# Archive for September 9th, 2011

### Stepper Dynamometer: High Speed Pullout Torque

Posted by Ed in Electronics Workbench, Machine Shop on 2011-09-09

Just to see what happened, I reversed the stepper dynamometer and drove the larger (480-ish mN·m) stepper in 1/8 step mode while recording the short-circuit current from the smaller (anonymous) stepper. Slowly cranking the step frequency upward produced this trace when the stepper stalled:

The bottom trace shows 30.4 rev/s = 191 rad/s = 1824 rev/min: a pretty good speed for a loaded stepper! The rotor began slowing just before the last sync pulse, but hadn’t lost any appreciable speed.

The current scale is 0.5 A/div (set on the Tek probe amp), which makes the winding current (500 mA/div × 10.4 mV_{pk} / 10 mV/div) = 520 mA_{pk}. The scope’s computed rms value includes the waveform after the stall, which isn’t helpful.

The small stepper has a 2.1 Ω winding resistance, so a short-circuit load dissipates (0.52 A)^{2} × 2.1 Ω = 567 mW_{pk} in each winding. The rms equivalent is half of that, so the total rms power is about half a watt, essentially all internal to the motor.

The pull-out torque depends on the peak torque load, not the rms and not the sum of the two windings, so it’s 0.6 W / (191 rad/s) = 3 mN·m, which doesn’t sound a lot for a 480 mN·m motor until you consider the screaming 6000 full step/s speed: pretty much off-scale high on most of the torque-vs-speed graphs you’ll see. Not much torque left out at that end of the curve, indeed.

In order to stall the motor at lower speeds, the load stepper must generate enough voltage into the load resistor (here, the winding resistance) to push the power/speed ratio (the torque!) above the drive motor’s ability. That implies the load stepper must always be larger than the drive stepper, which means I must conjure up a bracket for that NEMA 23 motor that’s been holding down a stack of papers on my desk…

Incidentally, the voltage required to produce that load current is 0.52 mA × 2.1 Ω = 1.1 V. The 0.58 v/(rev/s) open-circuit generator constant for the smaller motor predicts 0.58 v/rev/s × 30.4 rev/s = 17.7 V. Obviously, you can’t get from the open-circuit unloaded generator constant to the short-circuit loaded voltage… Lenz’s Law gets in the way.

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