# Do I need a buck converter?

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?

• yes and yes if the arduino can support the BT transmissions. Feb 4, 2020 at 12:03
• The bluetooth transmissions are separate from the arduino, it's just about powering the two components. Feb 4, 2020 at 12:17
• 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? Feb 4, 2020 at 12:17
• 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. Feb 4, 2020 at 12:38
• Why design a buck converter, rather than just buying one off-the-shelf? Feb 4, 2020 at 12:58

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.

• 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. Feb 4, 2020 at 12:51
• @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. Feb 4, 2020 at 12:57
• I know, I have the circuit around the MC as well Feb 4, 2020 at 16:47