Re: Solving the DC coil mystery

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Greg Leyh by way of Terry Fritz <twftesla-at-qwest-dot-net>" <lod-at-pacbell-dot-net>

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Original poster: "S & J Young <youngs-at-konnections-dot-net>

[snip]

So, list, if I use an inductor instead of the 500 watt resistor for
increased efficiency, what inductance should it have to charge a 19 NF
tank
cap with a break rate of 500 BPS max?  How should I calculate the value?
Greg Leyh's formula for his DC powered coils was BPS(max)/2 = 1/[2pi *
SQRT(Lreact/Cpri)].  Solving for L it is
1/((pi x BPS)^2 x C) = about 21 Henry.  Is that right?  And I end up
with
the tank cap charged to about twice the power supply voltage - right?
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Yes, 21H would be correct.  But your peak charging current 
would only be Vhvdc / SQRT(L/C), which I would guess is less
than 0.5A (3 joules).  A 3J reactor is easily of hand-held size.

At 1kW, you might want to consider a switching cap charger instead?
--


-GL
www.lod-dot-org