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>
 >
 > Antonio:
 > Is it that the negative slope of the V I curve occurs from when the gap
 > first breaks down until the voltage across the gap has dropped?

Supposing that the curve is fixed (it varies at least with temperature),
the negative resistance area is from when the gap breaks down until
a current when the voltage drop is minimum. If you force the current
across the gap, the voltage initially jumps to a large value, then
drops, reaches a minimum, and then rises again.

 > So if the arc were looked at as an arc that has already broken down the
 > V I curve would be the normal slope you would see with a resistor if
 > voltage supplied to it were varied?

Yes. If you force the voltage over the gap, initially the current is
low, but after reaching the breakdown voltage it jumps to a high value,
where the incremental resistance is positive again.

 > But the act of breaking down is like changing the gap area from an
 > infinite resistance into a lower resistance and that is when this
 > negative resistance characteristic takes place?

Imagine what happens in a Tesla coil primary: Initially the maximum
voltage of the capacitor appears across the gap, reaching the point
where the curve becomes vertical (breakdown). The current then starts
to rise, limited by the primary inductance. The voltage across the
gap drops as the current increases, following the curve. If the current
rises enough, the voltage starts to rise again (in a good gap it doesn't
reach this point). The current reaches a maximum, and then decreases.
The gap voltage rises as the voltage drops. When the current changes
direction, the same things occur with opposite polarity.
Why the gap conducts for several cycles? If the curve were static,
the voltage would not be enough for a second breakdown after a first
half-cycle swing, because not all the starting energy would return
to the capacitor. The curve is not static, but varies to smaller
voltages as the temperature of the gap rises, or ions accumulate in
the hot air. Quenching occurs when the remaining energy is not enough
for a breakdown at the current breakdown voltage.

 > Side note and question:
 > With positive resistance the current will drop when the voltage across
 > it drops.
 > With negative resistance the current will rise when the voltage across
 > it drops.
 > This is the concept of negative resistance right?

Better to say the the voltage drops when the current rises. You can't
force the gap to stay in the negative resistance area by forcing a
voltage over it. Note that a vertical line below the breakdown
voltage intercepts the curve in three points. Two (low current and
high current) are stable. The other (negative resistance) is unstable.
Stray capacitances and inductances would move the current to one of
the stable values.

 > Doesn't a normal switch display negative resistance?
 > With the switch open the voltage across it is high and the current is
 > low.
 > With the switch closed the voltage across it is low but the current
 > through it would be high (assuming there is a load present).
 > So with a switch when the voltage dropped the current went up.  That is
 > the opposite of what a resistor would do.

If the switch is controlled by the voltage over it, as in a relay with
the coil in parallel with its normally open contacts, you
can say that it shows "negative resistance" (put a resistor in series
with the contacts, if you want to try). In this case you can't bias
the device in the negative resistance area not even by forcing current.

  |i  *
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 > But since we are used to looking at standard electronic components
 > (caps, res, ind, diode, etc.) we try to compare it to what we know, and
 > it looks most like a resistor but backwards so we dub the term negative
 > resistance?

A resistor put backwards is a resistor... The inclination of the v x i
curve is what matters. If you bias a negative resistance device in the
negative resistance area and connect it to a series LC tank, or just
to a capacitor, the resulting circuit is unstable and oscillates,
initially as if a true negative resistor (that doesn't exist) were used.
Look at a Tesla coil primary circuit: A current source (the current-
limited power transformer) feeds a device with a negative resistance
region (the gap) in parallel with a series LC tank. This is an
oscillator.

 > So is it really that there is negative resistance or is it a change of
 > state kind of thing that is displayed?  Are we just calling a change of
 > state negative resistnace?

If the change is controlled by the voltage or current in the device,
the device is a nonlinear resistor, that can easily exhibit negative
resistance.

 > Sort of like trying to freeze water.  You can start cooling water down
 > and it will drop in temperature linearly down to about 32 degrees f.
 > once at that temp you need to keep pulling more and more heat out of it
 > to get it to turn to ice.  But while pulling out this heat the
 > temperature never dropped. All the heat removed left the temp the same
 > but made the water solid (change of state).  If we did not know how the
 > change of state verses btu's thing worked with water we would be a bit
 > stymied as to what was happening.  Not sure what we would try to call it
 > but you get the drift..........
 > Ignore this last paragraph if you think I am rambling.  :)

This is just another form of nonlinearity.

Antonio Carlos M. de Queiroz