PhysicsProf Steven E. Jones circuit shows 8x overunity ?

[futuristic]

Here is some more info about this: http://freeenergylt.narod2.ru/vladimir_pantiuhov/

[gyulasun]

I don't know Russian but what i understood from the video was that, he first shunted the capacitor on the left then he charged the capacitor using the battery and left the circuit to run alone without battery. As he does this, he shows the waveform of the pulses on the scope so that we can see its amplitude. When the circuit starts running alone, the amplitude decreases (maybe because the capacitor on the right is charging) and then increases again and stabilizes.

If my understanding is correct and there is no trick, then this should mean the circuit is self running with at least n = 2 since capacitor charging dissipates 50% of the energy by joule effect.

This seems to be an interesting circuit to compare with. It uses a separated coil to collect the energy and with less turns than the primary coil so that the pulses have lower amplitude and greater current. This should make it easier to feedback. The 'collecting coil' also has a middle connection that allows full wave rectifying of the pulses with only 2 diodes instead of 4 which means less voltage drop (0.7V instead of 1.4V).

Regards,
Jaime

Hi Jaime,

Thanks for your answer. I fully agree with your observations as happening in the video but let me quote this from you:

"If my understanding is correct and there is no trick, then this should mean the circuit is self running with at least n = 2 since capacitor charging dissipates 50% of the energy by joule effect."

While I believe there is no trick involved I think we have to elaborate on what your n=2 could exactly mean. Is it a COP of two?  (COP= coefficience of performance when you compare total output power to the input power you furnished in)  But if COP=2 then it should mean the puffer capacitor would not be discharged ever, right?  [Years ago I found about a half an hour run for such blocking oscillator (it was from Naudin TEP circuits if I recall correctly), then the 4700uF puffer cap gradually got discharged.]

Or we should introduce a so called temporary COP term where such circuits like this extend the total discharge time for the puffer cap when their ouput power is fed back to the input as an additional supply voltage?  Because this is what I believe happens here: run time is extended and your n=2 may mean run time doubles? 

So it is ok that this circuit self-runs but the big question is: for how long? for hours? days or weeks? more? 

By definition a COP of 2 means the run time is theoritically endless (in practice it boils down to the first component failure which can be many months or years).

This is why I think a temporary COP ought to be established lol to characterize such circuits, assuming of course what I think that this circuit stoppes working after some definite time like half an hour, an hour etc.

Thanks,  Gyula

[jmmac]

Hi Gyula,

About the 50% energy dissipation when charging a capacitor, it happens when you charge a capacitor using a resistor, for example. I'm not sure anymore if it happens in this case. Maybe someone else can say something about this.
If there is no energy loss, then my conclusion is wrong!

n (COP) is a ratio between the energy that enters and leaves the system by unit time (or power). In this case there is feedback and in order to maintain the system going forever, n must be >1. Since i was considering a 50% loss in the capacitor charging, the system had to compensate that with at least n=2 (EDIT: Here I was not counting the capacitor as belonging to the "system").

I don't have much experience with this kind of circuits but i'd guess the energy dissipation in the diodes alone, would be sufficient to lower the capacitor voltage and the amplitude of the oscillations in a small amount of time. Maybe i'm wrong.

Regards,
Jaime

Hi Jaime,

Thanks for your answer. I fully agree with your observations as happening in the video but let me quote this from you:

"If my understanding is correct and there is no trick, then this should mean the circuit is self running with at least n = 2 since capacitor charging dissipates 50% of the energy by joule effect."

While I believe there is no trick involved I think we have to elaborate on what your n=2 could exactly mean. Is it a COP of two?  (COP= coefficience of performance when you compare total output power to the input power you furnished in)  But if COP=2 then it should mean the puffer capacitor would not be discharged ever, right?  [Years ago I found about a half an hour run for such blocking oscillator (it was from Naudin TEP circuits if I recall correctly), then the 4700uF puffer cap gradually got discharged.]

Or we should introduce a so called temporary COP term where such circuits like this extend the total discharge time for the puffer cap when their ouput power is fed back to the input as an additional supply voltage?  Because this is what I believe happens here: run time is extended and your n=2 may mean run time doubles? 

So it is ok that this circuit self-runs but the big question is: for how long? for hours? days or weeks? more? 

By definition a COP of 2 means the run time is theoritically endless (in practice it boils down to the first component failure which can be many months or years).

This is why I think a temporary COP ought to be established lol to characterize such circuits, assuming of course what I think that this circuit stoppes working after some definite time like half an hour, an hour etc.

Thanks,  Gyula

[yfree]

@jmmac, @gyulasun,

My understanding of Russian is limited, self-taught.
jmmac, you are correct in understanding the video. The experiment has it's own thread. It starts somewhere here: http://www.001-lab.com/001lab/index.php?topic=1056.2800  , unfortunately it is in Russian. You will notice there that the schematic was updated with the capacitor in parallel with the collector coil. This tunes the ringing of the oscillator to the natural frequency of the ferrite. Somewhere in the thread he, Tiger2007, explains how to identify this natural frequency: a coil is wound on the ferrite and driven with the square-wave, the ringing occurring during the transitions of the waveform is at the natural frequency of the ferrite.

Best regards,

yfree

[gyulasun]

Hi Gyula,

About the 50% energy dissipation when charging a capacitor, it happens when you charge a capacitor using a resistor, for example. I'm not sure anymore if it happens in this case. Maybe someone else can say something about this.
If there is no energy loss, then my conclusion is wrong!

n (COP) is a ratio between the energy that enters and leaves the system by unit time (or power). In this case there is feedback and in order to maintain the system going forever, n must be >1. Since i was considering a 50% loss in the capacitor charging, the system had to compensate that with at least n=2 (EDIT: Here I was not counting the capacitor as belonging to the "system").

I don't have much experience with this kind of circuits but i'd guess the energy dissipation in the diodes alone, would be sufficient to lower the capacitor voltage and the amplitude of the oscillations in a small amount of time. Maybe i'm wrong.

Regards,
Jaime

Hi Jaime,

My understanding of this circuit shown in the Russian video with respect to the right hand side capacitor charging is that it is not a direct cap to cap discharge-charge scenario (where the 50% loss occurs if done directly): the left hand side (puffer) cap feeds the circuit and a circuit's component the coil's collapsing field charges up the right hand side cap. So a 50% loss in this charge transfer cannot occur as it does with a direct cap-to-cap setup: the energy comes from collapsing magnetic field when no energy is taken from the input (puffer) capacitor. This is how I think this works.
So there is some energy loss in this process but nowhere near the 50%.

The energy dissipation in the diodes can be minimized by using Germanium types like 1N34A types, also the transistor's saturation voltage could also be minimized by a good switching type bipolar transistor.
Question is still valid: how long can such a selfrunning circuit run? 

Thanks,  Gyula