RE: Gap Question

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

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

Antonio:
Yes I know a resistor placed in the circuit backwards is still a
resistor. :)
I meant something reacting in the opposite way a resistor reacts.

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

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?

Bear with me I think this is starting to soak in but have a few more
questions (I think).

Thanx

Luke Galyan
Bluu-at-cox-dot-net
http://members.cox-dot-net/bluu

-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Tuesday, February 24, 2004 6:52 AM
To: tesla-at-pupman-dot-com
Subject: Re: Gap Question

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  *
   |   *
   |  *
   |  *
   | *************
   |         ***
   |      ***
   |   ***          v
   +***--------------

  > 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