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

cnc_4_me wrote:

>I think you hit the nail on the head...That formula probably deals
>with AC motors...
>
>I googled around some more looking for answers on the torque verses
>voltage question for DC motors. I found about a half a dozen torque
>verses voltage curves on the bodine-electric site. Every curve was
>the same...
>
>If you half the voltage, you half the torque...Direct relationship...
>
>
>Wally
>
>

True, but that's still for a constant drive voltage. It doesn't
necessarily apply to a chopper-driven motor.

I'll give an example of the calculations. My motor specs (from the
nameplate) are:
Max voltage: 150V
Max speed: 2500RPM
Voltage constant: 40V / kRPM
Torque constant: ~3 in-lb/A
Continuous current: 9.2A (27 in-lb continuous, but close enough)
Maximum current: 37A (~ 111 in-lb)

Winding resistance (measured): about 1 ohm.

So, to pump max current (37A) at max speed (100V back EMF), I need a
137V driver.
I bought Geckos, so I'm limited to 80V - let's see what happens:
If I want amximum peak allowable torque, then I need to apply 37V plus
the back EMF for whatever speed the motor is at. This is 37V for my
motor (37 A * 1 ohm). So, I can calculate the maximum speed at which I
can still get maximum torque: 80V - 37V = 43V back EMF. 43 / 40 = 1.075
kRPM at full peak maximum intermittent
don't-do-this-for-long-or-the-motor-will-smoke torque.

If I do the same calculation with the (much more reasonable) continuous
current rating, then I get 80V - 9.2V = 70.8V. 70.8 / 40 = 1770 RPM.
Using an 80V supply, I can run all day at 1770 RPM, with full 27 in-lb
torque.
At no load, I can get the motor up to 2kRPM (80V / 40V/kRPM = 2kRPM)

Not bad, for driving a 150V motor at 80V :) (you could argue that it's
actually a 109.2V motor, but it's still pretty good performance with the
Gecko)

- Steve