Re: NST power rating -- another perspective

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Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net> 

Arrrrrrg!   My fonts didn't go thru.   I'll Repost*

<<*The Tesla list only sends out pure ASCII text.  Fancy or HTML fonts and 
special characters will get lost when the fancy stuff gets converted to 
ASCII. - Terry>>

 > Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net>
 >
 > A different perspective:
 >
 > I have a 4500 Vrms 22ma transformer (100VA).  The primary resistance is
 > 17ohms.  The secondary resistance is 22.1Kohms.
 >
 > 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 (n) 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 205Kohms.  I will transform the
primary
 > resistance to the secondary 17ohms * n^^2 or 23.9Kohms and add this to the
 > secondary resistance of 22.1Kohms for a total thevenin resistance of
 > 46Kohms.  The primary leakage flux has already been accounted for by the
short
 > circuit current measurement.  Next, I will decompose the 205Kohms into its
 > reactive and resistive component. Xl is 200Kohms (or 530 heneries) and Rs
 > is 46Kohms.  Therefore, the simplified NST model will be:
 >
 >      Vs = 4500 Vrms
 >      Ls = 530 h
 >      Rs = 46Kohm
 >
 > 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/46Kohms 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 now 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
 >
 >