From: "Terrell W. Fritz"
Date: Sat, 14 Sep 2002 21:22:08 -0600
Subject: Re: [TSSP] Top Voltage
Hi All,
Being a little frustrated with the fiber probe tonight, I whipped out the "good
o'l trusty" plane wave antenna. I hooked the pinger to the coil with a TEK
6015 HV probe and got the plane antenna calibrated (1V = 88.816kV). I then set
the firing voltage at 200VDC which is where the small ball sometimes breaks out
and sometimes does not this particular night...
http://hot-streamer.com/temp/PaulArc/020914/Tek00006.gif
I figured out the current directions tonight too and decide just to set things
so "they look right". I know what winding direction the secondary winding is,
but with a one turn primary.... Aruugh!! Darn Pearson! "Positive output
for electron flow in direction of arrow"... Of course electrons flow in the
"wrong" direction... I think Ben Franklin did that one... Oh well... I
decided just to "make it look right" :-)))
So, here is some real fun!
For the input:
200 VDCfire
Cpri = 47uF
ECpri = 0.94 Joules
Secondary peak voltage (no breakout):
164kV
Csec = 35pF
ECsec = 0.471J
Secondary peak current(no breakout):
1.68 A
Lsec = 0.492H
ELsec = 0.694J
I can see having a bunch of loss from primary to secondary, but the secondary L
and C energies "should" match. But Paul's work tells us the L is really less
and I assume C is really more in the resonant case. so I used:
1/2 x C x V^2 = 1/2 x L x I^2
and
Fo = 1 / (2 x pi x SQRT(LxC)
Fo=38300Hz
as "two equations and two unknowns" (I almost hurt my brain here! :o))) to
find:
C = 42.57pF
and
L = 0.4056H
Ldc = 0.492H and Cdc = 35pF
So Leq/Ldc = 0.8244 and Ceq/Cdc = 1.216. I may have the eq or ee or whatever
terms mixed up. Have to review Paul's paper for the proper terms... But
perhaps these numbers match what they should be as sort of an independent
experimental confirmation.
So with the new values, I get a secondary energy of 0.5725 Joules. So out of
the original 0.94 Joules, 61% made it to the secondary's peak power. 39% "got
lost" even with a fast k=0.25 ring up.
I also compared a breakout to no breakout energies (165kV vs. 131kV) and found
a breakout event consumes 0.213 Joules in this case. That is about 25% of the
original primary energy and 1/3 of the available secondary energy for just a
bare minimum breakout. Assuming the event time is 100nS and the voltage is
150kV, we get a current of 14.2 amps. That suggests that just a "little"
breakout is a 2,000,000+ watt instantaneous event!!!
Also note that the 1 inch diameter sphere's radius is 1/78.7 of a meter. That
suggests that the breakout voltage is 3MV / 78.7 = 38100 volts.
HAHAAHhahaaohoho.... Seems to be taking 150kV!!!!
So that's what I know tonight... Many more questions than answers... I guess
if this stuff was easy, everyone would be doing it :-))) I did order up that
nice P5205 differential probe. So at least I got a new toy coming :-)))
BTW - Paul mentioned about divider caps for the Jennings probe. They
apparently "did" make "one" to go to 120kV. I guess a 3pF 60kV shielded cap
that goes in series with the probe. I think Jennings made like "two" and the
recipe died with Joe Jennings there... The present staff at Jennings could
probably look the details up in their "paper room" (realizing, they hate me...)
but trying to find one, I would think would be impossible...
Cheers,
Terry
Maintainer Paul Nicholson, paul@abelian.demon.co.uk.