Thank you for the response. I'm thinking about load matching also.

Where the video, you're talking about, can be downloaded from?

And one more question: does anybody have an idea why such a huge
secondary voltage is required (TB in the patent claims that it can be
lowered with smaller secondary windings, but I didn't see anything
done with low secondary voltage)?

--- In MEG_builders@yahoogroups.com, Kent Andersen <sci@...> wrote:
>
> 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.
> >
> >
>