Re: Re: OT: help with motor run capacitor sizing

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: <a1accounting@xxxxxxxxxxxxx>

Hi Dave

I thought I would check again oops I forgot to subtract the L from 
the C winding

so C = L/(R^2 - (2*pi*f*L)^2) where f is the frequency.

I thought it would have a 2*pi*f in it.
I noticed the problem with the two extremes
if R> ZL(impedance of the L) in the direct winding, the current is in 
phase with the voltage so the winding with the C must have most of 
the supply voltage across the C to get the phase shift so you will 
need a lot more volts to get the right current.
if ZL>R in the direct winding the current is lagging 90deg with the 
voltage so the winding with the C must cancell the L to leave mostly 
R so the current is in phase with the voltage. The voltage will have 
to be reduced significantly to avoid over heating.

This suggests if ZL is near R you may get away with the same volts on 
both windings.

Bob
>
> From: "Tesla list" <tesla@xxxxxxxxxx>
> Date: 2006/06/21 Wed PM 03:09:27 EDT
> To: tesla@xxxxxxxxxx
> Subject: Re: OT:  help with motor run capacitor sizing
>
> Original poster: David Speck <Dave@xxxxxxxxxxxxxxxx>
>
> Bob,
> Thanks very much.  That's just what I was looking for.
> Dave
>
> Tesla list wrote:
> >Original poster: <a1accounting@xxxxxxxxxxxxx>
> >
> >Hi David,
> >
> >I stand corrected. There is a relatively simple formula assuming you
> >need a phase shift of 90deg between the currents of the windings.
> >C=(R^2)/L where C is the capacitance to be put in series in Farads,
> >R is the series resistance of the coil in ohms and L is the series
> >inductance of the coil in Henries.
> >
> >I did double check it, very quickly, because no 2*pie*f  is odd, 
it looked ok.
> >I don't know if it produces practical values of C.
> >
> >Note the current will be different in the winding with C compared to
> >the winding without the C (assuming identical windings) so you may
> >have to change the power supply voltage to compensate.  Assuming the
> >impedance of L > R then the voltage must be reduce to avoid over heating.
> >
> >Bob
> > >
> > > From: "Tesla list" <tesla@xxxxxxxxxx>
> > > Date: 2006/06/20 Tue PM 01:46:45 EDT
> > > To: tesla@xxxxxxxxxx
> > > Subject: Re: OT:  help with motor run capacitor sizing
> > >
> > > Original poster: <a1accounting@xxxxxxxxxxxxx>
> > >
> > > Hi David,
> > >
> > > I don't  think anyone replied to you so.
> > >
> > > There is probably no one formula is it would be dependent on the L R
> > > ratio of your winding.
> > >
> > > You could try putting a diode in series with each winding. The idea
> > > being that each winding is powered by alternative half cycles. You
> > > may have to do this at a lower supply voltage because the L of the
> > > windings will not limit the current.
> > >
> > > Bob
> > >
> > >
> > >
> > >  >
> > >  > From: "Tesla list" <tesla@xxxxxxxxxx>
> > >  > Date: 2006/06/18 Sun AM 12:41:02 EDT
> > >  > To: tesla@xxxxxxxxxx
> > >  > Subject: OT:  help with motor run capacitor sizing
> > >  >
> > >  > Original poster: David Speck <dave@xxxxxxxxxxxxxxxx>
> > >  >
> > >  > Terry,
> > >  >
> > >  > May I ask the EE types on the list for a little help with a not
> > > quite TC item.
> > >  >
> > >  > I have a number of two coil motors, some of them steppers, that I
> > >  > would like to use as simple AC motors by wiring them as  capacitor
> > >  > run motors.  One winding would run directly across the power supply,
> > >  > and the other winding would run in series with  an AC cap to provide
> > >  > a phase shift.
> > >  > Is there a formula that would suggest the optimum run capacitor value
> > >  > based on the inductance of the windings, or the current draw through
> > >  > one winding at their rated voltage?
> > >  >
> > >  > Thanks in advance.
> > >  >
> > >  > Dave
>
>
>
>