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I'm working on a project where there is no external 5V voltage regulator available but only 6V. I can not power it through the power jack because via this way I do not have enough current and get thermal shutdown.

Can I power it via 6V over USB? at the moment I'm trying this out and it works but what are the long term risks I'm prone to?

  • yes, it is bad. – Juraj Jul 28 at 9:04
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The Absolute Maximum Operating Voltage of the ATMega2560 is 6.0V. The same applies to the ATMega8u2 that does the USB to Serial on the Arduino Mega board.

Recommended maximum voltage is 5.5V

So yes, you can run it at 6 Volts, but you are right on the edge, so I wouldn't recommend it.

Possible risks are a decreased lifespan of the Arduino, and there might be some stability issues.

If the application isn't safety critical, and you don't mind having to replace the board after it failed in, say, two year of operation, it could be an acceptable risk (if there are no easily available alternatives)

PS you might want to measure you power supply. It may say it's 6V but could give out 6.2V. The output voltage of some supplies also vary with the load that's across it.

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    The incomming voltage is actually indeed 6.2 voltage – Roel Jul 28 at 9:38
  • put a diode (or two) on the incoming rail to drop it 0.6 (or 1.2)v. – dandavis Jul 28 at 19:50
  • We actually have a din rail inside the project. So we were planning on doing just that. – Roel Jul 29 at 10:12
  • Voltage drop across a diode depends on the current, so it won't give a very stable voltage (like a regulator does). But it will probably be fine. – Gerben Jul 29 at 12:48
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Is it bad to power a arduino mega with 6V via USB?

Yes, the Arduino Mega is not designed to be powered that way and you are probably abusing the USB specifications. USB can supply higher voltages than 5V but only if it is properly negotiated with the device.

The schematics (see Documentation tab) show a 5V regulator powered from the 2.1 mm Power-connector or from the "Vin" header pin. If the regulator specs say it can work from 6V, you should probably use that instead. However although the 1870_LD1117S50CTR seems to be a low-dropout (LDO) regulator, it has a dropout voltage of 1V. So I wouldn't power it from a 6V source - that seems equally marginal.

In short, I think you really need to find a more suitable power source.

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