I encountered a mass produced device that has a power circuit like this:

   5V from USB ==[boost converter]==> 13V ==[buck converter]==> 12V

and it confused me. I don't see why you would need a buck converter there.

The 13V rail is connected only to the buck converter, and nothing else. I think you can just set the boost converter to output 12V and omit the buck converter altogether, because power input is from USB and there's no need to drop voltage down (under normal circumstances). Or, if noise is a concern, you might use an LDO in place of the buck converter. However in case of this device the 12V rail is fairly noisy.

What is advantage(s) of this boost+buck configuration, over a single boost converter or a boost+LDO configuration?

EDIT: adding schematic which I reverse-engineered from the PCB, picture follows:


simulate this circuit – Schematic created using CircuitLab

enter image description here

Note: the other side of the PCB is a ground plane. No traces there.

  • 1
    \$\begingroup\$ What brings you to the conclusion that the power unit is as you describe? It does seem like a strange configuration. Post the schematic. \$\endgroup\$
    – Kartman
    Commented Mar 21, 2022 at 5:42
  • 4
    \$\begingroup\$ Maybe the manufacturer changed the product and now the 13V is unused and it was cheaper to not redesign it? We can't know the reasons the product ended being like that. \$\endgroup\$
    – Justme
    Commented Mar 21, 2022 at 5:52
  • 1
    \$\begingroup\$ At a theoretical level there’s no advantage but practically there could be reasons, for example there might be an option to run from a >12V supply rather than USB, or perhaps the designer wanted a fast time to market and used some known-good circuits rather than save on the BOM cost. We can only speculate. \$\endgroup\$
    – Frog
    Commented Mar 21, 2022 at 5:53
  • \$\begingroup\$ @Kartman Sure, added the schematic and a picture. \$\endgroup\$
    – q61org
    Commented Mar 21, 2022 at 7:35
  • \$\begingroup\$ Apart from the very common cause that @Justme indicated, I'd like to add that some designers think that buck is safer. That is because if the main switch in a boost converter fails (becomes short circuit by 99% chance) it'll short the supply afterwards. I really can't understand this because the same failure can cause catastrophic effects in a buck since the load which supposed to be supplied from a lower voltage will be connected to a higher voltage. But I don't want to think that this is the main idea behind the design of your circuit. \$\endgroup\$ Commented Mar 21, 2022 at 8:45

1 Answer 1


I’d say one possible reason is trust of the boost converter. Anecdotally I know people who see boost converters as a necessary evil, they are unstable and noisy.

These days it isn’t always the case, but you do see output ripple as quite high sometimes on boost converters compared to a similar buck converter. If a 12V rail needs to be very stable, it may be that they are boosting it to over 12V (13V in this case) in a rough and ready manner, and then brining it down to a very stable 12V in order to have the required cleaner supply rails.

We can’t know the real reason without being the designers, but it could be that the cost of getting a stable/clear enough 12V rail from a boost converter alone was higher than using the combination of a boost followed by buck.

  • 4
    \$\begingroup\$ The thing is you have much closer control over a buck converter. If the voltage drops, you switch on, if the voltage gets too high, you switch off. With a boost converter, you're basically flying blind in the phase where the inductor is discharging. If the load drops in this phase, your output voltage will overshoot, potentially by a lot. \$\endgroup\$
    – mow
    Commented May 24, 2022 at 11:12

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