0
\$\begingroup\$

I am trying to run this stepper motor (XY42STH34-0354A), but I am encountering two odd issues with its performance: slowing of the motor in 1/32 microstep mode, and a "soft cap" of the RPM where the speed of the motor barely changes despite increasing the RPM parameter.

Here are the motor specs:

Rated voltage: 12V

Current/phase: 0.35A

Resistance/phase: 34ohms

Inductance/phase: 33mH

Holding Torque: 20N-cm

I'm currently using a DRV8825 stepper motor driver with the motor using a 30V supply and am controlling it with an Arduino Mega, using this stepper library I found online. I believe that all my wiring is correct, and I've properly set the current limit on the DRV8825 to the rated current limit of 0.35A following the instructions on the Polulu website. The motor runs smoothly within a certain RPM range at all microstep settings.

When I use the 1/32 microstep mode of the DRV8825, it is noticably slower (~20%) than that of the other modes (1/16, 1/8, 1/4, 1/2, 1) for the same RPM setting, but other than that, works great (smooth, no rattling or other odd behavior). I think the motor may speed up slightly as the microstepping setting gets more coarse (towards fullstepping), but it's small enough that I really can't tell.

The second thing I noticed was that despite increasing the RPM setting, my motor hits a maximum of about 4-5rps. There is a range of RPM settings (300-900RPM) in which changing the RPM value won't result in any noticable difference in the speed of the motor, which is what I refer to when I say "soft cap". Exceeding that limit causes the motor to jitter.

Could anyone help me understand why the motors are behaving this way? Looking at some other NEMA17 stepper motors online, I see that they have a lot lower voltage ratings (2-3V) but much higher current/phase (1-2A). Could it be that the driver I chose isn't correct for my stepper's specs, or that the lower-voltage higher current types run faster or better? Thank you.

\$\endgroup\$
7
  • \$\begingroup\$ Lower voltage motors will also be lower inductance. Use a high voltage input to the chopper, and you can get good performance from them even at high step rates. \$\endgroup\$
    – Chris Stratton
    Commented Mar 6, 2018 at 19:53
  • \$\begingroup\$ See also \$\endgroup\$
    – Trevor_G
    Commented Mar 6, 2018 at 20:44
  • \$\begingroup\$ smaller microsteps reduce torque and thus limit acceleration and load limits max RPM/V, no microsteps maximizes torque , acceleration and velocity. I used Gcode on CNC shield with GRBL Panel windows software with serial to Uno for maximum performance on large servo sweeps and microstep for smooth smaller motions. 1m/s velocity on NEMA17. As I recall step rate max was 100Khz \$\endgroup\$
    – D.A.S.
    Commented Mar 6, 2018 at 20:49
  • \$\begingroup\$ The key to stepper servo performance to define max acceleration, velocity and mass to move with low noise and stiff holding torque but not overheating driver. Start with good specs. What are yours? \$\endgroup\$
    – D.A.S.
    Commented Mar 6, 2018 at 20:56
  • \$\begingroup\$ This was almost optimized and just doing a 4 corner seek. youtube.com/watch?v=I7DCYEM3dW8 \$\endgroup\$
    – D.A.S.
    Commented Mar 6, 2018 at 21:04

1 Answer 1

Reset to default
0
\$\begingroup\$

With high step count stepper motors the inductance of the coils plays a big part in limiting the speed you can drive it at a given terminal voltage.

schematic

simulate this circuit – Schematic created using CircuitLab enter image description here

Notice with the circuit above that the coil current has dropped to under a third of the nominal holding current at 12V resulting in less torque. Further there is a significant phase shift in the current which, in a closed loop system, will result in a further difference in applied torque if you do not correct for it in the switching angle.

This reduction of torque translates into a lesser top speed for any given shaft load.

If your stepper driver is current regulated, using a larger supply voltage will reduce this effect and allow the motor to run faster for a given load.

Your other alternative is to source a motor with significantly less inductance.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks - so because of the high inductance, a high step frequency causes the current in the coils to get attenuated too much and results in poor performance at higher speeds. A question about the drivers: I read that they apply a much higher voltage and then "chop" the supply when the current in the coil exceeds the set limit of the driver. You show in your diagram that the coil current has dropped to less than 1/3 holding current, but, if the driver is only chopping the current when the set limit is exceeded, shouldn't the coil current be the same as the holding current limit? \$\endgroup\$
    – sanlefa
    Commented Mar 6, 2018 at 22:27
  • \$\begingroup\$ @sanlefa the driver will only start regulating the current when it reaches the level you dictate/set to be the max drive current. At high step frequencies, that level is never reached. Using a higher voltage causes the current to grow faster. \$\endgroup\$
    – Trevor_G
    Commented Mar 7, 2018 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.