This is an initial presentation of how I theorize transformer induction takes place based on my charge separation research. I welcome any and all criticism! For those not familiar with my charge separation work, please read the attached pdf "Dielectric Charging via the E-Field".
Referring to the "Transformer Induction" pix, we have a basic ferrite toroid core that has a 3 turn primary and a 1 turn secondary. We now apply a 3v pulse to the primary and we will have a current increase in the primary depending on it's inductance as would be expected. We are not interested in this current but rather the voltage potentials that are present.
By experimental evidence, we will measure the potentials as seen on the primary and secondary points in the drawing with 1V/turn applied to the primary. Each primary wire segment that exists between the top and bottom of the toroid core surfaces will measure 1v for a total of 3v which is equal to the applied voltage to the entire primary. IOW, the remaining wire for each turn around the outside of the core is simply a connection for each said primary segment. We are ignoring the slight voltage drops across these outside connecting wires which will depend on the wire resistance and the mean primary current.
In like consideration, we will measure 1v across the secondary wire segment between the top and bottom core surfaces as shown. The actual voltage in practice will be slightly lower due to the primary resistance and the coupling between the primary and secondary. However, the overall point to be made is that the main voltage potentials exist between the wire segments in the toroid core.
IMO, charge separation occurs in the secondary via the E-Field generated in the primary. This secondary emf then is capable of producing usable power when loaded.
So, what theory supports this action? IMO it is the power flow or Poynting vector designated as S=EXH or IOW, S=E-Field crossed with the H-Field where H= 𝑵i/ 𝒍 . Even with an unloaded secondary, a very small H_Field exists horizontally in the core center due to a very small leakage flux generated by the primary magnetization current. With a loaded secondary this H-Field can become quite large which allows the generation of appreciable energy in the load due to the S flow. The core window area appears to act as a waveguide for the E and H Fields as the primary E-Field appears to be within this core area. Edwards and Saha (paper attached below) promote the use of the Poynting vector to be instrumental to transformer induction however, they show the E-Field extending outside the core window area. I say it doesn't and the only measurable E-Field outside the core window will be that generated by the small voltage drops in the outside connecting wires.
Regards,
Pm
While I disagree with Partzman's analysis I think there is the chance of something good here. My first image below shows the classical interpretation of his circuit where the primary inductance is being charged with current where the rising current is causing a rising field in the core thus inducing voltage both within the primary turns and the secondary turns. I have shown the voltages at a fixed point in time as batteries accounting for what Partzman calls voltage drop. The induction comes from the E field which I also show.
My next image is a huge parallel plate capacitor almost filling the space within the toroidal core. The significant feature of this is the capacitor gets charged without any external current to it, so Partzman is right to focus on this dielectric charging as something unique. Of course the displacment of electric charge within the dielectric is a form of current flow, so the primary does see that, the energy gained in the capacitor comes from the 3V input.
Here is how this feature can change the world. If the capacitor is initially charged to a voltage V by an electrical connection its energy is CV2/2. Disconnect that connection and now do this dielectric charging to double the voltage. The energy supplied via the primary input is CV2/2 since the transformer sees the same C. Total input energy CV2. Output energy stored in the capacitor is 2CV2. COP=2.
It would seem this will work also for conventional connection via a secondary. WHY HAS THIS NOT BEEN DONE? Is my math correct?
Smudge
On further reflection connecting a charged capacitor to a secondary would result in it discharging back into the transformer if there were not already an induced voltage there, so the dielectric charging is simpler to achieve. But it needs the internal construction of the capacitor to match with the inducing E field. That is multi parallel plate electrodes normal to the E field. Cylindrical electrodes won't work. Using a secondary connection to any form of capacitor will work with the added complication of timing the connection within the primary waveform.
Smudge
Quote from: Smudge on 2025.09.06, 08:48:29
Cylindrical electrodes won't work.
Neither will spirally wound capacitors...
Notice that Partzman observes cap's dielectric polarization even without a diode.
This is unusual because the induced voltage is of one polarity when the flux in the core increases and of the other polarity when the flux decreases.
Normally, these voltages*t cancel over the entire cycle and should not leave the cap's dielectric polarized in one direction (charged).
However, if the dielectric is not perfect and exhibits large dielectric hysteresis then the remanent electric field could end up non-zero at the end of the cycle.
Is so, then the voltage observed across the capacitor would be very dependent on the capacitor's technology - especially choice for its dielectric materials..
Quote from: verpies on 2025.09.06, 12:40:16
Neither will spirally wound capacitors...
Bifilar coils will.
Quote from: Centraflow on 2025.09.06, 12:46:37
Bifilar coils will.
...but we are not discussing coils in this subtopic. We are discussing capacitors.
Quote from: verpies on 2025.09.06, 13:21:24
...but we are not discussing coils in this subtopic. We are discussing capacitors.
Exactly that, coils turned into capacitors, charge the core of the transformer is one way, or turn the first coil into a pure plate and not an inductor.
This is the thing nobody seems to see, you have to experiment with this, an new "type" of induction using charge separation. I can asure you it works.
All wound as coils but work as capacitors. By using coils and not plates you change the way the charges separate.
As the capacitance charges, the electric field creates a magnetic field which I have explained before, the curl field, and this changes along the length of the coil in the for of a sine wave.
One coil shorted and the other normal, one gas no inductance and the other inductance and capacitance, this causes the changing wave.
Many years of experimenting until I had it right.
From there you then do charge separation using two supply sources, two shorted coils, and two normal coils.
There is a lot more to it than that as you have to charge up two coils on a common ferrite using these two supplies, electrons from the two creating a common flux in the ferrite, and then discharge this "saturated" ferrite into the coil/capacitors.
You will just have to imagin the circuit as it is not for open source.
Like here
Re: Tariel kapanadze's Energy Generator « Reply #455 on: 2025-09-04, 17:26:07 »
Quote from: partzman on 2025.09.05, 20:58:57
So, what theory supports this action? IMO it is the power flow or Poynting vector designated as S=EXH or IOW, S=E-Field crossed with the H-Field where H= 𝑵i/ 𝒍 . Even with an unloaded secondary, a very small H_Field exists horizontally in the core center due to a very small leakage flux generated by the primary magnetization current. With a loaded secondary this H-Field can become quite large which allows the generation of appreciable energy in the load due to the S flow. The core window area appears to act as a waveguide for the E and H Fields as the primary E-Field appears to be within this core area. Edwards and Saha (paper attached below) promote the use of the Poynting vector to be instrumental to transformer induction however, they show the E-Field extending outside the core window area. I say it doesn't and the only measurable E-Field outside the core window will be that generated by the small voltage drops in the outside connecting wires.
Regards,
Pm
Would this also mean that inserting a cylindrical electret (permanently charged electrostatic gradient) into a toroidal core should result in some miniscule-but-detectable potential or current in a primary/secondary coils on the toroid?
Also, same question but with an electret oscillating into and away from the toroidal core.
Quote from: Hakasays on 2025.09.07, 13:33:32
Would this also mean that inserting a cylindrical electret (permanently charged electrostatic gradient) into a toroidal core should result in some miniscule-but-detectable potential or current in a primary/secondary coils on the toroid?
Excellent question! An electret (charged or not) inserted within the core area will exhibit a potential change nearly equal to the volts/turn of the primary. This potential change will sum with any electret charge already present. The polarity will depend on the connections relative to the primary. If the electret is is loaded or conducting current, this current will be seen by the primary due to the Lenz effect. The electret will lose energy in this case depending on the load's mean current unless the electret was charge saturated. This is another subject altogether!
Quote
Also, same question but with an electret oscillating into and away from the toroidal core.
If I understand your question correctly, in order for any object to be charge separated and remain as such, it must be withing the confines of the toroid's center hole volume. If taken outside or placed outside, no charge separation will result.
Although I have not tried any biological entities, I have not found any object that will not charge separate up to and including non-conductive ferrite.
Pm
I was wrong to say the multi-plate capacitors are OK. In these the stored E field in the dielectric alternates in direction as you move up the stack so not OK for displacement by an external E field. Here is a quick write up of my scheme. May need a dedicated bench.
Smudge
Quote from: Smudge on 2025.09.07, 15:31:56
I was wrong to say the multi-plate capacitors are OK. In these the stored E field in the dielectric alternates in direction as you move up the stack so not OK for displacement by an external E field. Here is a quick write up of my scheme. May need a dedicated bench.
Smudge
What kind of geometry do you think would best highlight the
Quote from: partzman on 2025.09.07, 14:15:14
If I understand your question correctly, in order for any object to be charge separated and remain as such, it must be withing the confines of the toroid's center hole volume. If taken outside or placed outside, no charge separation will result.
I was thinking either a capacitor that could oscillate in+out of the center ring assembly, or a wheel-and-hub arrangement where a ring of dielectric could be spun continuously through it.
What would be the best arrangement to highlight or take advantage of the toroid-charge-separation relationship? Would a a very large diameter toroid, thicker, large volume, stacked capacitor plates, or just higher overall voltages work better?
Is the charge separation throughout the entire center, or is it 'stuck' closer to the surface of the ferromagnetic toroid?
Quote from: Hakasays on 2025.09.08, 02:35:53
What kind of geometry do you think would best highlight Partzman's quote "in order for any object to be charge separated and remain as such, it must be withing the confines of the toroid's center hole volume. If taken outside or placed outside, no charge separation will result."
Partzman thinks no E field exists outside the center hole volume, and that is not the case. The magnetic vector potential A field forms closed lines around the core flux and it is dA/dt that creates the E field. Certainly the A field is stongest inside that hole so the majority of the volts per turn occurs there. You can't use that outside E field in any outside circuit because in any such closed system the voltage integrates to zero. That also makes measuring that outside small E field very difficult. I think you also have to get clear in your mind what is meant by charge separation. The polarisation within a dielectric is a form of charge separation but no charge actually leaves the dielectric. In a capacitor the act of charging it causes charge separation where electrons move through the conductor creating charge separation within the circuit. I suspect Partzman's charge separation is the latter where he is correct, any outside circuit (no part enclosing the core flux) can obtain charge separation within the circuit. I have used the term "dielectric displacement" for the polarisation of the dielectric. In normal use that polarisation is driven by the charge accumulaion or deficit on the electrodes as given by the time integral of the current. The electrode charges create the E field within the dielectric to polarise it. In my paper the dielectric is already polarised and it gets additional polarisation from the magnetically derived E field. The optimum geometry is thus a disc of dielectric with electrodes covering the two flat surfaces.
QuoteI was thinking either a capacitor that could oscillate in+out of the center ring assembly
Without seeing how you connect to the capacitor I cannot really comment. If the circuit moves with the capacitor it doesn't enclose flux then there will be no current flow to make use of.
Quote, or a wheel-and-hub arrangement where a ring of dielectric could be spun continuously through it.
I can't see a continuous ring being useful without some means of utilising the dielectic displacement occuring only at a small region of the ring, in the rotating frame where the ring is stationary this is like a wave moving round the ring. A ferromagnetic band attached to the ring could obtain flux due to that wave passing wave passing through it, so a coil around that moving ring core could obtain induced voltage. But now you need means to connect to that moving coil.
QuoteWhat would be the best arrangement to highlight or take advantage of the toroid-charge-separation relationship? Would a a very large diameter toroid, thicker, large volume, stacked capacitor plates, or just higher overall voltages work better?
Stacked plates will not work as the upper and lower fields from the inner plates point in opposite directions. I have not yet derived the optimum aspect ratio of the disc capacitor but higher capacitance requires larger diameter and thinner disc. Higher voltage demands greater core area and higher frequencies.
QuoteIs the charge separation throughout the entire center, or is it 'stuck' closer to the surface of the ferromagnetic toroid?
The E field is not uniform across the hole, it has a minumum value at the center but it certainly is not a surface phenomenon.
Smudge
Thanks for all the clarifications, Smudge O0
I find the debate concerning simple capacitors interesting.
Some of my prior work with charges and E-fields relates to the work of Viktor Schauberger, Philipp Lenard and the Lenard (waterfall, spray electrification) effect.
Here we are not dealing with a simple plate capacitor but millions of complex fluid ones. For example, we have a single water drop or sphere with a surface area of A=4πr^2 and a given surface charge density. However if our drop is broken into two pieces we have two curved hemispheres 2πr^2(half of the sphere's curved area) plus the now exposed inside parts(the flat circular base of each 1/2 sphere) πr^2. Shown as O = ᗡ + D. As such the surface area increases by 50% and the charge density/E-field decreases. This also applies to two droplets or spheres which merge into one increasing the surface charge density/E-field or ᗡ + D = O.
Of course this is a very basic example. Suppose we had 100 droplets and each could split. The possible combinations of whole and split droplets is then ~1.27 × 10³⁰ or one nonillion, two hundred seventy octillion. This number is about 1.3 million times larger than the number of known stars in the universe. Which explains why no person or super computer has, even remotely, the capacity to predict what routinely happens in nature unless they use gross generalizations.
I like the water droplet example because it puts a new spin on "capacitors". It also helps to tackle the most complex problems which makes the simple ones like plate capacitors seem much easier.
Coincidentally, I found relative humidity or water vapor in the air can have substantial effects on real circuits, more so ones having higher frequency and voltage. Many people forget we live in the real world not an imaginary mathematical construction. In fact, I estimated one HF/HV circuit was losing over 40% efficiency to leakage and RH. How many people do you know who even considered this fact?... the number is very low imo.
AC
Quote from: Smudge on 2025.09.07, 15:31:56
Here is a quick write up of my scheme. May need a dedicated bench.
I have not observed that a plate capacitor placed in the hole of a toroidal core affects the primary current any more than this capacitor in series with the equivalent air capacitor that completes the secondary circuit.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=54067)
Also, there is a problem with the following statement:
Quote from: Energy_Gain_using_Dielectric_Displacement_by_Smudge on 2025.09.07, 15:31:56
Here we show the case for an uncharged capacitor (Figure 3) where the E field has been instantaneously switched off...
...as this will create infinite dA/dt and infinite -E.
Practically, the reversed polarity of -E integrated over time will discharge the capacitor unless some dielectric hysteresis keeps the charge.
Quote from: Smudge on 2025.09.07, 15:31:56
I was wrong to say the multi-plate capacitors are OK. In these the stored E field in the dielectric alternates in direction as you move up the stack so not OK for displacement by an external E field. Here is a quick write up of my scheme. May need a dedicated bench.
Smudge
Smudge,
Thank you for your paper, however I have a few comments.
First, I'm curious to know about the basis for your denying that the electrodes in your Fig 3 would have no measurable potential on their outside surfaces?
I understand your theory of doubling the charge with a pre-biased capacitor as you depict in FIg 4. For example, we charge a cap to 4vdc and then apply 4v/turn on the primary. Our cap now reaches a potential of 8v (due to charge separation) with the proper polarities applied . This new 8v on the cap will remain only as long as the primary is kept at 4v/turn. In fact, this charge separated voltage will follow the v/turn of the primary in magnitude, polarity, phase, and shape. So, in this case if the v/turn is returned to 0v/turn, the potential on the cap will instantly return to 4vdc if no energy is removed from the cap. I'm sure you're already aware of this but many reading here may not be.
In Fig 5, the "R" used as a load to capture the increased energy in the cap will also be charge separated when the primary 4v/turn is applied. This means the overall potential drop across "R" will be 4v in this case during the charge separated discharge cycle. However, one must be very careful here as the actual charge separated voltage across "R" will depend on the length of "R" compared to the height of the E-Field window. If it is dhorter for example, there will be charge separated voltage drops in the connecting leads.
Interestingly, as "R" pulls a current from the capacitor, this is not reflected back (no Lenz effect) to the primary. IOW, the core flux is not affected by any closed loop secondary as the current loop is contained in the core window. So, if our primary drive circuitry allows the primary to return the primary charge current to the supply voltage, we will only have the core magnetization current to account for our input energy consumption. Also during this primary current return to the supply, the primary voltage will reverse to -4v/turn and the voltage across the cap will be~0v during this time.
Regards,
Pm
Quote from: verpies on 2025.09.08, 19:52:15
I have not observed that a plate capacitor placed in the hole of a toroidal core affects the primary current any more than this capacitor in series with the equivalent air capacitor that completes the secondary circuit.
But what size was the capacitor in the hole? I suspect it was a bought component and not almost filling the hole with high K dielectric.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=54067)
QuoteAlso, there is a problem with the following statement: the E field has been instantaneously switched off
...as this will create infinite dA/dt and infinite -E.
Practically, the reversed polarity of -E integrated over time will discharge the capacitor unless some dielectric hysteresis keeps the charge.
Since the E field is E=-dA/dt, creating E=0 demands dA/dt=0. That is the opposite of your infinite dA/dt catastrophe.
Quote from: Smudge on 2025.09.09, 06:54:16
But what size was the capacitor in the hole? I suspect it was a bought component and not almost filling the hole with high K dielectric.
Self-made: 20cm
2 and ~10µm of fine barium titanate powder.
Quote from: Smudge on 2025.09.09, 06:54:16
Since the E field is E=-dA/dt, creating E=0 demands dA/dt=0. That is the opposite of your infinite dA/dt catastrophe.
OK but this means that the current in the primary cannot fall ...because if it falls then |dA/dt| > 0
Quote from: partzman on 2025.09.08, 19:57:15
Smudge,
Thank you for your paper, however I have a few comments.
First, I'm curious to know about the basis for your denying that the electrodes in your Fig 3 would have no measurable potential on their outside surfaces?
I did not say that, you have interpreted my statement (that there will be no excess or diminished charge within the electrode material) to mean no measurable potential. I stand by my statement as there is no conducting circuit for charge to flow through. There could be measurable potential as there will be charge separation within the electrode material, the top and bottom surfaces will have opposite surface charge. If you attemped a measurement what you get will be affected by the type of instrument and how it is connected or brought close to the surface. It won't be the same potential as that measured on the capacitor conventionally charged to a known voltage.
QuoteI understand your theory of doubling the charge with a pre-biased capacitor as you depict in FIg 4. For example, we charge a cap to 4vdc and then apply 4v/turn on the primary. Our cap now reaches a potential of 8v (due to charge separation) with the proper polarities applied.
No! This is not a capacitor charged to 8V yet. We have a situation where the dielectric has been stressed to be equivalent to what it would be under normal charging to 8V, but the CV value of charge for 8V is not present on the electrodes. We have energy stored in the dielectic that is double what it was, but the charge in the electrodes has not changed. I am not aware that this situation has ever been looked at before.
QuoteThis new 8v on the cap will remain only as long as the primary is kept at 4v/turn.
I would rephrase that to "This new state of dielectric stress will remain only as long as the primary is kept at 4v/turn. But we don't do that see comments below.
QuoteIn fact, this charge separated voltage will follow the v/turn of the primary in magnitude, polarity, phase, and shape. So, in this case if the v/turn is returned to 0v/turn, the potential on the cap will instantly return to 4vdc if no energy is removed from the cap. I'm sure you're already aware of this but many reading here may not be.
You are treating potential on the cap classically as relating to charge removed from one plate and delivered to the other. The dielectric stressing, storage of energy, has not done that. See later comment below.
QuoteIn Fig 5, the "R" used as a load to capture the increased energy in the cap will also be charge separated when the primary 4v/turn is applied. This means the overall potential drop across "R" will be 4v in this case during the charge separated discharge cycle. However, one must be very careful here as the actual charge separated voltage across "R" will depend on the length of "R" compared to the height of the E-Field window. If it is dhorter for example, there will be charge separated voltage drops in the connecting leads.
I am suggesting that the extra energy stored in the dielectric which was put there over a period of time cannot dissappear instantly. The same can be said for the E field creating that 4V/turn. It may be possible for the E field to get to zero much faster than the energy can return to the primary, in which case R gets connected when E is at or near zero.
QuoteInterestingly, as "R" pulls a current from the capacitor, this is not reflected back (no Lenz effect) to the primary. IOW, the core flux is not affected by any closed loop secondary as the current loop is contained in the core window.
Agreed, in which case it is not a secondary loop.
QuoteSo, if our primary drive circuitry allows the primary to return the primary charge current to the supply voltage, we will only have the core magnetization current to account for our input energy consumption.
No, during the core magnetization energy is transported into the dielectric so there is also a primary load current component accounting for that energy. If the dielectric charging is done resonantly the peak E will be reached while A is passing through zero so also the primary magnetizing current is passing through zero. Turning off the E simply means holding that current at that zero level so dA/dt becomes zero. The excess energy stored in the dielectric will want to dissipate but it can't do that instantaneously. Hopefully connecting R to it at that time will allow the full energy stored to dissipate in R.
QuoteAlso during this primary current return to the supply, the primary voltage will reverse to -4v/turn and the voltage across the cap will be~0v during this time.
No, the primary current is at zero so the return to the supply has already happened.
Smudge
Pm
[/quote]
Quote from: verpies on 2025.09.09, 12:38:36
Self-made: 20cm2 and ~10µm of fine barium titanate powder.
That 10µm thickness suggests the dielectric did not occupy much of the hole volumeand would receive only a small portion of the single turn volts. Any chance of creating a longer cylinder of dielectric that will fill the hole?
QuoteOK but this means that the current in the primary cannot fall ...because if it falls then |dA/dt| > 0
With resonant charging the current is already at zero when we keep dA/dt at zero by preventing the current from going negative.
Quote from: verpies on 2025.09.06, 12:40:16
...
However, if the dielectric is not perfect and exhibits large dielectric hysteresis then the remanent electric field could end up non-zero at the end of the cycle.
Is so, then the voltage observed across the capacitor would be very dependent on the capacitor's technology - especially choice for its dielectric materials..
I agree, and I think that this is the key point for capacitor charging; its technology is certainly decisive.
The electric field polarises the dielectric, aligning the electric dipoles of the material, and it is difficult to see how this would be possible without a non-linear effect of the dielectric, triggered or not by an asymmetry effect of the rising edge of the signal relative to the falling edge.
In any case, one thing can be predicted: no effect should be observed with an air dielectric capacitor.
The second observation is that there is no reason for the polarisation to persist if the dipoles are simply mobile in the field, because unlike a normally charged capacitor, we do not have excess charges on either plate.
A threshold effect is necessary for them to remain in their oriented position when the field stops, like an electret, which implies that the electric field must work to make them cross this threshold. We should therefore see this through an effect on the current that supplies this energy. My opinion is that if we do not see it, it is because it is insignificant compared to the magnetic or ohmic losses in the circuit, as is certainly the case with the charge of a 47nF capacitor.
I am thinking of a way to verify this.
Quote from: Smudge on 2025.09.09, 13:56:32
That 10µm thickness suggests the dielectric did not occupy much of the hole volumeand would receive only a small portion of the single turn volts. Any chance of creating a longer cylinder of dielectric that will fill the hole?
