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For a project I'm working on, I need to run a motor for a few seconds (somewhat rarely). The motor requires 24V@1A. The rest of the circuitry runs on 3.3V@300mA. My general thought is to use a 3.7V LiPo battery to act as a "current booster" (since it can discharge at a much higher rate than it charges) and add a buck regulator to get my 3.3V. A 5V@1A USB power supply would both charge the battery and power the 3.3V regulator.

That circuit is understood, using a LiPo charger with power-path circuitry to allow the buck regulator to run off the USB input or the battery when no USB power is available. And as a bonus, this also creates a built-in "UPS" that will run off the battery for some time if the USB fails or is disconnected.

But the second part (and the main reason for the battery) is the 24V requirement. This is easily done using a boost regulator to produce the 24V@1A from the LiPo (there are plenty of "high discharge" LiPo cells that will easily provide the necessary current). But I'm not sure how to use the LiPo as the power source without disconnecting the USB power and running only off the LiPo while the motor is running. This actually works, but it seems like there should be a more elegant solution. It's a different problem than the 3.3V buck regulator, since either the USB or the LiPo can provide the power, but with the 24V boost regulator, only the LiPo can provide the power.

I've done a fair bit of online research, but I haven't found anything that really covers this use-case, so maybe I'm not searching for the right thing. Any pointers?

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    \$\begingroup\$ If the cell can deliver the power, and you can design the boost, then the rest is easy. Just connect the boost directly to the battery. No need to disconnect anything. Just power the motor when you need to. \$\endgroup\$
    – mkeith
    Jul 15 at 3:44
  • \$\begingroup\$ With this approach, I'm concerned about overloading the 5V supply when the 24V boost regulator starts pulling 5A+. \$\endgroup\$
    – eric
    Jul 15 at 19:31
  • \$\begingroup\$ The battery is supplying those amps. Not the 5V supply. You can't just connect a 5V supply to a battery. There is a charge controller in between the 5V supply and the battery. The boost needs to be connected to the battery, not the 5V supply. When the battery starts boosting, it could cause battery voltage to sag and trigger the charger into a charge mode. But it shouldn't cause the 5V rail to crash. \$\endgroup\$
    – mkeith
    Jul 16 at 1:43
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Certainly you'll want to run the boost converter directly off the cell, not via the 3.3V regulator. One option would be to add a FET that could disconnect the 5V supply when you want to enable the boost converter; that's probably the simplest solution.

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  • \$\begingroup\$ Yes, of course; the 3.3V and the 24V regulators are both powered from the battery. A FET that disconnects the 5V supply from the battery is my current solution. I was just wondering if there was a better way. \$\endgroup\$
    – eric
    Jul 15 at 13:10
  • \$\begingroup\$ I can think of a couple of alternatives: 1. Leave the 5V supply active and hope that it current-limits gracefully; 2. Add a current-limit circuit so the 5V isn’t overloaded. I doubt that the former would receive much popular acclaim on those site. The latter is an option though- a 3R resistor would drop 900mV at 300mA so possibly above the dropout of the regulator, but when loaded down to 3V the 5V would supply no more than 600mA, well within its rating. \$\endgroup\$
    – Frog
    Jul 15 at 17:59
  • \$\begingroup\$ The second option is interesting. \$\endgroup\$
    – eric
    Jul 15 at 19:30
  • \$\begingroup\$ A simple solution is still a solution. \$\endgroup\$
    – Frog
    Jul 15 at 19:37

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