A new requirement came such that we need to power using an external 5V supply and use USB communication at the same time.
This says that if you are connecting an external 5V to the 5V pin of Arduino Nano you can connect a Schottky diode at the output of the external 5V regulator.
My question is if we power from an external 5V regulator with the help of a Schottky diode how Arduino determines which source to take.
Below is my schematic.' ARD_5V_IN' is the external 5V supply to Arduino.
It is hard to say, and it will vary between setups. The USB outputs 5VDC Which goes through a protection circuit that prevents power from going back into the USB connector. A simple version of that is a diode. If you and the USB are supplying ~5VDC it depends on the forward drop of the protection circuits and the actual 5VDC voltage. If everything is close it may share but a very small difference will cause it to pull from the higher voltage. If that starts to pull down it will also start using energy from the other source.