I don't have that much barium titanate powder.
Even if I did, keeping a large thickness of it squeezed without spilling would require tight side walls.
Quote from: Smudge on 2025.09.09, 13:56:32
With resonant charging the current is already at zero when we keep dA/dt at zero by preventing the current from going negative.
But the current still falls from some max to zero and when that happens, the induced E reverses. ...unless you keep it at max on purpose outside of the resonant regime.
ChatGPT, to whom I explained the problem, advised me to obtain the maximum effect.
Ferroelectric dielectric capacitors
🔬 Typical materials:
BaTiO₃: Barium titanate
PZT: Lead zirconate titanate (Pb[ZrₓTi₁₋ₓ]O₃)
BiFeO₃, KNO₃, etc.
⚙️ Characteristics:
Persistent (remanent) polarisation after field
Hysteresis effect (as in a magnetisation cycle)
High dielectric constant
Non-linear response (good for sensors or memory)
Quote from: F6FLT on 2025.09.09, 17:23:55
KNO₃
Are you sure? The stuff is hygroscopic and conductive when even slightly damp.
Quote from: verpies on 2025.09.09, 17:09:42
But the current still falls from some max to zero and when that happens, the induced E reverses. ...unless you keep it at max on purpose outside of the resonant regime.
Let's be quite clear about this. When the primary current is maximum (at the top of a sine wave) E is zero so no energy stored in the dielectric. Taking that as a starting point E reaches a maximum when the current has fallen to zero (E follows cosine wave) and the dielectric has now gained energy. That is the point where we wish to quickly make E=0 (i.e. stop current from going negative) and connect R. So I do not understand what you are querying here.
Smudge,
I having trouble understanding what you are proposing here!
For example:
I said-
I understand your theory of doubling the charge with a pre-biased capacitor as you depict in FIg 4. For example, we charge a cap to 4vdc and then apply 4v/turn on the primary. Our cap now reaches a potential of 8v (due to charge separation) with the proper polarities applied.
You said-
"No! This is not a capacitor charged to 8V yet. We have a situation where the dielectric has been stressed to be equivalent to what it would be under normal charging to 8V, but the CV value of charge for 8V is not present on the electrodes. We have energy stored in the dielectic that is double what it was, but the charge in the electrodes has not changed. I am not aware that this situation has ever been looked at before."
What means are we using to stress this dielectric? Is this a poled dielectric that we then assemble with the top and bottom plates and insert into the core? Let's assume yes at this point.
So now we have "the dielectric has been stressed to be equivalent to what it would be under normal charging to 8V" as you stated above. (I don't understand why we wouldn't start with 4v!) Now we apply 4v/turn to the primary. In your case as stated, we would expect the voltage stress across the dielectric to now be 12v and the energy required to accomplish this change in charge would come from the energized primary. Here is where I totally disagree! In all my experimentation with this type of charge separation, at no time have I ever seen energy taken from the primary for charge separation in any open circuited object. IMO, the energy required for this charge increase comes from the aether as I've stated many times before. Why is your example any different?
We can now apply the R load to said capacitor during the time 4v/turn is applied to the primary and we will see a drop in the cap voltage of 12v commensurate with the value of R. If we wait to apply R until the 4v/turn is removed from the primary, we will then see 8v across the cap and will proceed to discharge with R.
I have already run tests on your basic scheme and the results were COP<1. However, I must say that I used vertically oriented plates for the capacitor and you state that these types would not work!
Regards,
Pm
Edit: I see that you are assuming a resonant condition from your last post. Is this resonance between the primary and the stressed dielectric? All my references above were using a pulsed primary.
I found the perspective often depends on whether the person is a logical experimenter or simply uses generalized math and equations.
This problem is similar to the capacitor paradox imo. Charged cap A when connected to uncharged cap B equalizes it's voltage/charges with B. Half the voltage/charges end up on each cap A and B but somehow we lost 1/2 the energy. Most get this problem wrong because they generalize. Energy is lost because a set number of concentrated charges on A have redistributed themselves over twice the surface area of A plus B lowering the total charge density. This is the cause of the measured energy loss in A-B and any heat or radiation from the connection wire is simply an effect from the redistribution of charges. So in my opinion the supposed paradox is most people confusing cause and effect.
As always, I'm interested in your research, Jon... but having trouble following ATM.
Perhaps in part because I'm concerned about the stability of our world ATM... hoping we avoid WW III in the next months...
Quote from: verpies on 2025.09.09, 17:34:09
Are you sure? The stuff is hygroscopic and conductive when even slightly damp.
You're right, there's serious doubt. It is an ionic material. It would have to be very dry. I don't think ChatGPT's advice would be a good choice in practice.
There is a major flaw in the reasoning I see here and there. We are not dealing with a simple capacitor. It is never charged in the sense that free electrons would have accumulated on one plate while we would have a deficit on the other. It is impossible without contact with the plates.
It is the dielectric that is polarised and attracts positive or negative charges depending on the plate. Even if it has a certain capacity, the so-called capacitor is in fact closer to the functioning of an electret. While we are talking about capacitors, it is the 'electret' effect that is the focus of discussion, and it is not the same thing.
The electric dipoles are not free to follow the direction of the electric field almost instantaneously as in a simple capacitor. They are subject to constraints in the material, which can keep them in a particular orientation, constraints that the electric field has to overcome in order to polarise them.
This means that if we 'discharge' the capacitor in a short time, we do not recover the 'charge'. We recover a small part of the charge, the only free electrons from the plates that the polarisation of the dielectric had attracted, and this does not instantly cancel out the internal orientation of all the dipoles of the dielectric inside the insulator, with which the plates are not in contact except at the surface. The so-called capacitor will partially repolarise after a certain time following its discharge.
Such a component no longer responds to the classic equation I(t)=C.dU(t)/dt. A factor dependent on polarisation, which has a memory effect, must be added. It would be more like C.dU(t)/dt + (U(t)−Um)/Rm, where Um and Rm are related to the 'memory', with Rm including the attenuation of the repolarisation.
Quote from: Allcanadian on 2025.09.11, 16:26:07
Charged cap A when connected to uncharged cap B equalizes it's voltage/charges with B. Half the voltage/charges end up on each cap A and B but somehow we lost 1/2 the energy.
But the cap-cap energy transfer doesn't have to behave like that. If the transfer is conducted through an inductor then theoretically 100% of energy in cap A will be transferred to cap B. In practice ~90%.
the inductor makes use of the transfer. just cap to cap, the transfer happens without doing any work by way of the transfer. the fifty percent loss occurs because we didnt do anything with the tramsfer other than moving the electrons till the potential pressure is equalized.
mags
Quote from: Magluvin on 2025.09.12, 03:55:21
the inductor makes use of the transfer. just cap to cap,...
There is no such thing as cap-to-cap energy transfer in practice. The effect of the interconnecting component cannot be ignored ...be it a resistor, inductor or a compound of the two.
Quote from: Magluvin on 2025.09.12, 03:55:21
just cap to cap, the transfer happens without doing any work by way of the transfer.
There is energy conversion in any real transfer, e.g. conversion of current to heat in CRC transfers or current to magnetic flux in CLC transfers.
Energy conversion is equivalent to work.
Quote from: Magluvin on 2025.09.12, 03:55:21
the fifty percent loss occurs because we didnt do anything with the tramsfer other than moving the electrons till the potential pressure is equalized.
No, moving electrons are electric current and flowing any current through a resistor results in its conversion into heat and flowing this current through an inductor results in its conversion into magnetic flux. Thus conversion always happens during an energy transfer.
Practically you cannot connect two capacitors with a wire that does not have a resistance nor an inductance. For rigorous treatment you have to account for both phenomena.
Quote from: partzman on 2025.09.10, 15:14:35
Smudge,
I having trouble understanding what you are proposing here!
I start with the fact that in non-electrolytic capacitors there is never any flow of charge into or out of the dielectric. When we charge the capacitor by applying current the electrodes gain or lose electron charge and thus create an electric field in the dielectric which becomes polarised; at the surfaces in contact with the electrodes that polarisation appears as though these surfaces have charge on them. Those induced surface charges create electric fields that oppose the flow of current onto the electrodes, so our charging system uses energy to supply the current. The current source is seeing a back emf. This is a slightly different viewpoint that to charge a capacitor you apply voltage and the capacitor demands current. Note that the charge separation in the dielectric is driven by the charge separation on the electrodes,
but there are two different charge separations; they normally have the same value. To differentiate between these, I used the term dielectric displacement for the charge separation in the dielectric (maybe I shouldn't since displacement
D is a recognised EM vector alongside
E). The electrode charge separation is given by the integral of the current with respect to time. Within our circuit, when we do the math, that integral relates to voltage and we think of the capacitor being charged to that voltage.
Turning to the uncharged disc capacitor filling the hole of a ring core having no electrical connection to electrodes. The dielectric displacement will be driven by the
E field due to the changing flux in the core, not by an
E field from charge on the electrodes since there is no means for charge to get there, and that is a different ball game. The randomly orientated electric dipoles in the dielectric are aligned by that ring core
E field, but we cannot consider it to be a charged capacitor yet (there has been no conduction current so there is no electrode charge). If the
E field can be removed faster than the dielectric can relax and we put R across the electrodes the dielectric will quickly drive current to get the electrodes charged, losing a small amount of stored energy in doing so. That current spike ceases when the magnitude of charge on each electrode equals the opposite polarity bound surface charge on the dielectric. Thereafter we do have a conventional charged capacitor discharging exponentially through R with a reversed current. The loss of dielectric energy to get charge onto the electrodes is small so the voltage on this capacitor after that current spike is close to the earlier
E field potential difference across it.
Turning to the situation where we previously charge the capacitor (to 4V) and your specific queries.
QuoteWhat means are we using to stress this dielectric? Is this a poled dielectric that we then assemble with the top and bottom plates and insert into the core? Let's assume yes at this point.
I thought this was quite clear. There is no primary current and no flux in the core. We simply charge the capacitor conventionally (to 4V). This can be done outside the core and the charged capacitor then placed into the core, or it can be done with the capacitor in place. The dielectric is now stressed, has charge displacement, has internal charge separation held there by the charge on the electrodes.
We then energise the core to create an E field that doubles the stress in the dielectric. It has not doubled the charge on the electrodes, that charge Q=CV remains at the Q=4C value. The magnetic vector potential
A field penetrates the dielectric to add its
E=-d
A/dt field to the field already there from the +- Q on the electrodes.
QuoteSo now we have "the dielectric has been stressed to be equivalent to what it would be under normal charging to 8V" as you stated above. (I don't understand why we wouldn't start with 4v!)
We did start with 4V charge on the capacitor.
QuoteNow we apply 4v/turn to the primary. In your case as stated, we would expect the voltage stress across the dielectric to now be 12v
No. We have already done that, we don't do it twice. The dielectric was stressed to 4V (and the electrodes carried the Q=4C value), now it is stressed to 8V (while the electrodes still carry Q=4C).
QuoteIn all my experimentation with this type of charge separation, at no time have I ever seen energy taken from the primary for charge separation in any open circuited object.
With respect I am not aware that you used a disc capacitor as shown. In all your experiments I see a small purchased capacitor of unknown internal structure.
QuoteWe can now apply the R load to said capacitor during the time 4v/turn is applied to the primary and we will see a drop in the cap voltage of 12v commensurate with the value of R.
No. We quickly turn the E field off so the 4V/turn on the primary is turned off. With resonance the primary current could be passing through zero at that voltage so this is not difficult, we hold the current at zero. Then we apply the R load.
QuoteI have already run tests on your basic scheme and the results were COP<1. However, I must say that I used vertically oriented plates for the capacitor and you state that these types would not work!
What capacitor did you use? Did it nearly fill the core hole with dielectric?
Smudge
Quote from: Smudge on 2025.09.12, 14:07:41
[snip]
What capacitor did you use? Did it nearly fill the core hole with dielectric?
Smudge
No, not even close! It was a flat wound polyester film cap placed vertically that probably occupied 5% of the hole area or less.
Pm
Edit: I have enough Barium Titanate to fill a toroid with a hole that is ~19 cm^3 in volume. I will do the test after fabrication over the next several days.
Edit: With the given physical dimensions, the capacitance should be ~1300nfd using k=5000 for the dielectric.
Quote from: partzman on 2025.09.12, 20:38:08
I have enough Barium Titanate to fill a toroid with a hole that is ~19 cm^3 in volume. I will do the test after fabrication over the next several days.
Are you going to sinter it or compress it ?
I'd like to see whether the di/dt in the primary changes as the dielectric is inserted while all other things are being equal.
Quote from: verpies on 2025.09.12, 21:54:34
Are you going to sinter it or compress it ?
I'd like to see whether the di/dt in the primary changes as the dielectric is inserted while all other things are being equal.
The presence of that dielectric will make the apparent inductance of a coil wound on the core change value. So maybe simple before and after inductance measurements would suffice.
Edit. If you give me the core details I will calculate the inductance change for uncompressed powder. Electrodes would not be needed for this simple test, just fill the hole with the powder. Of course to take things further it requires the electrodes to be present and the means to connect to them without the conductor encircling the core.
Quote from: verpies on 2025.09.12, 21:54:34
Are you going to sinter it or compress it ?
I'd like to see whether the di/dt in the primary changes as the dielectric is inserted while all other things are being equal.
I am going to just compress it as I have no means to sinter it. I hadn't thought about making the assembly removable but maybe that would be useful.
Pm
Quote from: Smudge on 2025.09.13, 06:51:31
The presence of that dielectric will make the apparent inductance of a coil wound on the core change value. So maybe simple before and after inductance measurements would suffice.
Edit. If you give me the core details I will calculate the inductance change for uncompressed powder. Electrodes would not be needed for this simple test, just fill the hole with the powder. Of course to take things further it requires the electrodes to be present and the means to connect to them without the conductor encircling the core.
The core I will use is has a center hole measuring 31mm in diameter x 25mm high. My powder is 99.9% pure ground to 0.5-3.0 micron.
As for any connecting electrode being inside the core hole, it will exhibit charge separation resulting in an overall null measurement.
Pm
Edit: Change of plan- I will use an already available insert that measures 25mm I.D. X 21mm in height. This will allow the assembly to be removed and will also provide room for a load resistor and connecting lead.
Quote from: Smudge on 2025.09.12, 14:07:41
I start with the fact that in non-electrolytic capacitors there is never any flow of charge into or out of the dielectric. When we charge the capacitor by applying current the electrodes gain or lose electron charge and thus create an electric field in the dielectric which becomes polarised; at the surfaces in contact with the electrodes that polarisation appears as though these surfaces have charge on them. Those induced surface charges create electric fields that oppose the flow of current onto the electrodes, so our charging system uses energy to supply the current. The current source is seeing a back emf. This is a slightly different viewpoint that to charge a capacitor you apply voltage and the capacitor demands current. Note that the charge separation in the dielectric is driven by the charge separation on the electrodes, but there are two different charge separations; they normally have the same value. To differentiate between these, I used the term dielectric displacement for the charge separation in the dielectric (maybe I shouldn't since displacement D is a recognised EM vector alongside E). The electrode charge separation is given by the integral of the current with respect to time. Within our circuit, when we do the math, that integral relates to voltage and we think of the capacitor being charged to that voltage.
Turning to the uncharged disc capacitor filling the hole of a ring core having no electrical connection to electrodes. The dielectric displacement will be driven by the E field due to the changing flux in the core, not by an E field from charge on the electrodes since there is no means for charge to get there, and that is a different ball game. The randomly orientated electric dipoles in the dielectric are aligned by that ring core E field, but we cannot consider it to be a charged capacitor yet (there has been no conduction current so there is no electrode charge).
...
We are in complete agreement. That is what I was saying earlier in a different way in my reply #29.
Quote from: partzman on 2025.09.13, 14:23:09
The core I will use is has a center hole measuring 31mm in diameter x 25mm high. My powder is 99.9% pure ground to 0.5-3.0 micron.
As for any connecting electrode being inside the core hole, it will exhibit charge separation resulting in an overall null measurement.
Pm
Edit: Change of plan- I will use an already available insert that measures 25mm I.D. X 21mm in height. This will allow the assembly to be removed and will also provide room for a load resistor and connecting lead.
OK, I finished the assembly described in my edit above. The sleeve was a 3D printed cylinder which had a round copper electrode CA'd to one end. I then packed the inside with BT physically until no more could be forced into the sleeve and the BT was slightly above the upper surface of the sleeve. The top copper electrode was then put in place and forced to completely close with a vise. Numerous rounds of narrow tape was then used to hold the BT in what I think is a compressed state. The resultant capacity-a very disappointing 17pf! Applying manual physical pressure results in ~33 pf! This is essentially a failure IMO.
Obviously, the BT must be a compressed and sintered block.
Pm
Smudge and F6FLT,
So you both think that when I charge separate say a film capacitor as I demonstrated many times in the E-Field area in the center hole of a toroid, we are not really seeing a true charging of the capacitor? This belief arises from the fact that there is no apparent connection to the electrodes to supply the charging current, correct?
If so, then how do you explain the fact that a charge separated cap is able to supply energy commensurate with a given capacitance and charge separated voltage, to a load that is connected during the time the primary is supplied with a given V/turn?
Pm
Quote from: partzman on 2025.09.13, 20:26:11
The resultant capacity-a very disappointing 17pf!
Well, that's because the capacitance of a two-plate capacitor is inversely proportional to the thickness of the dielectric and directly proportional to its permittivity and the areas of the plates.
Quote from: partzman on 2025.09.13, 20:26:11
Applying manual physical pressure results in ~33 pf!
This means that the effective permittivity of your dielectric is ~2x higher when compressed.
Quote from: partzman on 2025.09.13, 20:26:11
This is obviously a failure IMO.
Not if your goal is to obtain the maximum dielectric polarization work instead of the maximum capacitance between plates.
If the polarization of that dielectric takes a lot of work then it should affect the di/dt of the primary winding.
Quote from: verpies on 2025.09.13, 21:37:56
Well, that's because the capacitance of a two-plate capacitor is inversely proportional to the thickness of the dielectric and directly proportional to its permittivity and the areas of the plates.
It is because the effective permittivity of the powder is 82.2 (for the 17pF value) and not the 5000 that is expected for the solid version. Goes to show that touching surfaces have an effective air gap, you need the chemical bond to eliminate that. Same happens in magnetic material, two C cores clamped together never reach the reluctance of an equivalent uncut ring core.
Quote from: partzman on 2025.09.13, 20:26:11
This is essentially a failure IMO.
Not necessarily. Taking your core measurements and assuming the core has square cross section, and assuming a relative permeability of 1000, the inductance of 10 turns is 0.384mH. This will resonate with 17pF at 1.9MHz. Without that C the coil will have some self resonance that may be much higher. Dumping the lump of dielectric in the hole should have the same effect as connecting a 1 turn secondary to an external 17pF capacitor and achieving that resonance. If I had all the core details I could refine this math (as you could also) to see whether looking for resonance is a viable method to tell us the wanted effect is there.
Quote from: partzman on 2025.09.13, 20:37:44
Smudge and F6FLT,
So you both think that when I charge separate say a film capacitor as I demonstrated many times in the E-Field area in the center hole of a toroid, we are not really seeing a true charging of the capacitor? This belief arises from the fact that there is no apparent connection to the electrodes to supply the charging current, correct?
Yes
QuoteIf so, then how do you explain the fact that a charge separated cap is able to supply energy commensurate with a given capacitance and charge separated voltage, to a load that is connected during the time the primary is supplied with a given V/turn?
Correct me if I am wrong, but there your circuit connecting the load encloses the flux in the core so that V/turn is present within the circuit.
Quote from: Smudge on 2025.09.14, 08:01:08
Not necessarily. Taking your core measurements and assuming the core has square cross section, and assuming a relative permeability of 1000, the inductance of 10 turns is 0.384mH. This will resonate with 17pF at 1.9MHz. Without that C the coil will have some self resonance that may be much higher. Dumping the lump of dielectric in the hole should have the same effect as connecting a 1 turn secondary to an external 17pF capacitor and achieving that resonance. If I had all the core details I could refine this math (as you could also) to see whether looking for resonance is a viable method to tell us the wanted effect is there.
The toroid core I am using is metglas and with a 10T primary, the inductance is 3.4mh. Therefore, the resonance frequency with 17pf of dielectric in the hole should be ~662kHz. I have tried driving the primary with a 50 ohm source using sine waveform and also a series connected 1k resistor with the primary. I see no resonance at any time manually sweeping the frequency spectrum. I then removed the dielectric and placed a single secondary turn with 22pf capacitance and again found no resonance with a sweep.
I also tried single and multiple pulses with the same results. The primary self-resonance is ~50MHz.
Pm
Edit: I also tried a ferrite core with a 10T primary and 608nH inductance. Resonance would be ~1.56MHz but I still found no resonance as above in all conditions.
Quote from: Smudge on 2025.09.14, 08:11:46
Correct me if I am wrong, but there your circuit connecting the load encloses the flux in the core so that V/turn is present within the circuit.
That has been the case with most of my past experiments however, thanks to you and this latest dielectric test, I have produced the following results seen below.
Basically, the ferrite core has a 4V/T primary with a 1.22ufd cap in parallel with a 90uh inductor both of which are in the center core hole. All leads are in the core center hole exposed to the E-Field. The resonance frequency is ~15.1kHz as seen in the scope pix below. CH1(yel) is the gate drive for the primary mosfet devices, CH2(blu) is the power supply, CH3(pnk) is the voltage across the LC network, and CH4(grn) is the LC current.
ICR1 shows the differential at the charge separation switching to be 6.15v. Note this is larger than the 4V/T of the primary.
ICR2 shows the RMS values for the LC voltage and current. This LC resonance is not seen by the primary.
Also note that this test is running continuously but can also be demonstrated with a single pulse as well.
It is my opinion that my tests with loads outside the core are also operating in this fashion but via Lenz, they affect the primary energy.
So, I leave you with the question, what is supplying the energy to this resonance circuit?
Pm
Quote from: Smudge on 2025.09.14, 07:28:59
It is because the effective permittivity of the powder is 82.2 (for the 17pF value) and not the 5000 that is expected for the solid version. Goes to show that touching surfaces have an effective air gap, you need the chemical bond to eliminate that. Same happens in magnetic material, two C cores clamped together never reach the reluctance of an equivalent uncut ring core.
The capacitance of my cap, that I described here (https://www.overunityresearch.com/index.php?topic=4797.msg116301#msg116301), is 4.2µF. I attribute this difference to the ~10µm thickness of the dielectric.
According to the formula C = επr
2/d the capacitance of this cap should be 8.7µF but because my dielectric powder is not sufficiently compressed or bonded, its permittivity is ~2x less than the advertised 5000.
