Sounds like you have done quite a bit of homework. This seems to be a
typical result from MEG builders
In a video that I watched where bearden goes into detail on the MEG's
operation it is more of a L/C
type of circuit. from what he said in that video the input waveform is
like /`\ ramp up and ramp down.
as you can tell there are some sweet spots between pulse width and
frequency of the drive signal.
since its a L/C circuit you must design a load match transformer that
will allow you to pull power out of it.
simply putting a resistive load across the output will detune the
circuit and throw it out of its "balance"
I have not done this myself yet as I have been short on time but that is
my next step to design a, more or
less a balun or unun to decouple the meg from the effects of the load.

(Kent)

lichtrov wrote:
>
> Hi all!
>
> I'm an EE and was intriqued by MEG because of simple circuitry
> required to operate. I've built my own replication of the MEG - also
> unsuccessfully (I mean I didn't obtain COP > 1). In my controller I
> can independently change both frequency and duty cycle of the gate
> driving voltage. My MEG has taps on both primary and secondary
> windings and I can play with turns ratio. While working without load,
> I obtained output voltage with close to sine waveform and amplitudes
> around reported by Bearden and Naudin. However, after loading the
> output (even lightly with 100kOhm resistor), output voltage decreases
> significantly to a few volts.
>
> It's pretty hard to debug the device since I don't understand how it
> should work. I started to dig into Berden's patent and I have a
> question: does anybody have an idea how Bearden measured the current
> in both primary (Fig 6D) and secondary (Figs 6G,H)?
>
> The question arised because the primary current he described has
> extremely small duty cycle - around 1 microsecond for both rise and
> fall. It supposes duty cycle with around 500 nanoseconds of active
> driving voltage - nothing comparable can be seen in Figs 6A,B. Also I
> tried to build high side current measurement circuit and (even with
> most recent chips!) it has around 500kHz bandwidth and cannot provide
> such a sharp waveform. Thus I presume that Tom measured a voltage on
> a small low side resistor. The voltage produced by such measurement
> can have short spikes as presented in Fig 6D because of capacitive
> coupling from adjacent circuitry and even from the ground ripple
> itself.
>
>