CAD CAM EDM DRO - Yahoo Group Archive

Re: 8-wire stepper: series is 1/2 the amps.Strange but True

Posted by jeffalanp
on 2003-06-02 11:26:48 UTC
Hi Mariss,
Thanks for the info. It appears as though the inductive reactance
puts a damper on the higher speed stepping current. My calculation
ended up being about 11% too fast (but then I didn't take any
resistance into account at all)
I looked at your chart("MS234A24V.wmf"). I guess the X axis
heading "RPM" would be Hz, and 552Hz would be sine cycles, or 4
steps, thus the drop off step rate would be about 2208 PPS?

Thanks,
Jeff


--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Mariss Freimanis"
<mariss92705@y...> wrote:
> First off, it's pi/4 or .785 (instead of 71%) of the area under the
> curve for a sine function compared to a square wave.
>
> To compensate for that a good drive changes its reference from sine
> to square wave (full-step sequence) once the speed is high enough
(4
> revs/sec) and the benefits of microstepping has disappeared.
>
> The second part: Torque begins to "drop off" past the speed where
> inductive reactance, not drive set current current begins to limit
> phase current.
>
> Inductive reactance (Xl) = 2*pi*F*L
> Inductive current (torque) = V / Xl
>
> Solving for F: F = V / 2pi*I*L
>
> Because the step rate is 4 times F (coil frequency),
>
> Full steps per sec = 4*V/2pi*I*L
>
> This is agreeably close with dyno data taken from "square" step
> motors. See the file "MS234A24V.wmf" in the files section.
>
> Mariss
>
>
>
>
>
>
> --- In CAD_CAM_EDM_DRO@yahoogroups.com, "jeffalanp" <xylotex@h...>
> wrote:
> > Hi Dave,
> >
> > > so, the question how does one determine where the torque starts
> to
> > > drop off when on is not using a single coil (on an 8 wire) but
is
> > > using less than nameplate on a series wired motor ?
> > >
> > > Dave
> >
> > I think torque will start to drop off when there isn't enough
> time
> > to charge the coil to required amperage before the next step
pulse
> > comes in.
> >
> > A microstepping drive, when set to full step mode, will generally
> > power each coil at about 71% of rated power. For full stepping
you
> > would get the following (4 steps shown):
> > Ph.A Ph.B
> > +71% +71%
> > +71% -71%
> > -71% -71%
> > -71% +71%
> > and so on. As you can see, at each step, one of the coils is
> > reversing it's polarity. You can figure out how long it will
take
> > the coil to go from + to - (or - to +).
> >
> > I think the formula is going to be something like:
> > 1 / (((Coil Amp *.71) / (Supply Voltage / Inductance in H))) * 2)
> >
> > So a 4 amp motor with 4mH inductance driven with 40V would be:
> > 1/(((4 amp motor *.71) / (40V/.004H)) * 2)
> > 1/((2.84/10000)*2)
> > 1/(.000284 * 2)
> > 1/.000568
> > 1760 steps per second
> > before it starts loosing torque.
> >
> > I'm sure one of the other members out there can (and hope they
> will)
> > scrutinize this and let us know if this turns out to be to case,
> > close, or way off.
> >
> > Jeff
> > http://www.xylotex.com

Discussion Thread

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