@Partzman: To cold bond the BaTi powder, you can find some shop with 80 ton press and ask them squeeze to powder for you into a pellet. 80 ton press is not an exotic piece of equipment (here (https://youtu.be/_Dqu5YtFiw4) is a video of a 500 ton press)
Quote from: Smudge on 2025.09.10, 14:36:42
Let's be quite clear about this. When the primary current is maximum (at the top of a sine wave) E is zero so no energy stored in the dielectric.
Yeah, the derivative is zero at the top of the sine wave, but getting to that top requires a period where the current changes and E is nonzero.
Quote from: verpies on 2025.09.14, 22:59:22
The capacitance of my cap, that I described here (https://www.overunityresearch.com/index.php?topic=4797.msg116301#msg116301), is 4.2µF. I attribute this difference to the ~10µm thickness of the dielectric.
According to the formula C = επr2/d the capacitance of this cap should be 8.7µF but because my dielectric powder is not sufficiently compressed or bonded, its permittivity is ~2x less than the advertised 5000.
Your compression got you within a factor of 2 to the 5000 permittivity. Using the same formula for patrzman's capacitor that would be 1nF at that 5000 value and only chieved 17pF his permittivity down by a factor of over 60.
QuoteYeah, the derivative is zero at the top of the sine wave, but getting to that top requires a period where the current changes and E is nonzero.
Agreed but I assumed a dielectric with no hysteresis so the previous history (an opposite polarity E rise and fall) would not affect this experiment.
Quote from: partzman on 2025.09.14, 16:24:42
That has been the case with most of my past experiments however, thanks to you and this latest dielectric test, I have produced the following results seen below.
Basically, the ferrite core has a 4V/T primary
The V/T depends on the rate of flux change so I guess you assume the voltage on the primary would always achieve that rate, hence you have a 10T primary against your 40V supply.
Quotewith a 1.22ufd cap in parallel with a 90uh inductor both of which are in the center core hole.
A circuit would help here as it is not clear whether the paralleled 90uH L and 1.22uF C are also in parallel with the primary or in series (my guess is in series). And how it is driven by the mosfet.
QuoteAll leads are in the core center hole exposed to the E-Field. The resonance frequency is ~15.1kHz as seen in the scope pix below. CH1(yel) is the gate drive for the primary mosfet devices, CH2(blu) is the power supply, CH3(pnk) is the voltage across the LC network, and CH4(grn) is the LC current.
ICR1 shows the differential at the charge separation switching to be 6.15v. Note this is larger than the 4V/T of the primary.
But does it change value if the circuit is pulled out of the hole?
QuoteICR2 shows the RMS values for the LC voltage and current. This LC resonance is not seen by the primary.
Are you saying the primary coil is seeing current impulses related to the pulsing with no indication of this LC resonance?.
QuoteAlso note that this test is running continuously but can also be demonstrated with a single pulse as well.
It is my opinion that my tests with loads outside the core are also operating in this fashion but via Lenz, they affect the primary energy.
So, I leave you with the question, what is supplying the energy to this resonance circuit?
It is from the 40V supply that is seeing current impulses that integrate to a non-zero value. In some respects this can be seen in the LC current waveform where the two sudden steps are of slightly unequal value. I think you could pull the circuit out of the hole and get exactly the same results.
Quote from: partzman on 2025.09.14, 15:23:26
The toroid core I am using is metglas and with a 10T primary, the inductance is 3.4mh. Therefore, the resonance frequency with 17pf of dielectric in the hole should be ~662kHz. I have tried driving the primary with a 50 ohm source using sine waveform and also a series connected 1k resistor with the primary. I see no resonance at any time manually sweeping the frequency spectrum. I then removed the dielectric and placed a single secondary turn with 22pf capacitance and again found no resonance with a sweep.
But there must be a resonance there of the 22pF with the secondary inductance. How are you looking for the resonance, a dip or increase in the primary voltage?
Quote from: Smudge on 2025.09.15, 07:50:16
Your compression got you within a factor of 2 to the 5000 permittivity.
Wouldn't that mean that the formula C = επr
2/d is incorrect and d does not affect the capacitance ?
r is the radius of the plates , d is the thickness of the dielectric and ε is its permittivity.
Quote from: Smudge on 2025.09.15, 08:34:59
The V/T depends on the rate of flux change so I guess you assume the voltage on the primary would always achieve that rate, hence you have a 10T primary against your 40V supply.
Yes. Any delay in the dispersion of flux within the core appears to be negligible.
Quote
A circuit would help here as it is not clear whether the paralleled 90uH L and 1.22uF C are also in parallel with the primary or in series (my guess is in series). And how it is driven by the mosfet.
I've posted the schematic below. The input is driven by the same 3/4 bridge circuit I've been using on the other charge separation circuits.
Quote
But does it change value if the circuit is pulled out of the hole?
See response below.
Quote
Are you saying the primary coil is seeing current impulses related to the pulsing with no indication of this LC resonance?.
Yes that is correct.
Quote
It is from the 40V supply that is seeing current impulses that integrate to a non-zero value. In some respects this can be seen in the LC current waveform where the two sudden steps are of slightly unequal value. I think you could pull the circuit out of the hole and get exactly the same results.
First, when the LC circuit is pulled from the hole, there is no voltage or current measured. It must be under the influence of the E_Field to achieve the measurements shown. There is also no change in the primary current.
Next, the primary current is non-symmetrical in it's rise and fall due to the nature of the 3/4 bridge. The current increase is created by the primary being switched between +40v and ground with relatively low loss mosfets. The collapsing current however, is switched between ground and +40 with the addition of one Schottky diode drop. This means the collapsing current sees a net voltage of ~40.4v. Therefore, the time for the collapsing current to reach zero is slightly shorter that the time for the increasing current. Since the generator is producing a symmetrical square wave, we see a slight pause in the negative portion of the resonant waveform during the charge separation polarity change.
Now, what has not been noticed is the fact that I'm able to get a potential difference between L1 and C1 to create the resonant waveform in the first place! If L1 were to be replaced by a resistor R1, the potential between R1 and C1 would be zero thus rendering zero circuit current. So how does this work? I have found that an inductor even with a precise layer wound construction, does not charge separate as expected. It is always lower in value than expected and at this time I do not have an explanation. This then provides the means for the potential difference between L1 and C1 and also explains why the charge separation differentials are not 2x the V/T as they should be.
Pm
Quote from: Smudge on 2025.09.15, 08:45:08
But there must be a resonance there of the 22pF with the secondary inductance. How are you looking for the resonance, a dip or increase in the primary voltage?
I looked for resonance changes in both primary and secondary voltages and currents and couldn't find any measurable changes.
Quote from: partzman on 2025.09.15, 14:23:07
I've posted the schematic below. The input is driven by the same 3/4 bridge circuit I've been using on the other charge separation circuits.
Thanks for that. Could I ask for more clarification, how was the scope probe ground lead connected. Was it as A or B in image Mod1 below? Same question when the LC circuit is pulled out of the hole, A or B in image mod2?
Quote from: Smudge on 2025.09.15, 18:56:58
Thanks for that. Could I ask for more clarification, how was the scope probe ground lead connected. Was it as A or B in image Mod1 below? Same question when the LC circuit is pulled out of the hole, A or B in image mod2?
It was A in Mod1 and B in Mod2.
Smudge,
In your configuration Mod1 with the ground in position B, the ground lead is charge separated and creates havoc with the overall results. So, I have run a test with the resonance circuit measured in a differential mode as shown in the schematic below.
The resultant scope traces are shown in "ICR Differential". CH3(pnk) is connected to the top of L1/C1, CH2(blu) is connected to the bottom of L1/C1, and the difference between them is measured and shown in the Math(red) channel. The scope ground connection is made in the 3/4 bridge primary drive circuitry.
With the complete L1/C1 assembly moved outside the core, no hint of resonance is seen.
Quote from: Smudge on 2025.09.15, 18:56:58
Could I ask for more clarification, how was the scope probe ground lead connected. Was it as A or B in image Mod1 below?
The answer to this question is very significant.
Quote from: partzman on 2025.09.15, 20:26:59
Smudge,
In your configuration Mod1 with the ground in position B, the ground lead is charge separated and creates havoc with the overall results. So, I have run a test with the resonance circuit measured in a differential mode as shown in the schematic below.
The resultant scope traces are shown in "ICR Differential". CH3(pnk) is connected to the top of L1/C1, CH2(blu) is connected to the bottom of L1/C1, and the difference between them is measured and shown in the Math(red) channel. The scope ground connection is made in the 3/4 bridge primary drive circuitry.
With the complete L1/C1 assembly moved outside the core, no hint of resonance is seen.
Thanks for doing that. I think this is quite significant as it shows that the E field in that core hole can excite the LC circuit. IMO this is not the dielectric displacement that I have been talking about but merely induction into components that have dimensions (including their connecting wires) that act like E field antenna. An L by itself will not carry significant AC current. Neither would a lone C. But put the two together and you get that 1A rms. That circulating 1A rms sinusoidal current can't create sinusoidal AC flux in the core. I think it would be interesting to put a series R in the LC circuit that dissipates significant sinusoidal AC power as shown in the image below. Then check whether that power comes from the primary power source via the difference in the current impulses at the leading and trailing edges of the pulse.
Quote from: partzman on 2025.09.15, 19:50:55
It was A in Mod1 and B in Mod2.
That was what I expected you to say. Your new differential measurement appears to settle the conflict.
@Partzman,
In anticipation of you showing genuine OU in this endeavour, and if the inductor produces the field shown in Mod4 image below, I am offering the possibility that this field crossing that within the core coheres the Larmor precessions of the atomic dipoles in the core for that to be the source. Wild I know but you have to start somewhere.
Smudge
Quote from: Smudge on 2025.09.16, 06:36:26
Thanks for doing that. I think this is quite significant as it shows that the E field in that core hole can excite the LC circuit. IMO this is not the dielectric displacement that I have been talking about but merely induction into components that have dimensions (including their connecting wires) that act like E field antenna. An L by itself will not carry significant AC current. Neither would a lone C. But put the two together and you get that 1A rms. That circulating 1A rms sinusoidal current can't create sinusoidal AC flux in the core. I think it would be interesting to put a series R in the LC circuit that dissipates significant sinusoidal AC power as shown in the image below. Then check whether that power comes from the primary power source via the difference in the current impulses at the leading and trailing edges of the pulse.
Up to this point I have been using short clip leads to connect L1 and C1 together but in the first scope pix below, these leads have been soldered. The improvement in both resonant voltage and current can be seen when compared to the results in my post #57. I haven't tried to measure the Q of this circuit yet but it appears to be quite high!
The next two pix show the circuit with a 1 ohm resistor placed as you show in your schematic using soldered leads. The second pix shows the measured primary current stored in R1(wht) with the circuit in place in the core. Also seen is the differential resonant voltage on the Math(red) channel.
The third pix shows the primary current in CH4(grn) overlaid on the R1 stored current with the circuit removed from the core. Essentially there appears to be no difference. Note that all these scope measurements are taken in the hi-res mode which is 16 bit full deflection resolution.
Quote from: Smudge on 2025.09.16, 13:35:19
@Partzman,
In anticipation of you showing genuine OU in this endeavour, and if the inductor produces the field shown in Mod4 image below, I am offering the possibility that this field crossing that within the core coheres the Larmor precessions of the atomic dipoles in the core for that to be the source. Wild I know but you have to start somewhere.
Smudge
You could be correct with your speculation however, at present the L1 core axis is 90 degrees to your diagram. I do expect to be able demonstrate OU with this concept but not quite there yet.
Quote from: verpies on 2025.09.14, 22:59:22
@Partzman: To cold bond the BaTi powder, you can find some shop with 80 ton press and ask them squeeze to powder for you into a pellet. 80 ton press is not an exotic piece of equipment (here (https://youtu.be/_Dqu5YtFiw4) is a video of a 500 ton press)
Better late than never. I do have a 20 ton "H" press and actually thot about using it but talked myself out of it when I considered how I would get the pellet out of a metal sleeve without destroying it. Any thots?
Quote from: partzman on 2025.09.16, 14:28:06
I do have a 20 ton "H" press...
20 ton is not enough at your plate area. You need at least 300MPa to bond the powder without heating.
Quote from: verpies on 2025.09.16, 14:44:07
20 ton is not enough at your plate area. You need at least 300MPa to bond the powder without heating.
OK, thanks!
I thought I would post a pix of the core I've been using for these past tests as in the past there was some question about the toroids I was using.
This core is assembled with 2 complete sets EC-70 core in N27 material with the center legs removed and stacked as shown. The AL=4.93uH/T^2 .
Quote from: partzman on 2025.09.16, 14:22:04
You could be correct with your speculation however, at present the L1 core axis is 90 degrees to your diagram. I do expect to be able demonstrate OU with this concept but not quite there yet.
So its magnetic field gets into the core, for one half of the core mostly parallel to the core field there, but for the other half of the core it would be anti-parallel? I'll ponder on that.
Could you please show us a picture of the LC in the hole at its position where you took your measurements? Thanks.
Quote from: Smudge on 2025.09.16, 18:20:33
So its magnetic field gets into the core, for one half of the core mostly parallel to the core field there, but for the other half of the core it would be anti-parallel? I'll ponder on that.
Could you please show us a picture of the LC in the hole at its position where you took your measurements? Thanks.
Attached is a pix of L1 and C1 in the core. The position of L1 in the plane shown is sensitive to it's location relative to the core. There is a large H-Field in L1 that could be coupling to the EC-70 core assembly although this doesn't appear to show up in the primary current. I have done some tests with L1 tightly coupled to the EC-70 core but need to do more testing before posting.
Quote from: partzman on 2025.09.16, 19:49:23
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=54105)
Rememeber that the core does not have an infinite permeability and the flux leaks out of it like this:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=51995)
Gaps or loaded secondary windings only exacerbate this leakage.
Quote from: partzman on 2025.09.16, 19:49:23
Attached is a pix of L1 and C1 in the core. The position of L1 in the plane shown is sensitive to it's location relative to the core. There is a large H-Field in L1 that could be coupling to the EC-70 core assembly although this doesn't appear to show up in the primary current. I have done some tests with L1 tightly coupled to the EC-70 core but need to do more testing before posting.
Thanks for that, it gives me a clear view of the coupling between L1 and the core, something for my brain to work on.
Quote from: partzman on 2025.09.16, 14:17:02
Up to this point I have been using short clip leads to connect L1 and C1 together but in the first scope pix below, these leads have been soldered. The improvement in both resonant voltage and current can be seen when compared to the results in my post #57. I haven't tried to measure the Q of this circuit yet but it appears to be quite high!
And presumably the scope channels are as stated in your post #57
QuoteThe next two pix show the circuit with a 1 ohm resistor placed as you show in your schematic using soldered leads. The second pix shows the measured primary current stored in R1(wht) with the circuit in place in the core. Also seen is the differential resonant voltage on the Math(red) channel.
I think you didn't mean "primary current stored in R1". The current through R1 has to be the sine wave that was CH4 green but will now be at a reduced level. I see you have CH4 rms value as 960.2mA compared to 1.519A previously but the CH4 trace is not there. Is that really the new rms current flowing round the LCR resonant circuit? If so it represents almost 1W dissipation that must come from somewhere. You have a primary current trace (white) and I see it is a triangular waveform, not what I expected, so I must rethink my analysis.
QuoteThe third pix shows the primary current in CH4(grn) overlaid on the R1 stored current with the circuit removed from the core. Essentially there appears to be no difference. Note that all these scope measurements are taken in the hi-res mode which is 16 bit full deflection resolution.
I see CH4(grn) overlaid on the differential resonant voltage Math(red) channel.
Edit. I guess R1 is a scope trace that is previously recorded and not an R1 resistor, so your third pix is really showing two traces overlaid (green on white). Would still like to know if the LCR circulatory current through the 1 0hm resistor is that 960mA.
Quote from: Smudge on 2025.09.17, 09:02:27
And presumably the scope channels are as stated in your post #57
Yes, that is correct.
Quote
I think you didn't mean "primary current stored in R1". The current through R1 has to be the sine wave that was CH4 green but will now be at a reduced level.
Sorry for the confusion. The "primary current stored in R1" should have read "primary current stored in Ref1". IOW, the previously measured primary current with the resonant L1/C1 in place and operating.
Quote
I see you have CH4 rms value as 960.2mA compared to 1.519A previously but the CH4 trace is not there. Is that really the new rms current flowing round the LCR resonant circuit? If so it represents almost 1W dissipation that must come from somewhere. You have a primary current trace (white) and I see it is a triangular waveform, not what I expected, so I must rethink my analysis.
I see CH4(grn) overlaid on the differential resonant voltage Math(red) channel.
Edit. I guess R1 is a scope trace that is previously recorded and not an R1 resistor, so your third pix is really showing two traces overlaid (green on white). Would still like to know if the LCR circulatory current through the 1 0hm resistor is that 960mA.
I will have to re-run the test and actually measure the current through R1. It will be considerably less than the resonant current of 1.519A without the resistor in the circuit but it will have some value that will produce an energy level that is not accounted for normally. The 960.2ma current measurement on CH4(grn) in the second scope pix is an rms measurement of the increase and collapse of the primary current. In the 3rd pix, the 953.3ma is again an rms measurement of the increase and collapse of the primary current with L1/C1 and R1 removed from the core.
I suspect the circuit anomaly could be due to what I call a lack of follow through.
For example, many suppose all of the magnetic field is contained within the core based on hearsay. That is, the textbooks show a closed field confirmed by many simulations. Of course this is complete nonsense and in the real world, measured with real hall effect sensors and arrays, some of the field around the coil always curls back on itself outside the coil and core.
Based on real world experience here is the first thing I see when looking at the circuit partzman posted.
1)The core coil is near a corner which increases magnetic field leakage. In effect it partially closes the field path.
2)The core coil is next to a nub extending from the middle of the inner core increasing magnetic field leakage. In effect it partially closes the field path
3)The inner coil L1 is literally on the nub extending from the middle of the inner core where I would expect leakage.
4)The inner coil L1 is aligned with the nub, near the core coil end where leakage should be the greatest.
Whenever we do experiments we should always test every premise. So if we suppose all the magnetic field stays within the core we should test it. This is literally the first test which should have been done once the coil was wound on the core because all the other premises are dependent on it.
I can also say with near absolute certainty the core field model Verpies posted is incorrect. Verpies coil is in the middle of the core, covering most of the leg, not in the corner like the actual circuit. There is no nub in the middle of the leg extending inward at the end on the core coil. It's obvious the model is nothing like the actual circuit.
It's not rocket science, buy a $2 linear hall effect sensor, attach it to your DSO, energize the core coil and map the direction and magnitude of the actual magnetic field on a big piece of paper with a picture of the coil and core.
AC
Quote from: Allcanadian on 2025.09.17, 15:56:48
It's obvious the model is nothing like the actual circuit.
It is very similar. Just because it does not mirror the actual circuit 100% does not mean it is "nothing like".
Quote from: partzman on 2025.09.17, 14:49:00
Yes, that is correct.
Sorry for the confusion. The "primary current stored in R1" should have read "primary current stored in Ref1". IOW, the previously measured primary current with the resonant L1/C1 in place and operating.
I will have to re-run the test and actually measure the current through R1. It will be considerably less than the resonant current of 1.519A without the resistor in the circuit but it will have some value that will produce an energy level that is not accounted for normally. The 960.2ma current measurement on CH4(grn) in the second scope pix is an rms measurement of the increase and collapse of the primary current. In the 3rd pix, the 953.3ma is again an rms measurement of the increase and collapse of the primary current with L1/C1 and R1 removed from the core.
You can get a good estimate from the measured AC voltage across the LC circuit and knowledge of the inductor value and frequency since i=V/wL. Without the R the unloaded Q is showing 40V pk to pk which with 90uH and 15.46KHz gives 1.62A rms against your measured 1.52A rms. Taking that 1.52 value as correct, with the 1 ohm resistor the voltage drops to 15V pk to pk which yields a current of 0.57A rms. That is 324mW dissipating in the R.
Quote from: Allcanadian on 2025.09.17, 15:56:48
I suspect the circuit anomaly could be due to what I call a lack of follow through.
.............
It's not rocket science, buy a $2 linear hall effect sensor, attach it to your DSO, energize the core coil and map the direction and magnitude of the actual magnetic field on a big piece of paper with a picture of the coil and core.
You are misjudging the knowledge of the people here, we know about flux leakage and we have means for getting that data. The clever thing is explaining how that leakage produces the anomaly, would you like to tackle that problem for us?
Quote from: Allcanadian on 2025.09.17, 15:56:48
I suspect the circuit anomaly could be due to what I call a lack of follow through.
For example, many suppose all of the magnetic field is contained within the core based on hearsay. That is, the textbooks show a closed field confirmed by many simulations. Of course this is complete nonsense and in the real world, measured with real hall effect sensors and arrays, some of the field around the coil always curls back on itself outside the coil and core.
Based on real world experience here is the first thing I see when looking at the circuit partzman posted.
1)The core coil is near a corner which increases magnetic field leakage. In effect it partially closes the field path.
2)The core coil is next to a nub extending from the middle of the inner core increasing magnetic field leakage. In effect it partially closes the field path
3)The inner coil L1 is literally on the nub extending from the middle of the inner core where I would expect leakage.
4)The inner coil L1 is aligned with the nub, near the core coil end where leakage should be the greatest.
Whenever we do experiments we should always test every premise. So if we suppose all the magnetic field stays within the core we should test it. This is literally the first test which should have been done once the coil was wound on the core because all the other premises are dependent on it.
I can also say with near absolute certainty the core field model Verpies posted is incorrect. Verpies coil is in the middle of the core, covering most of the leg, not in the corner like the actual circuit. There is no nub in the middle of the leg extending inward at the end on the core coil. It's obvious the model is nothing like the actual circuit.
It's not rocket science, buy a $2 linear hall effect sensor, attach it to your DSO, energize the core coil and map the direction and magnitude of the actual magnetic field on a big piece of paper with a picture of the coil and core.
AC
I am well aware of the so called leakage flux or H-Field which is negligible in an unloaded core which is what we have with the resonant L1/C1 in the core window. However, what we do have is an interaction between the H-Field of L1 during resonance with the host core. This interaction is complex due to positioning, coil aspect ratio, coupling factor, and lead lengths and positions of L1. Everything in the core window charge separates!
Pm
partzman
QuoteI am well aware of the so called leakage flux or H-Field which is negligible in an unloaded core which is what we have with the resonant L1/C1 in the core window.
The field leakage is not negligible and the moment the core coil is energized near the corner leakage occurs. I know this from mapping the fields in many core geometries under different conditions. This is why we need to do real tests to determine what is fact and what is not so much. It's similar to the hearsay that a current follows the shortest path between two points. In fact, proven by experiment, it never follows a straight line if not constrained by the conductor. Which begs the question how 99% of people could get something so simple so wrong?. We know why... they didn't check their premise.
