Is joule thief circuit gets overunity?

[ltseung888]

@ 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

I shall let poynt99 answer your posts.  He already received one of the oscilloscope test ready boards and he has a 4 channel high end oscilloscope.

[ltseung888]

Result from tester for Board 88.  Two Atten 1052CA oscilloscopes were used.  One for Input and one for Output.

COP = -4.29
Average Input Power =-0.02133 watts
Average Output Power = 0.091479 watts
One experimental error was seen on the Input BMP file.  The CH2 (current) Invert function should have been turned ON.  The power of the Oscilloscope Analysis is - such error can be caught....

[poynt99]

@ 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

Go back and re-read the posts I made explaining to you how power is computed by these scope measurements.

It makes absolutely NO DIFFERENCE what type of component is being measured, the computed power will be correct, in a perfect world that is. There are various subtle factors that will skew the measurement making it erroneous, if one is not aware of them and corrects for them.

[totoalas]

red LED says:         2013/02/24 at 09:17What is the purpose of the 0 ohm resistor?  Wouldn't a direct connection be the same?Reply

   
• admin says:         2013/02/24 at 09:45Yes, for measuring the lux, I shorted the 1 ohm resistor so that the LED output would be the actual light output.  For measuring the LED current, I removed the jumper from the 1 ohm resistor.  For each millivolt I measured across the 1 ohm with the DMM, it was the same as 1 milliamp LED current.  After the measurement, the short can be used for full light output or the 1 ohm resistor can be removed and the 0 ohm jumper put in its place.   THIS IS A QUOTE FROM RED LED 

[xee2]

poynt99
The equation "watts = amps x volts" is only valid for resistors in DC circuits. See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepow.html
This is the calculation the scopes are making. It is only valid when measuring power across resistors. If you doubt this, replace the LED with a resistor and you will see that there is no OU.