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Overunity Machines Forum



Joule Thief

Started by Pirate88179, November 20, 2008, 03:07:58 AM

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xee2

#15245
September 05, 2011, 02:48:29 AM

Joule thief powered by LEDs used as solar cells >> http://www.youtube.com/watch?v=STTenDq4_ws

rensseak

#15246
September 05, 2011, 04:20:32 AM
Quote from: xee2 on September 05, 2011, 02:48:29 AM
Joule thief powered by LEDs used as solar cells >> http://www.youtube.com/watch?v=STTenDq4_ws

Did you try also to focus the sun light beam with a lens?

xee2

#15247
September 05, 2011, 09:18:57 AM
Quote from: rensseak on September 05, 2011, 04:20:32 AM
Did you try also to focus the sun light beam with a lens?

No. This was just to see if it would work. It takes several minutes to start when placed in sunlight since the large 10,000 uF capacitor needs to charge up to at least 0.5 volts before Joule thief starts working. But then it will run as long as the LEDs are in direct sunlight. Would probably be better to use a smaller value capacitor.



freepow

#15248
September 05, 2011, 09:30:57 AM
@ anyone,    I did an experiment lighting a LED with used AA batterys, look at attached file called "current"

You probably know this anyhow, but I measured exactly 0.173 mA coming out of the LED,
and exactly 0.173 mA going into the LED, so I ask you... how much current is used up in doing the work of lighting the LED ????
The answer therefore should really be none !!!
The battery would get run down even If the LED was to be left out of the circuit, so why could'nt we after lighting that LED then use that 0.173 mA to go directly into a charge battery at the same time ????

Am I making sense or not ?

xee2

#15249
September 05, 2011, 03:20:34 PM
Quote from: freepow on September 05, 2011, 09:30:57 AM
@ anyone,    I did an experiment lighting a LED with used AA batterys, look at attached file called "current"

You probably know this anyhow, but I measured exactly 0.173 mA coming out of the LED,
and exactly 0.173 mA going into the LED, so I ask you... how much current is used up in doing the work of lighting the LED ????
The answer therefore should really be none !!!
The battery would get run down even If the LED was to be left out of the circuit, so why could'nt we after lighting that LED then use that 0.173 mA to go directly into a charge battery at the same time ????

Am I making sense or not ?

Current is the same everywhere in loop between positive and negative battery terminals. Look up Kirchhoff law.

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