RE: NST TEST GRAPHS (was NST power test)

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

Original poster: "John H. Couture by way of Terry Fritz <teslalist-at-qwest-dot-net>" <couturejh-at-mgte-dot-com>


Jim -

I agree that about 5% for the losses is a good guess for the graph shown at

       http://www.mgte-dot-com/tesla/rc-neon-curve.pdf

The voltage at the .011 uf capacitor was 6200 volts and the current was 22.4
ma giving 139 VA. At 5% the losses would be about 7 watts.

   Power Factor = W/VA
                = 7/139 = .05 x 100 = 5.00%

The phase angle would be arc cos(7/139) = 87.11 degrees.

  The Q factor can be found by

    Q Factor = tan(PF) = tan(87.11) = 19.8

This is close to the 20 you estimated.

The input (primary) circuit also gave some interesting numbers. For the .011
uf capacitor the VA was 96 and watts 40. The PF was 41.7 percent with a
65.35 phase angle. The 40 watts would include  the losses in both the
primary and secondary circuits. Note on the graph that for the .011 uf cap
the 96 VA of the primary is less than the 139 VA of the secondary as shown
above.

The graph was similar to the above for a NST load that consisted of a .01623
uf capacitor in series with increasing resistance. Some of the resistances
gave output VA greater than input VA. The curve for the input VA was non
linear and showed a maximum at about 100 Kohms. This would be an optimum
operating condition for this NST with this type of load. This would also be
the point of maximum overall efficiency. It is obvious that graphs of this
type would be very useful to coilers for finding the most efficient NST load
combination. But how do you relate this load combination to Tesla coils?

John Couture

-------------------------------


-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Monday, January 20, 2003 7:57 PM
To: tesla-at-pupman-dot-com
Subject: Re: NST TEST GRAPHS (was NST power test)


Original poster: "Jim Lux by way of Terry Fritz <teslalist-at-qwest-dot-net>"
<jimlux-at-earthlink-dot-net>

  > Original poster: "Jim Lux by way of Terry Fritz <teslalist-at-qwest-dot-net>"
  > <jimlux-at-earthlink-dot-net>
  >
  > Does any of this testing show that a simple transformer model (leakage
  > reactance, series resistance, magnetizing inductance, loss resistance)
  > doesn't adequately describe the NST operating in it's linear (non
  > saturated) regime?
  >
  > ------------------
  > None of my testing indicates that this changes. However, the tests did
show
  > that if the load is only a varying  resistance the sec voltage is linear
but
  > the sec wattage is non linear.

Which you would expect.. power transfer is a peak at matched resistance, and
zero with either a short or infinite resistance... Has to be nonlinear


  >
  > When the secondary VA is greater than the primary VA, it just means that
  > there is reactive power circulating in circuit formed by the secondary
  > leakage inductance, resistance, and the external capacitor.  In fact,
it's
  > a good indication of resonant rise in the secondary, since that would
  > nicely explain it.
  >
  > 0.01 uF at 60 Hz would be about +j265K ohms...
  > 7.5kV open circuit with 30 mA short circuit output current would imply
that
  > the leakage inductance has a reactance of about -j250K... (neglecting the
  > iron and copper losses, which are probably around 5-10%)
  >
  >
  > So, the 0.01 uF case is pretty darn close to resonant.  What was the
  > voltage across the capacitor in this case?
  >
  > -------------------
  > Unfortunately, I did not have a variable uf capacitor to look for
resonance
  > around .0106 uf. There was no voltage rise at .008 uf and the sec voltage
  > was 5700 volts. There was also no voltage rise at .011 uf and the sec
  > voltage was 6200 volts. If there is a voltage rise between .008 and .011
I
  > would expect that the tuning would be very sharp.

Given the high resistance (secondary winding resistance, if nothing else), I
don't know that the Q is all that high.  Ballparking it, I'd expect that the
resistance of the secondary is such that the loss in normal operation (i.e.
lots of current, low voltage) is around 5% of rated power. For a 7.5/30
(275VA) something on the order of 10W? At 30 mA, that's (R= P/I^2) around
11kOhm.. The reactance is (as calculated before) around 265K, so the Q( =
X/R) is going to be around 20. 5% might be optimistic.. 10% losses would
equate to a Q around 10.. To get a change in voltage of .707 (= 3dB) would
imply a change in C of around 20%...(say, from .01 to .012)

Perhaps Terry will make some current phase measurements (hint, hint) and
you'll see resonance clearly, as the phase flips through zero.