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I'm using two 28BYG-48 stepper motors rated for 12V. These are unipolar motors with 5 wires. 1 Common wire and 4 wires to control the 2 phases. The motor that is getting hot has around 76-77ohm the common pin and the other pins, while the motor that is cool has around 78-79 ohms.

Both motor are wired the same way through a ULN2003 driver. For my current tests, I was running the motor very slowly. Almost 1 step per second, I'd expect both motors to be cool as the current rarely switches direction.

Is it possible that one motor is getting more power than the other? That would explain why one is getting hot.

When I switch motor positions, the motor that was cool gets hot and the motor that was hot stays cool. My guess is it's a problem with the ULN2003 chips.

One other thing, the ULN2003 connected to the hot motor is quite cool compared to the ULN2003 connected to the cool motor. When there is no activity (no motor switching direction), one of the two motors gets hot progressively.

  • How can I fix this?
  • What could be actually causing one motor to get excessively hot?

BreadBoard Schematic

12v Motor

Note:

After taking pictures, I realized I mixed up the namings, those are ULN2003 chips. I have L293D chips but they're for other motors.

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  • \$\begingroup\$ Measure their power consumption. \$\endgroup\$
    – PlasmaHH
    Commented Sep 26, 2017 at 12:56
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    \$\begingroup\$ Since the heat stays with the driver, either your driver wiring is messed up or there may be an issue with the driver itself. \$\endgroup\$
    – Trevor_G
    Commented Sep 26, 2017 at 12:57
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    \$\begingroup\$ @LoïcFaure-Lacroix Visual inspection for shorts and bad connections is your first thing to try. Use your ohm-meter to see if things are connected the way you expect. I don't know what test equipment you have. A scope would help, but failing that you need to measure applied voltages and currents to see what is different between the two drivers. Note: Do a bunch of testing with the motors disconnected first in case you fry the drivers. You should know how it SHOULD work! Study the data sheet. \$\endgroup\$
    – Trevor_G
    Commented Sep 26, 2017 at 13:19
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    \$\begingroup\$ Your apparent belief that heat would only be generated when current "swtiiches direction" is fundamentally mistaken. If your control software deactivates (or PWM reduces) the driver once the motor is in position, that would indeed reduce heating - but it would also mean no (or at least lesser) holding torque. \$\endgroup\$
    – Chris Stratton
    Commented Sep 26, 2017 at 20:37
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    \$\begingroup\$ @Trevor I checked for shorts already, maybe it was the connection on the breadboard that wasn't right. I ended up converting the motor into bi-polar motors and using the l293d I had instead of the uln2003 chips. As a result, I can run those motors at 15v without them getting hot. Those l293d aren't getting hot either so I believe I might have damaged one of the uln2003 in a way the motor worked fine but thought my ohm-meter couldn't detect any short. \$\endgroup\$
    – Loïc Faure-Lacroix
    Commented Oct 2, 2017 at 18:46

1 Answer 1

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The L293D is only a driver that switches the supplied voltage to the load (motor). It has no current control, like chopper,... So basically when the motor is at standstill the current is limited only by its winding resistance. If it is spinning then the current is lower, since the back EMF voltage builds up: $$V_{supp} = I\cdot R+V_{BEMF} $$ $$I=\dfrac{V_{supp}-V_{BEMF}}{R}$$ $$V_{BEMF}=k_e\cdot N_{RPM}$$

therefore at N=0:

$$I=\dfrac{V_{supp}}{R}$$

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  • \$\begingroup\$ Moot point since the op is using the wrong type of driver. \$\endgroup\$
    – Jack Creasey
    Commented Sep 26, 2017 at 18:10

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