When choosing a power supply for a stepper motor, e.g. this one below:

Amazon Link NEMA 23

This one has: Rated Current: 4.0A Operating Voltage: 24-50VDC

The Stepper Driver I will be using is:

Stepper Driver

Voltage Range:20-50VDC Current Range: 1-4.2A

So would my power supply need a output Current of 4.0A?

And a Output Voltage of 24-50VDC, how is the performance of the motor affected if I use 24VDC vs 50VDC Power Supply?

Thank you in advance!


2 Answers 2


Normally you'd need to put a stepper motor driver between the stepper motor and the power supply. Nowadays the stepper motor is based on chipsets that regulate the current and the voltage depending on dip switches (in the older days bridge circuits were used and then the power supply was more important).

To my undertstanding for hobbyist projects (and maybe not only) depending on whether you use something like

enter image description here enter image description here
A4988 pololu Stepper motor driver

in both cases the ranges of input voltage and incput current are specified and the output current is usually specified by the user.

the following is from the dma 542 datasheet.

enter image description here

The input voltage that you need to connect to the stepper driver is anything between 20 and 50 VDC (36 is the typical Voltage). It shouldn't matter what is the recommended voltage of the motor.

You'd only need to flip the right SW switches (see below) to get a peak output current (to the motor) of 4.2 Amps.

enter image description here

If you have a desktop lab supply connected to the driver, you should be able to see how much current is being drawn from the driver. I'd expect that if you set the voltage of the power supply to 24V you'd get approximately double the current being drawn from the driver compared to if you had set 48V. However, I'd expect that in both cases the stepper motor should perform exactly the same.

  • $\begingroup$ Yes, thank you for your reply again!! So the stepper motor drover that I will be using has a voltage range of 20-50VDC, and current range of 1-4.2A. So, would that mean my power supply will need to have at 48VDC and 4.0A? $\endgroup$
    – user36991
    Commented Feb 19, 2022 at 10:40
  • $\begingroup$ I want to get the full 3Nm torque from my motor. $\endgroup$
    – user36991
    Commented Feb 19, 2022 at 10:42
  • $\begingroup$ Could you update the question and add the details of your stepper driver? $\endgroup$
    – NMech
    Commented Feb 19, 2022 at 10:58
  • $\begingroup$ Yes no problem, i’ve added the details of the driver now $\endgroup$
    – user36991
    Commented Feb 19, 2022 at 11:12
  • $\begingroup$ Thank you! This is very helpful $\endgroup$
    – user36991
    Commented Feb 19, 2022 at 19:46

For static and low-speed applications, it doesn't matter. If you need full torque at speed, it's nice to have a datasheet that provides a graph like this:

torque-speed-voltage image from: nanotec.com; note semilog scale

Roughly speaking, the effect of increasing the supply voltage from 24V to 48V would roughly double the RPM which is the "corner" on that graph.

The main reason is winding inductance. Magnetization is controlled by current, and rate-of-change of current is limited by max voltage. At the high end of the speed range, the lower supply voltage cannot increase the current fast enough.

For example, let's say you have a 200step/rev motor, 4A per winding, 2mH per winding, and 24V supply.

Max dV/dt = V/L = 24/.002 = 12000 A/s.

A very crude approximation of max rate needed is a triangular current profile (in reality the profile will be sinusoidal for effective high speed motion, but let's neglect it here for simplicity). This triangular profile has the current swinging from +4A to -4A and back to +4A, at each step = 8A swing per half step = 16A/step.

Thus the roughly highest rate at which a triangular profile is possible, is (12000 A/s)/(16A/s) = 750 steps/s = 3.75rev/s = 225RPM. This a crudely reckoned limit of where torque would start to fall off.

In real-world applications, at higher speeds it's also good to have excess torque capacity beyond what's "used" by the load, to let the system accelerate/decelerate and cope with resonances. Driver details would play a role there.

  • $\begingroup$ Thank you for your help! $\endgroup$
    – user36991
    Commented Feb 19, 2022 at 19:46

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