Bifilar pancake coil overunity experiment

[gyulasun]

Hi F6FLT,

I think the 'poles' of the fields are so close to each other, both for the case of the bifilar at 28 kHz or for the one turn coaxial loop, that they (i.e. the poles) are able to close themselves magnetically so that little stray fields remain to the outside world. The case is similar to closed magnetic circuits: if you have two identically sized horse shoe magnets and you let their facing unlike poles join to each other (to form a kind of horse race track), then the strong fields near the earlier unclosed ends reduce to much weaker fields.
The electric field for the case of the coaxial cable loop should behave similarly: the shield is connected to the generator BNC body output i.e. to a very low voltage potential and the E field is mainly confined inside the coax dielectric. Very likely the magnetic field is also able to close itself mainly inside the coax.
The E field for the bifilar coil also remains inside its structure, I think.

If you feed a DC current into the bifilar coil, where do you think the static poles would appear?

Would like to ask whether you already had time to read my reply to you on the measurements of coils self capacitance? (reply #145, previous page)

I partially agree with ayeaye in that the power that remains to feed the coil or the coax at resonance is small but still I maintain what I have written above.
Ayeaye wrote this: "All current just goes through it, it is not consumed."
I would say: Any time current is able to flow, it creates losses (in the form of heat and radiation), so it is consumed to a certain degree (worst case is 100%).

Thanks,
Gyula

[ayeaye]

An ignorant thinking perhaps, but is it resonance, or is it frequency above which it just cannot do induction any more? Resonance i guess should be when the input power is the greatest, then i think also the output power is the greatest. Like the length of pulse when it's pulsed, this i think is the only thing that matters when it's pulsed.

[F6FLT]

Hi F6FLT,

I think the 'poles' of the fields are so close to each other, both for the case of the bifilar at 28 kHz or for the one turn coaxial loop, that they (i.e. the poles) are able to close themselves magnetically so that little stray fields remain to the outside world. The case is similar to closed magnetic circuits: if you have two identically sized horse shoe magnets and you let their facing unlike poles join to each other (to form a kind of horse race track), then the strong fields near the earlier unclosed ends reduce to much weaker fields.
The electric field for the case of the coaxial cable loop should behave similarly: the shield is connected to the generator BNC body output i.e. to a very low voltage potential and the E field is mainly confined inside the coax dielectric. Very likely the magnetic field is also able to close itself mainly inside the coax.
The E field for the bifilar coil also remains inside its structure, I think.

Hi Guyla,

I understand what you're saying about the fields that are confined. It's surely one reason, in particular for the electric field. But I just found an additional explanation that I completely missed.
Remember that the input signal drops at resonance from 20v pp to 600 mv (and the current increases proportionally) and nevertheless I observe the same voltage from the open test probe feeling electric fields or with the "looped probe" feeling magnetic fields. From this I conclude that each field, the electric field and magnetic field that the probe detect, don't depend neither on the input voltage alone nor on the input current alone but depend on both. How is another question.

It's funny to see the strong resonance without the least effect on the amplitude but with a strong effect on the phase.
At resonance the signal probe is 90° out of phase with the input signal from the generator. When we change the frequency by only 2.5%, the signal probe becomes in phase or in opposition with the input signal (depending on the frequency lower or higher than the resonant frequency).

If you feed a DC current into the bifilar coil, where do you think the static poles would appear?

I don't understand the question in the context, no current is possible here in DC.

Would like to ask whether you already had time to read my reply to you on the measurements of coils self capacitance? (reply #145, previous page)

I partially agree with ayeaye in that the power that remains to feed the coil or the coax at resonance is small but still I maintain what I have written above.
Ayeaye wrote this: "All current just goes through it, it is not consumed."
I would say: Any time current is able to flow, it creates losses (in the form of heat and radiation), so it is consumed to a certain degree (worst case is 100%).

Thanks,
Gyula

I had read your reply on the measurements but didn't yet take time to study enough the method.

I remain troubled by the question of the probed magnetic field that doesn't increase at resonance while the current is increasing. The current goes the same direction in both wires and the magnetic field is proportional to the current only. So we should have an increase. May be it's a question of phase difference between the wires...
...to be continued

[F6FLT]

Well, when the current increases, and the voltage on the coil decreases, then the coil consumes less energy. All current just goes through it, it is not consumed. It would be weird to expect the magnetic field to increase in that case, it would be weird if it doesn't *decrease*.

It would be interesting to see if you do input and output power calculations at resonance. Does the input power decrease at resonance? Ah yes then, but what means resonance when the input is pulse. Pulse length when the voltage on the coil is minimal?

The "looped" probe detect the magnetic field. V=dΦ/dt and the magnetic flux Φ crossing the probe loop area is proportional to the flux of the bifilar inductance L, i.e. to -Ldi/dt. If i is increasing, V should increase proportionally. It's only a question of current in the bifilar coil, not a question of voltage. The current generating the magnetic field of the bifilar coil is the sum of the current in each wire. These currents go in the same direction. I will try to measure them and to look at their phase relation.

[ayeaye]

At resonance the signal probe is 90° out of phase with the input signal from the generator.

And what does that 90° there mean, as much as i understand it means that then the back-emf is almost equal to the forward emf, means that the net power is zero, though induction and power is likely at its greatest. Now when there is overunity, it should go over 90°, which means that then back-emf generates more power than the forward emf provides. One video shows a bifilar pancake coil charging a battery that provides the power, this is what should happen at overunity. Unfortunately nothing bout the battery voltage is measured there, so not enough evidence for overunity.

How comes, it's when it's sine, power does two cycles during one voltage cycle. It is that power can be both positive and negative. Negative is the power generated by back-emf. So in that case it's seen that the negative power is equal to the positive power. Means the average power is zero.