Re: Gap Question

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

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:

 > Original poster: "Luke" <Bluu-at-cox-dot-net>

 > Any way:
 > Ok as the gap breaks down the current starts to rise and the voltage
 > then starts to drop.  You mention that the current will rise to a
 > maximum and then decrease.  What causes that?

The oscillation of the LC circuit.

 > Why does the current not
 > increase to the maximum and stay there?

I was considering a charged capacitor in series with an inductor in
parallel with the gap. When the gap starts to conduct, the system
starts to oscillate.

 > Are you refereeing to the
 > capacitor being the voltage source to breakdown the gap and will give
 > less energy to the gap at later times in the discharge cycle?  Or do you
 > mean, even with a gap that has a steady voltage/current feeding it and
 > that source causes the breakdown of the gap that the current will rise
 > to a max and then decrease?

In the oscillations, the maximum voltage in the capacitor progressively
decreases, due to the energy lost to the gap (or transferred to the
secondary circuit, if there is one).

 > Also you mention in a good gap you never reach this point.  Is this due
 > to the fast quenching of a good gap?  Does a good gap quench (stop
 > arcing) before the current reaches its maximum?

If the gap is large enough to allow the expansion of the spark channel
mentioned in other posts, the voltage continues to decrease as the
current increases. If some geometrical factor impedes the expansion,
the voltage rises.

About your doubts about what is a negative resistance, you can consider
two interpretations:
A "true" negative resistor: v=-R*i. This is just a mathematical
abstraction that can't exist. Yo can even build an electronic circuit
that operates in this way, but always only for a limited range of
voltages and currents.
A device with "incremental negative resistance" delta v = -R* delta i.
Many physical devices show this behavior for a range of voltages and
currents, as a spark gap. The v*i product, that is the power dissipated
in the device, is always positive. It has to be biased by an external
source to stay in the negative resistance region.

Antonio Carlos M. de Queiroz