stepper motor wiring, and potentiometers adjustment

Hi All,
Mostly down with the hardware part of my built, now working on electronics!

1. My steppers are Linix 42BYGHD444-01H, and I can’t find the spec sheets…

As they come with six wires, I used the Ohmmeter technique to identify the wires:

Black, Green, Yellow >> Bk-G (15 Ohms )>> Bk-Y (7.5 Ohms) >> G-Y (7.5 Ohms)
Red, Blue, White >> R-B (15 Ohms) >> R-W (7.5 Ohms) >> B-W (7.5 Ohms)
So, if my understanding is right, I ignore the Yellow and the White, and work with the Bk-G/R-B pairs ?

2. Stepper drivers potentiometer adjustment:

In the instruction Ryan says "0.7 V or lower based on your stepper spec sheets…
As I have only the specs at the back of the motors, what should I consider as an acceptable adjustment ?

Current/Phase : 1.2A
Resistance/Phase : 7.0 Ohms
Inductance/Phase : 8.2mH
Holding torque : 0.4N.m


Thanks for your help/advises!


i don’t know how to wire your steppers. The once i used have only 4 wires (backl/green & red/ blue).

But i think that your tork is not high enough!

The steppers i use have

Current/Phase : 2,0A

Holding torque : 0,59 Nm

regards Stephan

Yup, looks that way.

Depends on your drivers and this number.

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Drivers are 8825 on ramps, how do i make the maths?

There are many YouTube videos on this subject, here is one:

For the DRV8825, it’s basically 2*Vref NOTE: most folks reduce the current to keep the heat down while maintaining enough torque to do the job.

I see a potential problem if you’re power supply is 12V and your steppers are series wired. To make the math easy, if you set for 1A drive current, your steppers will require 7V coil voltage to maintain 1A. This means 2 in series will require 14V.

I agree with Erwin, you might have a problem if these motors are designed for higher voltage and lower current. This can be inferred from the resistance, that it has a larger number of windings of finer wire. The good news is the required current would be lower.

A rough way to estimate the windings on two motors of the same size is to aim for similar resistive power dissipation. So for example suppose there is another motor of about the same size at 2.3 ohms rated at 1.5A, and say you would drive at half the rating. You would use 750 mA at 1.725 V for about 1.3 W power dissipated in resistance.

Your motor is 15 ohms so to achieve 1.3 W would require a current of only 0.29 A and voltage of 4.3 V. For two motors with the same amount of copper, with the only difference being a larger number of finer windings, the magnetic field and torque will be equivalent.

OK…sorted now! Thanks!

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