Re: Solving the DC coil mystery

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

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


> Original poster: "S & J Young <youngs-at-konnections-dot-net>
> Hi Rick & others,
>
> Yes, you are right.  See below.
>
<snip>

> > 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:
>
<snip>
>
> 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.