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*To*: tesla-at-pupman-dot-com*Subject*: Re: Directions for tesla coil research*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Mon, 06 May 2002 21:04:06 -0600*Resent-Date*: Mon, 6 May 2002 21:23:45 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <RnBSVB.A.u7H._i018-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "rob by way of Terry Fritz <twftesla-at-qwest-dot-net>" <rob-at-pythonemproject-dot-com> Tesla list wrote: > > Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-qwest-dot-net>" <evp-at-pacbell-dot-net> > > > How does a bigger top load increase Q. if Q = Xl / R and Xl = > > 2 * pi * f * l. > > > > from these equations, one would assume that if frequency goes down, due to a > > shift in resonant frequency, then Q decrease because Xl decrease. I know > > I'm missing something important here, right? > > > > Shaun Epp > > I think so. The discussion was about reducing the frequency by > increasing the top loading, but you have the wrong expression for Q in > this case. The Q is the parallel resistance divided by the reactance > so, for a given resistance (same power in the streamers) the Q will > increase. Your expression is for an inductor with a resistance in > series with it. If you look at the equivalent circuit you will see that > increasing the parallel resistance does lower the series resistance, > with the same effect of increasing the Q a bit. For coils with a Q much > over 10, as are all the ones under discussion here, the product of > series resistance and parallel resistance equals the square of the > reactance. > > Rp Rs = X^2 > > if I didn't make another typo. > > Ed > > Ed Something I forgot to mention about all of this. There is unloaded Q, and loaded Q. I assume that once the air starts to break down, you are in the realm of loaded Q. There is a very good discussion of these types of Q in the Amateur Radio Handbook, and probably lots elsewhere on the web. For example, when I design matching networks for LNAs at work, its best to have a circuit with a low loaded Q. The reason is that the Q factor multiplies the current in the inductors (circulating current), increasing the I*R loss. Thus high Q circuits can in some cases greatly magnify the losses. Still, though, the circuit with the highest unloaded Q elements has the lowest loss. I hope I am not being confusing :) Rob. -- ----------------------------- The Numeric Python EM Project www.pythonemproject-dot-com

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