QuoteHowever, what we do have is an interaction between the H-Field of L1 during resonance with the host core.
I agree, because L1 is literally right next to the core coil confined in the corner. I would expect a great deal of interaction considering the proximity of the circuit elements. However if you had a circular core larger in diameter and the elements not so close to one another much less interaction would occur. Of course, I would expect you already did simple comparative tests like this. I like to use an iron wire core which is easy to bend into different geometries when field mapping.
QuoteThis interaction is complex due to positioning, coil aspect ratio, coupling factor, and lead lengths and positions of L1. Everything in the core window charge separates!
I would agree the interaction is very complex considering the proximity of the core and circuit elements. More so when the field leakage can move both through and/or sweep across part of the L1 coil. It relates to my prior post on the possible number of combinations in a small number of variables. Four variables each having 1000 possible values has 1 quadrillion possible combinations.
AC
FYI ,an AI confirmed my suspicions...
Question, if a coil is wound in the corner of a square transformer core how much can the magnetic field leakage increase as a percentage?.
Minimal effect (<~5%) when the corner is gently rounded, most turns sit on the straight legs, and the core is large/high-µ.
Typical effect (≈5–25%) for a coil that places a significant fraction of its turns in a sharp corner on a normal laminated or ferrite square core.
Large effect (≈25–100% or more) if many turns are forced around a tight sharp corner, the corner has a small radius, or the core is small/low-µ — in extreme cases leakage can become comparable to or exceed the original leakage.
AC,
Most all my experiments in charge separation have involved the use of toroid cores! No corners! Same results! My results show that there is not so much an effect on L1 from the host core, but rather from L1 to the host core.
Pm
Quote from: Allcanadian on 2025.09.18, 17:52:03
Minimal effect (<~5%) when the corner is gently rounded, most turns sit on the straight legs, and the core is large/high-µ.
Typical effect (≈5–25%) for a coil that places a significant fraction of its turns in a sharp corner on a normal laminated or ferrite square core.
Large effect (≈25–100% or more) if many turns are forced around a tight sharp corner, the corner has a small radius, or the core is small/low-µ — in extreme cases leakage can become comparable to or exceed the original leakage.
Why does a sharp corner gives us more leakage inductance ? It looks like a magnetic flux has inertia. similar centrifugal force. :o
Quote from: Allcanadian on 2025.09.18, 17:52:03
FYI ,an AI confirmed my suspicions...
Question, if a coil is wound in the corner of a square transformer core how much can the magnetic field leakage increase as a percentage?.
Wrong question, you have not asked what percentage of the core flux leaks out which is what we need to work with. AI has given you a percentage increase from some dubious typical leakage figure. It is clear that AI has done that from its final statement
Quotein extreme cases leakage can become comparable to or exceed the original leakage.
And the original leakage is what?
Maximum leakage occurs when you have a small primary coil opposite to a small secondary coil where you find the leakage is greatly affected by the load current in the secondary, AI doesn't mention that.
Quote from: Smudge on 2025.09.19, 07:30:27
Maximum leakage occurs when you have a small primary coil opposite to a small secondary coil where you find the leakage is greatly affected by the load current in the secondary,...
...and greatly affected by any air gaps in the core ...especially on the secondary side.
Quote from: partzman on 2025.09.18, 21:05:44
My results show that there is not so much an effect on L1 from the host core, but rather from L1 to the host core.
And your measurement of the resonant LC current through L tells you what the waveform of the flux in the core from that LC current will be, for comparison with the primary flux from the primary current. Multiplying the induced sinusoidal voltage into the primary coil with the triangular current yields an energy exchange over a full cycle, but we don't have enough data to establish what the two polarities are, whether this yields energy gain or energy loss. This can be established by more experiments and perhaps the most useful could be to drive the primary with sinusoidal AC and not pulse, have your 1 ohm load there to give measured power out for comparison with measured power in.
If we will be use such a core, I don't know what it's called ?
And place your capacitor inside the core, on top of the winding.
This can be done even by using a thin insulator and winding the capacitor in many layers. 8)
Quote from: chief kolbacict on 2025.09.20, 10:57:22
If we will be use such a core, I don't know what it's called ?
Pot core
I have stated earlier on this thread that an inductor (cored or not) when placed in the center hole of a toroid did not charge separate properly. IOW, the measured voltage across said inductor did not reach the primary V/turn potential. This is not correct! My most recent testing shows an inductor will reach the expected V/turn of the primary.
I used this faulty logic to rationalize that this potential difference between and L and C therefore allowed a resonance condition to exist in the center hole. This is not the case!
L/C resonance can only exist in the center hole if L is affected by the host core. Not in the manner that AC described but rather whether the core has reached even a small level of saturation.
These are the results of my current tests but more work is needed to exactly define the conditions.
Pm
Quote from: partzman on 2025.09.23, 20:13:14
My most recent testing shows an inductor will reach the expected V/turn of the primary.
Measured like this ?
Quote from: verpies on 2025.09.23, 21:32:00
Measured like this ?
No! With the probe ground lead inside the E-Field area, it will charge separate and the resultant reading at the scope will be zero volts. The probe ground lies outside the core. With this arrangement, the resultant voltage will be the V/turn of the primary.
Quote from: partzman on 2025.09.23, 20:13:14
L/C resonance can only exist in the center hole if L is affected by the host core.
In your post #53 you showed some measurements for the set up in your image repeated here below. Clearly this LC was in resonance since your hall probe CH4(grn) was showing over 1 amp AC current. IMO the drive for this was not the volts for a single turn, that would imply the drive current coming through your CH3(pnk) scope probe impedance (like 10pF and 10 megohms?). IMO it was the capacitor C1 acting as an antenna to the E field there. If you disconnected CH3 I say you would still see that current, maybe you could do that to check it out. If you still get that resonance it is possible that capacitance from the wire where the hall probe is clamped, through the hall probe and then via the scope ground which passes outside the core and brings in the volts per turn could be the driver, but I suspect not. Is there any data for that leakage capacitance?
Smudge
Quote from: Smudge on 2025.09.24, 15:07:37
In your post #53 you showed some measurements for the set up in your image repeated here below. Clearly this LC was in resonance since your hall probe CH4(grn) was showing over 1 amp AC current. IMO the drive for this was not the volts for a single turn, that would imply the drive current coming through your CH3(pnk) scope probe impedance (like 10pF and 10 megohms?). IMO it was the capacitor C1 acting as an antenna to the E field there. If you disconnected CH3 I say you would still see that current, maybe you could do that to check it out.
You are correct in that the V/turn is not creating the resonance between L1 and C1 that is, if the primary is operating on the linear portion of the BH curve. If this is the case, there will be little to no resonance between L1 and C1.
In post #53 however, there is a great deal of saturation in the primary and we therefore have a large amount of flux outside the boundaries of the core. This flux is the primary driving source for L1 that creates the resonant condition. There is also a reciprocal condition in that L1 supplies an amount of H-Field back to the host core and the effects depend on the phasing and position of L1. Very interactive!
Yes, when C3 is disconnected, the resonant current current is still there.
Quote
If you still get that resonance it is possible that capacitance from the wire where the hall probe is clamped, through the hall probe and then via the scope ground which passes outside the core and brings in the volts per turn could be the driver, but I suspect not. Is there any data for that leakage capacitance?
Smudge
The current probe is a Tek TCP0020 and I could find no capacitance data for it. However, with the current probe removed and CH3 disconnected, the resonance condition still exists as can be seen by CH3 being placed close to the circuit.
If C1 is biased by an external voltage source and L1 is switched to ground during the time that voltage across C1 is increased by the V/turn of the primary, we will have resonance even with a linear primary as discussed previously.
Pm
Quote from: partzman on 2025.09.24, 20:55:17
In post #53 however, there is a great deal of saturation in the primary and we therefore have a large amount of flux outside the boundaries of the core. This flux is the primary driving source for L1 that creates the resonant condition. There is also a reciprocal condition in that L1 supplies an amount of H-Field back to the host core and the effects depend on the phasing and position of L1. Very interactive!
OK. Just to recap in case readers are lost, in post #47 you showed the waveforms and in post #53 you showed the circuit. You now have an explanation for how the LC circuit gets driven which makes more sense than my drivel. However in post#47 you said the resonance was not seen by the primary. Is there some form of non-reciprocity here that could be used to advantage?
Smudge
Quote from: Smudge on 2025.09.25, 08:58:02
OK. Just to recap in case readers are lost, in post #47 you showed the waveforms and in post #53 you showed the circuit. You now have an explanation for how the LC circuit gets driven which makes more sense than my drivel. However in post#47 you said the resonance was not seen by the primary. Is there some form of non-reciprocity here that could be used to advantage?
Smudge
Yes, there does appear to be conditions that indicate non-reciprocity. This being the orientation of L1 to the host core. See below for example.
In the attachments below, "Pri Curr1" shows two different primary currents stored in Ref1 and Ref2 that were the result of different physical placements of L1 relative to the host core with the circuit in resonance. In both cases, the axis of L1 is horizontal and L1's ferrite core is touching the host core. The maximum trace is one position and the minimum trace is with L1 rotated 180 degrees. Obviously we have a primary current increase from one position to the other that is apparently determined by the flux directions in L1 and the host core.
"Pri Curr2" now shows the primary current in CH4(grn) (that is superimposed over Ref2) that was measured with the resonance circuit removed from the host core. It appears this particular orientation of L1 does not affect the primary current. I know there is missing data here so I'll run a more complete demo of this if there is interest.
Pm
""Pri Curr2" now shows the primary current in CH4(grn) (that is superimposed over Ref2) that was measured with the resonance circuit removed from the host core. It appears this particular orientation of L1 does not affect the primary current. I know there is missing data here so I'll run a more complete demo of this if there is interest.
Pm""
I'm interested, although not following everything you're doing...
Question: is there any way for you to develop what you're doing into a SELF-RUNNER? even at very low power...
Quote from: PhysicsProf on 2025.09.25, 18:16:46
""Pri Curr2" now shows the primary current in CH4(grn) (that is superimposed over Ref2) that was measured with the resonance circuit removed from the host core. It appears this particular orientation of L1 does not affect the primary current. I know there is missing data here so I'll run a more complete demo of this if there is interest.
Pm""
I'm interested, although not following everything you're doing...
Question: is there any way for you to develop what you're doing into a SELF-RUNNER? even at very low power...
Well, that is the goal but nothing to report at the moment!
Pm
Quote from: partzman on 2025.09.25, 14:12:20
... It appears this particular orientation of L1 does not affect the primary current.
...
This situation is not unusual. Since the secondary circuit is coupled to the primary circuit, the load and losses are distributed across both circuits and may vary in each circuit depending on the mutual coupling, while the sum remains unchanged, as does the energy supplied by the primary circuit.
Quote from: F6FLT on 2025.09.26, 16:34:35
This situation is not unusual. Since the secondary circuit is coupled to the primary circuit, the load and losses are distributed across both circuits and may vary in each circuit depending on the mutual coupling, while the sum remains unchanged, as does the energy supplied by the primary circuit.
It seems quite significant to me that the primary current would remain the same with the resonant circuit removed and also with the resonant circuit inserted in the core in a particular orientation, assuming the resonant circuit is excited by this orientation.
Jon I got Grok to check out the paper that Hakasys posted in the chat and how it related to your setup. I'm not smart enough to know if it's slop or not but it's how I'm following along.
"This recent paper seems highly relevant to the ongoing discussion here on E-field charge separation, displacement currents, and the apparent lack of energy draw from the primary in open-circuit scenarios. It's titled "Nonlocal or Possibly Superluminal Maxwell Displacement Current Observed in the Near-field of a Spherical Capacitor" by Markoulakis, Walker, and Antonidakis, published in IRECAP Vol. 14, Issue 4 (2024, revised Feb 2025). Full text available at: https://papers.ssrn.com/sol3/papers.cfm?abstract_id=4790873 or DOI: 10.15866/irecap.v14i4.24903.
In summary, the authors conducted experiments with a large spherical air-dielectric capacitor (1.5m poles separated by 1.5m) pulsed at high voltage. They claim to observe the displacement current (Maxwell's μ₀ε₀∂E/∂t term) behaving nonlocally in the near-field—meaning instantaneous action-at-a-distance between the poles, with no measurable propagation delay (implying possibly superluminal speed, >>c). Delays were only from conduction currents in wires/electrodes, not the displacement itself. They argue this confirms Maxwell's original prediction for near-fields, where polarization of space (aether-like) allows instant effects without violating relativity for far-fields. Confidence level: 80% from stats on 100+ runs.
How does this tie into our thread?
Quoting partzman's OP:
IMO, charge separation occurs in the secondary via the E-Field generated in the primary. This secondary emf then is capable of producing usable power when loaded.
So, what theory supports this action? IMO it is the power flow or Poynting vector designated as S=EXH... The core window area appears to act as a waveguide for the E and H Fields as the primary E-Field appears to be within this core area.
And Smudge's Reply #1:
My next image is a huge parallel plate capacitor almost filling the space within the toroidal core. The significant feature of this is the capacitor gets charged without any external current to it... Of course the displacment of electric charge within the dielectric is a form of current flow, so the primary does see that, the energy gained in the capacitor comes from the 3V input.
The paper's nonlocal displacement could explain why charge separation/polarization happens "instantly" in the core window without apparent propagation time or energy reflection back to the primary (no Lenz if non-flux-enclosing, as discussed). It aligns with partzman's aether claims in Reply #25:
In all my experimentation with this type of charge separation, at no time have I ever seen energy taken from the primary for charge separation in any open circuited object. IMO, the energy required for this charge increase comes from the aether...
If displacement is nonlocal/instant in near-fields, it might act like an "aether polarization" that transfers info/energy without finite speed, explaining the no-draw observation. This also resonates with Allcanadian's cap paradox in #26 (redistribution as cause) and verpies/F6FLT's hysteresis emphasis (#20, #22, #28)—the paper notes bound charges in dielectrics enable this without free electron flow.
More directly, in partzman's LC resonance test (#50-53 quotes):
ICR1 shows the differential at the charge separation switching to be 6.15v. Note this is larger than the 4V/T of the primary... This LC resonance is not seen by the primary... So, I leave you with the question, what is supplying the energy to this resonance circuit?
(Smudge reply): It is from the 40V supply that is seeing current impulses that integrate to a non-zero value...
If the displacement current driving the LC is nonlocal, it could sustain the resonance instantly via E-field polarization, without the primary "seeing" it as a load—matching the paper's instant signaling between capacitor poles. No finite delay means no phased opposition, potentially amplifying effects like the >4V/T voltage.
Also echoes Centraflow's bifilar coil-caps (#4-6) and Hakasays' electret ideas (#7,10,12)—nonlocal effects might enhance curl fields or remnant polarization in ferroelectrics.
Worth testing? Maybe replicate their setup on a smaller scale in the toroid hole—pulse one "pole" (e.g., a plate), probe the other for zero-delay response. Could explain why vertical plates gave COP<1 but disc/LC shows promise."
Quote from: JimBoot on 2025.10.06, 02:47:17
Jon I got Grok to check out the paper that Hakasys posted in the chat and how it related to your setup. I'm not smart enough to know if it's slop or not but it's how I'm following along.
"This recent paper seems highly relevant to the ongoing discussion here on E-field charge separation, displacement currents, and the apparent lack of energy draw from the primary in open-circuit scenarios. It's titled "Nonlocal or Possibly Superluminal Maxwell Displacement Current Observed in the Near-field of a Spherical Capacitor" by Markoulakis, Walker, and Antonidakis, published in IRECAP Vol. 14, Issue 4 (2024, revised Feb 2025). Full text available at: https://papers.ssrn.com/sol3/papers.cfm?abstract_id=4790873 or DOI: 10.15866/irecap.v14i4.24903.
In summary, the authors conducted experiments with a large spherical air-dielectric capacitor (1.5m poles separated by 1.5m) pulsed at high voltage. They claim to observe the displacement current (Maxwell's μ₀ε₀∂E/∂t term) behaving nonlocally in the near-field—meaning instantaneous action-at-a-distance between the poles, with no measurable propagation delay (implying possibly superluminal speed, >>c). Delays were only from conduction currents in wires/electrodes, not the displacement itself. They argue this confirms Maxwell's original prediction for near-fields, where polarization of space (aether-like) allows instant effects without violating relativity for far-fields. Confidence level: 80% from stats on 100+ runs.
How does this tie into our thread?
Quoting partzman's OP:
IMO, charge separation occurs in the secondary via the E-Field generated in the primary. This secondary emf then is capable of producing usable power when loaded.
So, what theory supports this action? IMO it is the power flow or Poynting vector designated as S=EXH... The core window area appears to act as a waveguide for the E and H Fields as the primary E-Field appears to be within this core area.
And Smudge's Reply #1:
My next image is a huge parallel plate capacitor almost filling the space within the toroidal core. The significant feature of this is the capacitor gets charged without any external current to it... Of course the displacment of electric charge within the dielectric is a form of current flow, so the primary does see that, the energy gained in the capacitor comes from the 3V input.
The paper's nonlocal displacement could explain why charge separation/polarization happens "instantly" in the core window without apparent propagation time or energy reflection back to the primary (no Lenz if non-flux-enclosing, as discussed). It aligns with partzman's aether claims in Reply #25:
In all my experimentation with this type of charge separation, at no time have I ever seen energy taken from the primary for charge separation in any open circuited object. IMO, the energy required for this charge increase comes from the aether...
If displacement is nonlocal/instant in near-fields, it might act like an "aether polarization" that transfers info/energy without finite speed, explaining the no-draw observation. This also resonates with Allcanadian's cap paradox in #26 (redistribution as cause) and verpies/F6FLT's hysteresis emphasis (#20, #22, #28)—the paper notes bound charges in dielectrics enable this without free electron flow.
More directly, in partzman's LC resonance test (#50-53 quotes):
ICR1 shows the differential at the charge separation switching to be 6.15v. Note this is larger than the 4V/T of the primary... This LC resonance is not seen by the primary... So, I leave you with the question, what is supplying the energy to this resonance circuit?
(Smudge reply): It is from the 40V supply that is seeing current impulses that integrate to a non-zero value...
If the displacement current driving the LC is nonlocal, it could sustain the resonance instantly via E-field polarization, without the primary "seeing" it as a load—matching the paper's instant signaling between capacitor poles. No finite delay means no phased opposition, potentially amplifying effects like the >4V/T voltage.
Also echoes Centraflow's bifilar coil-caps (#4-6) and Hakasays' electret ideas (#7,10,12)—nonlocal effects might enhance curl fields or remnant polarization in ferroelectrics.
Worth testing? Maybe replicate their setup on a smaller scale in the toroid hole—pulse one "pole" (e.g., a plate), probe the other for zero-delay response. Could explain why vertical plates gave COP<1 but disc/LC shows promise."
Excellent work Jim, a good read over my coffee this morning.
The question is always, where and when did the external energy entre into the circuit
This was my problem in my work. I found that simple positioning of the dut would enable it to work or not, all other things being the same.
I am still open to how it works, and I am still looking on with great interest with Jons work.
Yes, I concur.
Where? Between 2 conductors in a magnetic field.
When? At greater than C and beyond our current ability to measure.
With these 2 assumptions .We know how to create a local magnetic field , we are in one all the time
Also assume that there is instant charge from nowhere between conductors as long as we don't short them out at the wrong time .
It is understandably too much to swallow for some .
Mind blowing stuff !
Here is a simple charge separation experiment that first charge separates C1 and then after 1us, a 60uH coil L2 is shorted across the output of C1. L2 and it's circuitry are located outside the core.
The first pix is the schematic of the test. The basic drive is from a 3/4 bridge shown in block form on the left of the schematic with the charge separation circuitry is shown in detail on the right. Mosfet M1 is switched on 1us after the application of 64 volts to the primary L1.
The first scope pix shows the bridge gate drive pulse on CH1(yel), the voltage across C1 on CH3(pnk), and the current in L2 on CH4(grn). Notice the ~1MHz sine wave current on CH4 for the first 1us of charge separation. Also notice the voltage drop in C1 of 789mv in CH3 as L2 charges.
The second scope pix shows the start and finish currents in L2. C1 is able to apparently supply energy to charge L2 to a peak current of 178ma which is the point of this demo. I will not bore everyone with my comments nor measurement analysis but rather leave this all up to you!
PM
Edit: Errors corrected regarding L1 as Smudge pointed out in his post #106.
Edit: Removed L1 strike throughs.
In the previous post #102, the capacitor used for C1 is a wound film type. It is placed vertically in the core so the plates are oriented as seen in the toroid cross section attached below. With the polarity of the E-Field as shown, C1 will charge separate with the polarities shown.
Using the hi-Z scope probe for a sniffer (not physically connected but laying close to a terminal), it appears that if the negative terminal of C1 is grounded, there will be no appreciable potential difference between the grounded terminal of C1 and the + terminal of C1. IOW, no apparent charge separation. So, this means the only charge path for C1 is thru any and all outside connections between the + terminal and ground. In this case we have the mosfet M1 with a Coss=100pf and a 3pf scope probe with 10Meg of resistance plus any stray capacitance. We can ignore the resistance and assume another 100pf of stray capacitance. This 203pf is the load to C1 before M1 conducts.
Now we look closely at the rise time of the charge separated voltage across C1 at the very start of the cycle and conservatively call it 100ns. From the experimental evidence above, it appears that C1 is fully charged to ~2.9v within this 100ns time period. If this is true, this would require di=dE*C/dt or di=2.9*.461e-6/100e-9=13.3A . However, our external conductive load is ~203pf so our available current to C1 is 2.9*203e-12/100e-9=6ma.
Obviously our outside path does not appear to be capable of supplying the current necessary to fully charge C1 as it appears.
So, does anyone wish to explain the physics that the experimental evidence shows?
Pm
Quote from: partzman on 2025.10.08, 20:51:58
In the previous post #102, the capacitor used for C1 is a wound film type. It is placed vertically in the core so the plates are oriented as seen in the toroid cross section attached below. With the polarity of the E-Field as shown, C1 will charge separate with the polarities shown.
NO!! The electric field will drive charge from one end of the capacitor plates to the other. The result will be as depicted in my addition to your image shown below. Although this looks like a charged capacitor the + and - signs denote the surface charge on each plate, and each plate has positive charge at one end and negative charge at the other. IOW at any point along the plate the dielectric is not stressed as at that point all plates carry the same polarity of surface charge. Your instrument measuring the voltage across the capacitor is initially treating the capacitor like a single conductor within the core and shows the open circuit voltage that you would expect to see with such a conductor placed there.
