Re: [CAD_CAM_EDM_DRO] Declining motor torque with lower voltage.

cnc_4_me wrote:

>I found this statment googling around today.
>
>"It is common motor knowledge that motor torque declines by the
>square of the change in supply voltage."
>
>First...Is this true.
>
>

I'd say that output power declines relative to the square of the voltage
change, but not torque. The motor will run proportionally slower, and
there will be proportionally less current (torque). Since power =
torque * speed, you get a square law decrease in power.

>And second if true would this be the proper way to appy this.
>
>1) You have a treadmill motor rated 130VDC, 2.5HP 6700RPM, and
> 373 oz-in continuous.
>
>2) The motor is run from a 80VDC power supply.
>
>Well, the change in voltage is 130VDC - 80VDC = 50VDC.
>The square root of 50VDC is aprox 7.
>So the motor torque declines by 373 0z-in / 7 = 53 oz-in
>
>???? Is this right ????
>
>Wally
>
>

Probably not :)

It should be a relative change, so you would take the ratio of actual
vs. rated voltage:
80 / 130 = 0.615
Then square that to get 0.378

This number then gets multiplied by the original torque specification:
0.378 * 373 = 141.25 oz-in

If you chose to run a 1000V motor at 950V, you wouldn't expect the 50V
difference to divide the output power by 7 (would you? :) ).

Note that this only applies to a DC motor that's attached to a constant
voltage, not one that's being driven by a chopper drive. Since you'll
usually be running the motor at much lower speeds than it's capable of,
the voltage available to create torque will be higher since the voltage
"used" to overcome back EMF will be low. Remember that the total
voltage across a motor is the sum of back EMF (proportional to speed),
and IR drop (proportional to torque). The IR drop is usually pretty
low, since it's proportional to the winding resistance, which is
generally a few ohms or less. (note that if you run through the power
calculations, this motor looks like it has a 9 ohm resistance, which is
a bit high (the calculation is 2.5HP * 746W/HP = 1865W. R=P/V^2 = 1865W
/ (130*130) = 9.06 ohms))

Let me know if I've succeeded in confusing you further :)
- Steve