PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[yssuraxu_697]

BTW it may just be an electromagnetic flywheel. In this case it is no problem to record "OU" in the flywheeling part but attempts to extract from there at greater rate than input will fail. At least when attempting to extract exact same form of energy.
Unless there is "unconventional" input from material itself in transistor, core or cap.

[Hoppy]

Hoppy,

If you eliminate the 1 Ohm resistor and keep the 1K load, then you can just measure the voltage drop in the load and calculate the current and power. You should have more accurate values this way.

Jaime

Yes, that's the way I would normally measure but I'm just comparing the two. 

Hoppy

[JouleSeeker]

Glad people are doing replications and tests. 
I just wrote an email to someone beginning a replication that may be of use to others as well:

....very glad you're looking at this little circuit.

I would ask that on your scope you look at the Power waveforms, Pin and Pout, and then tune the circuit as well as you can to minimize Pin.  Pin waveform should appear with a strong AC component, fluctuating around zero.  "Tuning" means adjusting the variable resistors in the circuit -- and the resistor to the transistor base -- so as to try to get the MEAN value of Pin to be close to zero.

The Pout waveform should show spikes of power, which remain on "one side of the zero line" when the Pin waveform is adjusted to average to near-zero.

That's what I've observed, and that's how I have evidence for (not "proof" of) super-efficiency, Pout/Pin > 1.

Thanks for taking a look at this!

Steven

[Montec]

Hello JouleSeeker
Measuring output power using current is one way to measure power. It does not matter whether the current is DC, pulsed DC or AC. The same equation I²*R=P holds true. The trick is to split a current into two equal currents. Taking an output across a load resister and passing it through a FWBR and charging a capacitor will give a maximum voltage across the capacitor. Using a variable resistor across the capacitor you can drain the energy (current) in the capacitor to a steady state voltage reading (across the variable resistor) that equals 0.707 times the max voltage you first measured. This is a half power measurement. The power dissipated by the variable resistor is equal to the power dissipated by the load resistor. The load resistor dissipates power from a non-sinusoidal current and the variable resistor dissipates power from a (nearly) DC current. A larger capacitor will make the DC smother at the expense of a longer measuring time. (Charge and discharge times increase.)


 

[jmmac]

Prof. Jones,

I'm trying to replicate your circuit without success. Can you please give some informations in order to help me?

- What's the voltage drop in your LED (in a dc circuit)?
- Did you use a normal ferrite toroid?
- I don't have 2N2222 transistors. Can you tell me if your circuit works as well with a BC547, BC547A or 2N3904 ?

Thank you. Hope you're having a nice time.
Regards,
Jaime

Glad people are doing replications and tests. 
I just wrote an email to someone beginning a replication that may be of use to others as well:
Steven