Re: [CAD_CAM_EDM_DRO] Re: My stepper getting hotter

mariss92705 wrote:

> Jon,
>
> That is not true. If it were, there would be no maximum limit on the
> power supply voltage!

OK, I was trying to simplify things, and oversimplified it, I guess.
Obviously, motor insulation and power transistor breakdown voltage
are additional limits, here.

> Step motors get hot for two reasons, copper losses and iron losses.
> Copper losses are due to current passing thru a resistance, (I^2times
> R) and are independent of power supply voltage in switch-mode drive.

(Well, there is some eddy current loss in the copper, too, especially as the
wire size gets bigger.)

> Iron losses are primarily eddy current heating of the motor
> laminations. These losses go up with the square of the power supply
> voltage. Inductive reactance currents are proportional to the supply
> voltage and eddy current losses are proportional to the square of
> this AC current.

I suppose it depends on the chopping circuit's characteristics. I have
seen some drivers that maintained the current (at idle) with a very
small delta-I, and the losses in the motor were mostly copper. Some
other drivers have a lot of hysteresis and let the current fall quite a bit
before turning it back on, and would cause more iron heating.

> Iron losses range from inconsequential at low voltages to being the
> majority cause of heat at high supply voltages.

Given a motor running at a speed which can be reached with a lower
DC voltage supply, but where the chopping action was not being
exercised much, would there be much difference between, say 48 and
64 V, with respect to heating? (clearly, there'd be a significant difference
in torque!)

What I was trying to get across to the original poster, was that I suspected
that his motor heating problem (if it was a problem at all) might not have
a lot to do with the 64 V power supply, but that he might be running more
current than he should be.

Jon