It just occurred to me that isolating the diodes or piezos from AC electrical noise (Johnson noise) is probably not going to work in terms of getting past the 10pA level. The parallel capacitance of my piezo is ~ 30nF. And the resistance of this piezo at such low current levels is in the giga ohm region. Therefore, the parallel capacitance is sufficient to short Johnson noise down to at least 0.01 Hz! So it does not seem to be related to AC electrical noise.
Obviously two piezos that are not connected to each other will *each* produce 10pA. IOW, they will not effect each other. So if it's only due to electrical isolation, then it must be DC current isolation. Hmm, actually that's easy to test! :-D Just place a low leakage capacitor between (not across) each element. This will block DC current between each element. This is actually based on an old theory of mine, where the DC voltage fluctuates between elements over time, and this alone might disturb each other to some degree and prevent the net current from going beyond 10pA DC. I don't know, it's an old theory that I no longer care for. Anyhow, if it's true that each element needs to be electrically DC isolated, then it will be unfortunate, as it could be difficult to isolate each component DC wise and still collect the 10pA from all parts.
So this is tested by placing a low leakage capacitor between each element. Of course you can't get more than 10pA in totality, but each element hopefully will still produce it's 10pA of current. If it does not, then it's not AC or DC electrical current. If that turns out to be the case, then what is it? I mean, surely they can still be in the same room and each produce 10pA.
Created on 2009-10-27 17:09:27 by EnergyMover
10pA effect, FE diodes, FE Misc devices, FE piezos, Free energy, Free energy devices, Homemade diodes, Science, 10pA, Diode, Free energy, homemade diodes, Piezo