40MHz Spark Gap Behavior

From:  Jim Lux [SMTP:jimlux-at-earthlink-dot-net]
Sent:  Monday, April 13, 1998 11:37 AM
To:  Tesla List
Subject:  Re: 40MHz Spark Gap Behavior

Terry, I don't know if your energy/power calculation is valid.

> 	I have examined the energy and lengths of the burst and have concluded
> following:
> 	I believe the gap is dissipating the vast majority of it's power during
> these bursts.  The voltage and current clearly are in phase and energy is
> being lost.  Heat, light, and RF radiation must be huge during these
> To take a guess at the numbers for the initial spike, if the current is
> 4000A at 2000V for 25ns the energy lost is very roughly 0.2 joule!!  This
> a pretty high rough calculation for a system that only has 0.034 joule to
> begin with.  The energy being lost in the gap at other times seems very
> insignificant!  This would explain why primary systems seem to loose so
> energy on the first oscillation.  Obviously, my number of 0.2 joules is
> correct but it does give an indication that the energy lost in these
> could be very high.

For an arc less than a cm long, I can't imagine the voltage drop at a
current of 4000A being as high as 2000V. It is more likely that the voltage
across the gap drops pretty dramatically (and very quickly, nanoseconds) as
the current comes up.  You might be able to calculate another number for
the energy by figuring out how much energy it takes to heat up a cylinder
of air some 0.1 cm long and .1 cm in diameter (i.e. the spark) to 10000K.
Somehow, I don't think it is in the tenths of joules range. 

I get a volume of  0.0007853975 cc. Make a (erroneous, but within an order
of magnitude) assumption that the specific heat of the air is 1.2 kJ/(kg K)
(at 3000 K) (the specific heat of gases changes pretty significantly as
they get really hot). Then, to heat our 1e-9 kg (= 7.8E-4 cc * 1.29 kg/m^3
* 1e-6 m^3/cc) of air some 10,000 degrees, it will take: 1.2 kJ/(kg K) *
1E4 K *1E-9 kg = 1.2E-5 kJ = 1.2E-2 J = .012 Joules

Some errors here are: The specific heat of a gas changes a lot (it gets
bigger) as the gas gets really hot. The arc is probably not .1 cm in
diameter (lightning -at- 50 kA is only .2-.5 cm) but more like .01 cm in
diameter (reducing the heat energy by a factor of 100).

So then, where else might the energy go? Into light? No, the light comes
from the gas cooling from an incandescent 10000K (neglecting ionization and
photon emission as the atoms drop back to ground state). Into the magnetic
field of the arc? (it is a conductor carrying current) This is pretty low.
Figuring using 1 uH/meter, our .1 cm arc is about 1E-9 H. The stored energy
would be .5 *L*I^2 = .5 * 1e-9 * 4E3^2 = 8E-3 joules. And of course, we
would get that back when the current stopped (it would go into either the E
field (capacity) or induce a current in the rest of the circuit).

Anyway you look at it, it seems that the gap isn't all that lossy. You
might want to carefully look at the relative time of the voltage and
current waveforms, since that has a huge influence on instantaneous power.
A few nanoseconds phase shift ( a few feet of coax...) could turn a 8 MW
pulse (4 kA -at- 2 kV at the same time) into a 8 W pulse (not at the same

As for other techniques for measuring current which work better for fast
pulses, how about a Rogowski coil? or, measuring the magnetic field next to
the conductor with a Hall Effect sensor of some sort, or, using the Faraday
effect (wrap that glass fiber around the conductor a bunch of times and
measure the polarization shift.)

Great work, Terry... you are raising the right sorts of questions.