# Re: Coil design questions.

```to: Larry

Dig out some older editions of ARRL's Radio Amateur's Handbook and do some
reading on resonant freq section --- it explains it very well.  In short,
you need to achieve only critical coupling.  Once you pass this point the
resonant peak starts to split into two separate frequencies --- one above
and one the same distance below the desired resonant freq.  You end up
pumping power into two freqs that then start beating against each other and
begin doing all sorts of destructive things to your sec coil and if
reflected into the primary your caps and xmfrs.  In most small and medium
size TCs critical coupling is reached around 0.18 to 0.22.  The archive
section details how to check your coupling.  Adjustment is by raising the
sec coil until you hit the right value.  A flat primary spiral provides
almost perfect critical coupling and you never need to use a cone unless
your coil is 2 inches in dia or less.  You never want to maximize coupling
in a classic Tesla oscillator circuit.

DR.RESONANCE-at-next-wave-dot-net

----------
> From: Tesla List <tesla-at-pupman-dot-com>
> To: tesla-at-pupman-dot-com
> Subject: Re: Coil design questions.
> Date: Friday, December 04, 1998 6:18 AM
>
> 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

```