Re: unipolar or bipolar?

Guy,

Thanks. You can think of half-stepping as a 2 microstep / step drive.
Everything that applies to higher resolution microstepping holds here
as well. The idea is to keep a constant vector sum of the phase
currents, i.e. sine and cosine (sin^2 + cos^2 =1).

Here's the math. You want your two "microsteps" to lie evenly between
the full-step locations. If a full step is 90 degrees, then these
locations are 22.5 and 67.5 degrees respectively. Solve for the sine
and cosine pairs for these angles and you get .383, .924 for 22.5 deg
and .924, .383 for 67.5 deg.

Next, normalize the amplitude by multiplying both pairs by 1.082 (
1 / .924 = 1.082). You now have .414, 1 and 1, .414 which are your
phase current ratios. As an example, say you have a 1A motor. Your
phase currents for all 8 half-steps will be:

+.414, +1.00
+1.00, +.414
+1.00, -.414
+.414, -1.00
-.414, -1.00
-1.00, -.414
-1.00, +.414
-.414, +1.00

These values give the smoothest possible motor motion because there
is no torque ripple (theoretically). This is better than the classic
half-step sequence of "weak step, strong step".

Mariss

--- In CAD_CAM_EDM_DRO@y..., "Guy Sirois" <guys@w...> wrote:
> Hi Mariss,
>
> thanks for the wonderful tutorial on steppers.
>
> I was recently designing drivers for my motors, using SGS L297 &
298 and
> incorporated the "half-step with shaping" method you mention, where
the peak
> current rises to 1.41 x the nominal. Now I think I will limit my
maximum
> current to the nominal value...
>
> However , I think the advantage of using this method lies in its
ability to
> be smoother than the standard half-step method, because the output
currents
> are more sinusoidal. Could you confirm this?
>
> I would then just run lower currents but still keep the shaping on
the
> pulses.
>
> Thanks
> Guy Sirois
>
> -----Original Message-----
> From: mariss92705 [mailto:mariss92705@y...]
> Sent: Thursday, February 21, 2002 11:41 PM
> To: CAD_CAM_EDM_DRO@y...
> Subject: [CAD_CAM_EDM_DRO] Re: unipolar or bipolar?
>
> Snip---
> There is a common misperception some users make regarding current
set
> with microstep drives. It has to do with the fact microstepping
> drives the motor with sine and cosine weighed currents. When the
> current in one winding reaches its maximum the other winding current
> is at zero. From this some surmise that it should be possible to
> increase the set current to 141% of rated since the motor power
> dissipation would then be identical to a full step drive. That is
> correct as far as power dissipation goes but it neglects a very
> important point most are not aware of.
>
> The magnetic flux path is not shared by the windings. This means the
> current in one winding does not contribute by adding or subtracting
> to the to the magnetic flux of the other winding. Though the
> dissipation may equal a full step drive by running the current at
> 141% of rated, the iron magnetic saturation is dependent on the
> current of one winding entirely.
>
> The result is the iron in the motor saturates at the higher current
> and degrades the motor's linearity. Consequently the microstep
> placement is not optimal anymore and the motor will exhibit greater
> low speed resonance than it otherwise would have.
> Snip---