Is joule thief circuit gets overunity?

[ltseung888]

The fact that the capacitor voltage decreased over time proves that the board is consuming more power than it is returning to the power source (capacitor). Thus your "OU" is due to measurement errors. A capacitor is rechargeable. Any current flowing back into it will recharge it.

Please check your Physics before making such bold statements.....  In particular, check pulsing current into a super capacitor.....

[xee2]

Please check your Physics before making such bold statements.....  In particular, check pulsing current into a super capacitor.....

If you think that this does not apply to your capacitor, then test with a normal capacitor. I have a degree in physics and, if you will check, you will find that pulsing current into capacitors will charge them up. I am only trying to be helpful. You are not making your measurements correctly.

[ltseung888]

http://www.overunityresearch.com/index.php?topic=1516.msg29450#msg29450

Some of you may say that the Output Power with the 2n2222 board is low (in the 0.0x range).  The noise may be a factor.  The measurements cannot be trusted.

The above link shows the result of the 2n3055.  The Average Output Power is in the range of 0.5W.  That should be way above the noise level.

@xee2
There are at least 10 other testers receiving the oscilloscope test-ready boards.  Is it likely that they all make errors???  Let us wait for their results.

*** One tester in Canada just picked up the FLEET package with the oscilloscope test-ready board.  Let us see when the other nine get theirs. 

[xee2]

@ ltseung888
The main problem with your measurements is that you are not computing the power going into the LED correctly. For a resistor the power used is amps through the resistor times voltage across the resistor. However, the LED is not a a resistor so this will not give the correct amount of power. If you do not believe this, then substiture a resistor for the LED in your circuit and you will find that your measurements no longer produce OU. I made the following video to try to explain this. I hope it helps.
https://www.youtube.com/watch?v=bTcQxC46pyw

[MileHigh]

Xee2:

Sorry you made a valiant attempt to cover the issue of the power dissipated in an LED including making a YouTube clip but you made a mistake.

You model the LED as a voltage source in series with a resistor.  You state that the power associated with the voltage source is zero because there is no resistance.  In fact that voltage source represents more power being dissipated in the LED.  Your model for the LED voltage source represents a voltage drop and any voltage drop times current flow means power is being dissipated.  So the voltage drop times the current through that voltage drop is what is also dissipated in the LED.

Total watts used by the LED = (Vled x i ) + (i x R)

If you use a DSO like Lawrence is doing then you can measure the power dissipated in the LED.  It's simply the current measurement that you get from measuring the voltage across a current sensing resistor times the voltage across the LED.  So there is no real difference in measuring the power dissipated in a resistor or an LED, you can use essentially the same technique.

However, there are limitations in what you can do with a DSO because of the sampling rate.  If you are trying to measure narrow spikes that could be a problem.  The DSO because of its sampling rate "sees" narrow spikes as a series of thin rectangles one stacked next to the other.  I don't know if this issue applies to Lawrence's case or not, but one should be aware that a DSO may have serious problems making measurements on very narrow spikes, whether they be voltage spikes or current spikes.

MileHigh