So far in all of Tom's, CMB's, and my diode arrays produce ~ 10pA, hence formulation of my diode equation, Vout = Rzt * 10pA.
Here's another equation to calculate the voltage charge across a capacitor each day. The total seconds per day is 60 * 60 * 24 = 86400 seconds.
The coulomb equation is C = A * s. So the increase in charge per day for a 1uF capacitor is 10pA * 86400 s = 8.64e-7 C.
The coulomb voltage equation is V = C / F. So the increase in voltage across a 1uF capacitor per day is thus 8.64e-7 C / 1.0uF = 0.86 volts. Of course, if it's loaded down to 100Mohms, then the maximum charged voltage would be 10pA * 100Mohms = 1mV. To achieve 2 volt requires 200Gohms. It could take the mylar capacitor some time to obtain 200Gohm parallel resistance. For this reason, if you want to obtain at least 2 volts charge on the 1uF mylar capacitor then it's best to have very clean components, no hand prints, dust, etc., which could decrease the parallel resistance below 200Gohms. You can use a cotton ear swab to clear everything. Once the 1uF mylar is charged to 2 volts, then it's not a big deal. Just pick it up in a relatively dark room, appropriately connect it to a red LED, and you'll definitely see a bright burst of light, if it doesn't pop the LED from initial excess current. :-)
Created on 2009-04-26 23:38:00 by EnergyMover
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