Is joule thief circuit gets overunity?

[gyulasun]

Hi Neo-X,

It surely depends on the excitation current flowing in the coil but I do not know by heart what relationship governs the vibration amplitude to the current strength when the AC or pulsed current is at the core's mechanical resonant frequency. It needs some studying of the technical literature of the 50s and 60s when ferrites were extensively researched.

Gyula

[xee2]

Out of the 20 boards from Mr. Zhou, 18 showed Average Output Power greater than Average Input Power as measured on the Atten Oscilloscope. 

ltseung888
How is average output power being measured? The scope probably assumes that load is resistive. If the output measurement includes the LED then the scope will not compute the output power correctly (LED does not obey ohms law).

[ltseung888]

ltseung888
How is average output power being measured? The scope probably assumes that load is resistive. If the output measurement includes the LED then the scope will not compute the output power correctly (LED does not obey ohms law).

The Output Power Computation is based on:
Instantaneous Power = Instantaneous Voltage x Instantaneous Current.

It does not matter whether the power is DC, AC or pulsed.  The Output Voltage Measurement is across the load.  The load can be LED, small motor or resistor.  It also includes the 1 ohm resistor.  The Output Current Measurement is across the 1 ohm resistor.  The Output Voltage Waveform can be seen from the top curve.  The Output Current Waveform can be seen from the bottom curve.  The Output power Waveform comes from the analysis of the CSV file.  The Average Output Power value comes from the CSV file.

Thus with the captured CSV file, we do not need to worry about whether the load is non-resistive.  Hope that answers your question.  That is why all the testers have digital oscilloscopes so that they can do such analysis.

[xee2]

The Output Power Computation is based on:
Instantaneous Power = Instantaneous Voltage x Instantaneous Current.

ltseung888
This is only true for a resistive load. When there are diodes in the load the voltage is increased by the diode junctions as well as the current. The power consumed by a diode is current squared times the forward resistance of the junction, not current times voltage across diode.

[ltseung888]

Just received another 20 oscilloscope test-ready boards.  Will do the double and triple checks in Hong Kong.  The boards for the three forums are already in the mail.  According to the Hong Kong post Office - arrive within 10 working days.  Thus independent testers with their oscilloscopes can do the detailed analysis.

Let the experts do their job and post their findings.