From Farmhand: Mr Clean shows the 1 Farad capacitor Negative connected to the circuit ground in my opinion I believe this is correct, according to the hand drawn schematic from Kurt. P.S. if the system is truly self running the 1 Farad capacitor should increase in charge or stay the same while running the load, This is partially correct, but even if the system is not self running, and the oscillator is starved for current and voltage, the 1F capacitor voltage will be seen to rise and continually power the circuit.... for a while until all soakage is bled out. To get the effect again, Mr. Clean will have to connect the battery to the 1F capacitor for a long time to get original conditions of dielectric absorption before disconnecting. This has all been stated in my earlier posts and can be demonstrated on the bench with even simpler circuitry. 72 Joules is a lot of energy. Discharge a 1F capacitor that has been fully charged for an hour or more. Now watch the voltage climb back, enough to light even a small incandescent lamp for a few seconds. How much easier it is to light a high intensity white led at much lower power for a much longer time. Note that the led's are not being driven to full 3 watts, but as Kurt says on his drawing are "dimly lit". If the 1F comes back to 20 % of it's initially charged value, it can supply over3 Joules of energy, possibly enough to power the circuitry in a starved mode. But if Kurt has not discharged the 1F fully, but has allowed the normal stored energy of 72 Joules to bleed down to the "sweet spot" this scenario is even more realistic. In my opinion there is no free energy here, just a lack of understanding of large electrolytic capacitors and a jumping to conclusion by the builder.
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