I am working on a flight control board inspired in this Sparkfun board:


It is basically an Arduino UNO with a FTDI chip instead Atmega8u2 as USB-UART bridge. It manages power stage in order to allow both USB and Jack connector at same time. In my device I want to allow both USB and ESC power options. My question is, can I replace the Jack connector and the LDO and just connect the 5V output of my ESC in the 5V wires of that schematic?

Something like this:


I actually don't know how smooth ESC's output voltage is. Maybe it is a good idea to let the LDO and connect the ESC over the jack input.

  • \$\begingroup\$ What is ESC? I am not familiar with that acronym and I saw no mention of it in the link. Where is it in your schematic? Also, please describe in your own words the purpose of Q1 (what you think it is for and what you think it is going to do). The way it is connected now does not make sense to me. If you explain what it is supposed to do, maybe I will come to understand or be able to suggest an alternative configuration. \$\endgroup\$ – mkeith Jan 13 '15 at 21:09
  • \$\begingroup\$ Hi, sorry I didn't specify it. \$\endgroup\$ – zapeitor Jan 14 '15 at 9:00
  • \$\begingroup\$ An ESC is Electronic Speed Controller. Very used in RC to controll motors. en.wikipedia.org/wiki/Electronic_speed_control They has 3 pins. GND 5v Signal. You control motor speed over signal. It is the same control signal than a servo. 5v is an output pin what can provide power from the battery. \$\endgroup\$ – zapeitor Jan 14 '15 at 9:09
  • \$\begingroup\$ Actually I don't understand either the purpose of Q1, but all Arduino boards I've seen have one like that. Its purpose is supossed to be mannage power stage with both Jack connector and USB power options. But I don't understand how that circuit works, that's why I just replaced the LDO with my 5v output from the ESC. I am not sure at all. I think I'm going to add a jumper to my board so you can choose between USB and ESC as power input, but not both at same time. Thank you. \$\endgroup\$ – zapeitor Jan 14 '15 at 9:09

Given that the 5V supplies are feeding a 3V3 LDO, then why not just use two Schottky diodes.

Put one in series with each supply. That will allow both to work, and the higher voltage will 'win'. Schottky diodes will only drop 0.1-0.2V in conduction.

Here is a 2A one I use often: http://www.digikey.com.au/product-detail/en/CDBB2100-G/641-1109-1-ND/1559070

  • \$\begingroup\$ It is a good idea. I'm going to try it. Looks more simple than this weird transistor. \$\endgroup\$ – zapeitor Jan 14 '15 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.