# RE: NST power rating -- another perspective

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

Gerry -

I believe if you do the tests with graphs (resistance vs wattage) of a NST
that you will find exactly what I found. Some of your comments below will
apply. You will need for the primary a voltmeter, ammeter, and wattmeter.
For the secondary you will need a HV voltmeter, ma meter, and a bunch of
power resistors and HV capacitors. Record all of the data, at least 9
columns, and make a graph of resistor load (x) and watts (y) to find the
maximum input and output power (watts). The curve will be a hump type at a
certain resistive load. The ma meter in the sec circuit will give you the
current. You can then find the sec power output by the equation
Output watts = R x I^2

You will find that this is not the standard power transfer problem. The
problem is what NST voltage times current will give you the maximum power
output? The power output refers to real power output not VA output.

You may also be interested as I was in making graphs for resistance load vs
sec volts, resistance load vs pri amps, pri VA vs sec VA, etc. Then add
capacitors to the loads and you will become an expert on NST operation with
Tesla coils.

John Couture

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-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Sunday, October 05, 2003 11:33 AM
To: tesla-at-pupman-dot-com
Subject: NST power rating -- another perspective

A different perspective:

I have a 4500 Vrms 22ma transformer (100VA).  The primary resistance is
17W.  The secondary resistance is 22.1KW.

The issue of power transfer, I believe, can better be understood by
starting from a simple model and progressing from there.  I will thevenize
the transformer to start.

First the turns ratio is 4500/120 or 37.5.  The open circuit output voltage
is 4500 Vrms and the short circuit output current is 22ma.  The thevenin
output impedance is 4500Vrms/22ma or 205KW.  I will transform the primary
resistance to the secondary 17W * n^^2 or 23.9KW and add this to the
secondary resistance of 22.1KW for a total thevenin resistance of
46KW.  The primary leakage flux has already been accounted for by the short
circuit current measurement.  Next, I will decompose the 205KW into its
reactive and resistive component. Xl will be 200KW (or 530 heneries) and Rs
is 46KW.  Therefore, the simplified NST model will be:

Vs = 4500 Vrms
Ls = 530
Rs = 46KW

The next step is to add a linear load (a pure ideal C with no esr).  I will
not (at this time) add a sparkgap nor a load resister.  I will pick C to
resonate with Ls at 60 hertz.  Xc and Xl will cancel and the current will
be limited only by Rs.  The output current will be 4500V/46KW or ~100ma
(note much greater than the 22ma rating).  Question: what is the power
across C.  Answer:  Zero real power, it is all reactive power.  At peak
voltage, there is zero current.  At zero voltage, there is peak
current.  If one were to calculate the instantaneous power vs time, it
would go positive and negative and the average power would be zero. Now,
lets get maximum real power transfer.

Realizing in a linear circuit, the only time one can get real power is with
resistance (I^^2 R). The maximum real power transfer will be with Rl
matched to Rs.  If we add Rl (=Rs) in series with C, the reactances will
again cancel and the power transfered into Rl will be 1/2 * V^^2 / (Rs +
Rl).  For this case, the power into Rl is 110 watts (higher that the VA
rating of the transformer because we are running at resonance - not what it
was spec'd for).

Now comes the complexity.  We add an ideal sparkgap that we can control
"the when" and "how long" it fires.  We remove Rl and neglect the TC
primary inductance for charging purposes.  The only components for this
consideration is Vs, Ls, Rs, C, and the sparkgap (standard topology).  The
TC primary in reality, will control the discharge rate of C and affect the
energy transfer time.

This is a non linear circuit that often results in a lot of confusion
(myself included so don't fret).  First, we realize that a charging
interval (at 60 Hz) is 8.3ms, so we will fire the ideal sparkgap every
8.3ms when Vc reaches peak.  For my coil, the energy transfer time is 18us
(to 1st primary notch) and we realize this is much less than a percent of
the charging time.  Lets assume all of the 1/2 CV^^2 energy goes thru the
sparkgap and into the secondary never to be seen again.  Otherwords, lets
open the sparkgap 18us after it fires.  Now we have a pseudo linear circiut
that would normally not have any real power delivered (remember we removed
the R in the load) but now we are removing real energy at a rate of 1/2
CV^^2 times the break rate (BPS).  This real energy transfer rate is REAL
POWER and has to be replaced by the charging circuit.  The hard question is
how much is this REAL POWER, how do you optimize for it, and what other
constraints do you need to consider (like not overvolting the transformer
at resonance).  I'm still learning the answer to this but believe the best
way is by simulation.  I DO believe the real metrix for spark length is the
REAL POWER transfered thru the sparkgap and not the VA rating for the
transformer (at least NST types).  This REAL POWER will certainly be
porportional to VA (everything else being the effectively the same), will
vary with Cp, the sparkgap setting, and the resultant BPS (static
gaps).  It will probably be very close to the POWER measured in the line
cord using a WATT meter (I^^2 R loses in the transformer would need to be
factored out).  The actual line cord VA could be significantly larger than
the VA rating of the transformer and will ultimately depend on your chosen
operating point.

Hope this adds some clarity to a very muddy subject.

Gerry R
Ft Collins, CO

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