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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