Re: Coil design questions.
Power is potential (voltage) x current (amperage). Therefore,
15,000 volts x .030 amps = 450 watts (would be twice that with 60 ma
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
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
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.
> From: Tesla List <tesla-at-pupman-dot-com>
> To: tesla-at-pupman-dot-com
> Subject: Re: Coil design questions.
> Date: Wednesday, December 02, 1998 3:14 PM
> Original Poster: Travis Tabbal <bigboss-at-inquo-dot-net>
> On Wed, 2 Dec 1998, Tesla List wrote:
> > Original Poster: "D.C. Cox" <DR.RESONANCE-at-next-wave-dot-net>
> > to: Travis
> > The 60 ma drive current will give you much better performance as the
> > won't be starving for recharge current. If you have to stick with the
> > ma drive, then reduce your cap to around .005 or .006 MFD and use
> > turns on your primary inductor to acheive resonance. At the 6 inch dia
> > would strongly suggest a flat spiral primary as an inverted cone would
> > begin to produce overcoupling and forced resonance which can lead to
> > problems especially with neon sign transformer (NST) powered systems.
> > you have the means to check the coeff. of coupling shoot for a value
> > between 0.18 and 0.20 which seems to work very well with small and
> > size systems.
> Thanks for the responce. After more reading I was beginning to think the
> flat primary would be a better choice. I can probably get more
> transformers, the shops arround here are kind of a pain but it's worth
> How much current does a 15KV/30ma neon pull anyway? I want to make sure I
> don't burn up the varriac. :)
> I probably can't check coupling, I'm not sure how I would go about that.
> Is there a coupling meter? ;) How about an equation?