I need to power an Ardumoto Shield to drive two 12V DC motors. I was planning to connect a 12V battery to the Max 18V input on the Ardumoto and a 9V battery to the barrel jack input of the Arduino Uno. Can you tell me if this should be ok or if it would cause any damage to the Arduino?
It mentions in the Ardumoto hookup guide:
There's no mention of the Max 18V input on the Ardumoto.. Advice appreciated, thanks.
2 Answers 2
Do not supply power to both the Arduino barrel jack input and VIN on the shield!
There goes the simple answer. Choose one.
If you power the system through arduino barrel jack input, you will have polarity protection, but the voltage output on VIN pin (which also used by Ardumoto shield) is slightly lower, about 0.6V lower than the input voltage because of the voltage drop on the protection diode.
If you have want to differentiate the supply for Ardumoto and Arduino (as described on your plan), you have to tamper the Ardumoto shield to isolate both supply. Although I never try it myself, it seems possible if you understands how L298 works.
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I'll probably just use the 18V Max Vin on the Ardumoto shield. Or maybe get rid of the Vin header pin on the Ardumoto that connects to the Arduino Vin.somers– somers2018年02月05日 15:11:41 +00:00Commented Feb 5, 2018 at 15:11
The L298 used in the Ardumoto shield has a maximum supply voltage of 46V. Hence 18V should be sufficient to power it as long as the motors have enough power too.
The Arduino voltage regulator, on the other hand, won't be able to handle such a high voltage. Hence you will have to remove the Vin pin that connects the Arduino & the shield in order to isolate the power supplies of the two.
Only then should you attempt plugging in the shield & then powering the Arduino through the 9V battery.