QuoteUsing the hi-Z scope probe for a sniffer (not physically connected but laying close to a terminal), it appears that if the negative terminal of C1 is grounded, there will be no appreciable potential difference between the grounded terminal of C1 and the + terminal of C1. IOW, no apparent charge separation. So, this means the only charge path for C1 is thru any and all outside connections between the + terminal and ground. In this case we have the mosfet M1 with a Coss=100pf and a 3pf scope probe with 10Meg of resistance plus any stray capacitance. We can ignore the resistance and assume another 100pf of stray capacitance. This 203pf is the load to C1 before M1 conducts.
Now we look closely at the rise time of the charge separated voltage across C1 at the very start of the cycle and conservatively call it 100ns. From the experimental evidence above, it appears that C1 is fully charged to ~2.9v within this 100ns time period.
No for the reason stated above. C1 is not charged. If you bring C1 outside the core while its connecting wire remains inside the core you will still get that 2.9V, the only difference being the electric field is driving charge along the wire and not along the capacitor plates.
QuoteIf this is true, this would require di=dE*C/dt or di=2.9*.461e-6/100e-9=13.3A . However, our external conductive load is ~203pf so our available current to C1 is 2.9*203e-12/100e-9=6ma.
Which over the first 1uS creates 13mV charge. You could bring C1 outside the core keeping the circuit connections the same and do a differential measurement of voltage across it to check that.
QuoteObviously our outside path does not appear to be capable of supplying the current necessary to fully charge C1 as it appears.
So, does anyone wish to explain the physics that the experimental evidence shows?
I hope I have done that.
Smudge
Smudge,
Thank you for your concise response! I will respond, but I first would like to see if there are any other thoughts on the experimental evidence of this circuitry.
Pm
Further to my reply #104 above, in reply #102 Pm shows CH4(grn) as being the current in L1 which is the primary. He also says a 60uH coil L1 is connected across the output of C1 which is incorrect, it is L2 that is connected there. We must assume these are simply typos and indeed the current rise to 178mA was that measured in L2. That seems reasonable as L2 resonates with C1 at 30KHz and that current rise over 4uS is reasonable for the 3.2V drive in the single turn. Also reasonable is the 789mV drop seen on CH3(pnk) as that is actually C1 charging (not discharging) via current through L2. It appears like a discharge because the scope is initially seeing the notional +3.2V for a single turn (not the capacitor charge) and when current flows through L2 it charges the capacitor +ve at the bottom (it flow in there) and -ve at the top (it flow out there) (the polarity is opposite of what PM shows) hence it subtracts from that initial value.
Smudge
Quote from: Smudge on 2025.10.09, 14:41:32
Further to my reply #104 above, in reply #102 Pm shows CH4(grn) as being the current in L1 which is the primary. He also says a 60uH coil L1 is connected across the output of C1 which is incorrect, it is L2 that is connected there. We must assume these are simply typos and indeed the current rise to 178mA was that measured in L2. That seems reasonable as L2 resonates with C1 at 30KHz and that current rise over 4uS is reasonable for the 3.2V drive in the single turn. Also reasonable is the 789mV drop seen on CH3(pnk) as that is actually C1 charging (not discharging) via current through L2. It appears like a discharge because the scope is initially seeing the notional +3.2V for a single turn (not the capacitor charge) and when current flows through L2 it charges the capacitor +ve at the bottom (it flow in there) and -ve at the top (it flow out there) (the polarity is opposite of what PM shows) hence it subtracts from that initial value.
Smudge
Yes, you are correct regarding my typos with L1 and I have made the corrections in my post.
As to your other comments here, I will still wait a bit longer to comment to see if others are willing to chime in!
Pm
Ok Iwill have a go.
if i understood Smudge correctly, this would men that a ceramic cap would also seem to charge.
How does the peak current in L2 compare when C1 is removed and replaced with a simple conductor?
It would seem according to your schematic that you likely still have a rapidly changing flux in L1 which could account for the energy transfer to L2 when M1 is switched on. C1 discharge would be a combination of the EMF induced from the transformer action as well as any charge in the dielectric. It just doesn't seem like we could call this strictly a dielectric discharge alone.
Dave
Quote from: 3D Magnetics on 2025.10.09, 23:26:22
Ok Iwill have a go.
if i understood Smudge correctly, this would men that a ceramic cap would also seem to charge.
No. I think the misunderstanding all hinges on the perception that there is an E field within the hole in the core and none elsewhere. That is wrong, the E field forms closed circles around the core flux. The E field is not uniform along those E field closed lines and certainly the field in the core hole is greater than that outside. The whole closed line integral of that E field is the volts/turn from the changing flux. Any closed circuit that encloses the flux obtains that voltage. A closed circuit that does not encircle the flux (like one that is wholly within the core hole) obtains zero induced voltage. That does not stop a small electric antenna (a short piece of isolated wire) within the core hole being driven by the E field there to create its own electric dipole field from oscillating charge appearing at its tips. That aslo applies to a lump of dielectric within the hole and that is true delectric displacement. Pm's C1 capacitor has conductive electrodes along the E field so they get charge separation and his isolated device does act like a short length of conductor electric dipole. If he put RF into his primary he could use a radio receiver to discover the radiation from his isolated capacitor that is acting like a small antenna. But his 0.461uF capacitor is not being charged and discharged. His apparent voltage seen by the scope is due to the scope probe connection forming a closed circuit around the core flux which gets that single turn voltage induced into it. As I pointed out if that induced voltage is allowed to drive current to charge the capacitor it would get the opposite polarity to the observed voltage.
3D, Dave, Smudge, and other readers,
Please do not think I'm being rude, but I'm waiting for anyone else that might wish to comment on all of this before responding to any and all comments.
Regards,
Pm
Thanks Smudge ,
You may be right but I doubt that Jon would have missed that obvious explanation.
Like it only there because of the measurement taking place .
It aligns with Mikes work is some respects and his cap is surely being charged.
More opinions either way will help Jon to explain it so c`mon guys, poke some stick about!
I must have missed something because I no longer understand the point.
In view 1, as the circuit is not closed, no current flows along the plate. A plate is a conductor. Consequence: the same potential at the top and bottom.
In view 2, this is what is interesting. The field induces polarisation of the capacitor's dielectric. We will therefore measure a voltage at its terminals and use it.
Isn't that the point?
Smudge,
Well, it looks like all that are going to respond have, so here is my general response to your physics on my charge separation circuitry. In general I mostly disagree with your analysis and I hope to show why with this post.
First, let's take your image in post #104 showing the charges on the capacitor plates. I have already been through this thinking process and I too logically concluded that there would be no apparent energy available in C1 if this were the case. However, there is real energy available in C1 as will be demonstrated below! So, there are only two explanations to resolve this. One, your image depicts the correct polarities and there is some other mechanism that requires further investigation to help resolve this dilemma or two, the unconnected ends of the plates do not have the charges as shown! I have attempted tests to prove this point one way of the other but they are too unreliable at this point to have any solid conclusions. I hope to be able to provide this in the future however.
So, that brings us to what I feel is proof that C1 does contain real energy after charge separation. The last pix is the schematic and shows that this circuit has been modified from the original shown in my post #102 in that L2 has been replaced with R2 and D2. R2 provides a load which simply dissipates energy instead of storing energy and D2 prevents R2 from conducting thru the substrate diode of M1 during the negative half cycle.
The current probe CH4(grn) is placed as shown and take note of the arrow indicating the direction of the conventional (+) current flow. This arrow is printed on the head of the current probe so there can be no confusion. At this point I will remind the reader that if a positive current is taken from or exiting the positive terminal of a charged capacitor, the capacitor is being discharged or losing energy. If a positive current is entering the positive terminal of a charged capacitor, the capacitor is being charged or gaining energy. If this is confusing, I'll be happy to demonstrate!
The first scope pix shows the circuit operating with no connection to the R2/D2 network. CH3(pnk) shows the voltage across C1 and CH4(grn) shows the current in L1. In theory there should be no difference between the start and finish voltages across C1 but due to the 3/4 bridge drive, there is a some.
DCE1 pix shows the circuit in operation. CH1(yel) is the gate drive to the 3/4 bridge, CH2(blu) is the supply voltage, CH3(pnk) is the voltage across C1, and CH4(grn) is the current thru R2/D2. R2/D2 is switched across C1 after a 4us delay of the rising edge of CH1. The important measurements to note here are cursors A and B that show a differential of 4v from prior to start to charge separation before R2 is connected to the circuit.
DCE2 pix now shows the time interval when the load resistance of 20 ohm-1% R2 is connected across C1 to ground via M1. Here we see a voltage drop in C1 of 1.44v and an average current thru R2 of 125.7ma. We can now calculate the voltage drop across C1 using dE=di*dt/C or dE=.1257*12.43us/1.07e-6=1.46v. So, we have measured and calculated voltage drops that are reasonably close for our charged capacitor connected to our resistive load over a given period of time!
DCE3 shows the differential voltage of 8.08v after C1 is discharged by R2 until the C1 reaches the most negative value via charge separation. This is commensurate with the fact that we have a 4v/turn primary. If this is confusing, remember that the primary L1 first falls to zero volts during the first half of the bridge transition (first 4v/turn) and then the bridge reverses the voltage across L1 (second 4v/turn) for a total of 8v across C1.
DCE4 now shows the very real voltage left on C1 after the cycle is complete when the bridge is no longer conducting and the voltage across L1 has returned to zero. Here we see the difference via the A and B cursors from a start of 0v across C1 to a finish of 1.48v across C1. This is reasonably equal to the voltage drop during the first phase discharge of C1,
IMO, this indicates that C1 is nearly instantly charged to it's full energy level by the E-Field in the center of the core and this energy can be manipulated. I would welcome any other explanation.
Regards,
Pm
Quote from: partzman on 2025.10.13, 15:37:11
Smudge,
Well, it looks like all that are going to respond have, so here is my general response to your physics on my charge separation circuitry. In general I mostly disagree with your analysis and I hope to show why with this post.
First, let's take your image in post #104 showing the charges on the capacitor plates. I have already been through this thinking process and I too logically concluded that there would be no apparent energy available in C1 if this were the case. However, there is real energy available in C1 as will be demonstrated below! So, there are only two explanations to resolve this. One, your image depicts the correct polarities and there is some other mechanism that requires further investigation to help resolve this dilemma or two, the unconnected ends of the plates do not have the charges as shown! I have attempted tests to prove this point one way of the other but they are too unreliable at this point to have any solid conclusions. I hope to be able to provide this in the future however.
So, that brings us to what I feel is proof that C1 does contain real energy after charge separation. The last pix is the schematic and shows that this circuit has been modified from the original shown in my post #102 in that L2 has been replaced with R2 and D2. R2 provides a load which simply dissipates energy instead of storing energy and D2 prevents R2 from conducting thru the substrate diode of M1 during the negative half cycle.
The current probe CH4(grn) is placed as shown and take note of the arrow indicating the direction of the conventional (+) current flow. This arrow is printed on the head of the current probe so there can be no confusion. At this point I will remind the reader that if a positive current is taken from or exiting the positive terminal of a charged capacitor, the capacitor is being discharged or losing energy. If a positive current is entering the positive terminal of a charged capacitor, the capacitor is being charged or gaining energy. If this is confusing, I'll be happy to demonstrate!
The first scope pix shows the circuit operating with no connection to the R2/D2 network. CH3(pnk) shows the voltage across C1 and CH4(grn) shows the current in L1. In theory there should be no difference between the start and finish voltages across C1 but due to the 3/4 bridge drive, there is a some.
You would get the same scope traces if C1 were replaced by a piece of wire. The scope is seeing the voltage across a single turn. You would not claim the piece of wire is charged to 4V. If you keep the piece of wire there but bring C1 outside the core you would get the same scope pics. There you could scope both sides of C1 to see the same voltage as you would expect. With such a fast rise time C1 doesn't have time to gain charge. Same happens with C1 in the core hole so it acts just like a piece of wire and your scope is seeing induced voltage, not charged voltage.
QuoteDCE1 pix shows the circuit in operation. CH1(yel) is the gate drive to the 3/4 bridge, CH2(blu) is the supply voltage, CH3(pnk) is the voltage across C1, and CH4(grn) is the current thru R2/D2. R2/D2 is switched across C1 after a 4us delay of the rising edge of CH1. The important measurements to note here are cursors A and B that show a differential of 4v from prior to start to charge separation before R2 is connected to the circuit.
Not charge separation as in charging of C1 but the same induced charge separation you get along a length of wire.
QuoteDCE2 pix now shows the time interval when the load resistance of 20 ohm-1% R2 is connected across C1 to ground via M1. Here we see a voltage drop in C1 of 1.44v and an average current thru R2 of 125.7ma. We can now calculate the voltage drop across C1 using dE=di*dt/C or dE=.1257*12.43us/1.07e-6=1.46v. So, we have measured and calculated voltage drops that are reasonably close for our charged capacitor connected to our resistive load over a given period of time!
That is not a voltage drop of a charged C1, it is C1 starting at 0V charge enabling the induced 4V to be seen by the scope, i.e. initially acting like a piece of wire, but not being a piece of wire it then receives charge by induced current flow so it gains charge and also voltage, voltage that subtracts from the induced constant 4V yielding exactly what you see.
QuoteDCE3 shows the differential voltage of 8.08v after C1 is discharged by R2 until the C1 reaches the most negative value via charge separation. This is commensurate with the fact that we have a 4v/turn primary. If this is confusing, remember that the primary L1 first falls to zero volts during the first half of the bridge transition (first 4v/turn) and then the bridge reverses the voltage across L1 (second 4v/turn) for a total of 8v across C1.
That 8V (+-4V) is induced voltage, not induced charge. What you see there is exactly as expected.
QuoteDCE4 now shows the very real voltage left on C1 after the cycle is complete when the bridge is no longer conducting and the voltage across L1 has returned to zero. Here we see the difference via the A and B cursors from a start of 0v across C1 to a finish of 1.48v across C1. This is reasonably equal to the voltage drop during the first phase discharge of C1,
It is the charge received by C1 during the first phase.
QuoteIMO, this indicates that C1 is nearly instantly charged to it's full energy level by the E-Field in the center of the core and this energy can be manipulated.
I disagree. A fast voltage step will be passed by any C at zero charge of C as you well know. Only after the step will C get charged by current flow. During that fast rise of voltage C acts just like a piece of wire. That same reasoning applies when C is within the core hole, it initially acts like a piece of wire. There the tiny initial flow of current along a wire to get one end at the induced voltage level relative to the other also applies to C, it is so tiny that we ignore it. And it is not C being charged.
Smudge
Quote from: 3D Magnetics on 2025.10.10, 14:39:19
Thanks Smudge ,
..........It aligns with Mikes work is some respects and his cap is surely being charged.
Mike Nunnerley's work certainly deserves some attention. His toroidal bifilar coils on top of each other used as capacitors raises questions on how the electron gas on the surface of a negative electrode at high voltage can act like a ferromagnetic core where the otherwise randomly orientated magnetic dipoles get aligned in an applied B field. With his layout those surface charges do form a ring so there could be a closed magnetic reluctance there that is not taken into account. Indeed a reluctance that is switched on and off, something that doesn't exist in our EM theory but could open the door to exotic effects.
Smudge
Quote from: partzman on 2025.10.13, 15:37:11
Smudge,
Well, it looks like all that are going to respond have, so here is my general response to your physics on my charge separation circuitry. In general I mostly disagree with your analysis and I hope to show why with this post.
[...]
IMO, this indicates that C1 is nearly instantly charged to it's full energy level by the E-Field in the center of the core and this energy can be manipulated. I would welcome any other explanation.
Regards,
Pm
Can you not pull the capacitor out after the intial "charging" and measure its voltage? So we can see if it is holding a real charge or it was only some induced voltage seen on the scope?
Regards
Quote from: Smudge on 2025.10.14, 08:42:32
You would get the same scope traces if C1 were replaced by a piece of wire. The scope is seeing the voltage across a single turn. You would not claim the piece of wire is charged to 4V. If you keep the piece of wire there but bring C1 outside the core you would get the same scope pics. There you could scope both sides of C1 to see the same voltage as you would expect. With such a fast rise time C1 doesn't have time to gain charge. Same happens with C1 in the core hole so it acts just like a piece of wire and your scope is seeing induced voltage, not charged voltage.
This is not correct as is experimentally shown below. The schematic remains the same except that C1 has been replaced by a piece of wire.
"DCE1 wire" shows the scope traces with the same labeling as before. Especially note CH3(pnk) has no change in voltage potential at the time M1 turns on. Why? Because the piece of wire is just that and no more! It has little inductance, capacitance, and resistance, but it is charge separated to 4v and therefore represents a reasonably stiff voltage source for R2. This is in contrast to C1 which definitely is a capacitor that has been fully charged in nanoseconds and exhibits a voltage drop commensurate with the 20 ohm load over the 12.43us time period.
"DCE2 wire" shows the voltage across the wire in CH3 and the resulting current thru R2 and D2 to be 3.913v and 172ma respectively. Considering the voltage drop in D2 to be .4v, the resultant current in R2 would be (3.913-.4)/20=175.6ma which is reasonably close to the scope measurement.
Regards,
Pm
Edit: Replaced "DCE2 wire" with new version that has the proper current polarity thru R2/D2.
Quote from: 3D Magnetics on 2025.10.09, 23:26:22
Ok Iwill have a go.
if i understood Smudge correctly, this would men that a ceramic cap would also seem to charge.
I have not found any type of capacitor that will not charge separate!
Pm
Quote from: web000x on 2025.10.09, 23:51:37
How does the peak current in L2 compare when C1 is removed and replaced with a simple conductor?
The peak current in L2 would be much greater in this case for the same reasons stated in my post #117.
[/quote]
It would seem according to your schematic that you likely still have a rapidly changing flux in L1 which could account for the energy transfer to L2 when M1 is switched on. C1 discharge would be a combination of the EMF induced from the transformer action as well as any charge in the dielectric. It just doesn't seem like we could call this strictly a dielectric discharge alone.
Dave
[/quote]
The changing flux in L1 is based on the simple turns ratio of secondary to primary and the secondary load as in any transformer. Lenz is still in full effect here. IOW, the discharge current in C1 is reflected back to the primary but we also can see that the energy lost in C1 is equal to that gained in L2 within reason.
Pm
Quote from: Frederik2k1 on 2025.10.14, 10:14:50
Can you not pull the capacitor out after the intial "charging" and measure its voltage? So we can see if it is holding a real charge or it was only some induced voltage seen on the scope?
Regards
With the capacitor placed in the E-Field of a core, it will follow the voltage of the primary in magnitude and polarity. The magnitude is reduced somewhat by the effective coupling factor of the the primary to capacitor. If the capacitor is removed from the core, all E-Field influence is gone.
Pm
Grok's take.
"Strong evidence for partzman's OP—E-field polarizes dielectric, creating usable EMF without traditional coupling. But conservation holds in calcs (output << input, ~0.1% efficiency if primary E ~10–20µJ). Anomalies could be parasitics (probe loading ~10pF adds C, altering rise). Why no primary draw? Displacement confines to window (waveguide analogy), minimizing reflection." and "No overunity yet; gains likely losses elsewhere (eddy, radiation). Exciting for unconventional transformers (e.g., pulsed HV).
If testing: Replace C1 with air/ceramic (#111) for pure displacement; scope primary VA with/without load."
Quote from: partzman on 2025.10.14, 14:32:20
This is not correct as is experimentally shown below.
With respect I say the experiment shown below exactly supports my view.
QuoteThe schematic remains the same except that C1 has been replaced by a piece of wire.
"DCE1 wire" shows the scope traces with the same labeling as before. Especially note CH3(pnk) has no change in voltage potential at the time M1 turns on.
If you compare your DCE2 wire.png with DCE2.png of your post #117 the capacitor also shows no change of voltage at that point.
QuoteWhy? Because the piece of wire is just that and no more! It has little inductance, capacitance, and resistance, but it is charge separated to 4v and therefore represents a reasonably stiff voltage source for R2. This is in contrast to C1 which definitely is a capacitor that has been fully charged in nanoseconds
There we have the difference in our perceptions. You are prepared to accept the wire gets "charge separated" (quantity of electrons driven from one end to the other) for the scope to see the 4V/turn potential. What quantity of charge? It has to be that which charges the scope probe capacitance to 4V. With C1 there we get that same small quantity of charge into the scope probe but now you claim C1 has magically become fully charged to 4V. I say that is wrong, at that point C1 has only seen a tiny change of charge so virtually still uncharged. The scope probe has 4V charge and that is what is seen. After M1 turns on you say (with C1 there)
Quoteand exhibits a voltage drop commensurate with the 20 ohm load over the 12.43us time period.
You see it as a voltage drop but I see it as a voltage rise as C1 gets charged. What the scope sees is the initial 4V outout falling as C1 gets charged. May I suggest you bring C1 outside the core as shown in your modified image below. I think you will get exactly the same scope traces as your DCE2.png, and my explanation for those traces will be validated. There you can't claim C1 is charged to 4V in a few nanoseconds. Perhaps that result will persuade you that C1 initially acts like the piece of wire in transposing the 4V into the scope, then gets charged.
Quote"DCE2 wire" shows the voltage across the wire in CH3 and the resulting current thru R2 and D2 to be 3.913v and 172ma respectively. Considering the voltage drop in D2 to be .4v, the resultant current in R2 would be (3.913-.4)/20=175.6ma which is reasonably close to the scope measurement.
As expected.
I think your perception of charge separation within a dielectric placed on the core hole has some merit, but only for a large lump of dielectric almost filling the hole having just two electrodes, top and bottom. IMO using a large value capacitor in that hole will not exhibit such behaviour.
Smudge
Smudge,
OK, here are the test results of your modified circuit shown below with C1 placed outside the core and replaced with a wire in the core.
Yes, as you say, the first and second scope pix appear to look identical to the equivalent pix with C1 in the core. We see a change in C1 of 1.48v which is commensurate with 130ma over 12us, plus we see a change in start to finish of -1.40v in C1 as previous.
In the third scope pix Ch3(pnk) is across C1 and CH4(grn) is the current thru C1. Here we see C1 being charged to a negative potential by the negative current. I agree with all this for this circuit. What I don't agree with is that these results are equal to what is happening with C1 in the core!
Why? Because in this case it is quite obvious that the current thru R2/D2 will charge C1 from an initial 0v state. However, how do we logically justify the fact that with C1 in the core while being charge separated to ~4.00v, we see a positive current in R2/D2 that is drawn from C1 which will logically discharge C1, not charge it? Do you not agree?
Regards,
Pm
This is the operation you explain for C1 being in the core
My take on what is happening.
When C1 is inside the toroid the charging electric field between the plates is equal to a wire, a single turn, and a magnetic field is created.
