PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[JouleSeeker]

Yes. See this example > http://www.youtube.com/watch?v=smOiVmKv9f8

Note that since the output is pulses, a capacitor is needed to sum the energy of the pulses. And a diode is needed to prevent capacitor discharging back through transistor when it turns on. Each pulse adds energy to the capacitor and increases its voltage until the capacitor input power and output power are equal.

If the Joule thief output is AC (not normal for a Joule thief), then a full wave rectifier should be used instead of the diode.

If you will look at the input Power waveform from the Tek 3032 (in my early posts), you will see that the instantaneous Pin has a large AC component.  The Tek provides V(t) * I(t), then over numerous cycles will calculate the mean input power.  I trust this method much more than  using a DVM to measure I (meter) * V (battery), given the AC component in the input power.

Indeed, measuring the input power is challenging, given the remarkable AC component observed... hence the suggestion to use a cap for the input in lieu of a battery.  Or a calorimeter for the measurements.

I'm looking for a capacitor that does not "leak" appreciably, something in the 0.5 F range would be great.

[xee2]

If you will look at the input Power waveform from the Tek 3032 (in my early posts), you will see that the instantaneous Pin has a large AC component.  The Tek provides V(t) * I(t), then over numerous cycles will calculate the mean input power.  I trust this method much more than  using a DVM to measure I (meter) * V (battery), given the AC component in the input power.

Indeed, measuring the input power is challenging, given the remarkable AC component observed... hence the suggestion to use a cap for the input in lieu of a battery.  Or a calorimeter for the measurements.

I'm looking for a capacitor that does not "leak" appreciably, something in the 0.5 F range would be great.

Yes this is true. I originally put a large capacitor between ground and the output of the amp meter to filter the noise. But I have found that all of my digital meters do a good job of computing the average current even with these pulses so I stopped adding the capacitor. All of my meters seem to give about the same reading without the capacitor as with it. However, this may not be true for all meters, so adding a large capacitor may help in some cases. I am also assuming that there is generally no need to get a super-exact measurement in order to determine if over unit exits, I have yet to find a device that was even close.

If your Tek meter is giving you 8 times more output power than in I think you should also try another method of testing since that is rather suspicious. I am sure that is what you told you students when they came up with questionable results.

I hope I do not seem too negative, I really do have an open mind. That is why I examine devices claiming over unity. However, after so many claims turn out to be wrong I guess I do get a bit suspicious when someone claims such a large COP.

[NerzhDishual]

Hi OU blokes,

I'm far from a skeptic, just the contrary.

OK. OK! F' measurements! The TEK gives OU ! God save the TEK.

My intuition (and also some experiments) tells me that all these kinda JT CCTs are, indeed, more or less, 'OU'.

But 'intuition' is not scientific. Is it?

My experiments was about a non charge conservation anomaly.
For ex: a replication of : http://jnaudin.free.fr/html/tepcoil.htm
Yes, there is non charge conservation if you use a kinda JT CCT.
A non charge conservation but not 'OU'. (1/2 * C * V *V - wise).

You know what?
IMHO, Nature does not like to be disturbed and "over" reacts.
This appends when you use square waves, resonance. and other shenanigans.
Of course, if you dismiss the existence of Aether you should experiment some psychological trouble.
Just IMHO.

Very Best

[xee2]

I'm far from a skeptic, just the contrary.

There is a lot of good evidence that T. Henry Moray had a working OU device. You may want to read up on him.

[poynt99]

If you will look at the input Power waveform from the Tek 3032 (in my early posts), you will see that the instantaneous Pin has a large AC component.  The Tek provides V(t) * I(t), then over numerous cycles will calculate the mean input power.  I trust this method much more than  using a DVM to measure I (meter) * V (battery), given the AC component in the input power.

Indeed, measuring the input power is challenging, given the remarkable AC component observed... hence the suggestion to use a cap for the input in lieu of a battery.  Or a calorimeter for the measurements.

I'm looking for a capacitor that does not "leak" appreciably, something in the 0.5 F range would be great.

I urge you to reconsider Professor.

When dealing with DC power sources, heavy averaging of both the battery voltage and current signals is the most reliable way to measure input power. You simply multiply the two DMM values together (taking the CSR value into account), and the result is an accurate net average INPUT power measurement.

.99