RE: NST power rating -- another perspective

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

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


Gerry -

It appears I misunderstood what you were trying to measure, something I am
prone to do. You mention maximum power transfer and the Thevenin theorem. I
have used the Thevenin theorem only for  bridge problems. What does (^^2)
mean? I cannot follow your Thevenin equations but maybe Antonio, Paul N, or
some other coiler can help.

It is obvious I did not make myself very clear when I said the maximum watt
seconds from a NST would be about 25% of the nameplate rating. Refer to my
reply to John Freau. My NST tests were not with Tesla coil type loads and
were to only show what a NST would do if a series of resistor loads were
attached to the NST. That is, the tests would show that a curve made from
the data is a humped curved with a maximum at about 25% of the NST nameplate
watt rating. If instead of resistors different types of TC loads are
attached to the NST the operation gets pretty complicated. I also made these
types of RC tests which produced a set of interesting curves.

John Couture

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


-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Tuesday, October 14, 2003 2:26 AM
To: tesla-at-pupman-dot-com
Subject: Re: NST power rating -- another perspective


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

Hi John,

  > Original poster: "John H. Couture" <couturejh-at-mgte-dot-com>
  >
  > The best way to show the TC "power" relationships is to draw the VA,
Watts,
  > and RVA values as a right (Pythagorean) triangle. The watts as the x axis
  > and the RVA as the y axis. With this triangle you can easily show how a
VA
  > can be greater than Watts. The secret is in the power factor.

Agreed

  > In your tests you should use 100 Watts instead of the 100 VA, etc.
Watt
  > seconds is energy and work done while VA seconds includes circulating
energy
  > that is returned to the source by the reactive components. To do proper
  > input metering you need a voltmeter, ammeter and wattmeter. This will
also
  > give you the power factor as a decimal
  >    PF = watts/VA

I agree, if this was what I was trying to measure.

  > The TC input circuit is similar to the NST test with a load that includes
a
  > resistance and a capacitor.

I also agree, except the resister in the TC circuit during charging is the
esr of the Cap and any wiring resistance (negligable compared to NST wiring
resistance) and these are values that we want to be as low as possible
(ideally equal to zero since they result in losses during high current
discharges).

In my experiment, I wasn't trying to correlate power to spark length.   THis
measurement would be necessary to correlate to John Freau's equation.  My
guess, based on simulation results from others, is that my PF is around .5
with the static spark gap and the value of Cp that worked best.  This also
seems to correlate with my measured spark lengths.  The 100VA resulted from
the measured output of the xformer [Vs(oc) x Is(ss)] and not measured in the
TC environment. My xformer was bought over 40 years ago and has no name
plate and I was only mentioning this for reference (could that be part of
the disconnect).


  > When a coiler tests a NST with different resistance loads he will find
that
  > the maximum wattage output that he will get from the NST will be about
25%
  > of the nameplate rating on the NST.

This could very well be for a pure resistive load, but for me this is not an
interesting case. The voltage transfer function for an L R RL circuit (where
L and R represent thevenin parameters of the NST and RL is the load
resister) is:

VL = Vsec(oc) [ RL/(sL + R + RL)]

the current in RL is:

   IL = Vsec(oc) / (sL + R + RL)

the power across RL is:

PL = Vsec(oc)^^2  x   RL / [(sL)^^2 + sL(R + RL) + (R + RL)^^2]

I'll let you solve this for your RL or set the derivative to zero and find
the optimum RL for maximum power transfer.  As for me, I prefer to focus on
a capacitive load since this is the TC application.  With an L R Cp RL
circuit,  the power transfered will be different than 25% of VA.  You could
even choose Cp = Cres and RL = R (with no spark gap) to get more power (and
I really mean watts) across RL than the VA rating of the transformer
(measured VA, of course, will always be equal to or higher than the power
draw).  I'm still not interested in this case since we want RL to be as low
as possible and not try to match it to the transformers R for TC's.

If you model an L R Cp circuit (again with no sparkgap) the only power
dissapated is in the R of the transformer.  When you add the sparkgap to
this circuit, then real energy is removed from the charging circuit making
its way to the secondary.

I really believe that we are mostly in agreement on these things.  I hope
you view it that way too.

Gerry R
Ft Collins, CO