Re: Running Unipolar steppers with a Bi-polar driver?

Mariss,
Thanks, that helps explain alot. So, Once I start charging up the
coil (lets assume parallel bipolar motor, 2mH of inductance and
2A/phase with a 40V PS, like you used earlier), it will take 100uS to
charge up to 2A. Now I decide I want to reverse polarity, not only
will it take 100uS to get the current back to 0 but another 100uS
before it is energized in the other direction for a total of 200uS
(true???) I'm guessing this is where the unipolar (with same L) has
a great advantage in that it just uses a different coil for going
the "other direction", while the originally charged coil will just
decay through free-wheeling diodes in the FET, thus saving a whole
100uS of time. Yes? Another question: how much 'drag' (for lack af a
better word), will the decaying coil put on the rotor, keeping it
from turning the way the newly energized coils is pulling, while it
decays? Anything significant?

Thanks again,

Jeff

--- In CAD_CAM_EDM_DRO@y..., "mariss92705" <mariss92705@y...> wrote:
> Jeff,
>
> Real close. The time constant for R and L is: T = L/R. A current
> source as you know has infinite source resistance, thus with R
> becoming infinity, L/R becomes zero. That is why I said the time
> constant does not matter.
>
> What governs then is simply V/L with respect to how rapidly current
> can be injected or removed in a coil.
>
> The same applies to an RC time constant. Assume you made R equal to
> zero. The RC time constant would then be zero as well. Now, attach
> the capacitor to a current-limited power supply. The cap will
charge
> at a rate of I/uF volts per second, not instantly as the time
> constant would suggest. This is a mirror image corrolary to the V/L
> instance.
>
> Interesting subject.
>
> Mariss