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So I have a stepper motor that's rated for 2.9V and 0.8A and I connected it to a L9110 Driver and I wanted to drive it with my arduino. But I don't know how to drive it safely without damaging the motor. I am using the stepper.h library and thought of connecting it to 3.3V pin on the Arduino but I am a bit concerned.. is it Ok to power it with 3.3V if no then What can I do?

Edit: the L9110 module I am using is a dual H Bridge.

Edit2: I was Indead Powering the Stepper and the Module directly from the arduino.. luckily the arduino survived.

Edit3: So I didn't have a 3V power supply so I used a Voltage Divider with a 9V battery and connected the L9110s module with the battery (after its voltage divided) and the module powers on.. but when the arduino starts driving the stepper the module's led starts Flickering instead of staying on and the stepper doesn't turn... I don't know what is the problem.

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  • I don't think the motor will be your problem. I think the arduino is in danger. If the motor is powered 0.4V higher than specified, who cares. But you try to draw 800 mA from the arduino 3.3V output. That's too much. Just use an external power to drive the motor, not the arduino. BTW: I assume you' ll use a L9110 Module with two driver chips. One L9110 alone is only useful for DC motor; not for steppers. – Peter Paul Kiefer Jan 29 at 16:17
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    It may just because I'm not completely awake, but I'm finding it difficult to follow your wiring description. A diagram or schematic would be better. – timemage Jan 29 at 16:17
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    It is unclear from your description / question if you are connecting a stepping motor directly to the pins on an Arduino. Further, we do not know which Arduino you are using. But, in general, an Arduino's outputs are likely directly connected to the processor. And there are no (or very few) processors that can survive getting 800mA pulled from one of their output pins. So do be very careful or you may burn out the processor. – st2000 Jan 29 at 16:20
  • Oops... I powered it directly from the arduino... but it's still alive(I think) – Mohamed Technology Jan 29 at 17:46
  • Is it fine if I used a 1n4007 diode to "protect" The Arduino UNO from The Motor? Or it will be useless?? – Mohamed Technology Jan 29 at 19:27
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Motors draw a LOT of current compared to CMOS circuits.

The 3.3V output on most Arduinos can only provide a very small amount of current. The Arduino Uno, for example, says it only provides a max of 50mA from its 3.3V regulator. Your motor driver probably needs >10X more than that.

The overload protection on the Arduino's voltage regulator will probably kick in and cause it to shut off. It might take a while to cool down before 3.3V output comes back. I've heard that the protection on Arduino knockoffs doesn't always work, so you could also fry the voltage regulator if you try to draw too much from it.

Bottom line: Don't do that.

Get a high current supply for your motor driver.

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  • I was powering the driver directly from the arduino. Thankfully the Arduino isn't Dead (yet) so I decided to Use a 9V battery with a Voltage divider to power the driver and the motor But now when the arduino Starts driving the stepper the driver's led starts flickering instead of staying constant. And now the motor doesn't turn (see my last edit for more details) – Mohamed Technology Jan 29 at 19:02
  • If you're talking about one of those little square 9V batteries, forget it. Those don't have enough juice to power an Arduino for long, never mind a motor. They have very low total energy storage (measured in mAh, 500 mAh is common) and very low peak current. (A little googling said "Estimate 350 mA peak for 10 minutes".) If your motor needs 800 mA, you have less than half the current you need to even run the thing. It might work for a few seconds when the battery is completely fresh, but it might not. – Duncan C Jan 29 at 20:23
  • Do you have any recommendation for any Readily Available power source for that Motor? – Mohamed Technology Jan 29 at 21:04
  • Nope. Look at the current requirements, and find a switching supply that provides at least 20% more current (so you have some "wiggle room".) – Duncan C Jan 29 at 21:19

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