Kapanadze Cousin - DALLY FREE ENERGY

[verpies]

The feedback to the 4046 PLL can come from an e-field antenna near the resonating secondary, or from inductively or capacitively coupled signal off of the bottom of the secondary. I've had good results from just a one-turn loop loosely around the very bottom of the secondary.

Got it.
Did you ever try to pick up the standing wave node with a neon bulb and a fiber-fed phototransistor ?  That would be cool!

[verpies]

Probably I'm not in your level of knowledge guys, but I'm doing some tests, a lot to say the truth, and I would like to share something that I noticed in one of my tests. Sorry if it looks like dumb, but to me it was interesting.
https://youtu.be/LM9H4UEa7Uk

This behavior can be explained using capacitive reactance X_C=1/(6.28*f*C)
where:
X_C is your capacitive reactance expressed in Ohms
f is your frequency in expressed Hertz
C is your capacitance between the sheet of aluminum and the cases of capacitors, expressed in Farads. 
This capacitance is non-zero even if the capacitors are 50cm away from the aluminum foil.  Diode junctions have capacitances, too.

[ariovaldo]

This behavior can be explained using capacitive reactance X_C=1/(6.28*f*C)
where:
X_C is your capacitive reactance expressed in Ohms
f is your frequency in expressed Hertz
C is your capacitance between the sheet of aluminum and the cases of capacitors, expressed in Farads. 
This capacitance is non-zero even if the capacitors are 50cm away from the aluminum foil.  Diode junctions have capacitances, too.

Thank you Verpies!

[Dog-One]

That is a good approach if your power amplifier can handle the frequency.

Yes, it can run well up to about 30kHz.  In light of your other comments though, it seems this may be an inferior approach.  I'm going to take another look at the example circuits you posted and see just how far off I am with my original design.

Yes, for maximum power transfer and minimum reverse energy flow, an inductive load should be counterbalanced with matching capacitance and the DC ohms of the load should be equal to DC ohms of the power amplifier's output.  All in accordance with MPTT.

Ah hah!  This is exactly what I thought at first.  Why I deviated from my instincts are unclear.

Thank you Verpies for setting me straight again.

[T-1000]

The current hurdle I'm faced with is the reaction of the toroid transformer when connected to the grenade and induction heater coils.  If you have the resonating capacitors incorrectly set, you get huge spikes that destroy the Class-D amplifier.

This is why there are protecting high voltage Shotky diodes in series to MOSFETs so the BEMF spikes cannot break isolation...

P.S> Turn your skype on when you have a chance.

Cheers!