Stepper motor ampage question

[Fivetide]

m_c I ment closed loop stepper sorry

- Closed-loop vector control technology ensures the high torque output, while ensuring that the motor does not lose step.
- With overvoltage, overcurrent, tracking error or over-error, and other protection features.
- Closed-loop driver can significantly improve the high-speed performance of the motor, increasing the effective torque of the motor by more than 30%; Effectively reduce the motor heating and motor vibration, the output current of the driver becomes larger or smaller as the load and speed increase or decrease; It can improve the processing speed and accuracy of the machine and reduce the energy consumption of the machine.

[m_c]

I suspected you really meant closed loop. Yes, they will offer more performance for a given size, and will recover more gracefully depending on settings (I forgot to mention, one big issue with using a stepper for a spindle, is if you do stall it, it will probably remain stalled until you stop/restart it).

If you want to run some comparisons, find the relevant speed/torque curves, then run the figures through the calc here - http://wentec.com/unipower/calculators/power_torque.asp

One thing I should of mentioned in my last post about DC/Induction motors, is the rated power is only at full rated speed. So if it's 200W at 4000RPM, it will only deliver 100W at 2000RPM.
With an AC motor and VFD, you do have the possibility of running the motor above it's rated speed, however you enter a power constant stage. I.e. if the motor produces 1Nm at it's rated speed of 1500RPM (works out to 160W), if you were to double it's speed to 3000RPM, you would only get 0.5Nm (which is still 160W)