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NST Power Rating Con

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RE: NST power rating con

Original poster: "John H. Couture" <couturejh-at-mgte-dot-com> Tom - There is only one test to find the maximum watts output of your NST. You need a HV voltmeter, ma meter, and a bunch of "power" resistors. Then make a graph of output watts vs the resistive load. I did this with a 7500 V, 30 ma, 225 watt NST. The maximum secondary watts output was 59.3 watts, 3900 volts, 15.2 ma. The input was 120 volts, 1.45 amps, 174 VA, 60 watts, 34.5% PF. The NST overall efficiency was 98.8% at maximum output!! The efficiency drops to 22% with an 8.0 Kohms load. Overall efficiency % = 59.3/60 = 98.8% Many other tests (1980's) gave surprising results. With certain combination loads of resistors and capacitors it was possible to make graphs with VA output greater than VA input! It is obvious the NST with a TC load can be an interesting challenge when it comes to creating an equation for output spark length. John Freau's equation is about as good as it gets. This all means that for TC's the NST nameplate rating must be used with the utmost caution. John Couture

Re: NST power rating con

Original poster: Neonglo-at-aol-dot-com In a message dated 10/2/03 8:55:45 PM Central Daylight Time, tesla-at-pupman-dot-com writes: >Original poster: Thomas <tom-at-pwrcom-dot-com.au> > >Ok, I can now see why there is a need to indicate the maximum secondary voltage and current even though both can not be supplied at the same time. Exactly. Into a purely resistive load, a typical NST at 120 volts can supply 50% to 55% of it's secondary voltage rating, while delivering 80% of the rated current. This means that a 15kV/30mA NST is actually meant to put out about 7500 to 8250 volts at 24mA on a continuos basis when 'properly' loaded with neon tubes. (The tubing in neon signs is properly loaded by selecting a transformer which will supply 80% of it's current ratings into the tubes in that circuit.) Of course, adding capacitance to the circuit changes everything. Just running the high voltage wire through grounded metal conduit can add enough capacitance to cause numerous problems to an NST. It's no wonder they are so fragile in TC use with large tank caps. That's why safety gaps and a Terry filter are so important to the safe operation of an NST in a TC. ------------------------------------------------- Tony Greer Special Effects Neon Lubbock, Texas

RE: NST power rating con

Original poster: "John H. Couture" <couturejh-at-mgte-dot-com>

For the watts rating of a NST refer to my post Oct 2,2003. The watts output of a NST is a humped curve on a graph of resistance vs watts. The maximum wattage is always much less than any nameplate rating. The 7500 KV, 30 ma, 225 watts(?) NST that I tested had a maximum output of 59.3 watts. This gives a ratio of max watts/nameplate watts = 59.3/225 = .2636 = 26.36%

However, the max test watts output to test watts input was watts output/watts input = 59.3/60 = .988 = 98.8% This is the main reason that using NST nameplate numbers for rating TC's doesn't make sense. John Couture

Original Message----- From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Thursday, October 02, 2003 1:29 PM
Subject: Re: NST power rating con
Original poster: Ed Phillips <evp-at-pacbell-dot-net> Original poster: Thomas <tom-at-pwrcom-dot-com.au
> > > The more I look into this the more proof I get that the actual power available from an NST is only half of the face plate values' product, i.e. P =(V x I)/2.It's the only way I could get this to work out:
 > > Also it gives an extremely close value (+10%err) for spark length when half the secondary VA is used for P in: L = 1.7sqrt(P) for my coil.
 > > I think the +10% length measured is due to the primary cap being resonant with the NST, and a slightly too wide spark gap. > > Why is this con(fidence trick) by NST manufacturers not mentioned on any Tesla coil design web sites (that I've seen)? > > It's almost as bad as the *peak music power* con used by cheap audio gear manufacturers
. > > Tom L. Not exactly correct. With a pure resistive load the power output limit is as you say. With a resistive/capacitive load as in TC applications the power can be very much higher than the product of name plate open-circuit voltage and short-circuit current. I've never seen an NST rated in watts. Has anyone else? Ed

RE: NST power rating con

Original poster: Harvey Norris <harvich-at-yahoo-dot-com> ---
Original poster: "John H. Couture"

Tom - > > There is only one test to find the maximum watts > output of your NST. You > need a HV voltmeter, ma meter, and a bunch of > "power" resistors. Then make a > graph of output watts vs the resistive load. I did > this with a 7500 V, 30 > ma, 225 watt NST. The maximum secondary watts output > was 59.3 watts, 3900 > volts, 15.2 ma. The input was 120 volts, 1.45 amps, > 174 VA, 60 watts, 34.5% > PF. The NST overall efficiency was 98.8% at maximum > output!! The efficiency > drops to 22% with an 8.0 Kohms load. > > Overall efficiency % = 59.3/60 = 98.8% This sounds like a verification of the maximum power transfer principle, which occurs when R(int) = R (load). When this occurs the voltage that developes across the load will be ~ half the voltage found at open circuit. (3900 volts with load vs 7500 volts open circuit) Since a shorted NST represents only R(int) as that load, if R(load) = R (int): this is twice the impedance, (or more properly resistance), on the total current loop, so we should expect ~ half of the measured short circuit current to exist. ( 15 .2 ma loaded current vs 30 ma short circuit current) Problematic with these definitions is how we define R(int). Things seem to make sense if we define it as the inherent impedance of the secondary, not merely its resistance. The limitations involved with ferromagnetic resonance are also shown by giving the secondary an equal capacitive reactance to R(int)'s reactance. A current equal to what would develope by ohms law never actually developes, and what does develope is only a small % of that value. Applying the same "maximum power transfer" principle to just reactances would mean that perhaps we should only expect HALF the current that would be found with ohms law to develope. I doubt if it even goes close to that. The limitations of what a "resonance" should yeild by book value calculations as an "ideal" component vs what it delivers as a "real" component seems to be quite a vast difference. A 1000 ohm multiturn 60 henry air core coil put into 60 hz resonanace only delivers 75-80% of its possible resonance. The same coil placed with 480 hz alternator inputs only delivers 5% of its possible resonance This figure can be doubled to 10% by allowing another magnetic field to exist in magnetic opposition to that coil. It is thought that internal capacity brought on by interwinding capacity is the culprit that limits the available amount of resonance. This also seems comparable since at 8 times the frequency, the limitations have been increased 8 fold for the best possible resonance obtainable for the input conditions. > > Many other tests (1980's) gave surprising results. > With certain combination loads of resistors and capacitors it was possible to make graphs with VA output greater than VA input! This happens to a great degree with certain alternator source frequency air core transformer resonant circuits. An ending load can have 10 times the resistance vs what is inputed as a primary input, and on the secondary side a 10 fold voltage rise can be obtained. With ordinary transformer principles we might expect an accompanying 10 fold reduction from the secondary currents, since the output voltage has risen 10 fold. Yet we can obtain the same amount of current on the secondary ending circuit as that inputed to the primary. The VA input vs VA output has increased 10 fold, even with the the loads having a 10 fold difference in resistance! HDN

