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

cnc_4_me wrote:

>--- In CAD_CAM_EDM_DRO@yahoogroups.com, Jon Elson <elson@p...> wrote:
>
>
>>cnc_4_me wrote:
>>
>>
>>>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.
>>>
>>>
>>>
>>>
>>I believe the treadmill motor in question is a PM motor. The
>>
>>
>torque is
>
>
>>proportional
>>to current, the speed is proportional to voltage. The speed would
>>
>>
>fall
>
>
>>to 61.5% or
>>4123 RPM. The torque available would still be 373 Oz-In.
>>
>>Jon
>>
>>
>
>I looked this over again and have a new conclusion…As you say torque
>is proportional to current and I agree with this. But I came to a
>different conclusion on the end torque with lower voltage.
>
>Since you are running the motor aprox 38% below rated voltage you
>cannot push the amps thru the motor to get full torque.
>
>I left out the current ratting before but here it is 18.5 amp.
>Armature resistance would be 130vdc / 18.5A = 7 ohm.
>

No, completely wrong. The effective resistance is 7 Ohms, but most of that
is not real resistance, but actually what is called "back EMF". This is
a voltage
generated by the moving windings in the magnetic field that acts against the
applied voltage. If there was no back EMF, the motor would accelerate until
it exploded. The true armature resistance is much more likely to be
around one
Ohm, possibly lower. This is what you would see in the case where the
motor is
not allowed to turn.

You can be sure this is what we are talking about by doing a power
calculation. If
the real resistance was 7 Ohms, the power loss would be I^2 * R = 18.5 *
18.5 * 7 =
2396 W or 3.2 HP. Since this equals the total power input to the motor
(130 V * 18.5 A = 2405W) there is no power left to come out the motor shaft.
Obviously a motor that absorbs 3.2 electrical HP and produces zero
mechanical
output is worthless.

Jon