Re: Coil design questions.

To: tesla-at-pupman-dot-com
Subject: Re: Coil design questions.
From: Larry Fennigkoh <nnttnn-at-mixcom-dot-com> (by way of "Terry Fritz (moderator)" <twf-at-verinet-dot-com>)
Date: Sun, 06 Dec 1998 14:40:00 -0700
Approved: twf-at-verinet-dot-com
Thanks to all for the great info regarding primary coil design.  Having
built my
first TC over 35 years ago (jr high school) from a Popular Electronics or
Science
article - and clearly not knowing what I was doing, I've had an urge - ever
since
- to build another.  I've just finished the secondary.    This has proven
to be an
extremely informative web site.  Thanks again.
Larry

Tesla List wrote:

> Original Poster: "Barton B. Anderson" <mopar-at-uswest-dot-net>
>
> 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