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*To*: tesla-at-pupman-dot-com*Subject*: Re: Q Factor and Overall Efficiency*From*: "Ed Phillips" <evp-at-pacbell-dot-net> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)*Date*: Sat, 02 Sep 2000 17:02:11 -0600*Delivered-To*: fixup-tesla-at-pupman-dot-com-at-fixme

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

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