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Re: Watts o' Problems

"Since the transformer is current limited, it will put out 9000VAC at
0mA or
30mA at 0VAC normally into a resistor.  They figure the most you can get
out of it is 4500VAC at 15mA if the source resistance is matched to the
load resistance.  Of course, that's only 67.5 watts which is really what
your transformer will put into a resistor.  Since these are normally run
at
low voltage but high current, they really don't care what the power is
an
they don't take a lot of time figuring it out ;-))  

If you short the output, you will get 30mA but the input current will be
0.030 x 9000 / 120 = 2.25 amps.  However, the current will be 90 degrees
out of phase so the power is low but the "VA" rating is 2.25 x 120 =
270VA.
 So it's anybody's guess how to rate these things ;-)

Fortunately you can count on getting 270 real watts out of each of your
NSTs in a Tesla coil because we charge caps with them which basically
removes their current limiting."

	Don't think all of this is correct; its even more complicated than
Terry states it.  The transformer short-circuit current is limited by
its internal leakage reactance, which is determined by the gap in the
magnetic shunts used for that purpose.  For instance, for the 9000 volt
30 ma transformer the internal leakage reactance is (9000/0.03) or
300,000 ohms.  As far as the output current goes, if you load the
transformer with a resistor, if the transformer resistance is neglected
the voltage across it will be 90 degrees out of phase with the drop of
the internal reactance.  Net result is that for the 9000 volt example
the output voltage will be about 0.666 x 9000 or 7790 volts at a current
of 15 ma and the load resistor will be about 519,000 ohms.  At a an
output voltage of 4500 the current will actually be about 26 ma and the
load resistance will be about 173,000 ohms.

	Now take the case where the transformer is connected to a "matched"
capacitor (capacitance is series resonant with the internal reactance). 
With no load connected the current which would flow if there were no
voltage breakdown of either the transformer or capacitor would be
limited only by the internal resistance of the transformer, which I
estimate as about 9000 ohms in this example.  The current would thus be
about 9000 volts / 9000 ohms, or one ampere.  The voltage across the
terminals would then "try to rise" to 300,000 volts.  This wouldn't
happen, due both to voltage breakdown and core saturation, but it
illustrates the principal and shows why you should never use a neon-sign
transformer without a protective gap.  With the capacitive load the
transformer can put out lots more power than with a resistive load.

	When you throw in the spark gap things become even more complicated, as
the load impedance the TC presents to the transformer is reactive,
non-linear, and time varying.

	Bottom line is that you can indeed get more power out of an NST than
you would calculate by multiplying the open-circuit voltage by the
short-circuit current (your 270 watts, for example).  The only sure way
to determine the power input to a TC is to put a real wattmeter in the
primary circuit.  Because the power factor will be less than 1, the
power will be something less than the product of the line voltage and
the RMS input current.  As a specific example, I have a small coil using
a 12 kV, 60 ma NST.  It runs with a capacitor slightly under the
"matched" value, and the input current is about 9 amps at 120 volts. 
This is 1080 volt-amperes (don't know the actual power, as I don't have
a wattmeter that large), which is greater than the product of 12000
volts and 60 ma.  If I open up the primary gap the transformer input
current (and spark length) increase accordingly.

	Interesting topic.

Ed