Re: NST power rating con

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

Original poster: Ed Phillips <evp-at-pacbell-dot-net> 

Tesla list wrote:
 >
 > Original poster: "John H. Couture" <couturejh-at-mgte-dot-com>
 >
 > Harvey -
 >
 > I don't believe this is a verification of the maximum power transfer
 > principle. The tests indicate that the maximum power available from a NST is
 > only about one quarter not one half of the nameplate rating. Note that one
 > half the voltage and one half the current gives one fourth the power. For
 > example the maximum output wattage for these tests was 59.3 and the
 > nameplate 225 watts or
 >
 >       Max power = 59.3/225 = .2636 = 26.36 %
 >
 > It appears that very little power is available from NSTs for their size. No
 > wonder they produce such short spark lengths compared to distribution,
 > instrument transformers, MOT, etc. Spark length equations should take this
 > into account. I have not heard of any other coilers who have made these NST
 > tests.
 >
 > The other problem with NSTs is that the TC input impedance is generally
 > never equal to the optimum needed to get the maximum power out of the
 > transformer. To my knowledge no coiler ever tests to find the optimum
 > impedance for the NST he is using and then designing his TC input impedance
 > to match. This is not a problem with distribution, instrument transformers,
 > etc.
 >
 > The fact is that NSTs appear to produce much longer sparks compared to the
 > power available. Are we missing something regarding TC operation with NST's?
 > My tests show that it is possible to produce VA outputs greater than VA
 > inputs with combination resistance and  capacitance TC loads. Does this
 > somehow provide an extra gain?
 >
 > John Couture

	A few points on which I'd welcome comments.  Assume an NST rated at
15000 volts OC and 60 ma SC.  The internal reactance will thus be
(15000/0.06) = 250,000 ohms.  If a 250,000 ohm resistor is connected to
the terminals the total secondary circuit impedance will be
sqrt(250000^2+250000^2), or 353,533.3906 ohms [good to may 2 decimal
places but the little calculator turns out nice long numbers].  The
total current flowing will be 15,000/353500 = 0.0424 amperes and the
total power dissipated in the resistor will be 0.0424^2 x 250,000 = 450
watts and is indeed half the nameplate rating.  Now change the situation
a bit and put an 0.025 ufd capacitor (-106100 ohms reactance) in series
with the resistor.  The net circuit reactance is now (250000-106100) =
143899 ohms and the total impedance will be sqrt(250000^2+143899^2) =
288,456 ohms.  The current flowing will be 15000/288456 = 0.052 amperes
and the power in the resistive loat will be 0.052^2 x 250,000 =676
watts.  Reducing the capacitance to 0.012 ufd (-221040 ohms) will result
in a net reactance of about 29000 ohms and the total circuit impedance
will be 251670 ohms; the current flowing will be 0.0596 amperes and the
power in the resistor will be 880 watts.  If the capacitance is reduced
to the "matched" (resonant) value of 0.0106 ufd the circuit reactance
will be zero and the voltage across the resistor will be 15000 and the
power will be 900 watts.  Think I got all those numbers OK but one can
always make mistakes.

	Bottom line is that with a load with capacitive reactance you can
indeed get an output power equal to the nameplate OC voltage and SC
current.  Since the leakage reactance is affected somewhat by the
current flowing and is not necessarily exactly 250000 ohms these numbers
are only approximate but they illustrate the principle.  Note that, if
everything were linear, the open circuit voltage of the transformer
would approach infinity (insulation would fail and/or core would
saturate first) and the short circuit current would be limited only by
the internal resistance of the transformer.

	In TC operation the load is not resistive of course, but same very
general principles apply.  I'm sure some of the amazing results reported
for systems using NST's are the result of near-resonant operation of the
secondary.

	As for VA outputs greater than VA inputs I think such results have to
be due to instrumentation errors.

Ed