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Re: Terry's GM HEI Coil Update



Ed,

    Coulombs, if his capacitor is bigger than yours in the spark gap, his
discharge is going to be hotter, and longer.  If his supply is at 60 mA, and
yours at 30mA, the value of capacitor he needs at the same frequency is
twice the value.  Impedance matching desides when the gap fires.  So the
output impedance = 15Kv/30mA  which is 500K ohms, but his output impedance
is 250k ohms.  Therefore F= 2(pi)RC, and for a spark gap C = (F/2(pi)R)/5
Because capacitor timing equations are based on 63% of their actual charge
time, and in order to get the maximum spark voltage, and length you cannot
use a normal frequency calculation.  If you were using the Universal Time
Constant Curve, then you would see how the spark gap could be calculated
using that equation.  If you work the equation for the two resistance values
you get to totally different answers, and one is weak compared to the other
at the spark gap.  Therefore, a Bigger capactor stores more electrons, and
has a higher number of watts out, although, that is pulsed, and in
microseconds..  So, I guess the better way of asking that question who has
the highest output over the time period in uS's.

James


<<<< I think the equation "F= 2(pi)RC" should be F = 1 / (2 x pi x R x C).
The other equation "C = (F/2(pi)R)/5" is from the fact that 5 RC time
constants are needed for a capacitor to charge (or drain) to within 0.6738%
of its final value.  F = 1 / T = 1 / (5 x 2 x pi x R x C) so the second
equation should be C = 1 / (5 x 2 x pi x R x F).  This would allow the cap
to reach over 99.3% of its full charge.  However, that is for a square wave
input.  A sine wave input is different where it would reach 98.06% of the
full value.  I don't think that applies since this assumes the transformer
is a pure resistance and does not account for the inductance and energy
storage characteristics of the actual circuit.  In that case the capacitor
value is really about twice what the first equation would suggest for
maximum power to the system (unless you use the first value with resonant
charging which risks blowing the neon to get the same power level).  The
second equation gives a far too small cap value.  But perhaps the moderator
is reading the mail too closely.... - Terry >>>>




>Original Poster: Ed Phillips <evp-at-pacbell-dot-net>
>
>"Original Poster: "James" <elgersmad-at-email.msn-dot-com>
>
>    Try checking out the value of capacitor that you were using, and
>compare
>it to the value he is using, in watts."
>
> What is the conversion factor from watts to farads?????????
>
>Ed
>
>