Quote from: partzman on 2025.10.15, 15:23:42
Smudge,
OK, here are the test results of your modified circuit shown below with C1 placed outside the core and replaced with a wire in the core.
Yes, as you say, the first and second scope pix appear to look identical to the equivalent pix with C1 in the core. We see a change in C1 of 1.48v which is commensurate with 130ma over 12us, plus we see a change in start to finish of -1.40v in C1 as previous.
In the third scope pix Ch3(pnk) is across C1 and CH4(grn) is the current thru C1. Here we see C1 being charged to a negative potential by the negative current.
It is the same current flowing around the closed circuit that you show as positive in the first and second scope pics.
QuoteI agree with all this for this circuit. What I don't agree with is that these results are equal to what is happening with C1 in the core!
Why? Because in this case it is quite obvious that the current thru R2/D2 will charge C1 from an initial 0v state. However, how do we logically justify the fact that with C1 in the core while being charge separated to ~4.00v, we see a positive current in R2/D2 that is drawn from C1 which will logically discharge C1, not charge it? Do you not agree?
No. That current logically charges C1 from 0V. I don't see C1 as being charge separated to ~4V, I see C1 showing charge separation identical to a piece of wire so you are seeing an induced voltage, not a charged voltage. I tried to demonstrate this with your image of the capacitor plates where I put + and - symbols against the conductive vertical plates. I try again here where I put colours on those plates to demonstrate you are seeing charge separation along a conductor, not charge separation across the dielectric. C1 is at 0V charge while showing 4V induction and then gains charge. The two scope pics (C1 in the core and out of the core) show identical situations.
Smudge
Quote from: Smudge on 2025.10.15, 17:52:23
It is the same current flowing around the closed circuit that you show as positive in the first and second scope pics.
No. That current logically charges C1 from 0V. I don't see C1 as being charge separated to ~4V, I see C1 showing charge separation identical to a piece of wire so you are seeing an induced voltage, not a charged voltage. I tried to demonstrate this with your image of the capacitor plates where I put + and - symbols against the conductive vertical plates. I try again here where I put colours on those plates to demonstrate you are seeing charge separation along a conductor, not charge separation across the dielectric. C1 is at 0V charge while showing 4V induction and then gains charge. The two scope pics (C1 in the core and out of the core) show identical situations.
Smudge
I of course disagree, so could you show the current flow along with potentials in the charge separated schematic that supports your claim that C1 is charging. I'm sorry but I just don't get it!
Regards,
Pm
Hello Partzman,
Could you briefly explain again what type of capacitor is used inside the toroid? Perhaps you could provide another detailed image showing the capacitor in the setup where it is actually used in the toroid.
I think I've lost track of what type of capacitor is used. Am I correct in assuming that it is not spiral wound?
Quote from: Frederik2k1 on 2025.10.16, 13:40:13
Hello Partzman,
Could you briefly explain again what type of capacitor is used inside the toroid? Perhaps you could provide another detailed image showing the capacitor in the setup where it is actually used in the toroid.
I think I've lost track of what type of capacitor is used. Am I correct in assuming that it is not spiral wound?
Yes, it is a polyester film capacitor that is wound in an oval shape as is seen in the pix below. I assume that this is spiral wound. In any case, the plates are vertical in a horizontally placed core.
Pm
Quote from: partzman on 2025.10.15, 19:52:11
I of course disagree, so could you show the current flow along with potentials in the charge separated schematic that supports your claim that C1 is charging. I'm sorry but I just don't get it!
Regards,
Pm
Smudge,
No reason to detail any more on what I requested above. I see what you are saying and I would agree at this point! Thank you for your patience!
Pm
Quote from: partzman on 2025.10.16, 14:44:02
Smudge,
No reason to detail any more on what I requested above. I see what you are saying and I would agree at this point! Thank you for your patience!
Pm
I started to compose a reply but it quickly developed into a deeper concern involving the difference between two types of E field, that from quasi static charge distrbutions (on conductors) and that from time changing magnetic fields. In particular I show that the virtual particles or virtual photons that carry these E fields do not annihalate when the two E field sum to zero at a region of empty space, only the effect annihalates. The virtual particles pass through that space unhindered. But if that region of space is not empty (like it is part of a conducting wire) then that region (and the rest of the wire) will contain its own sources of both types of virtual particle, sources that respond to the arriving virtual particles. I will eventually produce a document on this deeper issue but it will take some time.
Smudge
Smudge
Quote from: Smudge on 2025.10.17, 07:37:41
I started to compose a reply but it quickly developed into a deeper concern involving the difference between two types of E field, that from quasi static charge distrbutions (on conductors) and that from time changing magnetic fields. In particular I show that the virtual particles or virtual photons that carry these E fields do not annihalate when the two E field sum to zero at a region of empty space, only the effect annihalates. The virtual particles pass through that space unhindered. But if that region of space is not empty (like it is part of a conducting wire) then that region (and the rest of the wire) will contain its own sources of both types of virtual particle, sources that respond to the arriving virtual particles. I will eventually produce a document on this deeper issue but it will take some time.
Smudge
Smudge
Doesn't superposition play a part here, with fields combining at a point in space?
Quote from: Smudge on 2025.10.17, 07:37:41
I started to compose a reply but it quickly developed into a deeper concern involving the difference between two types of E field, that from quasi static charge distrbutions (on conductors) and that from time changing magnetic fields. In particular I show that the virtual particles or virtual photons that carry these E fields do not annihalate when the two E field sum to zero at a region of empty space, only the effect annihalates. The virtual particles pass through that space unhindered. But if that region of space is not empty (like it is part of a conducting wire) then that region (and the rest of the wire) will contain its own sources of both types of virtual particle, sources that respond to the arriving virtual particles. I will eventually produce a document on this deeper issue but it will take some time.
Smudge
Smudge
What is interesting, to me at least, is that if one views C1 to be instantly charged or not, the end results are the same!
Then there is the fact of transposition when bias is applied along with the resulting anomaly.
Pm
Quote from: Smudge on 2025.10.15, 17:52:23
No. That current logically charges C1 from 0V. I don't see C1 as being charge separated to ~4V, I see C1 showing charge separation identical to a piece of wire so you are seeing an induced voltage, not a charged voltage. I tried to demonstrate this with your image of the capacitor plates where I put + and - symbols against the conductive vertical plates. I try again here where I put colours on those plates to demonstrate you are seeing charge separation along a conductor, not charge separation across the dielectric. C1 is at 0V charge while showing 4V induction and then gains charge. The two scope pics (C1 in the core and out of the core) show identical situations.
Smudge
This is what I feel is happening in this circuit. The residual negative voltage on the capacitor is fairly conclusive evidence of this.
Dave
In reference to my 1st post on this thread, see the triangular shaped coil below with 30 turns with a split ferrite filter section placed over one leg of the coil. Also placed thru this ferrite section is a single turn with scope probes attached. Now, before looking at the scope pix below, how much voltage do you think will appear across the single turn wire when 30v is applied to the triangle coil?
Pm
Quote from: partzman on 2025.12.01, 20:57:55
In reference to my 1st post on this thread, see the triangular shaped coil below with 30 turns with a split ferrite filter section placed over one leg of the coil. Also placed thru this ferrite section is a single turn with scope probes attached. Now, before looking at the scope pix below, how much voltage do you think will appear across the single turn wire when 30v is applied to the triangle coil?
Pm
That ferrite section will supply the majority of the inductance of the 30 turn loop. You would expect the 1 turn to obtain about 1 volt and your scope shows this at about 0.93V. For 1 turn that is a dPhi/dt flux change of 0.93Webers/S. That flux change comes from the current rise at the application of the 30V that is almost linear over the 12.8uS pulse. With your measured input current rise to 150mA over that pulse time the effective Reluctance of your magnetic circuit calculates as 3.78E5 yielding an effective inductance of 2.381mH for your 30 turn coil. After the end of your pulse we see the coil current reducing back to zero so you must have some means for that current to flow after switch off, i.e. a resistor that is then receiving energy from the circuit. Your math channel that assumes that falling current is coming from the 30V supply is IMO misleading. But without a view of your actual circuit my analysis could be wrong.
Smudge
Quote from: Smudge on 2025.12.02, 10:07:06
After the end of your pulse we see the coil current reducing back to zero so you must have some means for that current to flow after switch off, i.e. a resistor that is then receiving energy from the circuit.
...or a diode.
Quote from: Smudge on 2025.12.02, 10:07:06
But without a view of your actual circuit my analysis could be wrong.
...and probe positions for the other channels.
Quote from: partzman on 2025.12.01, 20:57:55
... how much voltage do you think will appear across the single turn wire when 30v is applied to the triangle coil?
I would expect the turn ratio to dictate that. ½ : 30
Quote from: Verpies on 2025.12.02, 10:30:09
I would expect the turn ratio to dictate that. ½ : 30
Why 1/2:30? I would say 1:30
Quote from: Smudge on 2025.12.02, 13:03:20
Why 1/2:30? I would say 1:30
Because full turn is 360° around the core.
Quote from: Verpies on 2025.12.02, 13:13:51
Because full turn is 360° around the core.
The single conductor (through the ferrite cylinder) connected to the scope probe creates a full turn 360° around the ferrite core that is carrying nearly all the flux. There is no such thing as 1/2 a turn, either the secondary conductor passes through the core (coupling one turn or N turns) or it doesn't (coupling zero).
Edit. Of course you could drill a radial hole and pass one wire end through it so the coil only intersects half the flux and that would be like 1/2 turn.
Quote from: Smudge on 2025.12.02, 14:31:01
The single conductor (through the ferrite cylinder) connected to the scope probe creates a full turn 360° around the ferrite core that is carrying nearly all the flux.
Aha! but then you have to treat the scope probe as a part of the turn ...which is the main issue in this experiment.
Quote from: partzman on 2025.12.01, 20:57:55
In reference to my 1st post on this thread, see the triangular shaped coil below with 30 turns with a split ferrite filter section placed over one leg of the coil. Also placed thru this ferrite section is a single turn with scope probes attached. Now, before looking at the scope pix below, how much voltage do you think will appear across the single turn wire when 30v is applied to the triangle coil?
Pm
ill take a poke at it. i like this one.
30v in, 1v on the single wire out. all of the action is in the hole of the core. all the rest is additional resistance. if im correct, now add 2 more cores, 1 to each other line of the triangle and do the same, 1 wire out for each. should now have 1v for each single wire out. every turn in the core will read the same as every other turn.
i get what you are trying to prove here. i have different ideas on that, but let me know if i have the answer to the question correct first.
mags
Quote from: Verpies on 2025.12.02, 16:07:11
Aha! but then you have to treat the scope probe as a part of the turn ...which is the main issue in this experiment.
I don't see that as an issue (problem?). If you stick a resistive load there, is the resistor as part of the turn an issue? Would you still consider a 1/2 turn? I think not. Move that resistor away so that there are long wires connecting to it. Is that an issue? No, IMO you should take account of the effect of additional resistance, capacitance, inductance as
external to a single turn feeding them, not something that somehow creates a partial turn.
It is an issue when you stick a capacitor at the end of these probes in lieu of a wire.
Quote from: Magluvin on 2025.12.02, 20:53:51
ill take a poke at it. i like this one.
30v in, 1v on the single wire out. all of the action is in the hole of the core. all the rest is additional resistance. if im correct, now add 2 more cores, 1 to each other line of the triangle and do the same, 1 wire out for each. should now have 1v for each single wire out. every turn in the core will read the same as every other turn.
i get what you are trying to prove here. i have different ideas on that, but let me know if i have the answer to the question correct first.
mags
Very good!! There are several points to this exercise and you have hit on one. If two more cores are added with each having a single wire secondary, we now have three cores placed on the triangular coil. However, the output of each secondary will now be ~.33 volts for a total secondary voltage of ~1 volt. The reason for this is that we still only have a source of 1 volt/turn.
Pm
Edit: BTW, you are correct on the ~1 volt on the secondary.
thanks for that. is that triangle a setup done by you? if so, did you try 2 more cores for the other 2 angles with 1 wire output to confirm the .33v output?
mags
Quote from: Magluvin on 2025.12.03, 21:03:05
thanks for that. is that triangle a setup done by you? if so, did you try 2 more cores for the other 2 angles with 1 wire output to confirm the .33v output?
mags
Yes and yes! Below is a pix of three ferrite cores on the triangle coil with single wire secondaries all connected in series.
The scope pix shows the total voltage measurement of the secondaries on CH3(pnk).
If one begins to view transformer induction as a result of charge separation in the E-Field, even complex winding arrangements become easy to "see". For example, in any given "E" core with the primary and secondary wound on the center leg, what is the E-Field magnitude relative to the primary V/T in each window? Or again in an "E" core with windings on each leg, to induce the center leg from the outside windings, they must be in a bucking configuration. But what if I wish to configure the outer leg windings in an aiding configuration which would normally not induce the center winding. What sort of center leg winding configuration would I need to provide induction? Yes, it is possible.
Pm
good stuff O0
now if you can, measure one of the single turns with just the 2 additional cores in place, not connecting all 3 in series. does that bring the single turn down to .33v?
i like this thread. ;)
mags
Quote from: Magluvin on 2025.12.03, 22:09:11
good stuff O0
now if you can, measure one of the single turns with just the 2 additional cores in place, not connecting all 3 in series. does that bring the single turn down to .33v?
i like this thread. ;)
mags
its late and im thinking on this.
the reason i ask the above question is, if there is still 30v at the input, 1v per turn, the how does each of the 3 single wires not have 1v each? does what i ask make sense? has the application of the 3 cores split the E field of the 30 turns into sections of .33v of E field?
so this is a pulse driven input. not ac waveform. could it be that if it is a pulse, is there a time variation of the input pulse for each test? what im getting at is, the additions of more cores to the 30 turn triangle increases the induction value vs just 1 core i would have to believe. so for the tests of 3 cores vs just 1, i would think the pulse would need to be longer to reach the same max current value as it would the single core test. or am i missing it.
im still on the edge with the E field thought train. sorry. have been for a long time, im sure you know me by now. :D lol
here is the gist... if for each test, 1 core, 2 cores then 3 cores, the 30 turn coil reaches the same max current for each case, should not the E field in each 30 turn coil section be the same? or do those cores break that up into sections? thats why i asked above to measure just 1 single wire in 1 single core but with the 2 other cores in place. this is impotant to me and ill conduct my own experiments with you on this here. just thought since you have it setup, hopefully you could try these things. im diggin it.
im not interested i what others here have to say on this, for now. just me and you bud. please enlighten me. :)
mags
I would be most interested if PM would use just two ferrite cylinders and connect the two output wires in parallel so they form a closed circuit. Depending on how they are paralleled they either do nothing or they short out the induced voltages. For the shorted case the inductance of the series primaries is significantly reduced. Now put the two output wires across each end of a very long length of 50 ohm coax cable so that the shorting action is delayed. What do the waveforms look like then? Initially the two primaries see 50 ohm loads (adjusted by the turns ratio) but later that changes when the time-delayed shorting effect arrives. I can see some anomalous effects occuring at that instant in time.
Quote from: Magluvin on 2025.12.04, 05:38:37
its late and im thinking on this.
the reason i ask the above question is, if there is still 30v at the input, 1v per turn, the how does each of the 3 single wires not have 1v each? does what i ask make sense? has the application of the 3 cores split the E field of the 30 turns into sections of .33v of E field?
so this is a pulse driven input. not ac waveform. could it be that if it is a pulse, is there a time variation of the input pulse for each test? what im getting at is, the additions of more cores to the 30 turn triangle increases the induction value vs just 1 core i would have to believe. so for the tests of 3 cores vs just 1, i would think the pulse would need to be longer to reach the same max current value as it would the single core test. or am i missing it.
im still on the edge with the E field thought train. sorry. have been for a long time, im sure you know me by now. :D lol
here is the gist... if for each test, 1 core, 2 cores then 3 cores, the 30 turn coil reaches the same max current for each case, should not the E field in each 30 turn coil section be the same? or do those cores break that up into sections? thats why i asked above to measure just 1 single wire in 1 single core but with the 2 other cores in place. this is impotant to me and ill conduct my own experiments with you on this here. just thought since you have it setup, hopefully you could try these things. im diggin it.
im not interested i what others here have to say on this, for now. just me and you bud. please enlighten me. :)
mags
OK, here is the test result with three cores in place with a measurement taken on only one core's secondary.
If only two cores are placed on the triangular coil, then each secondary will measure ~.5 volts for a total of ~1 volt. I will show this in my next post in answer to Smudge's request for two paralleled secondaries.
Pm
Quote from: Smudge on 2025.12.04, 11:02:02
I would be most interested if PM would use just two ferrite cylinders and connect the two output wires in parallel so they form a closed circuit. Depending on how they are paralleled they either do nothing or they short out the induced voltages. For the shorted case the inductance of the series primaries is significantly reduced. Now put the two output wires across each end of a very long length of 50 ohm coax cable so that the shorting action is delayed. What do the waveforms look like then? Initially the two primaries see 50 ohm loads (adjusted by the turns ratio) but later that changes when the time-delayed shorting effect arrives. I can see some anomalous effects occuring at that instant in time.
OK, here is the test with two ferrite cores on the triangular coil. The pix shows the in-phase physical secondary connection only.
The first scope pix Tri Coil Tr4 shows the measurements with the secondaries connected in parallel and in-phase.
The second scope pix Tri Coil Tr4A shows measurements with the secondaries again connected in parallel but out-of-phase.
I'm still looking for a spool of coax cable I thot I had but so far no luck!
Pm
Quote from: partzman on 2025.12.04, 15:15:50
OK, here is the test result with three cores in place with a measurement taken on only one core's secondary.
If only two cores are placed on the triangular coil, then each secondary will measure ~.5 volts for a total of ~1 volt. I will show this in my next post in answer to Smudge's request for two paralleled secondaries.
Pm
so to my question, is it, in your opinion, that the 2 other cores are dividing the E field in the triangle to get that result? the reason for only testing for the voltage across 1 single wire in the core was to see if it is the other cores alone that are causing that division, and it apears so. id have to say that if the 30 turns coil were round, we can adjust what we read on that single wire if we move those other cores closer or further away from the one being read . does that sound right?
as it comes to the E field developed along the triangle coil without cores, would that E field be considered 1v??
mags
Mags,
Here is the response from ChatGPT on this subject only I used a round coil instead of a triangular one.
You said:
I have a round 30 turn coil that is 5" in diameter. I put a 1" pair of ferrite cores over the coil and insert a single turn. I now apply a 30 volt pulse to the 30 turn coil and I will measure what amount of voltage across the single turn?
ChatGPT said:
If the coupling is reasonably good, you can treat this exactly like a transformer.
Basic transformer relationship
Vs=Vp×NsNp
Vs=Vp×NpNs
Where:
Vp=30Vp=30 volts (your applied pulse)
Np=30Np=30 turns
Ns=1Ns=1 turn
Ideal case
Vs=30×130=1 volt
Vs=30×301=1 volt
Realistic case
The actual voltage depends on coupling coefficient kk, which in your geometry may be significantly less than 1, especially because:
The ferrite cores are only 1" while the coil is 5" diameter → not fully linking the field
The single-turn loop may not tightly couple to all flux
Pulse rise time matters (leakage inductance)
In practice:
Vs≈k×1 V
Vs≈k×1 V
Typical kk for such a poorly-fitting geometry might be 0.1 to 0.5, so the observed voltage could be something like:
0.1–1.0 volt
If the single turn is placed entirely inside the closed ferrite core
...and the 30-turn coil's flux fully passes through that core, then kk could approach 1, and you would indeed see very close to 1 volt.
If you want, tell me how the ferrite cores are arranged over the coil (e.g., clamshell? does it close fully around the conductor?) and I can estimate a more accurate coupling coefficient and real output voltage.
You said:
OK, now if I place an additional ferrite core set on the same 30 turn coil, how will this affect the single turn measurement on the first ferrite core?
ChatGPT said:
Short answer: The voltage on the first single-turn will stay almost the same, unless the second ferrite core significantly diverts flux away from the first one.
But let's explain why.
✔ What happens when you add a second ferrite core set to the same 30-turn coil?
You now have something like this:
One 30-turn primary coil
Two separate ferrite cores placed around different parts of the coil
Each core may contain its own secondary winding (your single turn)
This behaves like a transformer with multiple secondaries, except that the magnetic paths are not shared—they are local to each ferrite core.
Key Principle
The voltage induced in a secondary is proportional to the local magnetic flux through that core:
Vs=NsdΦdt
Vs=NsdtdΦ
Adding another ferrite core simply creates another local flux path, but it does not steal flux from the first core unless the coil's field is "flux-limited."
✔ What actually changes?
Case 1 — The ferrite cores are far enough apart and not interacting
Each core sees roughly the same field generated by the 30-turn coil.
The first core still sees nearly the same flux as before
The induced voltage in the first single-turn stays the same
The second core also produces its own induced voltage (also ~1 V ideal)
This is the most common situation.
Case 2 — The cores are very close together so that their flux paths compete
This is uncommon but possible if:
They are touching or nearly touching
Their magnetic circuits strongly interact
The ferrite partially saturates (unlikely with 30 V pulse unless pulse is long)
In this case, the effective permeance of the system changes, and the flux divides between the cores. However, even then:
The first core might see slightly less flux
The reduction is usually small unless the ferrite saturates
✔ Practical conclusion
Adding a second ferrite core will NOT significantly change the induced voltage in the first single turn.
You will still measure approximately:
Vs1≈3030=1 V×k1
Vs1≈3030=1 V×k1
where k1k1 is the coupling coefficient to the first core.
The second ferrite core will produce:
Vs2≈1 V×k2
Vs2≈1 V×k2
Both voltages exist independently.
✔ Exception: If your pulse is long and the ferrite saturates
Then inductance drops, current rises, flux changes, and both secondary voltages may fall as the ferrite saturates.
If you tell me:
Pulse duration
Pulse rise time
Ferrite material type (e.g., 3C90, N87, etc.)
Coil current
Placement of cores
...I can calculate whether saturation is possible and estimate the exact performance.
So, we seemingly have a dilemma!
