Re: Coil design questions.

To: tesla-at-pupman-dot-com
Subject: Re: Coil design questions.
From: "Barton B. Anderson" <mopar-at-uswest-dot-net> (by way of "Terry Fritz (moderator)" <twf-at-verinet-dot-com>)
Date: Sat, 05 Dec 1998 10:08:40 -0700
Approved: twf-at-verinet-dot-com

Larry,

A flat, conical, or vertical primary can ALL over-couple or under-couple. The
main advantage of a flat primary is it's "distance" away from the top
terminal,
thus, there is less likely a tendency to strike the primary (of course, it can
still happen). If one builds a vertical or cone primary in an attempt to
couple
more energy to the secondary LC, they will gain nothing over a flat primary. A
flat primary can couple just as much energy to the secondary and even
couple too
much. As most coils don't couple beyond 0.2, a vertical, conical, or flat
primary
can all provide this coefficient.

Bart

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