# Understanding stepper motor current draw

I have the following motor:

Manufacturer Part Number: 23HE30-2804S
Number Of Phase: 2
Step Angle: 1.8 deg
Holding Torque: 1.85Nm (261.98oz.in)
Rated Current/phase: 2.8 A
Phase Resistance: 1.13 ohms± 10%
Inductance: 5.4 mH ± 20%(1KHz)


Driven by a 24V, 6A PSU

(Side note: Which doesn't make much sense to me 24V/1.13ohms = 21.23A?)

When a stepper is engaged in a holding state under no load - Both phases are passing current, which would mean a load of 5.6A being drawn?

My PSU however only shows 1A being drawn, why?

In this question "Understanding Stepper Power" a user states that for a motor with a phase current of 5.8A

If you are using say a 24 V power supply, then the average current flow will be 1.6 A ...the peak current flow (which comes from the output capacitor) will be 11 A peak

How did they arrive at 1.6 and 11A?

I have seen people calculate general load by multiplying rated phase current by 1.4

The rated current is the maximum current for one winding while the other winding carries no current. If both windings are active the sum of the squares of the two currents should be equal or less to the square of the rated current (quadrature drive).

Thus with full steps a 5.8A setting will send ~4A to both windings.

Which again conflicts with the "Understanding Stepper Power"

How can I confidently calculate and verify my stepper motor draw current?

• what driver do you have between the 24 V PSU and the stepper? Does it control the current by any chance? Commented Jun 20, 2023 at 13:28
• Yes, I have a DM556T_V4 driver set to a current limit of 3.5A Commented Jun 20, 2023 at 14:24
• If it's a buck control of current, the 3.5 A to the motor will draw much less than that from the PSU. Commented Jun 20, 2023 at 15:02