1

When powering Arduino through Japan Jack at 12V USB communication becomes unstable and frequently interrupts/disconnects. Using Arduino Uno R3 and Linux on the connected computer.

Does anyone have any insight why this might be happening?

7
  • This is a guess but since 12V is at the upper edge of what the board can accept over its jack is your PSU stable? If it isn't your regulator should heat up. You could measure this by monitoring the output from the VIN pin (apparently its bidirectional). This link might help open-electronics.org/the-power-of-arduino-this-unknown Commented Feb 9, 2017 at 12:51
  • @Matt can you point me to where they say this? according to the arduino page 12V is the upper limit of the recommended voltages, but it can work up to 20V (in fact, the regulator has a maximum voltage of 20V, but there is also a diode). So... Why 12V is the maximum?
    – frarugi87
    Commented Feb 9, 2017 at 15:31
  • @frarugi87 the effective limit is not the maximum voltage the regulator can accept when supplying tiny currents, but rather the voltage at which the heat produced by the difference between accepting power to satisfy the actually needed current at a high voltage and emitting only a much smaller amount of power at a low voltage, is more than can be removed from the device while it remains at an appropriate temperature. So the actual voltage limit depends on current draw and cooling situation. Commented Feb 9, 2017 at 17:04
  • @Matt ok, so 12V is not the limit of what the board can accept. According to this blog, the uno draws 46.5mA when no load is present. The datasheet says that with a 1cm*1cm copper pad, the 1117 has a thermal resistance of around 100°C/W, thus saying that 1.2W can be safely dissipated without triggering the thermal protection. At 46.5mA and 20V the 1117 dissipates only 0.7W, so it is perfectly safe to give 20V to a board without external loads. The 1117 can deliver up to 1A, thus leaving about 950mA of loads applicable... continue
    – frarugi87
    Commented Feb 9, 2017 at 17:28
  • So when you have 950mA of current absorption (from the 5V) 1.2W mean that you can have at most a 6.2V voltage on the input of the regulator (about 7V on the input pin, since there is a diode). I'm not saying that you can power everything, I'm saying that claiming that 12V is at the edge of the board voltages is wrong, since it only depends on how much current you are drawing (up to 30mA of current draw even 20V are fine)
    – frarugi87
    Commented Feb 9, 2017 at 17:32

1 Answer 1

1

12V is a little on the high end for what the arduino can handle by itself. If I were you I would use a DC step down Buck Converter module to hold the voltage at somewhere around 6v.

http://www.ebay.com/itm/like/400673913185?lpid=82&chn=ps

They're super cheap and pretty easy to use. Simply solder the + and - leads from your 12v power supply to the +in and -in on the buck converter. Then the + out and -out to the + and - for your arduino. Turn the potentiometer until you get somewhere around 6V.

Edit:

Actually I was just reading online, and some people have trouble communicating with the arduino through USB while an external power source is connected. Internet says this problem is related to sharing voltages.

https://forum.arduino.cc/index.php?topic=194598.0

2
  • I disagree. According to the arduino page 12V is the upper limit of the recommended voltages, but it can work up to 20V (in fact, the regulator has a maximum voltage of 20V, but there is also a diode). So 12V is perfectly fine. 6V, on the other hand, is undervoltage: dropout of the regulator is around 1V, the diode has a 0.6V dropout, so the minimum voltage should be around 6.6V...
    – frarugi87
    Commented Feb 9, 2017 at 15:33
  • The thoughts are not bad in the general sense, but unlikely to be related to the observed problem. Setting the output to 6v however is erroneous advice - the Arduino's on-board regulator needs a higher input voltage that than, while 6v is too high to safely bypass the on-board regulator. Commented Feb 9, 2017 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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