Re: Directions for tesla coil research

["Tesla list" <tesla-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.

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