Re: 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>

Kevin,

Very interesting.  But how do you get much better than 50% efficiency?  That
does not agree with Bert Hickman's proof that half the power is dissipated
in the series resistor.  Why do you use 0.250 amp in your resistor power
loss equation?  If you meant 0.292 amps, that is the same mistake I made -
one can't use the average current to calculate the resistor loss.  If Bert
is correct, and I believe he most certainly is, then your resistors are
dissipating more like 1,750 watts if you are charging your tank cap through
them.  Maybe your circuit setup is different than I am imagining.  Can you
do an ASCII rendition of it or let us know where it is posted?

--Steve
----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Tuesday, July 24, 2001 12:35 PM
Subject: Re: Solving the DC coil mystery


> Original poster: "Kevin Ottalini by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <ottalini-at-mindspring-dot-com>
>
> Steve/All:
>       In my DC coil I currently use 5Kohm (4ea 20K, 200watt resistors in
> parallel, so 800 watt max power dissipation).
>
> My calculated loss is running around 425 watts at a sustained (average)
> power level or 3.5kw, or about 87.9% efficient.  The resistors just get a
> little warm at 3kw with a small fan to help.
>
> The calculation goes like this:
> --------------------------------------------------------------
>    average power utilization:  3500watts
> average DC output voltage:  12,000v
>                    average current::  3500watts/12,000v  = 0.292 amps
>                          peak current::  12,000v/5,000ohms = 2.4 Amps
>                              resistance:  5,000ohms
>   resistor I^2R power loss:  I^2R = (0.250A^2*5000ohms) = 425 watts
>                   Overall Efficiency: ((3500w-425w)/3500w)*100 = 87.9%
>
> The resistors are really intended for "softening" the current surge to
> keep from blowing the diodes.
>
> With 3500watts (minus the 425watts loss in the resistors), and say
> my tank cap is 0 .0625uFd, and given that 1 joule = 1 watt/second,
>  then each bang is 1/2*C*V^2:  0.5 * 6.25E-8 * 1.44E+8 = 4.5 joules
>
>                    so at 3075 watts:  3075w / 4.5j = ~683 BPS maximum
>
> Which is pretty close to the observed performance.
>
> I would rather see that 425watts go into the coil instead of heat,
> but I haven't had a chance to play with HV inductors yet
> (much more of an art than resistors).
>
> Kevin