Re: Electric Breakdown

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Dave Dockstader wrote:
> It seems that I don't understand ionization as well as I should.  I had thought
> that all the ions between electrodes would feel the same electric field and
> would be ionized simultaneously, much as all the electrons in a conductor
> move together.  Why don't they all ionize together?  Ignoring end effects,
> don't they all experience the same field?  Am I overidealizing the problem.

Certainly, ignoring any field inhomogeneities, such as end effects, the ions
*do* experience the same *initial* field.  Once a breakdown avalance gets
strongly underway in some region, the field in the deeply interior region of
high ion density weakens as the forming conductive ion/electron cloud moves in
a way which tends to neutralize the external field.  This weakened field 
strength means that the potential drop across the dense region of the
conductive cloud is less than average for its longitudinal extent.  Since,
presumably the electrode potentials are held (more or less) fixed this means
that an *inceased* potential drop occurs in the nonconductive region between
the conductive cloud and the electrodes.  This increased potential drop
increases the field strength in the region at the outer ends (along the
field direction) of the ion cloud.  This region of increased field strength
at the leading edges of the ion/electron cloud is what insures that the
breakdown chain reaction continues to propagate across the gap with ever
increasing vigor until the entire gap is filled with a high concentration of
conductive ions and electrons which then shorts out the external circuit of
the electrodes and the field strength collapses and a huge current flows.
The violence of these initially ionizing collisions which propagate the
region of ionization greatly heats the conductive cloud to a very high 
temperature, and this along with some recombination of electrons with ions
causes the photon emission we see as the glow of the spark.

Once the spark is struck the current flows easily with a greatly reduced 
overall field strength since the conductive ion channel has now been heated
to many thousands of degrees and this thermal agitation in the channel
prevents reneutralization of the channel and keeps it sufficiently ionized
to continue the conduction of the spark.  If the spark is oriented
horizontally then eventually convection processes push the hot spark channel
upward causing it to rise like a Jacob's Ladder demonstration or to bow
upward in the middle which stretches the conduction channel into a weaker
field region causing it to break open (like a lab prop in a Frankenstein
movie) and a new straight spark then forms directly across the electrodes
once the potential drop across the now newly high impedence medium rebounds.

Before the field is turned on the distribution of velocities of gas particles
is a Maxwell-Boltzmann distribution.  There will presumably always be a few
ions initially present (due to old tracks of background cosmic rays and other
ionizing radiation, left over exhaust remanents from previous chemical 
reactions, etc. if nothing else).  When the electric field is turned on
an initial ion accelerates in response to it until it collides with an
intervening gas particle and looses much of its accumulated kinetic energy to
the gas particle.  Eventually an ion-gas molecule collision occurs with such
vigor (due to the relative velocity of the constituent particles being in the
high end tail of the MB distribution) that the collision ionizes the gas
molecule.  Now just after the collision instead of one ion in this vicinity
we have two or more ions (and electrons) present, and these charged particles
accelerate in the field and collide with other neutral gas molecules.  This
is the beginning of a chain reaction cascade of ionization the sweeps across 
the region of high field strength between the electrodes.

> On a new plug the end of the center electrode has sharp edges so I presume
> the fields are strongest here and ionization occurs here first and propagates
> across the gap.  Am I getting close?

Sounds good to me.

> What about preferences for forming positive or negative ions?  I had expected
> that an electrostatic precipitator (used to remove dust from power plant
> exhaust) would strip electrons from the dust forming positive ions, which would
> then drift to negative plates.  Instead, I am told it is the other way around.
> Any body know why one way should be any better than the other.

Since the gas is electrically neutral I would expect an even mix of positive
and negative charges in the spark.  I'm not sure how many of the negative
charges will be in the form of bare electrons and how many of them will be
in the form of negative ions.  I expect the fraction of each is mostly 
determined by the electronic/chemistry properties of the gas particles.
Remember that when a bare electron collides with a neutral atom/molecule it
will either stick to it making a negative ion, bounce off possibly
electronically exciting the neutral particle, or knock off other secondary
electrons producing a positive ion and more electrons (which might then marry
up with other neutrals and make more negative ions).  Since the electron mass
is many thousands of times lower than the ion masses it is the bare electrons
that are colliding with other particles with the highest *speeds*.  *But* an
electron and a singly charged ion both accumulate the same kinetic energy
over the same distance in the same field.  I certainly don't know enough
about atomic physics and chemistry to know what the relative
(bare electron)/(negative ion) concentration ratio would be in the spark.

David Bowman
dbowman@gtc.georgetown.ky.us