Is joule thief circuit gets overunity?

[Kator01]

Condradelektro,

welcome.. thank you for listening and understanding. The time for misleading people with unproved claims is over. There has been time enough to learn the basics.

@tinselkoala: thank you, very well presented. The loop of the clip including the wire from the resistor add up at least to 20 nH. And yes, it is known by some electronic specialists that in high speed switching the ground-wire ( eg. source-connection of a mosfet to ground) poses the problem of high frequnecy ringing. I am searching for a document of a friend of mine who did a lecture on this topic

Regards

Kator01

[Kator01]

Now , here we are

these neagtive ringing even reflects back and causes an additional voltage-peak in a drain-inductor !
This was the reason, this investigation was presented

Regards

Kator01

[xee2]

Battery voltage increase does not prove that battery is charging. Put a capacitor in parallel with battery and then disconnect the battery. If circuit is over unity it will charge the capacitor. If the circuit is under unity the capacitor will discharge. The rate of discharge shows how close the circuit is to unity. See for expample: http://www.youtube.com/watch?v=8T9HQkDnIuU

[MileHigh]

Put a capacitor in parallel with battery and then disconnect the battery. If circuit is over unity it will charge the capacitor. If the circuit is under unity the capacitor will discharge. The rate of discharge shows how close the circuit is to unity.

This does not apply to a Joule Thief.  Be very careful about making generalized statements.  A Joule Thief does not return any power to the supply battery at all.  Like I already posted, the measurement for any possible over unity in the case of a Joule Thief circuit is typically done by comparing the average power drain from the supply battery with the average power output to the LED.

[totoalas]

in my joule ringer circuit,
the wire from the collector is looped to a ring magnet thru an output ( open circuit winding) of the transformer then to a diode back to the supply battery positive....  As I decrease the load current the input battery voltage increase...  so with no load, the extra charge from the coil increased the input battery but for a certain time until it lowers back by 0.2 highere than without the bypass....  so in my case   just an increase of 0.2 volts
thanks xee2   for the simple test set up for ou  at least we have some benchmark to see .....
totoalas