Re: FW: Re: Tesla Coil Efficiency Test

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

Original poster: "Paul Nicholson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <paul-at-abelian.demon.co.uk>

Simply accounting for the coupling is not enough.

John Couture wrote:

> With the pickup coil wire wound tightly around the bottom of
> the secondary coil the coupling would be high, something
> like 0.6 or higher. 

You might be surprised how much wire you'd need to wind in order
to couple that closely to the secondary!   But that's beside
the point anyway.  You could couple the lamp through a transformer
in the coil base lead, for example, or just put the lamp directly
in the base lead.  Either will give you a k=1 coupling.  

But just knowing the coupling is not enough. The two cases above
are both k=1, but the xformer presents the resonator's RF to the
lamp at a lower or higher impedance than the lamp would see when
placed in the base circuit.  Thus a given lamp would show a
different light output in the two cases, despite having the
same k to the same coil at the same power.

What you'd have to calculate at the very least is the reflected
impedance of the lamp (however you couple it), so that it can be
compared with the resonator's equivalent series resistance (ESR),
ie the notional coil base resistance that would by itself account
for the total rate of energy dissipation, including streamer load,
but excluding the lamp load. 

But then having calculated the reflected lamp resistance, you can
see that: if the reflected lamp resistance is small compared to
the coil's ESR, then you'll get a low lamp wattage.  If the lamp
resistance is higher, then you'll get a higher wattage at the lamp.
If you arrange the reflected lamp resistance to equal the average
reflected streamer load resistance, then the lamp wattage will
equal the average watts delivered to the streamer load.

I do see another way to employ a lamp test that might give you
the watts-to-streamer, and therefore the overall efficiency.

It requires two lamp tests, one just below breakout, and another
at the normal operating level.  This gives you a lamp wattage for
the unloaded and loaded conditions respectively. 

If the equivalent coil base resistances are:

  Rc  . . . the resonators total ESR
  Rl  . . . the reflected resistance of the lamp.
  Rs  . . . the reflected resistance of the streamers.

then the firing energy will be dissipated in the ratios of these
resistances. 
   
For the below-breakout lamp test, we measure an input power Pu
and a lamp wattage Wu, related by

  g * Pu = Wu * (Rc + Rl)/Rl,

where g is an unknown factor denoting the supply circuit
efficiency.  For the loaded test, an input power Pl gives a lamp
wattage Wl, and 

  g * Pl = Wl * (Rc + Rl + Rs)/Rl

Dividing these two disposes of the unknown g to give

 Pu * Wl / (Pl * Wu) = (Rc + Rl)/(Rc + Rs + Rl)

and if we make Rl small compared to the other loss resistances
(eg by making sure the streamers don't diminish significantly
when the lamp is inserted) we can approximate 

 Pu * Wl / (Pl * Wu) ~= Rc/(Rc + Rs)

so that 

 RF efficiency =  Rs/(Rc + Rs) ~= 1 - Pu * Wl / (Pl * Wu)

We don't get the overall efficiency here unless both Rc and Rl 
are known so that we can calculate the factor g.  Then the
overall efficiency would be 

 Watts-to-streamer/Supply-watts = g * (1 - Pu * Wl / (Pl * Wu))

One way or another, to get at this overall efficiency, we can't
seem to avoid having to measure a Q factor or an ESR somewhere
along the way.  However, obtaining just the RF efficiency using
only simple equipment like this would be very useful, since we
expect that most of the system loss occurs in the RF parts of
the system. 
--
Paul Nicholson
--