2

On many tutorials I have looked at, they say that you should power your Arduino with a 5 - 12V external power supply, but on the Arduino, there is a voltage regulator. Is there anything wrong with powering your Arduino with power outside this range? Solved

3
  • 1
    it won't work very well below 5 V ... the regulator has to dissipate too much power when the input voltage is high .... this sort of question has been asked before – jsotola Jun 10 at 23:38
  • Thank you for your answer. – Nathan Jiang Jun 10 at 23:46
  • 1
    it is 7 to 12 V – Juraj Jun 11 at 4:20
1

Let me just give you an example of what happens even with 12V: One time I was working on a claw machine that I made for the Arduino and when I plugged in 12V while the 5V USB port plugged into the computer, guess what happened. Well, I'll tell you: the Arduino burned up and it was broken. I couldn't even put in the code because it was so cooked. I don't think it turned on either. You see, 7-12V is just a range. You typically want to use what is in between the voltage rating. Even with 12V, you have to be careful. So, hopefully that clarifies. Basically, it will burn up your Arduino into hot plastic, sort of like a hot glue gun. The datasheets agree if you don't believe us. You can try it, but just be aware that I warned you not to. This is the datasheet for the Mega 2560 rev3 It's all because of voltage, which is a sort of pressure for electricity. You could say the same for the water specifications. What happens when you add too much pressure to a water container? It bursts, right? In a way, this can be applied to the Arduino. Thanks for reading, hopefully this helped.

4
  • 1
    Well, things are a bit more complicated. For the MEGA, if you use the barrel jack the power is fed by the LD1117 IC. It has a roughly 1x1cm heatsink, but in order to be more conservative I'll ignore it. Consequently the thermal resistance is 110°C/W. If the arduino remains at 30°C, the maximum power the 1117 can dissipate is about 860mW. Let's remove 10% to be a bit safer (so 780mW). When the mega is idle with no external devices, it draws 73mA; consequently the 1117 can bear a 10.5V drop, which puts the maximum input voltage at 15.5V. If, on the other hand, you [...] – frarugi87 Jun 11 at 14:58
  • 1
    [...] want to add two bright LEDs (20mA each), the current jumps to 113mA, lowering the voltage drop the 1117 can withstand to 6.9V (and so a maximum input voltage of 11.9V). As you can see, the maximum input voltage highly depend on what you are building... – frarugi87 Jun 11 at 14:59
  • 1
    This sounds more like a power supply or board issue; there should be zero issues with powering through the barrel jack and having USB plugged in at the same time; that's a very common scenario when, e.g., you're directly connected to a 3D printer or CNC machine through USB. – Dave Newton Jun 11 at 16:43
  • 1
    Well, it is a clone. I am not sure if that has to do with anything, if you want I can delete my answer, if it doesn't make sense. – Austin Jun 11 at 16:46
3

You could use a step-down converter to lower the voltage and also stabilize it to 12V or even lower and you wouldn't have any problem even if you power supply isn't really stable and outputs higher voltage.

(Instead of buying one you can even build it, it isn't really hard to make one)

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.