I have an original Arduino Uno R3. According to the specifications the recommended input voltage is 7 to 12 Volt.

However, I have a laptop adapter of 19V that I want to use to power my arduino because it offers more current than the 9V adapter I have as well.

I don't want to damage the Arduino or the DFRobot 2A Motor Shield I'm using. Can I safely plug in the adapter and will it still work correctly?

  • 1
    Input Voltage (limits) 6-20V. What kind of motors do you have, because these have to be able to handle 19v too.
    – Gerben
    Mar 16, 2015 at 14:59
  • The motors need to handle 19v if powered separately via PWRIN, but not if they use the 5v regulated supply from the Arduino via VIN. See my answer below for more details.
    – cjs
    Apr 1, 2017 at 10:41
  • 19V is too close to the absolute maximum rating (20V) for comfort. Any voltage surge could kill the regulator. And you'll have thermal problems as well. So I'd say powering the Uno at 19V is very bad advice! Stick to the recommended input voltage range instead! Apr 1, 2017 at 16:59

2 Answers 2


It's often better to supply separate power to high-current applications (such as motors) rather than taking power off the Arduino. This avoids problems with current and thermal limits of the Arduino on-board power regulator.

Looking at the manual for your board, the motors can be powered by a separate power supply hooked up directly to the motor shield. There's a six-pin header on the board marked VIN and PWRIN. When the jumpers are on the VIN side Arduino-supplied power is used to drive the motors; when the jumpers are on the PWRIN side the motors are supplied with power from the terminal block next to the jumper. (The rest of the board remains powered from the Arduino, but that's usually a negligible amount of current compared to your motors.) A photo in the manual demonstrates the pins set to PWRIN and an external power supply hooked up.

The documentation says you can supply any voltage between 4.8 V and 35 V DC, but be careful! What they don't make clear in the documentation (but is clear from the schematic and the L298N datasheet) is that this power is supplied directly to power side of the L298N and the motors without further regulation. The chip can handle up to 50 V DC, so it would be fine with a 35 V input, but you'll want to check that your motors can handle the voltage you want to give them.

If the motors can handle 19 V input, and your 19 V power supply is reasonably well regulated itself (which a laptop computer power supply generally would be), you can use that to power the motors directly through the motor shield and continue using your current power supply to power the Arduino.

(The ST L298 page has further information, including a slightly less readable copy of the data sheet and application notes.)


According to the page you linked:

The board can operate on an external supply of 6 to 20 volts. If supplied with less than 7V, however, the 5V pin may supply less than five volts and the board may be unstable. If using more than 12V, the voltage regulator may overheat and damage the board. The recommended range is 7 to 12 volts.

So, yes. You can use the 19V external power supply to power the board. However, be aware that the voltage regulator can overheat over time and damage the board.

  • 1
    If someone has a way of calculating how hot the voltage regulator would get, I'd greatly appreciate the knowledge. I'm also not sure if the regulator will get hotter quicker depending on how much current is being pulled through it (for the motor driver shield)
    – krol
    Mar 16, 2015 at 16:11
  • 1
    Also, there may be current issues - I found an analysis that says the current limit of the Uno is 1A ( electricrcaircraftguy.blogspot.com/2014/02/… ). TheDFRobot 2A Motor Shield might try drawing more than the Arduino can handle. I don't know how to calculate that
    – krol
    Mar 16, 2015 at 18:57
  • 1
    yes, the more current, the more heat. to calc the temp, you need to know the junction-to-ambient rating of the regulator in watts. To calc watts, subtract 5 from 19 (14) and multiply by the current. If we assume, say, 200ma; 14*.2=2.8w; way too much. divide watts by J2A to get deg C.
    – dandavis
    Apr 1, 2017 at 13:41
  • I agree with @dandavis. Here's the datasheet for the regulator: onsemi.com/pub/Collateral/NCP1117-D.PDF. A dissipation of 2.8W would raise the junction temperature by 188°C. If ambient was 25°C, then the junction temperature would be 213°C, far above the 175°C limit at which the regulator internally shuts off. So it won't work at that current (200 mA). Apr 1, 2017 at 16:51
  • Also, 19V is too close to the absolute maximum rating (20V) for comfort. Any voltage surge could kill the regulator. So I'd say powering the Uno at 19V is very bad advice! Stick to the recommended input voltage range instead! Apr 1, 2017 at 16:54

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