The light bulb would seem to be experiencing about 0.7 amp RMS of current in a part designed for 0.25 Amp, the math trace
shows about 11 VA applied and 7 VA returned leaving about 4 VA consumed or 4 Watts output. But the math trace doesn't
seem to jive with the calculated power.

EDIT: Mistake on the input power it was actually 12.7 volts at 0.4 Amp input or 5 Watts. 4 / 5 = .8 or 80% efficiency.

Yeah this setup can slay a few fallacies I think, there is the reduced input under load effect, the load not reflecting to the primary
effect, it can demonstrate some reactive power issues that get misunderstood, it can show resonant activity in the tanks that is
no where near 90 degree shift between voltage and current and it can show in phase inductive loads and out of phase inductive
loads.

I think I can get up to 77 % real efficiency end to end with some loads. Or very low efficiency when caused.

I think the output and input impedance can be varied for different loads.

I think I will apply the "Thevenin's theorum" test to try to determine the maximum and minimum output impedance.

I'll need to reduce the coupling so that the output short circuit doesn't max out my power supply or I can use a battery, but the
pulsing circuit is frequency sensitive to supply voltage so the circuit needs a regulated supply to stay stable on the right
frequency as well as a supply of 12.5 volts or so that's unrestricted for the primary coil. If it's an issue depends on if the
frequency rises with a reducing voltage or drops with a reducing voltage and how much it varies with 0.1 volt supply change.

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Post edited to rectify reported input power value.

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OK here's the result of a "moderate" input power level of about 12.5 volts and 0.09 A input open circuit
and 12.5 volts and 0.5 A input shorted, resulting in the following output conditions respectively from left to right.

So the Thevenin resistance Rth is the open circuit voltage Vab divided by the short circuit current Iab. Or - Vth / Ith = Rth .

Vth - 7.86 / Ith - 0.363 = Rth - 21.6 Ohms.

It can vary quite a bit, and with a small adjustment it maxes out the 1 amp current meter. So I think it can go well below 10 Ohms.

I wonder what will happen when I short it with a low resistance high inductance coil that has a core. 

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Here's some Lissajous patterns resulting from the 1:2 harmonics causing a double frequency on the HV tank while the LV tank
remains at about 300 kHz, the bottom right scope shot shows the LV tank when continued reduction of capacitance on the HV tank
begins to force the LV tank into a similar but less than perfect double frequency of almost 600 kHz. The HV double frequency is fairly
even voltage but low current.

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Quote from: Farmhand on August 06, 2014, 11:41:26 PM
Here's a shot of the tank voltage and the current in the ground line (voltage across a 0.8 Ohm resistor in ground line) the
elevated plate is about 1500 mm x 300 mm x 4 mm thick. I think it has some tens of pF capacitance, maybe 100 pF just going by the extra capacitance I need without the antenna connected. Maybe I should put a 100 pF capacitor in series with the antenna.

Not sure how to measure the voltage current and phase in the antenna wire itself.
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HI,
I don't find a formula for your elevated plate, but what import is the plate area and his height .
This capacitor have 3 components
  a) Capacitor plate relative to the Univers, C1.
  b) Capacitor plate relative to the Earth, C2.
  c) Capacitor of the Earth relative to the Univers, C3. (760uF, http://web.mit.edu/sahughes/www/8.022/lec06.pdf)
So your Plate capacitor, C, is the sum of C2+ C1 in serial with C3.
  C=C2+C1*C3/(C1+C3)  and if C3>> C1   C=C2+C1

Here is a formula for a disc in space, http://en.wikipedia.org/wiki/Capacitance , find "Circular disc".
       C= 8 * epsilon0 * a,   with epsilon0=8.84e-12 and a=rayon  in m
Your plate has an surface of 1.5*0.3=0.45m² and a disc of same surface has  a rayon r=sqrt(A/Pi) = sqrt(0.45/Pi) = 0.378m
so for  a=0.378m --> C1=8*8,84e-12*0,378 pF=26,7 pF
You must add the Plate to earth  capacitor C2=epsilon* A/d  with A=Area(m²) and d=disque to earth length in m.
C2=8,84e-12*1.5*0.3/d   but I don't know D!
  - if d=5m then C2=79.56pf
As C3>> C1  (760uF>>26.7pF) , your capacitor plate is
   C=C1+C2=79.6pf+26,7 pf=106.3pF
Don"t forget, this is a raw calculus but it's approximately your conclusion!

This is an error...