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Re: Capacitor C/Peek

Tesla List wrote:
> 
> >From rwstephens-at-ptbo.igs-dot-netThu Oct 24 22:57:34 1996
> Date: Fri, 25 Oct 1996 00:37:12 -0500
> From: "Robert W. Stephens" <rwstephens-at-ptbo.igs-dot-net>
> To: tesla-at-pupman-dot-com
> Subject: Re: Capacitor C/Peek
> 

<SNIP>

> Now take a 12000 volt RMS xfmer and charge a 0.125 mfd cap to 16968
> volts.  That's 17.99 Joules.  Bang this in 5 microseconds and your
> sinewave peak power is 5.088 megawatts!  That's approximately the
> spec of R. Hull's Nemesis coil and exactly those of my MTC system. My
> first half sinewave is 6.25 microseconds though (80 kHz), so my theoretical
> peak power is lower at merely 4.07 megawatts, and then taking K=0.2 and gap
> losses into account I'm down to maybe 15% of that (a measely 610 kilowatts)
> out the top?
> 
> I realize that the above is fast and simplified 'figurin' which
> ignores a lotta stuff so flames are welcomed if I'm out more than 10 db. : )
> 
> rwstephens

Robert,

Your post got me thinkin' about coil efficiencies again...

What if we look at this from a slightly little different light: How much
energy
do I start out with in the primary, and how much do I end up with in
the secondary after I've transferred all the energy I could from the
primary. The ratio of output vs input energy (transfer efficiency/bang) 
could be a meaningful way of measuring how well a coil could perform. It
should be measurable with either a 2-coil TC or a magnifying TC.  

Assume you've done "single bang" measurements, and are able to farily
repeatably set your primary gap breakdown voltage via a static gap.
Also, assume you've
estimated your peak secondary voltage indirectly (via low duty cycle
sparklength or via calibrated pickup plate/storage scope). Assume your
resonator is high Q, and is heavily capacitively-loaded so that
virtually ALL of the resonator's energy will be stored in the toroid's
capacitance at the point of peak secondary voltage. [Since most of the
secondary's self- capacitance is concentrated towards the bottom, we'll
choose to ignore it in this analysis - this only makes our analysis a
little bit more pessimistic]. 
Finally, lets assume that we quench the gap at the end of the first
beat.

We can compute the ratio of maximum secondary output energy versus
initial input energy per bang:
 
   Ep = Initial energy = 0.5CpVgap^2
   Es = Max Sec Energy = 0.5CtVsec^2

The Energy Transfer Ratio = Es/Ep per bang

Plugging in some specific parameters for my coil:
My "single shot" low duty cycle output spark length is about 20", which
I estimate to be about 400,000 volts, The toroid capacitance is about 26
pF. My
primary gaps are set for about 19 KV, with a primary tank capacitance of
0.0205 uF. The previously measured primary Q was about 11 with the gap
firing (no secondary), so my primary performance seems typical enough. 
Plugging in these values I get:

   Ep = 3.7 Joules
   Es = 2.1 Joules  
 
   Es/Ep =  56% (!)  (per-bang)

This is a much better than 15%...  No flames today Robert, you're
certainly within 10 db   :^) 

Now, I realize that I've made LOTS of assumptions above. Once we get
breakout, the secondary output voltage and Q will both decrease.
However, its not clear that these changes would cause a dramatic
fall-off in transfer efficiency from the single-shot case. In fact,
after breakout, when the secondary Q drops to a level closer to the
primary Q, the energy tranfer efficiency might actually improve! 

- Am I all wet?

- What do you get when you plug in _your_ coil's parameters?  

As usual, throw any extra Joules my way...

Safe, and maybe even higher power, coilin' to ya!

-- Bert --