So I have an Arduino and an L293D. The L293D logic is powered by the +5V output of the Arduino, while the DC motor is powered by an external power supply.

Here's the part I don't get. If I connect the Arduino and the power supply grounds together the circuit works, otherwise it doesn't. Can you explain me why?


1 Answer 1


While a schematic would have helped describe the problem statement better, one key concept might help in clarifying this matter:

  • A voltage is the potential difference between two points in a circuit, it is not an absolute value of any physical characteristic at a single point in a circuit. Thus, there is no absolute potential involved, it is relative value, a difference.

How this applies:

The control side of the L293D is actually powered by the +5V from the Arduino only when the +5V has a reference ground available, that corresponds to that particular +5V supply, in other words, the ground of the Arduino board.

The L293D does not have a separate drive-side ground pin, just the "Heat sink and Ground" pins, which are also the ground reference for the \$V_{cc1}\$ pin.

If you note the schematic on page 3 of the datasheet, the ground references for \$V_{cc1}\$ and \$V_{cc2}\$ are one and the same, the GND pin(s). Thus, that reference needs to be connected to the Arduino's reference ground.

  • \$\begingroup\$ Thank you for the complete answer, I get the problem now ! So do you think it's a good idea in that case to plug the L239D +5V to my external power supply too ? In that case I would only have the control pins connected to my arduino ? (I've notice that if for example I don't put capacitor on my DC motor inputs I get a lot of noises across the whole circuit, so I would like to decouple it from the arduino as possible) \$\endgroup\$ Dec 22, 2013 at 10:02
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    \$\begingroup\$ No, the +5V should not be connected to the external power supply at all - else you'll have either a blown regulator IC on the Arduino (most likely) or a blown power supply (not so likely). Just the ground pins are to be interconnected. \$\endgroup\$ Dec 22, 2013 at 10:23

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