I like Dumped schematic better, but regardless, Naudin is making a very interesting claim.
I think there is a mathematical principle here with any feedback system. You take a fraction of the output power and feed it back to the input. Thus, you've lowered the input requirement. Easy! But remember, not all the input power goes to a load, just loops around back to the input, so it is deceiving. In fact, if he turns off the lights, he can have maybe 99% of the power input to the cooker be recycled and only 1% coming in from the grid to offset losses. Do you see what I'm saying. Looping the power back like that proves nothing, it is just an extra unnecessary complication! So I must conclude that it is not over unity, because of the power meters, therefore, we must now ascertain how much the load is consuming. And measuring the power pulses is hard and inaccurate, but I'll give it a try.
edit: I added the second picture from Naudin, where he shows the power measured. Well judging from the current and voltage waveform he shows on his website, they are not perfect sinusoids, one side seems clipped, and as we know, the average power in a sinusoidal waveform feeding a resistive load is 1/2 V I, where V and I are the amplitude values, and he did not take this into account for this graph, so there is 1/2 penalty there that we need to correct, then because the waveform amplitude is modulated by 50 Hz, that's another 1/2 we need to take off, so his time averaged power is closer to 1000 watts not 4000 watts peak, however, because the waveforms are clipped on one side as I mentioned, I would take off another 30%, so that puts us at around 700 watts, which is what the input is from the grid.
So no extra power, just a show for the naive.
PS, I added the third image to bring the point home about recycling the power and why higher power to the cooker, while lower power is input from the grid, is really not anything special at all and certainly does not mean overunity operation.
« Last Edit: 2013-01-27, 01:37:57 by EMdevices »