So say you want to use transistors with your Arduino Nano to power components, which require greater than 12V. If your power supply is more than 12V, is there a way to safely use the same power supply to power the Arduino?


The voltage regulator (on the Vin pin) is supposed to be provided with 12V at max, since it is not cooled (it is a linear regulator, that dissipates the excess voltage as heat). If you have significantly more, you should at best buy a little switching regulator, to directly get a regulated 5V to feed to the 5V pin of the Arduino (that's what I suggest). Be sure to use a regulator, that can provide the needed current (which also depends on what else you provide over the Arduino).

If you really don't want to use an extra regulator and you don't have much more than 12V (maybe like 15V), you might get away by cooling the voltage regulator on the Arduino. The voltage regulator on a genuine Arduino will go into emergency shutdown, if it overheats, but it will not be damaged easily. Here you could try a bit. But be careful to not stress it too much. The clones from china often have different/cheaper regulators, that will mostly be fried if you overheat them. But they are very cheap.

  • If powered with 12V, do those output pins still output 5V?
    – BobaJFET
    Jan 31 '20 at 22:35
  • The output pins of the Arduino still use 5V, since the Arduino still works with 5V. The 12V only get regulated down, when you provide them to the Vin pin or the barrel jack. A dedicated voltage regulator takes the 12V and outputs 5V for the Arduino to run on. The only question is, if the voltage regulator can stand the power to dissipate, which depends on the provided input voltage and the drawn current.
    – chrisl
    Jan 31 '20 at 23:52

The recommended input voltage for the Arduino Nano is 7-12 V, the max. 20 V.

So you could use upto 20 V, but this is indeed not advisable.

What you can do, is to use a so-called step down/boost down converter which converts the voltage back to a voltage that your Arduino Nano can (easily) handle. The voltage drop will be changed into heat, but I guess considering you are using a higher voltage power supply, you are not using battery power.

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