- Hi All,
Here is my considered opinion on the internal workings of the MEG.
1. The secondary capacitance plays an important role. The output waveforms
are distorted sine waves, not square waves. If there were no capacitance
the outputs would be square pulses.
2. The circulating currents in the secondary tank circuits have a far
greater role in shuttling the flux back and forth than the primary current.
The usual view that the primary currents steer the flux is somewhat
erroneous (but see later remarks).
3. The main driving action is not from the current switched from the DC
power supply. The main driving action comes from the switching transitors
applying a short circuit to the primary coils.
4. When you short out a coil you apply a very high magnetic impedance in
series with the magnetic circuit at that physical point in the magnetic
circuit (i.e. just under the coil). This acts as a flux blocker. (In
simple terms if the flux tries to change value it creates a current in the
shorted coil to oppose that change). When the shorting transistor is
switched on it has the effect of open circuiting that magnetic limb to any
future flux change (the flux in the limb remaining constant).
5. You can now visualise the alternate shorting of the two primary coils as
acting like a single pole two way magnetic change-over switch. The magnet
is switched to each arm in turn by this method.
6. If you did this without tickling up the tank circuits the result would
be zero. The magnet supplies constant flux to both limbs, the alternate
shorting actions do nothing because there is no change of flux taking place.
7. But if the tank circuits are somehow energised, so that there are AC
flux changes going on, and the shorting action is synchronised with the AC,
then the magnet being switched magnetically from one arm to the other could
support the AC waveform. (This would make an interesting experiment. Don't
feed DC power to the drive circuit, just have the transistors acting as
short circuits. Externally shock excite the resonant secondary tank
circuit, watch the decay time constant (tau=2*Q/omega). Now repeat the
experiment but this time synchronously switch the shorting transitors, see
if the time constant increases to indicate energy being supplied by the
8. The small DC energy supplied by the DC power source is there to overcome
losses in the drive circuit. Note the DC level of the current waveforms in
the Patent and in Bearden's original MEG paper. With a normal push-pull
drive circuit you would expect unidirectional half sine current pulses. The
MEG waveforms are not unidirectional. The current (taken from a DC supply)
is definitely AC. During part of the cycle current is being fed back into
the power source. This is not normal push-pull flux switching action.
9. The point in the AC cycle where the shorting/switching takes place is
probably critical. It is wrong to expect to be able to drive in a square
wave and get the thing to work as most reseachers have found (this will be
more like normal push-pull action). It is probably necessary to control the
switching point with reference to the output waveform, and that requires
some extra (but not too sophisticated) electronic circuitry. Note the
position of the switching transients on the published output waveforms.
That phase delay is a forced situation. You won't get that with normal
10. The timing for the change-over action is also probably critical. Is
there a "make before break" action? Is there a small time gap? There is
plenty of scope for experimentation here.