Single AA battery to light WHITE LED for long-long time

[hartiberlin]

Try this circuit.
Saves 2 diodes losses and light output doubles.

Regards, Stefan.

@Groundloop,
please let your battery run on it and see, how low the voltage will go.
It will probably stay around 0.5 Volts for a long time and really suck up
environmental energy this way.

@All,
best will be to use a 3 wire parallel bifilar coil on the toroid.
Best would be to drill the 3 wires together before winding it ,
so the coupling between the 3 wires is highest,
like in a Bedini coil.

[Groundloop]

Stefan,

Yes, I will keep the circuit running until the LED goes dim.
I do not know how long time that will take. 
I will try your ideas but on a new circuit.

Groundloop.

[Groundloop]

@All,

Here is my 9VDC version. If I use a fully charged battery then the LED will light full bright.
With a depleted (5,29V) battery then the LED will still light dimly.

The resistor is controlling how much current to use. If you use a lower value then this circuit
will run from any voltage.

One funny thing, with my high Ohm resistor and very little current usage from the battery,
the LED will only light when connected ONE way. If I turn the LED around then it goes black.

Any ideas why this happens?

[EDIT] I changed the 2N3904 to a PN2222A. Much better transistor for this application.

Groundloop.

[Feynman]

@stefan
Thanks for the schematics

@groundloop, stefan, zon, amigo

My vote is for "negative resistor" theory, since it is consistent with QED and the work of Bearden and Bedini.  However, we cannot exclude any possibility at this point, to include hitting resonance with the Earth's magnetic field.  The one problem with Earth field theory is that we are not in the ELF frequency range, so it is difficult for me to see how this extra energy is entering the circuit at tens of kilohertz. But anything is possible.

My personal suspicion is that we are possibly hitting the internal resonance of the coils and toroid, and pulling energy from the local active vacuum which is deposited into the battery and/or capacitors. I suspect the sharp gradient of the switching circuit has something to do with this, but again, this is purely speculation on my part.

[magnetoelastic]

What happens when the battery is removed? 

The LED goes out

Therefore, the battery is providing the power for the LED

No mystery.