I am driving a few stepper motors using a A4988 chip, and my motors are unipolar. The driver I chose is bipolar, so, obviously, I've decided to drive it as a bipolar stepper by leaving out two of the leads. (Haven't decided which to do; one way will increase torque and another will increase speed.) Do I need decrease the current that the chopper driver can supply to prevent damaging my motor?

EDIT: I'm concerned that the current from energising double the amount of coil at one time that if I drive it at 1.5 A, then it will overheat because of resistance. Someone has said that since there is more resistance then you can technically run it on a lower current because of ohms law with the same voltage... blah blah blah. But since I have a chopper drive... please inform me on how much current is safe. Could I add an external heatsink somewhere to be safe?


1 Answer 1


The chopper will decrease the intensity for you, that's its all purpose.

The rate at what a magnetic field in a coil can change is V/L (voltage / inductance), to drive a stepper quickly you need to change the coils polarity fast and therefore you want to use high voltages (12Volts for a 3Volts motor for example).

High voltages generate high currents (V = R * I), and the coils of your stepper should be able to dissipate then this much power:

P = V * I

That gives you a big P and dissipating power is bad because we want to generate torque and not heat, that means we need to lower I and the only way to lower I is by PWM the voltage.

The chopper does exactly that, chops (PWM) the 12 volts to prevent the coils from burning at the same time it drivers the motors with 12 volts (that gives you a nice polarity change rate in your coils)

  • \$\begingroup\$ Thanks for the reply! So I would drive it as a normal bipolar stepper? The current is 1.5 A for unipolar, I think I've read that you need to decrease the current by 25% or something but I can't remember. EDIT: After looking online at diagrams it looks like unipolar has just an extra wire in the middle of each coil that you can use to not have to reverse polarity, so it would just be an extra wire sticking out in the middle that is useless. Am I correct? A concern I have is that, since I am going to be using 2x the coil as a unipolar would that I wouldn't have sufficient cooling. \$\endgroup\$ Commented Oct 28, 2013 at 1:24
  • \$\begingroup\$ My guess is that if your coils can dissipate 1.5A, putting 2 coils in series will still dissipate 1.5A, (if they were in parallel they could handle 3.0A) I would just ignore the extra wire and go with 1.5A. \$\endgroup\$
    – aguaviva
    Commented Oct 28, 2013 at 9:43
  • \$\begingroup\$ I see your point but since you double the amount of a coil energised at one time... \$\endgroup\$ Commented Oct 28, 2013 at 20:45
  • \$\begingroup\$ Not sure what you mean. Ultimately you can always start running your motors with a low intensity and then keep raising it until the motors start to be a bit warm :) \$\endgroup\$
    – aguaviva
    Commented Oct 29, 2013 at 10:09
  • \$\begingroup\$ That's what I'm going to do. What I mean: Coil [{Lead 1}--------{Lead 2}--------{Lead 3}]. That's my motor. As unipolar, I'd energize between 1 and 2 OR 2 and 3. As Bipolar I would energize between 1 and 3 always, but just switch the polarity. Longer coil = more resistance = more heat. \$\endgroup\$ Commented Nov 1, 2013 at 20:27

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