# Re: Q Factor and Overall Efficiency

```Tesla list wrote:
>
> Original poster: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
>
> 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.

The Boonton Q Meter circuits apply a voltage from a VERY low resistance
generator in series with the L and C.  This equation is indeed correct
under this condition.

> What is the different arithmetic of the double-tuned circuit for finding Vs
> other than sqrt(Cp/Cs), sqrt(Ls/Lp) and .5CsV^2 ?

The above equation is indeed exactly correst if all of the energy
stored in the primary capacitor is transferred to the capacitance of the
secondary circuit.  NOTE THAT Q DOESN'T ENTER INTO IT AT ALL!!!!!!!  The
only way that circuit losses (Q related) enter into things is if the
coupling is low and the losses are high, so that the primary oscillates
for many cycles before it transfers all of the [remaining] energy to the
secondary, which of course loses energy due to its losses.  In a real TC
the secondary Q goes way down as soon as streamers form, so that the
losses dominate the ultimate voltage in a way which I feel is impossible
to analyze.  IF you had a very good equivalent circuit for the impedance
of the streamers as influenced by time since ionization occurred,
current into the streamers, etc. then it would be possible to do a
pretty good Spice simulation (not analysis) to model the probable output
waveform.

>I am saying that because
> you know Vp and the Test Q you can find the secondary volts by
>     Vs = Vp * Qtest

I say this is incorrect in general, depending on how you arrive at Q
test.  In calculating the energy transfer in a double-tuned circuit you
have to include the effect of primary resonant frequency, secondary
resonant frequency (as loaded by the discharge, in the case we are
considering), and the coupling factor.  If you measured the secondary Q
with a Q meter for example, your equation would give a wildly optimistic
prediction of output voltage.

> Are you saying that the secondary voltage Vs does not increase when the
> primary voltage Vp is increased?

Of course not!  That's not the topic under discussion. I would say
that, in general, that once a discharge starts the secondary voltage
will not be proportional to the input voltage, but will increase more
slowly as the input voltage (voltage on the primary capacitor when the
gap fires) due to the decreasing impedance of the streamers with
increasing power into them.

>
> John Couture
>

Ed

```