Help wiring M60STH88-3008DF Steppers Motors to 752 Drivers

[Jonathan]

You should set the drivers to the setting closest to (but not greater than) 2.1 amps. If you had wired them in bipolar parallel (presumably you have your reasons for not doing this) then you would set the driver to 4.2A.

The reason Ohm's law does not immediately appear to apply when using 43v is the current is pulsed, not contsant. The drivers use PWM (Pulse Width Modulation) or chopping techniques to limit the current to the motor. If you did apply 43v to the motors constantly it would quickly destroy it as that is 62 times the rated power!

You are using a higher voltage since the inductance of the motor limits the rate of change of current through the motor, and increasing the voltage counters this. Using a low voltage will limit the maximum current that can get to the motor at high step rates (since there is only a finite time for the current to rise before it is cut off at the next step), thus limiting the torque.

For an ideal inductor v(t)=Ld(i)t/dt
Rearrange that for d(i)t/dt:
d(i)t/dt=v(t)/L

We want to maximise the rate of change of current, so clearly the formula above shows that increasing the voltage or decreasing the inductance achieves this. This is why you should buy motors with the lowest inductance you can find. Putting the motors in bipolar series increases the inductance, as inductors add in series ... so in parallel the value is a quarter of the series inductance. This may explain why the torque is slightly reduced at high speed with bipolar series, though I'm only guessing.

(There's a bit more to it than that in reality, but that's the general idea.)