I want to use a 2s Lipo for a autonomous car built with a Beaglebone and Motor Bridge Cape from Seeedstudio. The Motor Cape has an built in voltage regulator which steps down 7,2V to 5V system voltage. This allows supplying motors an Beaglebone from same source.

While charging a 2s LiPo charger is supplied with 12V. Is it possible to charge the LiPo under load? Or do I have to supply the Beaglebone from 12V source while charging?

This would mean that there has to be a switch between battery source and 12V source which selects the source automatically. That is not a complicated task, but supplying the system by battery all the time would be easier.

For example the swithing between LiPo and 12V source could be realized like this:


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ by "loading lipo" and "lipo loader" do you mean "charging lipo" and "lipo charger"? I think "charging lipo under load" is better understandable than "loading lipo under load" \$\endgroup\$
    – Wesley Lee
    Commented Apr 3, 2016 at 18:21
  • \$\begingroup\$ that is correct. It should be called "charging under load" \$\endgroup\$
    – Timm
    Commented Apr 3, 2016 at 19:16

1 Answer 1


Adding a load to the battery while charging means that the load is actually added to the charger. Since the charger must provide enough current to both charge the battery and supply the device there may be issues if the load is disconnected while the battery is being charged.

Microchip's AN1149 describes a method for converting a battery and charger into an offline supply that switches to wall power when the battery is being charged. Since the charger only ever has the battery as a load, the above problem is avoided.

  • \$\begingroup\$ Application Note AN1149 (ww1.microchip.com/downloads/en/AppNotes/01149c.pdf) helps me very well. Connection the load directly is not recommended. The note displays how automatic switching between battery and power supply can be implemented. \$\endgroup\$
    – Timm
    Commented Apr 3, 2016 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.