Solving the DC coil mystery

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

Original poster: "S & J Young by way of Terry Fritz <twftesla-at-qwest-dot-net>" <youngs-at-konnections-dot-net>

Hi Rick & others,

Yes, you are right.  See below.

----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Monday, July 23, 2001 6:35 AM
Subject: Re: Need help with mystery


> Original poster: "Rick W by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<rickwilliams404-at-earthlink-dot-net>
>
> Hey Steve,
> A thought. The short charge time can't be good as far as pulse current is
> concerned. Your tank cap charges awful fast with 67 ohms. A 1 millisecond
> charge time would still be quick enough for the BPS desired. 10K ohms *
19nF
> = .19ms then times 5 time constants = close to 1 ms. I don't know the size
> of the RSG disk or the rpm but 1ms is about 1/2" of travel on the outside
of
> a disk rotating at 1800rpm.
>
> What I'm getting at is try a higher resistance, a few K anyway. You
> shouldn't be hurt by increased charge time and it would help keep sudden
> pulse current down.
>
> Rick Williams
> Salt Lake City
>
Bert Hickman published the key to my mystery a few weeks ago.  I quote:

(Bert or others, please elaborate on the "It can be shown ...".  It has
taken me a while to decide this is true, and I would like to see the
calculations / reasoning that prove it.)

"It can be shown that if a capacitor is charged through a resistor, and if
the charge time is sufficient to virtually fully charge the cap (Tcharge >
3RC), then the series resistor will dissipate the SAME amount of energy as
the energy that ends up being stored in the capacitor. Knowing this
simplifies the problem a bit.

If the Energy/bang = 0.5*C*(V^2) Joules and the break rate = X
bangs/second, then, the power transferred to the TC primary circuit will be
0.5*X*C*V^2 Watts, and the power dissipated by the series (water) resistors
will be the SAME (for 50% efficiency by the charging system):

   Presistors = 0.5*X*C*(V^2) Watts

Because of the significant inefficiency (only 50% max), pure RC charging
systems are seldom used for higher power pulse work, and reactive (LC)
charging systems are preferred instead.

I didn't pay close enough attention to Bert's wisdom before.  If I am
pulling a KW out of my HVDC supply, than 500 watts of it is going into any
resistance in the tank cap charging circuit!  When I was using no physical
resistor, the 500 watts was being dissipated in the gap, wiring, and my DC
reservoir caps which got warm due to their internal resistance.  Not good!

So Rick, as you say, to limit charging current I should use as high a value
of power resistor as will fully charge the tank cap at max BPS.  My 8
rotating electrodes are #10 brass bolts at a radius of 3".  At 500 BPS the
RPM is 3750 and the time they overlap (about 1/4 inch) is about 212
microsecond.  At 19 NF & 3RC, R will need to be about 3.7K ohms.  This would
limit current peaks to about 3.2 amps.

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?

--Steve