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I have a chinese arduino nano that I'm planning to use to open a door using a numeric keypad.

My power source is 12V and I want it running 24/7. I have read that the first thing to fail on an arduino is the voltage regulator due to the heat of lowering the voltage to something the arduino can use.

I have think of buying two big resistors of around 1W each to build a voltage divider and only feed 7V to the Vin pin of the arduino, so it will increase its lifespan.

Is this a good idea?

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  • It's basically another load to the battery. Some low power DC-DC voltage regulator (12V -> 5V) will be much better.
    – KIIV
    Mar 10, 2018 at 20:06
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    that is a very good idea ... doing experiments with the Arduino is the best way to learn
    – jsotola
    Mar 10, 2018 at 20:30
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    READ THIS: hackingmajenkoblog.wordpress.com/2016/08/30/…
    – Majenko
    Mar 10, 2018 at 20:48
  • How much current do you expect to use? If you know that you can calculate how much power is converted to heat. If your circuit is only using a few tens of milliamperes, the voltage regulator will only need to dissipate a few tenths of a watt of heat.
    – Gerben
    Mar 10, 2018 at 21:23
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    @Majenko Nice off-site explanation! You could make it an answer! (Copy/paste, upload images).
    – Nick Gammon
    Mar 10, 2018 at 22:21

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Terminating voltage by dropping it over large power resistors is a bad idea, efficiency wise. There are better options.

Switch-mode Buck (=step down) converters are usually far more efficient than linear regulators. A linear regulator dumps the excess voltage in a resistor which converts the energy mainly into heat. A switch-mode converter works by switching on/off an inductance at the right time, which is far more efficient.

For step-up/step-down DC-DC converter theory look here, here and here. For a comparison for switch-mode against linear regulators look here.

A step-down converter module like the ones with a LM2596 should be good enough for you. See here on ebay.

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