Hi Ian

I think we're on the same page more or less with this. I understand that the driver does not induce a square wave of current. As I understand it, that's exactly why we use over voltage in order to try to get the leading edge of the current waveform as steep as possible.

I wasn't my intention to suggest that the

*output mechanical power* of the motor is VA, that's why I said "your motor will have the same

* available * power of 60W", meaning from the PS - which

*is* VA. Clearly some of this is not converted into mechanical power otherwise we'd have a 100% efficient motor and we'd all be rich

I agree that the output power of the motor effectively doubles as the PS voltage doubles - Mariss told me that too when he said this (I'm sure he won't mind me including it here.)

"

*Look at it this way: Doubling the power supply voltage doubles the mechanical power output of a step motor. The efficiency of the motor doesn't change, so twice as much electrical power has to be supplied to the drive. Electrical power is Volts times Amps. Because you doubled the voltage, you doubled the electrical power at the same amperage.*

Conclusion: Power supply current is independent of supply voltage for a given motor. It is the same value regardless of power supply voltage.

Mariss"

What I still don't really understand though is if Jim does indeed get more oomph by wiring in parallel as opposed to single coil, where is that extra oomph coming from? Like I said, there is 60W of electrical power available to drive the motor in both cases. As Mariss says - the efficiency of the motor doesn't change at the same amperage, and as you've said - not all of that will be converted into mechanical power, - but whatever proportion it is, is the same.

The only thing I can think of at the moment is prbably absolute rubbish but I'll throw it in anyway. We know that the current is induced in pulses. We know that the quicker we can get the current to rise to its 2.5A the better. Inductance of parallel is less than single coil. This presumably allows the current's leading edge to be steeper. Therefore the 2.5A is "present" in the motor for longer. Therefore the power is available for longer on each pulse than it is for single coil. - just a thought.

Ian

EDIT: As you correctly say - we're all learning. A thought just crossed my mind. A chopper will apparantly only draw about 2/3 of the rated current of the motor. Is this (I wonder) because as we know it's applying that current to the motor in pulses, it's only actually applying it for 2/3 of the time?