I am planning to build a circuit with a bluetooth audio receiver (MH-M18) and an Arduino Nano, connected to a DIN connector.

I have a wire with 12V supply readily available to connect to the circuit. The bluetooth module works with 5V and draws up to 20mA.

Is it a good idea to design a buck converter to step down the 12V to 5V to power both the bluetooth module and the Arduino, or should I just power the Arduino with the 12v and power the bluetooth module with the 5V output from the Arduino?

  • \$\begingroup\$ yes and yes if the arduino can support the BT transmissions. \$\endgroup\$
    – Sorenp
    Feb 4, 2020 at 12:03
  • \$\begingroup\$ The bluetooth transmissions are separate from the arduino, it's just about powering the two components. \$\endgroup\$
    – Arrrow
    Feb 4, 2020 at 12:17
  • \$\begingroup\$ It sounds doable. Could you please explain what made you think of the buck converter? Are you trying to shave off power dissipation from the Arduino LDO? \$\endgroup\$
    – eeintech
    Feb 4, 2020 at 12:17
  • \$\begingroup\$ The bluetooth module only works at 5v, and I hadn't realised that the arduino can just work at 12v, so I thought I needed it regardless. Only after ordering components for the converter I realised the arduino can just be powered with the 12v. An LM7805 would be an alternative, but the circuit is going in quite a confined space and it would generate quite a lot of heat so I figured that wouldnt be viable. \$\endgroup\$
    – Arrrow
    Feb 4, 2020 at 12:38
  • 1
    \$\begingroup\$ Why design a buck converter, rather than just buying one off-the-shelf? \$\endgroup\$
    – Simon B
    Feb 4, 2020 at 12:58

1 Answer 1


The Arduino has a linear regulator, which means the current in = current out, and the excess voltage translates into wasted heat in the regulator.

Your datasheet shows that the receiver module will use a maximum current of 20mA. Let's assume it uses this all the time.

If you use a buck converter, it will use something like 10mA from your power source (a battery?). 12V * 10mA = 120mW. 5V * 20mA = 100mW. The difference is 20mW which translates to heat in the buck converter. 10mA is just an example number that I have made up - if you have a particular regulator then the datasheet should allow you to estimate how much the input current will be.

If you use a linear regulator, it will use 20mA from the power source. 12V * 20mA = 240mW. 5V * 2mA = 100mW. The regulator will heat up by 140mW.

This question tells me you can get 900mA through that pin when using a 7V power supply, which is 1800mW power dissipation (900mA * (7V - 5V)). So 140mW should be no problem for the regulator.

You might want to use a buck converter if:

  • You care about wasted power (more power = less battery lifetime!)
  • You care about heat (maybe if this circuit is designed to work in the desert at noon, or if you want to put it in a tiny box with no airflow)
  • You want to experiment with buck converters

Otherwise, I don't see any reason not to use the Arduino's built-in linear regulator.

  • \$\begingroup\$ Thank you for your answer, Point 2 (it's going to be placed behind my car radio) and 3 are kind of relevant here, point 1 less so since the power source is a car battery. I was planning to use the MC34063 regulator. \$\endgroup\$
    – Arrrow
    Feb 4, 2020 at 12:51
  • \$\begingroup\$ @Arrrow I don't think it will be enough heat to be a problem for point 2, but you could test it. Also MC34063 is a control chip for a regulator, not a whole regulator. \$\endgroup\$
    – user253751
    Feb 4, 2020 at 12:57
  • \$\begingroup\$ I know, I have the circuit around the MC as well \$\endgroup\$
    – Arrrow
    Feb 4, 2020 at 16:47

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