# Re: Coil design questions.

```My TC experience is limited but I know a little about transients in coupled
resonant circuits. Here's my "theory":

Under ideal conditions you want all the energy in the primary resonator
transferred to the secondary resonator. This will take several cycles of RF to
complete for a "properly" coupled system. If the coupling is too tight,
some of
the energy transferred to the secondary will begin to come back to the primary
before it all has been transferred. The idea is to pump the system with
transients from the spark gap at a rate that is compatible with the
coupling (the
rate of transfer of energy). Make sense??

Tesla List wrote:

> Original Poster: Larry Fennigkoh <nnttnn-at-mixcom-dot-com>
>
> Greetings:
>
> A basic question for you seasoned coilers:  why is a the flat spiral
primary,
> and lower coupling coefficient, better than a vertical helix or inverted
> cone??  Seems sort of counter-intuitive . . . don't you want to maximize
> coupling?  thanks.
> Larry
>
> Tesla List wrote:
>
> > Original Poster: "D.C. Cox" <DR.RESONANCE-at-next-wave-dot-net>
> >
> > to: Travis
> >
> > Power is potential (voltage) x current (amperage).  Therefore,
> >
> > 15,000 volts x .030 amps  =  450 watts      (would be twice that with
60 ma
> > xmfr)
> >
> > P = E/I  , so   I = P/E , current = power / potential
> >
> > If you ignore losses (very slight in this case) you can assume you need to
> > put as much power into the xmfr as you will be getting out of the
secondary
> > side, so
> >
> > Pin   =   Pout
> >
> > Transposing,    Pout  =  Pin
> >
> > so, if these two are equal then you can assume you also have 450 watts on
> > the primary side of the xmfr, therefore,
> >
> > Pin  =  potential (volts) x current (amps)     again, same as secondary
> > side, so
> >
> > Pin =  E x I ,      I  =  P/E  =  450 watts / 120 volts  =  3.75 amps
> > primary current draw
> >
> > I would suggest a 5 or 6 amp fuse on the primary side for protection.
> >
> > Also, if you can't find a 15 kV, 60 ma xmfr just use a 12 kV, 60 ma xmfr.
> > They are much easier to find and the output difference will be very small.
> >
> > Hope this helps you out and now you can see how the math is done.
> >
> > DR.RESONANCE-at-next-wave-dot-net
> >
> > ----------
> snip

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