I am taking an old HP G62-407DX apart, and I have replaced the motherboard with a raspberry pi. With some help on Superuser, I figured out what the display used. Now, I would like to know how to limit the power output of my battery. I am trying to use the battery that the computer normally uses, but on the battery, it says "10.8V 4200 mAh." I really don't want to burn out my pi, so any answers on how to do this will be greatly appreciated.

closed as off-topic by Milliways, techraf, joan, Jacobm001 Jan 3 '18 at 0:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be specific to the Raspberry Pi within the scope defined in the help center." – Milliways, techraf, joan, Jacobm001
If this question can be reworded to fit the rules in the help center, please edit the question.


For this situation you are going to need an adapter that converts 10.8v down to the 5v needed by the Raspberry Pi. Typically you would use what is called a Buck Converter for this. These are readily available in online stores such as Amazon, ebay, etc.

Do a search for terms such as "12v to 5v dc-dc buck converter with usb output" to find one.

Note: For your input (10.8v) you will specifically want one that accepts input voltages less than 12v.

Good luck! Let us know what you find.

  • My RPi has been running for the last couple of years like this. I have a 12V battery connected to a battery charger that is designed to be permanently connected and then I have a pair of those low cost (only a few GBP each) buck-boost converters with an input range of 7 to 18 Volts DC - one provides 12V at up to 3A (IIRC) for things that need it (my electric front door lock and a wireless RC receiver) and the other a 5V 3A supply to my RPi (via an RPi specific UPS). – SlySven Dec 25 '17 at 11:11
  • @MrChips Thanks for the great answer. I actually took apart an outdated car charger, following my friends instructions, and one of the boards inside (I don't remember the exact name) is functioning as a voltage regulator. I will definitely look into your solution though if this one breaks. – Jmaxmanblue Dec 25 '17 at 20:26
  • @JaredGizersky Awesome! Glad to be of assistance. Merry Christmas! – MrChips Dec 25 '17 at 20:28

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