Re: Q Factor and Overall Efficiency

["Ed Phillips" <evp-at-pacbell-dot-net> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)]

Original poster: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
> >
> >
> > The Q factor tests have been discussed many times in past List posts.
> > However, these tests give the Q factor in a low voltage condition. This is
> > not the TC operating Q factor. The operating Q factor would give the
> > secondary voltage with the equation
> >     Vs = Vp * Q
> > where Vp is the primary peak voltage.

	Wrong!  If you had a simple series-resonant circuit the voltage across
the load would be Q times the voltage applied to the end of of the
capacitor or inductor.  Think that "magnifiers" work on that basic
principle.  The ordinary TC is a coupled double-tuned circuit, and obeys
different arithmetic.

> 
> I don't agree that this is valid for a disruptive TC and I
> have stated why many many times so I won't repeat myself yet
> again. Operating unloaded Q can easily be found by measuring
> the decrementing waveform on an oscilloscope. 

	Agree on that.

>I have found it
> is pretty much the same as the results I've obtained in low
> level tests.
> ----------------------------
> It is a well known fact that increasing the primary voltage (Vp) will
> increase the secondary voltage (Vs) of a disruptive TC with no other
> changes. This means the Q factor would also increase. 

	What leads you to say that?  Doesn't make any sense to me.  The only
way the Q could change would be if the load represented by the discharge
changed with power, and the only way I can see it changing is to a LOWER
impedance which, in shunt with the coil, would lead to a LOWER Q.
????????????????

Ed