RE: NST power rating con

 Date: Sat, 04 Oct 2003 05:11:22 -0600 - RE: NST power rating con

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

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


-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Friday, October 03, 2003 2:24 PM
To: tesla-at-pupman-dot-com
Subject: RE: NST power rating con


Original poster: Harvey Norris <harvich-at-yahoo-dot-com>


--- Tesla list <tesla-at-pupman-dot-com> wrote:
  > Original poster: "John H. Couture"
  > <couturejh-at-mgte-dot-com>
  >
  >
  > Tom -
  >
  > There is only one test to find the maximum watts
  > output of your NST. You
  > need a HV voltmeter, ma meter, and a bunch of
  > "power" resistors. Then make a
  > graph of output watts vs the resistive load. I did
  > this with a 7500 V, 30
  > ma, 225 watt NST. The maximum secondary watts output
  > was 59.3 watts, 3900
  > volts, 15.2 ma. The input was 120 volts, 1.45 amps,
  > 174 VA, 60 watts, 34.5%
  > PF. The NST overall efficiency was 98.8% at maximum
  > output!! The efficiency
  > drops to 22% with an 8.0 Kohms load.
  >
  >     Overall efficiency % = 59.3/60 = 98.8%
This sounds like a verification of the maximum power transfer principle, which occurs when R(int) = R (load) When this occurs the voltage that developes across the load will be ~ half the voltage found at open circuit. (3900 volts with load vs 7500 volts open circuit) Since a shorted NST represents only R(int) as that load, if R(load) = R (int): this is twice the impedance, (or more properly resistance), on the total current loop, so we should expect ~ half of the measured short circuit current to exist. ( 15 .2 ma loaded current vs 30 ma short circuit current)

Problematic with these definitions is how we define R(int). Things seem to make sense if we define it as the inherent impedance of the secondary, not merely its resistance. The limitations involved with ferromagnetic resonance are also shown by giving the secondary an equal capacitive reactance to R(int)'s reactance. A current equal to what would develope by ohms law never actually developes, and what does develope is only a small % of that value. Applying the same "maximum power transfer" principle to just reactances would mean that perhaps we should only expect HALF the current that would be found with ohms law to develope. I doubt if it even goes close to that. The limitations of what a "resonance" should yeild by book value calculations as an "ideal" component vs what it delivers as a "real" component seems to be quite a vast difference. A 1000 ohm multiturn 60 henry air core coil put into 60 hz resonanace only delivers 75-80% of its possible resonance. The same coil placed with 480 hz alternator inputs only delivers 5% of its possible resonance. This figure can be doubled  to 10% by allowing another magnetic field to exist in magnetic opposition to that coil. It is thought that internal capacity brought on by interwinding capacity is the culprit that limits the available amount of resonance. This also seems comparable since at 8 times the frequency, the limitations have been increased 8 fold for the best possible resonance obtainable for the input conditions.
  >
  > Many other tests (1980's) gave surprising results.
  > With certain combination
  > loads of resistors and capacitors it was possible to
  > make graphs with VA
  > output greater than VA input!
This happens to a great degree with certain alternator source frequency air core transformer resonant circuits. An ending load can have 10 times the resistance vs what is inputed as a primary input, and on the secondary side a 10 fold voltage rise can be obtained. With ordinary transformer principles we might expect an accompanying 10 fold reduction from the secondary currents, since the output voltage has risen 10 fold. Yet we can obtain the same amount of current on the secondary ending circuit as that inputed to the primary. The VA input vs VA output has increased 10 fold, even with the the loads having a 10 fold difference in resistance!

HDN

Date: Sat, 04 Oct 2003 18:33:22 -0600 - Re: NST power rating con

 Original poster: Ed Phillips

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
Date: Sun, 05 Oct 2003 17:27:31 -0600 - Subject: RE: NST power rating con
Original poster: "John H. Couture"

Ed - If you do the tests and make the graphs as I mention in my reply to Gerry I believe you will get the answers to all of your comments. Note in your calculations that as soon as the sec amps on the NST start the sec voltage reduces and the output wattage changes. The trick is to find the optimum conditions for maximum output power (watts). This requires a graph or other method because the power curve is a hump type. The tests will also show you that the VA out greater than VA in is not due to instrument error. - John Couture 
 
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