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

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

--- In CAD_CAM_EDM_DRO@y..., "jeffalanp" <xylotex@h...> wrote:
> 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
> >