# Tesla free energy collector

I'm trying to see how this circuit works:

This is taken from http://www.instructables.com/id/TESLA-FREE-ENERGY-COLLECTOR/. I tried the circuit with 1n4148 diode and I can see a voltage on the capacitor, meaning that the circuit can collect energy. But I don't understand why I don't see the same behavior when I replace the diodes with 1n5819 schottky diodes. Presumably a schottky diode will have lower drop of voltage and so the capacitor gets charged more. But the capacitor does not hold any voltage in this case. Can someone explain why? P.S. I really don't want to generate free energy, just want to understand how this circuit works.

• Schottky diodes also have higher leakage current – user253751 Mar 12 '18 at 5:11
• @jsotola: With what in mind? The OP understands how the rectifier works and is aware of the difference between the diode types. You can give better guidance than that! – Transistor Mar 12 '18 at 8:42

Measure the voltage across the capacitor. (Presumably your meter has a 10 Mohm impedance).

Now connect a 1 Mohm load across the cap, and note what that does to the voltage.

You can now calculate the current in that 1 Mohm resistor, and (from the voltage across both 1M and 10M loads) the effective source impedance, which will let you calculate the leakage current in those 1N5819s. Compare that with their datasheet spec.

Load the circuit with a resistor equal to that source impedance (see "maximum power transfer theorem") - that should collect the highest power you can get. Calculate that power...

TL/DR : free energy, but not much of it. For fun, how long will it take to pay back the price of 4 1N4148s at \$0.15/kWh?

It appears that you are really trying to understand why using different diodes results in different outcomes.
If you compare the specifications for each type of diode, you most likely will find that their leakage currents (reverse resistance) are different. The one with the higher resistance (low leakage) will not discharge the capacitor, while the one with the smaller resistance (higher leakage) will discharge it.