Re: simple stepper power question.

Mariss -

Thanks - I'll give that a try.

Regards,
Tony

--- In CAD_CAM_EDM_DRO@y..., "mariss92705" <mariss92705@y...> wrote:
> Tony,
>
> Yes, and I bet the L298 got even hotter! This method depends on a
> very efficient drive to work, where the drive's contribution can be
> neglected. Otherwise a refinement is necessary to eliminate the
> drive's contribution to the total power change. Here's how to do
that.
>
> Use a pair of ferrite-core inductors in the place of the motor. I
use
> two 600uH, 7A rated chokes. You can find these under the heading
> of "hash chokes" in the Digikey catalog.
>
> These chokes have very small losses (<1W) and can be used to
simulate
> a step motor on the test drive. This means all losses will be in
the
> drive.
>
> Run your drive with these chokes and graph the power supply Watts
> into the drive from 0 amps/phase to maximum amps/phase.
>
> Subtract this wattage from your no-load (zero torque) to full load
> (stall torque) power supply wattage readings. The result will be an
> accurate reading of the power absorbed by the motor.
>
> Mariss
>
> --- In CAD_CAM_EDM_DRO@y..., "tonyjeffree" <tony@j...> wrote:
> > Mariss -
> >
> > Tried your procedure yesterday evening on a small motor of
unknown
> > power (other than it was marked 1.1A/phase 4V, 6-wire motor) and
a
> > 25V supply, driven via L297/298.
> >
> > Measurements using 1/2 and full coil at rated amps yielded
> reasonably
> > sensible numbers - the motor looks to be about 60 oz-in. Just to
> see
> > what would happen, I upped the L297 current setting to 2 amps and
> re-
> > measured, still on full coil. (Yes, I know - the motor *did* get
> > warm <g>). The measured supply current under load actually
> *reduced*
> > relative to the supply no-load current under these conditions.
> >
> > Any theories?
> >
> > Regards,
> > Tony
> >
> > --- In CAD_CAM_EDM_DRO@y..., "mariss92705" <mariss92705@y...>
wrote:
> > > Dave,
> > >
> > > Yes there is. All you need is a multimeter and a calculator. Do
> the
> > > following:
> > >
> > > (1) Measure your DC voltage accurately
> > > (2) Set the meter to DC Amps
> > > (3) Put it in series with the power supply to the drive
> > > (4) Run the motor at a known speed (greater than 1 rev / sec)
> > > (5) Measure the power supply current at no load on the motor
> > > (6) Now the tricky part. Load the motor down and note the
current
> > at
> > > the moment the motor stalls. This may have to be repeated
several
> > > times to get an accurate reading.
> > > (7) Take the difference between the no-load current and the
stall
> > > current. Multiply the result by your power supply voltage. The
> > > product will be the motor output power in Watts mechanical.
> > > (8) The torque at that speed will be: oz-in = Watts * 4506 /
full
> > > steps per second
> > >
> > > Mariss
> > >
> > >
> > >
> > >
> > > --- In CAD_CAM_EDM_DRO@y..., "dave_ace_me" <davemucha@j...>
wrote:
> > > > is there a formula to determine power of a stepper ?
> > > >
> > > > I have a Slo-Syn NEMA 34, 1.7 volt, 4.5 amp
> > > > single stack.
> > > >
> > > >
> > > > Also, has anyone put together a simple test
> > > > for stepper power ? Seems like a weight,
> > > > and a long crank arm with holes drilled
> > > > each inch would allow one to keep moving
> > > > the wieght farther out until the motor
> > > > started missing steps.
> > > >
> > > > Dave