Powerflex 525 with 380V motor

Got an application where I have a spindle motor rated at 7.5HP, 380VAC, 11 amps (name plate ratings) but we have 600VAC supply in our panel.

So going with an Allen Bradley Powerflex 525 drive that is powered with 600VAC, I can change a parameter setting so that my output voltage is 380V.

So now the main question is what size Powerflex?

If I go with a 15HP drive (600VAC input) it gives 19 amps output.

I believe I have to calculate that at 19 amps (600VAC) it will give me about 12 amps (380VAC) and I should be good to run the spindle motor.

If I wanted to be cautious I could bump up to a 20HP motor but regardless I can't select anything less than the 15HP.


Does that all sound correct?

Does my speed (Hz) get affected since I need to run the spindle at 24,000 rpms?

Drives are really sized by amps, selected by volts. Torque and amps are essentially equivalent once the motor is at rated volts, so no, your calculations getting you to 12A are incorrect. Sizing the drive is simpler than that really, just size it to the full load amp rating on the motor nameplate, because that is how you will be using it. But that said, a 10HP 600V drive will be rated for probably 10A, so the next size up is 15HP anyway.

Don't know where you were going with the 20HP issue, but the above rule still holds, size the drive by the motor FLA.

As to the 24,000 RPM, totally separate issue. Assuming that your motor is a 2 pole induction motor, AND the nameplate of the motor says it is rated for 380V at 400Hz, then if you give it that, the speed will be what it should be. It has nothing to do with the incoming voltage. Think of the incoming voltage as just the raw material resource for creating the DC power that your VFD inverter section needs in order to run the motor, that's all it is.

So to that point, the only issue related to the higher incoming voltage that you need to be concerned about is that your DC bus, and thereby the height of the DC pulses in the PWM output to the motor, will be very much higher than the motor is designed for. Typical motor insulation is designed to handle 2 to 2-1/2 times the line voltage, 3-4X if it was designed for inverter use, as this spindle likely is. The reason for the higher amount is because the PWM pulses are at the DC bus level, not the AC RMS, and then can create much higher peak voltages as "standing wave" spikes, well known to damage older insulation. So even if, let's say, this one used 2000V winding insulation, the problem is that you are starting out with an artificially high line voltage, 600V RMS, so that becomes an 850VDC bus, which can result in standing wave spikes of 2400-2600V, far above the insulation even of a good quality inverter duty design. So if you are going to do this, I strongly suggest two strategies, either or. Since it's a good idea to protect a drive with a line reactor, and you get the same benefit from a transformer, eschew the reactor for a 600-380V transformer. Option two would be to add a good quality dV/dt filter between the VFD and the motor, let the filter clean up the output by slowing down the rise time of the pulses before they get to the motor. The filter would need to be sized for the motor current as well, but also make sure it too is rated for a 600V system, even though you are turning it down to 380V, the same pulse height issue exists for the filter too.

Best practice would be both. Put a transformers ahead of the drive, then put a (now 380V) dV/dt filter below it.

Thanks for the good info.

The 20HP thing was just if I wanted to upsize to give me some more breathing room with amp supply. From what your saying, that is way overkill anyways.

A 10HP drive (525-600V input) is rated for 12A (0-600V output) which sounds like it covers me for the 380V 11A motor. To get that breathing room I mentioned earlier, I would probably consider going to a 15HP to get 19A output for only a little more money.

The Powerflex 525 does give me some PWM and DC Bus parameters to play with. I'll have to investigate these a little further.

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So far this is all exploratory for me. My actual design currently has a 600V || 480V transformer and then I'm using a 7.5HP drive (380-480V input) rated for 13A (0-460V output). If I understand your reply correctly, this should be good for controlling the motor in question.

Just thought I could simplify by eliminating the need for the transformer and keeping a standard 600V drive on the shelf for a spare that I could use in several different machines.

Thanks again for the help!

The only thing I would add to jraef's sizing instructions is that you must size not only to the continuous running amps, in this case it would be the 11 amp FLA of the motor, but also the short-term overload amps desired or needed due to short-term overload torque required by the load.

For example, with the given 11 amp motor, the operating conditions of the load may require short-term torque of 160% of running torque. That translates to about 160% of FLA which would be 11 x 1.6 = 17.6amps. The drive you select would have to be rated for at least 11 amps continuous and 17.6amps short-term to fully cover the load requirements.

In the VFD world, short-term generally means for one minute or less.

And, yes, unless you are driving a fan or centrifugal pump, forget hp for sizing, for the reasons mentioned above.

Hope this helps, DickDV