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

Mariss,
OK, OK, I got it (I think). The inductive time constant for both
parallel and series motors will be the same (t=L/R, the time it take
to get to 62.3%), but to overcome the increased inductance and
resistance in a series wound motor (compared to parallel), I would
need to double the volatge to achiecve the same time as parallel.

A unipolar motor would have a time constant about half that of a
parallel (or series motor), but to achieve that rate I would need a
voltage 1.414 times the parallel voltage.

Since we use a constant voltage source (40V -unity- in your
example, and not 1.414 or 2 times the amount) for all motor types
(series, parallel or unipolar), time now becomes the variable that
must change because the voltage is now a constant, and the "inductive
time constant" becomes irrelevant (but not meaningless).

How does that sound, Closer?

Jeff

--- In CAD_CAM_EDM_DRO@y..., "mariss92705" <mariss92705@y...> wrote:
> Jeff,
>
> Right data, wrong conclusion; the "inductive time constant" has
> nothing to do with it.
>
> Rather what matters is V/L. V is the voltage applied across
> inductance L and is expressed in amps/second.
>
> Say you have a 2A/phase (parallel connection) motor with 2 mH of
> inductance and you are using a 40VDC supply. V/L is 20,000 amps/sec
> and it will take 100 uS to go from 0 to 2A in the winding.
>
> Re-connect the motor in series. It is now a 1A/phase, 8 mH motor.
The
> V/L is now 5,000 amps/sec and it will take 200 uS to go from 0 to
1A
> now.
>
> Mariss
>