[Prev][Next][Index][Thread]

Re: Coil design questions.



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