My main objective with this project was to control these two DC motors with an Arduino Nano (it is not a UNO), powered by a 3-Cell 750mAh LiPO battery (both the motors as the Arduino itself). So, I came up with something like this (it is a simplified version):

Simplified version of circuit.

If the load on the two mottors rise, could the Arduino be damaged by this current increase? Is the voltage too high for the Arduino handle? Is there any more issues that this setup can have?

When searching, I found that it would be wise to get a fuse between the power supply and the arduino (before reaching Vin). Is that necessary?

  • You're not controlling anything, they will run as long as the battery is connected. Yes, you need the fuse - Lithium Polymer batteries do not tolerate shorts. Mar 16, 2016 at 13:20
  • You mean shorts at the motor part? If it happens, it may damage the arduino? How the current would behave in case of a short before the first motor?
    – yZaph
    Mar 16, 2016 at 13:32
  • Any short of the battery can cause it to quickly overheat, burst, and even burn. It has nothing to do with the Arduino being damaged, rather the connection of prototypes modules like this increases the chances of wiring being - even very briefly - wrong either during connection or if something moves later. Mar 16, 2016 at 13:34
  • But how would a fuse between the first branch and the Vin be of any help?
    – yZaph
    Mar 16, 2016 at 13:40
  • 1
    The fuse should be between the battery and the rest of the circuit, so that in the case of a short it can hopefully disconnect the battery from everything entirely. Mar 16, 2016 at 13:41

1 Answer 1


If the load on the two motors rise, could the Arduino be damaged by this current increase?

No, the motor current doesn't pass through the Arduino. It is completely separate.

Is the voltage too high for the Arduino handle?

Borderline. The regulator (a lowly 7805) may well get very hot at that voltage (depending on the Arduino's current draw) and go into thermal shutdown.

Is there any more issues that this setup can have?

Maybe add a small buck regulator to the 5V pin instead of using the on-board linear regulator. That will not overheat and will be far more efficient making your batteries last longer. Often marketed as "UBEC" (Universal Battery Elimination Circuit) and used throughout the model vehicle industry these are very easy to get hold of and are ideal for powering things like Arduinos from batteries like this.

  • So besides the temperature, this setup will work, right? If I can't find these regulators, is there any possible home-made circuit to replace it?
    – yZaph
    Mar 16, 2016 at 12:26
  • You will be able to find them unless you live in Vanuatu. Even then you can probably get someone to fetch one in a canoe. Building buck regulators is a bit of a black art and not a very easy task - you need to do the PCB layout properly or you end up with a lot of stray EMI being generated. But yes, besides the heat, it should work. I assume the "simplification" has removed the control circuitry?
    – Majenko
    Mar 16, 2016 at 12:29
  • The arduino itself probably isn't going to draw enough current to justify a switching supply. Smaller RC BECs are linear anyway, and those large enough to be switchers may waste enough power themselves to negate the advantage for a light load. An intermediate regulator at 9v can move some of the heat off the board, but will require its own capacitors. Mar 16, 2016 at 13:39
  • @ChrisStratton It all depends on what the OP has connected to the Arduino. Add a bunch of LEDs and the load will rise.
    – Majenko
    Mar 16, 2016 at 13:40

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