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RE: Q Factor and Overall Efficiency
Ed -
The voltage at the end of the secondary inductor is
Vs = Vp * Q
This equation is used by the Boonton Model 260-A Q Meter.
What is the different arithmetic of the double-tuned circuit for finding Vs
other than sqrt(Cp/Cs), sqrt(Ls/Lp) and .5CsV^2 ? I am saying that because
you know Vp and the Test Q you can find the secondary volts by
Vs = Vp * Qtest
Are you saying that the secondary voltage Vs does not increase when the
primary voltage Vp is increased?
John Couture
-----------------------------
-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Tuesday, August 29, 2000 8:16 PM
To: tesla-at-pupman-dot-com
Subject: Re: Q Factor and Overall Efficiency
Original poster: "Ed Phillips" <evp-at-pacbell-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