Pm
Edit: Actually there is no dilemma as each added core reduces the triangular coil's current and thus the flux which in turn reduces the E-Field in each added core section.
ok. not an actual answer to some of my questions, but its a start..
firstly, it didnt provide any E field in the mix, only flux. so as you know im a firm believer in faraday flux cutting vs E field theory. so the ai pivots toward my beliefs so far.
on your triangle with 3 cores and only one with a wire being tested for voltage, does the voltage in the test change if you slide the other 2 cores say closer to the 30v input corner and then slide them closer to the 2 corners of the angle with the core and single wire being tested? all to see if the division changes. if not then i think the cores are just increasing inductance and we are not reaching same max current for each case.
i initially was thinking what the ai said that you would still get 1v from the single test wire when you applied the other 2 cores. and then it said that the pulse length would need to be different with the added cores.
if the ai did claim 1v still, then it is saying what i said before that the addition of the other 2 cores increases the inductance of the triangle/circle 30 turn winding, in which i believe you will need to change the pulse width of your input in order for each case to reach max current, same max current for each case, in order for the test to qualify accuracy in each case. so for example, i see in your scope shots different currents on the scope for each case you have shown. i think that needs to be addressed in order to qualify fair tesing here.
hopefully that makes sense. hate to be repetitive, but i need what im saying to sink in..
mags
also, we are not loading the test wire in any significant way. so reading that wire should not alter the functions of the whole to affect anything as the ai alluded to if it were loaded.
mags
I would like to take a guess on the separate ferrite cores and lower voltage. Don't worry, Your thread, not here to argue :)
My guess is, each extra ferrite cores added changes the reluctance in the primary magnetic path. As we decrease reluctance, the primary behaves with more and more reactance.
Usually lowering reluctance should result in more voltage output from a condensed field within the core, but the separate cores play a role, resulting in each separate core seeing less push from the primary because the choke effect is creeping in more and more as you add cores.
I bet if an inductance meter was used, you would witness the triangle getting more and more inductive every core that you add.
Just my guess
Edit- OOps I missed your last line.. "Edit: Actually there is no dilemma as each added core reduces the triangular coil's current and thus the flux which in turn reduces the E-Field in each added core section." I guess it was already explained
Mags,
The first scope pix below Tri Coil Tr5 is a test where the triangular coil has three sets of ferrite cores in place with the pulse frequency is lowered to achieve a peak coil current of 150ma. The voltage across a single secondary is now seen to be 352mv.
If the other two ferrite cores are moved closer to or away from the active core, no change is seen.
Tri Coil Tr6 is a measurement of all three secondaries connected is series.
Pm
here is a pdf that discusses what you are showing. the second to the last depiction is the same, but reversed in and out.
mags
Quote from: partzman on 2025.12.04, 15:34:02
OK, here is the test with two ferrite cores on the triangular coil. The pix shows the in-phase physical secondary connection only.
The first scope pix Tri Coil Tr4 shows the measurements with the secondaries connected in parallel and in-phase.
The second scope pix Tri Coil Tr4A shows measurements with the secondaries again connected in parallel but out-of-phase.
I'm still looking for a spool of coax cable I thot I had but so far no luck!
Pm
Thanks for doing that. Clearly the input inductance has reduced for the second case as expected. I think the reduction would be geater if each ferrite had more than a 1 turn secondary. I have amended my paper on transformers connected to a delay line to make things clearer, attached here. It is that sudden drop in flux that occurs after the delay and how it returns energy to the source that is of interest here for this two transformers case. Does it really happen?
Smudge
Quote from: Magluvin on 2025.12.05, 02:25:49
here is a pdf that discusses what you are showing. the second to the last depiction is the same, but reversed in and out.
mags
I had not seen that pdf before so thanks for sharing! It is exactly what I've been researching lately only I've used the basis of E-Field charge separation instead of core flux. Results are the same.
In the first pix below we see the physical setup that produced the following scope trace pix. Two identical toroid cores are physically connected together with a 12 turn winding and then each toroid receives an additional 12 turn winding as shown.
The two outside windings are connected in a series bucking mode and then a 24 volt peak pulse is applied. The center winding is the secondary.
The CH1(yel) trace is the input pulse to a 3/4 bridge drive circuit, CH2(blu) is the power supply, CH3(pnk) is the open circuit output across the secondary, and CH4(grn) is the primary current. Math(red) can be ignored
This configuration is basically the same as the last figure you show in the pdf only with two cores instead of three.
Here we see a primary consisting of 24 total turns that is supplied with a 24 volt pulse that is producing a 24 volt output pulse across 12 turns. This is certainly not normal transformer action. My explanation using the E-Field is that the primary voltage is split between the two primary windings with each having 12 volts applied. The coupling between these two primary windings is very low so we have the secondary winding subjected to two out-of-phase 12 volt E-Fields in each core window. These E-Fields therefore add in the 12 turn secondary to equal 24 volts. All voltage levels are approximate for clarity.
The primary and secondary inductances will be nearly equal but this is for another discussion.
Pm
Quote from: Smudge on 2025.12.05, 15:02:37
Thanks for doing that. Clearly the input inductance has reduced for the second case as expected. I think the reduction would be geater if each ferrite had more than a 1 turn secondary. I have amended my paper on transformers connected to a delay line to make things clearer, attached here. It is that sudden drop in flux that occurs after the delay and how it returns energy to the source that is of interest here for this two transformers case. Does it really happen?
Smudge
You're welcome. I agree that more secondary turns would result in a greater inductance reduction.
I think I might be able to simulate your delay line circuit if you can give an idea of how much delay to start with.
Pm
here is another pdf that explains how flux propagates through a toroid core transformer. it follows what i believe it to be.. im a flux guy. ;)
it is a compliment to the last pdf i uploaded. ill post both here.
Mags
Quote from: Smudge on 2025.12.05, 15:02:37
Thanks for doing that. Clearly the input inductance has reduced for the second case as expected. I think the reduction would be geater if each ferrite had more than a 1 turn secondary. I have amended my paper on transformers connected to a delay line to make things clearer, attached here. It is that sudden drop in flux that occurs after the delay and how it returns energy to the source that is of interest here for this two transformers case. Does it really happen?
Smudge
In that paper I show that plotting
primary current against
flux can show the two diiferent energies, load current energy and magnetizing current energy, as inputs. What I failed to do was show that plotting
secondary current against
flux shows that transfer of energy to the load as an output. Below I correct that combining the two currents onto one chart where the red areas are input and the green areas are output. The second chart is the case where the secondary current jumps from 1A to 2A and during that jump the flux reduces down to zero. Taking the rectangle areas as 1 unit of energy it is now clear that before that secondary jump the primary input is 1.5 units of energy where 1 unit has passed through the transformer to be stored in the delay-line and 0.5 units is lost to the core. If we then disconnect the delay line and immediately replace it with a current source that ramps up from 1A to 2A we get the situation shown in the second chart. That current generator pumps in 1.5 units of energy and if we somehow used the 1 unit stored in the delay line the total energy input so far is 1.5+0.5=2 units. The pumping action causes the primary to return 2 units of energy to its source so there is no OU.
Smudge
Quote from: Smudge on 2025.12.07, 09:45:39
In that paper I show that plotting primary current against flux can show the two diiferent energies, load current energy and magnetizing current energy, as inputs. What I failed to do was show that plotting secondary current against flux shows that transfer of energy to the load as an output. Below I correct that combining the two currents onto one chart where the red areas are input and the green areas are output. The second chart is the case where the secondary current jumps from 1A to 2A and during that jump the flux reduces down to zero. Taking the rectangle areas as 1 unit of energy it is now clear that before that secondary jump the primary input is 1.5 units of energy where 1 unit has passed through the transformer to be stored in the delay-line and 0.5 units is lost to the core. If we then disconnect the delay line and immediately replace it with a current source that ramps up from 1A to 2A we get the situation shown in the second chart. That current generator pumps in 1.5 units of energy and if we somehow used the 1 unit stored in the delay line the total energy input so far is 1.5+0.5=2 units. The pumping action causes the primary to return 2 units of energy to its source so there is no OU.
Smudge
Thanks for this update! I have already attempted to simulate this core topology with a delay line but was not successful. LtSpice can not and will not create a usable matrix with the 'k' factors or coupling factors of the actual core assembly and thus gives an error message saying basically that the core model is impossible. I'm not sure if using capacitor/gyrator modeling could yield a workable solution but it seems that you have answered the question of whether this idea could yield OU so I won't attempt a solution.
Pm
here is pics of what im going for.
each core will have 20 turns 30awg and all 8 will be in series.
once they are done ill mount them on a plexy square as shown below using small spacers. this arrangement will make for simple threading of the secodary through all 8 cores. being the sec will be longer for each turn, im using 26awg to reduce its resistance. going to make 3 separate sec windings of 10 turns each for ease of trying 1 winding as an output or up to all 30 turns in series.
the pdf uses ac but im going to try pulsing first. should have the transformer together tomorrow.
mags
Quote from: Smudge on 2025.12.07, 09:45:39
What I failed to do was show that plotting secondary current against flux shows that transfer of energy to the load as an output.
I told you so (https://www.overunityresearch.com/index.php?topic=4889.msg117558#msg117558).
Just FYI.
It appears to me in my continued testing of charge separation in various cores, that the E-Field magnitude is dependent on the amount of flux retained in the core material. IOW, any leakage flux outside the core creates a loss in the E-Field within the core window and attributes to the apparent E-Field outside the core. As a core saturates for example, more flux leaves the core and also creates a reduction in the measured E-Field within the core window.
Therefore IMO, a lossless core relative to it's flux magnitude, would result in zero E-Field outside the core.
Pm
Below are the mental gymnastics that I have to perform in order to understand the statement:
Quote from: partzman on 2025.12.18, 21:21:51
It appears to me in my continued testing of charge separation in various cores, that the E-Field magnitude is dependent on the amount of flux retained in the core material.
The word "retained" suggests something that remains after being kept. It is not far form there to the magnetic remanence and associated hysteresis losses.
But somehow I have doubts that the author has the BH curve and magnetic remanence in mind.
Quote from: partzman on 2025.12.18, 21:21:51
IOW, any leakage flux outside the core...
"leakage flux" suggests an issue with flux confinement and the Hopkinson's law.
Quote from: partzman on 2025.12.18, 21:21:51
As a core saturates for example, more flux leaves the core...
Yes, ferromagnetic saturation decreases the differential permeability, increases the associated reluctance and causes flux leakage ...so the focus is on flux confinement in the core - not on the ferromagnetic remanence.
Quote from: partzman on 2025.12.18, 21:21:51
Therefore IMO, a lossless core relative to it's flux magnitude, would result in zero E-Field outside the core.
Considering the above, the phrase "lossless core" does not refer to a core that has no hysteresis losses.
Rather the focus is on the flux confinement within the core and the flux distribution according to Hopkinson's law in a system where the reluctance of core's path increases with its saturation wrt to the reluctance of paths outside of the core.
Quote from: partzman on 2025.12.18, 21:21:51
Just FYI.
It appears to me in my continued testing of charge separation in various cores, that the E-Field magnitude is dependent on the amount of flux retained in the core material. IOW, any leakage flux outside the core creates a loss in the E-Field within the core window and attributes to the apparent E-Field outside the core. As a core saturates for example, more flux leaves the core and also creates a reduction in the measured E-Field within the core window.
Therefore IMO, a lossless core relative to it's flux magnitude, would result in zero E-Field outside the core.
Pm
I disagree. The image below shows the E field around a lossless core where the length of the arrows depict the field strength. There is no sudden change to zero field going from "inside the core" position to outside the core.
Smudge
Magluvin
Nice set of cores. Maybe suitable for this experiment ? https://www.youtube.com/watch?v=Dz0GUqxCPp4
How about parallel resonant tank circuit on primary ? Should it matter ?
Quote from: Smudge on 2025.12.19, 08:33:43
The image below shows the E field around a lossless core where the length of the arrows depict the field strength. There is no sudden change to zero field going from "inside the core" position to outside the core.
...and a couple of words about its loop integral, please.
Quote from: Verpies on 2025.12.19, 04:30:29
Below are the mental gymnastics that I have to perform in order to understand the statement:
The word "retained" suggests something that remains after being kept. It is not far form there to the magnetic remanence and associated hysteresis losses.
But somehow I have doubts that you have the BH curve and magnetic remanence in mind.
Yes, 'retained' is not the best choice of words here just simply the 'magnitude' of flux! The BH curve and remanence are not in consideration because we can determine results from a single unipolar pulse without any field collapse.
Quote
"leakage flux" suggests an issue with flux confinement and the Hopkinson's law.
Not being familiar with the complex (to me) Hopkinson's law, the simple answer is a matter of flux confinement that is, flux contained within a core verses flux outside that same core at any given time.
Quote
Yes, ferromagnetic saturation decreases the differential permeability, increases the associated reluctance and causes flux leakage ...so the focus is on flux confinement in the core - not on the ferromagnetic remanence.
Yes that is correct!
Quote
Considering the above, the phrase "lossless core" does not refer to a core that has no hysteresis losses.
Rather the focus is on the flux confinement within the core and the flux distribution according to Hopkinson's law in a system where the reluctance of core's path increases with its saturation wrt to the reluctance of paths outside of the core.
Again, a poor choice of words on my part and yes, you are correct.
Pm
Quote from: Smudge on 2025.12.19, 08:33:43
I disagree. The image below shows the E field around a lossless core where the length of the arrows depict the field strength. There is no sudden change to zero field going from "inside the core" position to outside the core.
Smudge
OK, can you give some field strength numbers for this lossless core verses a core with a minor loss of say 5%?
Pm
Quote from: partzman on 2025.12.19, 15:21:46
OK, can you give some field strength numbers for this lossless core verses a core with a minor loss of say 5%?
Pm
I will answer this when I get back from my Christmas break. I will be away from my computer for the next 9 days.
Smudge
Quote from: partzman on 2025.12.18, 21:21:51
...
Therefore IMO, a lossless core relative to it's flux magnitude, would result in zero E-Field outside the core.
...
Only if it had infinite permeability.
If it has finite permeability, then since the magnetic flux takes all possible paths, it also passes through the air, so E = -dΦ/dt is not zero.
The preferred path of the flux, with equal cross-section and length, is obviously the one with the highest permeability, in terms of the permeability ratio.
In the general case, the proportion is given by the ratio of reluctances R=µ.S/L, where S is the cross-section and L is the length of the path. In practice, the flux will never be zero outside the core.
Quote from: F6FLT on 2026.01.22, 10:50:07
Only if it had infinite permeability.
If it has finite permeability, then since the magnetic flux takes all possible paths, it also passes through the air, so E = -dΦ/dt is not zero.
The preferred path of the flux, with equal cross-section and length, is obviously the one with the highest permeability, in terms of the permeability ratio.
In the general case, the proportion is given by the ratio of reluctances R=µ.S/L, where S is the cross-section and L is the length of the path. In practice, the flux will never be zero outside the core.
Your u.S/L is not reluctance, it is the inverse, permeance.
Quote from: partzman on 2025.12.19, 15:21:46
OK, can you give some field strength numbers for this lossless core verses a core with a minor loss of say 5%?
Sorry about the late reply, I forget things these days. I think we need to go back to your earlier post #169
QuoteIt appears to me in my continued testing of charge separation in various cores, that the E-Field magnitude is dependent on the amount of flux retained in the core material. IOW, any leakage flux outside the core creates a loss in the E-Field within the core window
It certainly will create a reduction from the value that occurs without flux leakage, but that is not a loss in the normal meaning of the word (power or energy loss).
Quoteand attributes to the apparent E-Field outside the core.
I took your "outside the core" to mean outside the core window. You seem to consider voltage induction as occurring only on that part of the conductor within the core window (within the donut hole), and my reply post #171 shows that there is E field present everywhere external to the core, not just in the window. I hope you now have moved away from that E field being "apparent" and from it being connected to flux leakage.
QuoteAs a core saturates for example, more flux leaves the core and also creates a reduction in the measured E-Field within the core window.
Flux does not leave a ring core as it saturates. Increased driving current gives a smaller increase of flux. That is not flux leakage. E field everywhere and your measured voltage are proportional to rate of change of flux, so they are reduced but this is not due to flux leakage.
QuoteTherefore IMO, a lossless core relative to it's flux magnitude, would result in zero E-Field outside the core.
I hope I showed that to be nonsense.
Moving on to your latest request, what are the core dimensions and characteristics you want numbers for? What is the driving waveform? Do you want E field vector magnitudes at varying points. Or are you just looking for a single number, the relative change in magnitude between a lossless core and a 5% lossy core that will apply everywhere external to the core?
Smudge
so here im presenting 3 examples.
a toroid core. blue box is 4v ac input to 4 turn primary at the bottom of the core. at the top of the cutaway examples, 3 different secondaries. the yellow circle is a scope.
on the left the single turn is wound through the core.
middle just above the top of the core but same proximity to the core.
right, the secondary has close access to the top outer core and outer sides of the core.
what should we see in the 3 scopes? not concerned with phase differences of the middle and right scopes compared to the left.
then if the yellow circle is a load, which secondary would produce the most output?
If you say the e field is inducing all of the wire of a normally wound secondary, in the window and outer sides, would your answer to what you believe the scopes would show, all be the same results? again, not concerned with phase differences of the middle and right scopes compared to the left.
i know there looks like 5 turns for the primary.. lets say it is 4.
Mags
Quote from: Smudge on 2025.12.19, 08:33:43
I disagree. The image below shows the E field around a lossless core where the length of the arrows depict the field strength. There is no sudden change to zero field going from "inside the core" position to outside the core.
Smudge
if the core is lossless, why is the e field looking weaker around the outside of the core? or better to say, why is the e field looking stronger in the window of the core?
mags
Quote from: Magluvin on 2026.01.24, 03:58:42
so here im presenting 3 examples.
a toroid core. blue box is 4v ac input to 4 turn primary at the bottom of the core. at the top of the cutaway examples, 3 different secondaries. the yellow circle is a scope.
on the left the single turn is wound through the core.
middle just above the top of the core but same proximity to the core.
right, the secondary has close access to the top outer core and outer sides of the core.
what should we see in the 3 scopes? not concerned with phase differences of the middle and right scopes compared to the left.
then if the yellow circle is a load, which secondary would produce the most output?
If you say the e field is inducing all of the wire of a normally wound secondary, in the window and outer sides, would your answer to what you believe the scopes would show, all be the same results? again, not concerned with phase differences of the middle and right scopes compared to the left.
i know there looks like 5 turns for the primary.. lets say it is 4.
Mags
The left one is a 1 turn secondary with a 4 turn primary so with 4V ac input we get 1V ac output. That assumes the core has high permeability so flux leakage (because the primary only occupies a small part of the core) is negligible. With a poor low perm core the output will be less than 1V.
For the other two there will be zero volts except at high frequencies when capacitive coupling and RF coupling from source to scope take effect.
Smudge
Quote from: Magluvin on 2026.01.24, 05:05:02
if the core is lossless, why is the e field looking weaker around the outside of the core? or better to say, why is the e field looking stronger in the window of the core?
mags
The E field is due to the flux in the core changing with time. The flux lines are circular. For any small length of flux (that is changing with time) the E field reduces with distance from it. Within the circle the total contribution to the E field from the entire flux around the circle is greater than elsewhere. Maximum E field is within the window close to the inner surface of the core. Core loss plays no part in this geometric effect.
Smudge
Quote from: Smudge on 2026.01.24, 08:12:54
The E field is due to the flux in the core changing with time. The flux lines are circular. For any small length of flux (that is changing with time) the E field reduces with distance from it. Within the circle the total contribution to the E field from the entire flux around the circle is greater than elsewhere. Maximum E field is within the window close to the inner surface of the core. Core loss plays no part in this geometric effect.
Smudge
I find this statement confusing! Circular in the core?
OK, let's see if we can agree on some common truths with conventional theory! We apply a DC voltage to a primary on a toroid core. This results in a current ramp that starts from zero and increases to some peak value without reaching saturation of the core. During this current ramp, flux starts from zero and increases to a peak value in the core. During this time, zero flux appears inside the hole of the core or outside the core and therefore zero E-Field can appear around the core. We also have placed a single turn secondary completely around the core. We know that this completed turn will produce a voltage between it's open ends. Why? Because the surface bounded by the secondary loop cuts through the core where the flux B is varying with time.
If we agree thus far, then how does the potential across the secondary vary through it's perimeter?
Regards,
Pm
Quote from: partzman on 2026.01.24, 15:25:58
I find this statement confusing! Circular in the core?
You have a circular ring core and the B field forms circular lines within it. Flux is that B field times the area so the flux is the total number of lines forming a circular ring of flux (the old measure of flux was lines per square cm).
QuoteOK, let's see if we can agree on some common truths with conventional theory! We apply a DC voltage to a primary on a toroid core. This results in a current ramp that starts from zero and increases to some peak value without reaching saturation of the core. During this current ramp, flux starts from zero and increases to a peak value in the core. During this time, zero flux appears inside the hole of the core or outside the core and therefore zero E-Field can appear around the core.
There is zero flux external to the core material but there is the magnetic vector potential
A field there. And that
A field changing with time (d
A/dt) produces the
E field. I am surprised you say zero E field during this flux build-up because you know that V occurs while the flux is changing.
QuoteWe also have placed a single turn secondary completely around the core. We know that this completed turn will produce a voltage between it's open ends. Why? Because the surface bounded by the secondary loop cuts through the core where the flux B is varying with time.
Which it does through this build-up phase, producing the
E field that drives the electrons in the secondary wire to create the voltage.
QuoteIf we agree thus far, then how does the potential across the secondary vary through it's perimeter?
Ah, I think I can see where the problem lies. You consider the surface bounded by the loop is where the magic happens and that is where the voltage happens. That is incorrect. The flux passing through that surface doesn't magically induce voltage at its periphery, it is more complex than that. You have to integrate the
A vector tangential component along the conductor and take rate-of-change of that sum to get the voltage. That integral sum is equal to the flux passing through that surface, but that does not mean the electric field driving the electrons is equally distributed around its boundary. V=dΦ/dt hides the
E field necessary to create the voltage. My image shows the
A vectors, hence also the
E vectors, in their true colours.
Smudge
Quote from: Smudge on 2026.01.24, 17:41:15
You have to integrate the A vector tangential component along the conductor and take rate-of-change of that sum to get the voltage.
That's right.
A bit of warning, though: You cannot apply the Ohm's law to that voltage in order to determine the current flowing in the loop. i=ℰ/R does not work.
Quote from: Verpies on 2026.01.25, 06:01:07
That's right.
A bit of warning, though: You cannot apply the Ohm's law to that voltage in order to determine the current flowing in the loop. i=ℰ/R does not work.
Yes, because the load current alters the magnetic field in the core and that alters the magnetic vector potential and that alters the E field that you used in the first place. Many people find difficulty in getting out of this conundrum. Your i=ℰ/R suggests the use of the E field (ℰ field?) there and I would have used the symbol V for voltage.
Quote from: Smudge on 2026.01.25, 08:33:15
Yes, because the load current alters the magnetic field in the core and that alters the magnetic vector potential and that alters the E field that you used in the first place. Many people find difficulty in getting out of this conundrum .
Indeed
Quote from: Smudge on 2026.01.25, 08:33:15
Your i=ℰ/R suggests the use of the E field (ℰ field?)
ℰ: induced EMF
Also, I'd like to add that A=0 implies that B=0 ....but B=0 does not imply that A=0 since superposition of multiple non-zero A fields can yield B=0.
