Re: [CAD_CAM_EDM_DRO] Re: Stepper motor voltage VS amps.

On Tue, 20 May 2003 02:43:44 -0000, you wrote:

>
>>
>> >
>> >and that even though it may be connected to 40 volts, the actual
>> >voltage reaching the motor will be 2 volts ?
>>
>> no.. actual current averaged over time is the rated motor current.
>>
>> You boost the voltage so that the rise time of the current through
>the
>> motor winding is as fast as you can make it.. More torque that way
>on
>> shorter pulses...
>>
>> Harvey
>
>I got the desire for a FAST rise time, but the current is limited.
>therefore the voltage seems like it should be much less than the
>power supply. V=I*R would seem to use a fixed resistance in the
>motor, and a limit on the current?

as it averages, it is.

for the steady state, neglecting the rise time of the waveform, the
current is indeed based on the motor resistance, but it's an average
based on full on (supply) full off (ground) pulses.

>
>average voltage and average current would be back at nameplate ?

Yes.

>
>Maybe I'm missing something about the inductance?

You can't change the current through an inductance instantaneously.
So you apply a voltage, and the current "ramps" up to the maximum
value. During that "ramp up" time, the current through the windings
is lower than the steady state current. It tends to limit the
available torque, especially at high pulse rates (where the ramp up
time becomes a significant fraction of the step time).

The higher the voltage you can through across the motor, the faster
the current will build up. This is where the performance increase
comes with a higher supply voltage.

Because the chopper action limits the average dissipation to the
"proper" motor current, you see the proper motor current. The actual
voltage is (by a gedanken experiment) higher than the motor's rated
voltage.

The current and power dissipation are what's important.

Harvey

>
>Dave
>