RE: [CAD_CAM_EDM_DRO] Re: unipolar or bipolar?

Mariss,

thanks for that enlightening explanation. I had to take some time to really
get it and draw it on paper.

That's quite different than in the SGS application notes !
Your system always has a current flowing in the windings at any time, while
with the SGS technique, each winding spends 1/4 of its time unpowered.

And your current waveform looks more like a sinusoid while theirs is more
like a triangle. And you don't exceed the nominal motor current.

I'm a bit embarassed to ask you this but, do you know of some commercially
available chip that can do this?

If not, I guess I'll just stop trying to reinvent the wheel and just buy a
G201 drive. Speaking of which, is the drive always in 1/10 step mode or is
it possible to go down to 1/4 and 1/2 step? What I mean to say is does my
CAM program has to output pulses 5 times faster if I replace a common
half-step drive with a G201? This is assuming I want to keep the same speed
on my machine.

Thanks again for everything

Guy



-----Original Message-----
From: mariss92705 [mailto:mariss92705@...]
Sent: Friday, February 22, 2002 4:44 PM
To: CAD_CAM_EDM_DRO@yahoogroups.com
Subject: [CAD_CAM_EDM_DRO] 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