup\$ Jun 22, 2023 at 23:54

2 Answers 2


Why don't you make a unified power input that is compatible with either 5 volts from USB or 1.2 volts from the battery. You would then boost up to (say) 6 volts and buck down to 3.3 volts.

If you are worried about series conversion efficiency you can do this: -

  • Boost to 3.3 volts and, if on battery supply, bypass the buck
  • On USB the boost output will naturally be near enough 5 volts but...
  • Activate the buck regulator
  • \$\begingroup\$ I've added an update to my original question. I think option A (boosting the battery to 3.6 V and then dropping both supplies through the buck converter) might make the most sense. I initially didn't want to do that because of increased losses, but I like the idea of a cleaner power supply with the buck converter. I can just feed the 5 V in at the output of the boost converter, because that isn't susceptible to backfeeding. \$\endgroup\$ Jun 23, 2023 at 15:25

The back-feeding path is probably through mosfet body diode D2.
It happen if N-chan. or also P-chan. is used.
To avoid this put a D4 schottky in series with inductor.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ better put the diode in series with the FET (on drain side), to save on conduction losses. \$\endgroup\$
    – tobalt
    Jun 23, 2023 at 7:29
  • \$\begingroup\$ @tobalt Originally I was thinking about it, but I didn't know if it not going about chip version buck, where M1D1 is in one package. It is also possible to connect schottky in series with V1 or to drain, but then the input cap can be charged via schottky leakage and run the buck occasionally, so resistor parallel to input cap is needed to waist the leakage. \$\endgroup\$ Jun 23, 2023 at 12:37
  • \$\begingroup\$ Yes, this is a chip version-- it's the TI TPS62237. I also asked about this on the TI designers' forum, and they replied that it's fine to just have a Schottky on the 5V before the input, as most of the reverse leakage is through the input. This seems like one easy solution. \$\endgroup\$ Jun 23, 2023 at 15:06

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