I dont have tons of time atm. But Ive started a 4 core model. The 4 cores are from the 5 core model I had which fell some time back and 1 broke. tried to count the turns o the broken one but reverted to count turns of a good one. about 60 turns 30awg red. The green sec will be 70 turns of 26awg. Both old stock Radioshack magnet wire pack.
As we have seen with Partsmans 12 turn experiments, if we apply 12v to each of the outer primary windings, the middle secondary was 24v. The pdf claims that he sec currents can be no more than the currents of the input. Thats where I have chosen to boost the sec turns to 70. So if the sec is heavily loaded, we hope to get more voltage out, in ratio to the primary's and roughly similar output currents as given to the primary windings. Hopefully we get decent results..
Will have the sec wound soon.
Mags
Are you powering the primaries in a phased sequence ?
no. as per the pdf, it shows all primaries in series, of which the alternative could be parallel.
if it were sequential, i could imagine a polarized output on the sec. i need to test some things. like can we get field collapse currents back out of the primary while the secondary is loaded. the pdf suggests such as it claims that loading the secondary does not lead to increased primary current and will not stop resonance of the primary if it is set up for resonance.
mags
There is another field in transformer induction which is discussed little and that is the H-Field. This is a demo of that field.
The first pix shows the physical layout used in the test. The 2" OD toroid has a 12T primary and also has two 4uf film caps connected in parallel which are connected to a piece of wire that runs vertically in the core and placed close to the 12T primary. IOW, we have a piece of wire in parallel with an 8uf cap in the center hole of the toroid. One end of this parallel network is connected to ground and the other end has the CH3(pnk) scope probe attached. Also, a current probe is seen connected across the single piece of wire to measure the current in the network and is seen on CH4(grn) of the scope.
The next pix is taken from a paper by Edwards and Saha linked below which shows the H-Fields around a coil of wire. The single wire in close proximity to the primary is induced by the H-Field as well as being charge separated by the E-Field.
HF1 shows the charge separation measured at the top of the network. It is seen to be ~ +/-3.8v with 48v applied to the 12t primary.
HF2 shows a resonant current build-up to a peak level of 791ma.
HF3 shows the power measured with the current probe measuring the current to the primary winding. Here, CH1(yel) is the pulse to the switching network and CH2(blu) is the supply voltage.
Note the lack of resonant voltage across the 8uf film cap.
Pm
Edit: Corrected pdf file.
Quote from: partzman on 2026.02.26, 16:13:09
The next pix is taken from a paper by Edwards and Saha linked below which shows the H-Fields around a coil of wire.
It is not in that paper, it must be taken from some other.
Smudge
Quote from: Smudge on 2026.02.26, 19:21:31
It is not in that paper, it must be taken from some other.
Smudge
Ooops! My mistake! I have replaced it with the corrected pdf.
Thanks!
Pm
Quote from: partzman on 2026.02.26, 16:13:09
The next pix is taken from a paper by Edwards and Saha linked below which shows the H-Fields around a coil of wire.
It also states "Equivalent single turn current sheet" which is wrong because the equivalency is destroyed by the pitch component of the current flowing in a multi-turn helical winding (odd or single-layer).
Quote from: Verpies on 2026.02.27, 10:41:41
It also states "Equivalent single turn current sheet" which is wrong because the equivalency is destroyed by the pitch component of the current flowing in a multi-turn helical winding (odd or single-layer).
Would you elaborate please?
Pm
Quote from: partzman on 2026.02.28, 15:05:55
Would you elaborate please?
I can only guess which part deserves an elaboration, since my own statement is perfectly clear to me.
Is it the "the pitch component of the current" that is unclear to you ?
If "yes", then consider that a helical winding has two perpendicular dimensions, i.e.: a radius and a pitch. Pitch is the perpendicular distance progressed for one full revolution of the radius. Analogy: the "
Threads per inch (TPI)" specification of a machine bolt's thread is a measure of the pitch.
Most people consider only the direction of the current flowing in the circle swept by the end of a rotating radius segment anchored at one point (the center of the circle) while completely ignoring that this point also moves in the perpendicular pitch dimension ...and so does the current with it.
A current sheet does not have any pitch component, so it is not equivalent.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=55524)
The bottom choke does not confine the magnetic flux to the core at all ( the ratio of its toroidal flux to axial flux is zero ).
The top choke confines most of the flux to the core, but not all because its pitch component of the current is non-zero and is equal to all of the other chokes' pitch components of the current ( consequently, their axial fluxes are equal, too ).
( in all of these chokes, the current travels 359° around the axis of the core ).This has practical consequences: e.g. two coaxial full single-layer toroidal chokes have non-zero coefficient of mutual inductance because their currents have non-zero pitch components of the current.
This makes them behave like two windings of a 1t:1t transformer. Consequently every choke depicted above will induce EMF or current in the others.
If the pitch component of their currents is zero then so is the coefficient of their mutual inductance and no mutual induction takes place, since the entire magnetic flux generated by one winding is confined to one core (this is possible for windings spanning the entire circumference of the core in even number of pitch-reversing layers):
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=55528)
Two pitch-reversing layers spanning the entire circumference of a toroidal core.
The pitch component of the current in one layer cancels the same in the second layer.All of the above is also applicable to helical windings on solenoids, since a solenoid is just a fragment of a very large toroid.
Verpies,
Sorry to have caused you all the work of your previous post just to make your point. I understood what you meant, I just couldn't understand why you said "It also states "Equivalent single turn current sheet" which is wrong because the equivalency is destroyed by the pitch component of the current flowing in a multi-turn helical winding (odd or single-layer)."
Edwards' drawing clearly shows the vertical winds as parallel and at 90 degrees to the page. The only pitch seen is in the top and bottom portions of the windings. So clearly, his "equivalent single turn current sheet" for that portion of the windings is valid IMO.
Pm
Quote from: partzman on 2026.03.01, 15:15:06
Edwards' drawing clearly shows the vertical winds as parallel and at 90 degrees to the page. The only pitch seen is in the top and bottom portions of the windings. So clearly, his "equivalent single turn current sheet" for that portion of the windings is valid IMO.
In this illustration:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=55514)
All of the turns of the winding are slanted and this means that the current is progressing from top to bottom as well.
This progression happens in every turn - not just the top and bottom portions of the winding.
I have added the parts of the turns hidden under the page in red color:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=55530) The top to bottom current progression also happens in these hidden red halves of the turns.
Collectively, this current progression is equivalent to a current flowing in a straight wire from top to bottom that looks like this:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=55534)
Notice that this equivalent vertical current generates magnetic flux which is perpendicular to the page (I marked it as red circles).
Since Edward's original diagram does not account for that vertical current progression nor for the magnetic flux generated by that current - thus his Fig. 6b is not equivalent to Fig. 6a.
Feature E Field B Field H Field
Common Name Electric Field Magnetic Flux Density Magnetic Field Strength
Primary Source Charges(Q) Total Current (J total) Free Current(J free)
Physical Effect Pushes charges Deflects moving charges Magnetizes materials
SI Unit V/m Tesla (T) A/m
Utility: Engineers use it because it depends only on the driving current, making it easier to calculate in complex circuits.
mags
What is the difference between the magnetic H field and the B field?
H is a bit like the number of magnetic field lines and B kinda is how tightly packed they are. More amps/more turns/shorter core means more field lines (bigger H - Aturns/m), higher permeability (measure of how easily those field lines can "flow") means they can be packed tighter together in the core (larger B - more intense magnetic field).
mags
Quote from: Magluvin on 2026.03.01, 18:29:11
What is the difference between the magnetic H field and the B field?
Perceived origin.
In SI base units the B field is kg/A⋅s² and the H field is A/m so you need to divide the B field by Newton/Ampere² to obtain the H field ( incidentally N/A² = µ₀ ).
...and √(1/µ₀ε₀) = speed of light (c).
Quote from: Magluvin on 2026.03.01, 18:29:11
H is a bit like the number of magnetic field lines and B kinda is how tightly packed they are.
Φ is the number of magnetic field lines. B is their areal density ( or like you wrote: "
how tightly packed they are" ).
There is something important missing from this table that we all know and use, magnetic attraction or repulsion. I have added it in red using your word Pushes to mean applying force..
Quote from: Magluvin on 2026.03.01, 18:21:23
Feature E Field B Field H Field
Common Name Electric Field Magnetic Flux Density Magnetic Field Strength
Primary Source Charges(Q) Total Current (J total) Free Current(J free)
Physical Effect Pushes charges Deflects moving charges Magnetizes materials
Pushes magnetic dipoles
SI Unit V/m Tesla (T) A/m
Electrons have a magnetic dipole moment in addition to their mass and electric charge. Moving electrons are the life-blood of our electrical world so it is surprising that none of our devices use their magnetic property, the Bohr magneton, to achieve movement. There will soon be more on this subject in my
Energy from electron spin bench https://www.overunityresearch.com/index.php?topic=4307.msg99424#msg99424 (https://www.overunityresearch.com/index.php?topic=4307.msg99424#msg99424).
Smudge
Quote from: Verpies on 2026.03.01, 16:16:57
In this illustration:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=55514)
All of the turns of the winding are slanted and this means that the current is progressing from top to bottom as well.
This progression happens in every turn - not just the top and bottom portions of the winding.
I have added the parts of the turns hidden under the page in red color:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=55530) The top to bottom current progression also happens in these hidden red halves of the turns.
Collectively, this current progression is equivalent to a current flowing in a straight wire from top to bottom that looks like this:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=55534)
Notice that this equivalent vertical current generates magnetic flux which is perpendicular to the page (I marked it as red circles).
Since Edward's original diagram does not account for that vertical current progression nor for the magnetic flux generated by that current - thus his Fig. 6b is not equivalent to Fig. 6a.
What approximate amount of difference would the pitch make in a typical single layer H-Field measurement as compared to Edward's original diagram?
Pm
Quote from: partzman on 2026.04.09, 20:50:13
What approximate amount of difference would the pitch make in a typical single layer H-Field measurement as compared to Edward's original diagram?
The flux angle will change by tan⁻¹(1/N). The flux leakage will increase by that amount and cause induction outside of the core disproportionally to its µ
r. The normal component of the flux's magnitude will change proportionally.
To avoid all that, always use an even number of opposite-pitch layers that cover the entire core equally and uniformly.
In reference to my post #192 (https://www.overunityresearch.com/index.php?topic=4797.msg118488#msg118488) using the same layout and components but running the circuit continuously rather than eight pulses, yields the following data. This experiment utilizes both the H-Field of the primary and the charge separation inside the toroid core.
P1 shows the input power taken from the 48v supply. CH1(yel) is the pulse input to the 3/4 bridge driving the 12T primary winding, CH2(blu) is the supply voltage, CH3(pnk) is the voltage across the wire/capacitor, CH4(grn) is the current through the primary, and CHM(red) is the math channel showing the resultant watts with CH2*CH4.
P2 shows the measurements of the power returned to the 48v supply from the primary.
P3 shows the output power in the Math(red) channel with CH3*CH4 with CH4 now measuring the current in the wire connected across the 8uf capacitor.
P4 shows the increase in voltage across the 8uf cap produced by the average resonant current for the positive half cycle.
P5 shows the average current in the wire with the wire/cap rotated 90 degrees clockwise inside the toroid core in reference to the primary winding. This indicates that the H-Field induction to the wire is valid when properly oriented. With the wire and capacitor both charge separated to the same potential, no current would be expected to flow unless supplied by an outside source which is in this case, is the primary's H-Field on the inside of the core.
Regards,
Pm
Quote from: partzman on 2026.06.26, 14:40:55
In reference to my post #192 (https://www.overunityresearch.com/index.php?topic=4797.msg118488#msg118488) using the same layout and components
I will need the probe positions on the circuit that is being measured to reply to this meaningfully.
Quote from: Verpies on 2026.06.26, 18:42:01
I will need the probe positions on the circuit that is being measured to reply to this meaningfully.
I hope the following is sufficient to explain the scope connections. The indicated ground is the scope ground connection.
Pm
Quote from: partzman on 2026.06.26, 19:47:19
I hope the following is sufficient to explain the scope connections. The indicated ground is the scope ground connection.
Unfortunately it is not enough.
Please draw the path that Ch3 and Ch4 grounds take on that diagram.
If Ch4 is a contactless current probe then skip its ground.
Also, where in the bridge does Ch2 connect ?
Quote from: Verpies on 2026.06.26, 21:35:05
Unfortunately it is not enough.
Please draw the path that Ch3 and Ch4 grounds take on that diagram.
If Ch4 is a contactless current probe then skip its ground.
Also, where in the bridge does Ch2 connect ?
OK, is this complete enough?
Pm
Quote from: partzman on 2026.06.27, 16:23:21
OK, is this complete enough?
Pm
may not be.. C.C
hey pm
the cap in the core... i see that you have the cap with a wire from top of the cap to the bottom and you are measuring current at the top.. is that wire in the window of the core?
mags
Quote from: partzman on 2026.06.27, 16:23:21
OK, is this complete enough?
No. Is the path of Ch3 probe's ground lead like I have drawn with the green line ?
Quote from: Magluvin on 2026.06.27, 18:59:19
may not be.. C.C
hey pm
the cap in the core... i see that you have the cap with a wire from top of the cap to the bottom and you are measuring current at the top.. is that wire in the window of the core?
mags
Yes, the wire and the cap are both in the core window. There is a pix in my post #196 of this assembly with the current probe in position and if you look closely, you can see the wire just to the left of the probe and the two paralleled caps to the right. The important point is that the wire is as close to the primary winding as possible in order to receive the maximum amount of induction from the primary's H-Field as possible.
Pm
Quote from: Verpies on 2026.06.27, 19:12:59
No. Is the path of Ch3 probe's ground lead like I have drawn with the green line ?
Yes, that is correct.
Pm
Quote from: partzman on 2026.06.27, 20:43:18
Yes, that is correct.
So you are forming a shorted loop around the core. See the extended green line that marks this loop.
This loop is essentially a shorted secondary winding of your transformer. As in any shorted secondary winding, I expect a high current to flow in it.
The voltage measured by the Ch3 probe is determined by the voltage divider formed by the resistances of the two wire halves that form this shorted loop (the probe's ground wire and clip being one of them) and the voltage determined by the pri:sec turn ratio (I think that's 1:1 in this circuit).
Quote from: Verpies on 2026.06.27, 21:07:25
So you are forming a shorted loop around the core. See the extended green line that marks this loop.
Shouldn't that green loop include the probe input impedance (like 10 megohms and 10pF), i.e.not a short circuit?
Smudge
im not really concerned with loops outside the core or to say, outside the window of the core. so smudge is correct.
so, if the wire looping the top to the bottom of the cap is what is being induced, then why is the capacitor in the window also? its been a bit, but i thought this was all to show the induction effect of just the cap in the window of the core. are you trying to show that the that the capacitor currents being induced can oppose the currents of the induced wire?
mags
Quote from: Verpies on 2026.06.27, 21:07:25
So you are forming a shorted loop around the core. See the extended green line that marks this loop.
This loop is essentially a shorted secondary winding of your transformer. As in any shorted secondary winding, I expect a high current to flow in it.
The voltage measured by the Ch3 probe is determined by the voltage divider formed by the resistances of the two wire halves that form this shorted loop (the probe's ground wire and clip being one of them) and the voltage determined by the pri:sec turn ratio (I think that's 1:1 in this circuit).
No, this is not correct at all. The current loop is completely inside the LC network (wire and 8uf cap) that is physically inside the core window. In my actual test circuit, there is/was no ground lead connected to the CH3 probe and the scope common ground shown on the schematic was connected to the CH1 probe.
However, I connected a ground lead to the CH3 probe and measured the current in the ground lead with the CH4 current probe. The result is seen below and is showing essentially near zero current both with and without the CH1 ground lead connected.
Pm
Quote from: partzman on 2026.06.28, 14:34:30
The result is seen below and is showing essentially near zero current both with and without the CH1 ground lead connected.
Yes, the scopeshot confirms it .
Quote from: Smudge on 2026.06.28, 09:13:09
Shouldn't that green loop include the probe input impedance (like 10 megohms and 10pF), i.e.not a short circuit?
Smudge
Yes, however Ch3 still measured the induced voltage through the window of the core.
To measure the voltage only across C1, the Ch3 probing should be routed like the green line.
Routing it like the orange line will not produce the same results.
Quote from: Verpies on 2026.06.29, 12:36:19
Yes, however Ch3 still measured the induced voltage through the window of the core.
To measure the voltage only across C1, the Ch3 probing should be routed like the green line.
Routing it like the orange line will not produce the same results.
If I understand you correctly, the ground lead for CH3 should pass thru the toroid as shown with the green line, If this is the case then yes, the results will be different. The ground lead will now be charge separated equally with the capacitor which will result in a zero or near zero voltage differential at the probe's input amplifier.
If routed like the orange wire or if no ground lead is attached to the CH3 probe, then the probe's input amplifier has zero volts for one voltage input with the other receiving the voltage across the cap thru the probe's 10M resistance and 3pf capacitance yielding the correct voltage across the cap.
However, for a clear confirmation with no doubt, we can do a differential measurement of the voltage across the LC network when floating off ground. We first remove the ground connection from the LC network and place the CH2(blu) probe on this un-grounded connection with the CH3 probe still in place on the other LC network connection. Both CH2 and CH3 have no ground wires connected to the probes thus forcing the probes electrically to rely on the cable shield of each for their respective ground reference.
P7 shows the results of this differential measurement displayed between the cursors(pnk) as CH3-CH2 on the Math(red) channel.
Regards,
Pm
Quote from: partzman on 2026.06.29, 14:42:21
If routed like the orange wire or if no ground lead is attached to the CH3 probe, then the probe's input amplifier has zero volts for one voltage input with the other receiving the voltage across the cap thru the probe's 10M resistance and 3pf capacitance yielding the correct voltage across the cap.
No because of the EMF induced in the orange measurement loop according to the Faraday's law ℰ=-dΦ/dt, where Φ is the magnetic flux threading that loop.
The orange routing (https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=56571) generates the same measurement in Ch3 as the red routing (https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=56579) depicted below (assuming individual isolated grounds of other channels, which your scope does not have):
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=56571)
Quote from: Verpies on 2026.06.29, 15:03:43
No because of the EMF induced in the orange measurement loop according to the Faraday's law ℰ=-dΦ/dt, where Φ is the magnetic flux threading that loop.
The orange routing (https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=56571) generates the same measurement in Ch3 as the red routing (https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=56579) depicted below (assuming individual isolated grounds of other channels, which your scope does not have):
OK, so based on your above response, my question for you, Smudge, Mags, and anybody else that is following, does the wire in the center of the core experience any charge separation at all?
If so, to what potential level?
Pm
The significance of these measurements is the build up and decay of some kind of resonance shown in post #192 and continuous in post #206. What is this resonance? IMO it is due to the self inductance of the 8uF capacitor which appears to be about 0.114uH. it demonstrates that applying an electric field of a certain type (E=-dA/dt) across the shorted capacitor gets inside that closed circuit, and we should be concentrating on how that occurs and can it be put to good use. PM talks of charge separation and that is what the electric field does in the short passing through the core center. It must also do something to the series LC of the capacitor and it is obvious from these results that for rapidly changing E field the short and the LC combination do not exhibit the same charge separation. With the square wave applied E field PM gets this significant effect that AFAIK has not been seen before. I think PM is wrong to ascribe it to an H field, in my opinion it is a feature of the particular type of E field here, and would not occur in an E field between two electrodes. The pumped currents PM is measuring are quite significant, so I think he should put a small resistor in series with his short on the capacitor and determine the output power going into that R determined from the current measured by the current probe. It should be possible to orientate R so that it is at right angles to the E field. Will that give us OU? PM has all the expertise and equipment to find out, and I get the feeling that he has done something like this before. What is new is the realization that a magnetic vector potential derived E field can get inside a closed circuit within that E field, a circuit that does not enclose magnetic flux.
Smudge
Further to the previous post the E field in the core window is not uniform. It is maximum closest to the inside surface of the core and minimum at the center, but both pointing parallel to the core axis. This is probably why we get induction within that interior closed circuit. Maximum induction will occur wth the wire short touching the inner surface and the capacitor along the center. If PM used a rectangular shorted wire circuit instead of his shorted capacitor he will lose the resonance but should still see a square wave current flow within that closed circuit. It would then be interesting to observe whether the input sees that effect when the rectangular circuit is moved into the core or is switched from open to closed (but the switch has to be positioned there and controlled remotely). Then place a load resistor as described previously and measure power into the load and power into the driving circuit. Is it OU? The voltage driving the load is the difference between the two induced voltages along the axis-parallel sides of the rectangular circuit. We can''t measure this difference using two probes and a differential measurement because both probes will see the same flux change. If the differential could be taken within a single probe (are there such probes?) then maybe we could. I will produce a view showing the E field vector lengths using the FEMM axisymmetric simulation but deriving them for the H field within a current loop where the math is the same as the E field within a flux loop.
Smudge
Quote from: Smudge on 2026.07.01, 09:08:14
The significance of these measurements is the build up and decay of some kind of resonance shown in post #192 and continuous in post #206. What is this resonance? IMO it is due to the self inductance of the 8uF capacitor which appears to be about 0.114uH. it demonstrates that applying an electric field of a certain type (E=-dA/dt) across the shorted capacitor gets inside that closed circuit, and we should be concentrating on how that occurs and can it be put to good use. PM talks of charge separation and that is what the electric field does in the short passing through the core center. It must also do something to the series LC of the capacitor and it is obvious from these results that for rapidly changing E field the short and the LC combination do not exhibit the same charge separation. With the square wave applied E field PM gets this significant effect that AFAIK has not been seen before. I think PM is wrong to ascribe it to an H field, in my opinion it is a feature of the particular type of E field here, and would not occur in an E field between two electrodes. The pumped currents PM is measuring are quite significant, so I think he should put a small resistor in series with his short on the capacitor and determine the output power going into that R determined from the current measured by the current probe. It should be possible to orientate R so that it is at right angles to the E field. Will that give us OU? PM has all the expertise and equipment to find out, and I get the feeling that he has done something like this before. What is new is the realization that a magnetic vector potential derived E field can get inside a closed circuit within that E field, a circuit that does not enclose magnetic flux.
Smudge
Smudge,
You have no idea how thankful I am in regards to your response! I hope to respond a little bit later but right now I'm experiencing a third round of C-Diff! I think we have a solution now using pulsed antibiotics as I'm already feeling better.
In the meantime, you might wish to ponder the results when the LC network is rotated 90 degrees.
Regards,
Pm
Quote from: Smudge on 2026.07.01, 15:59:27
It is maximum closest to the inside surface of the core and minimum at the center, but both pointing parallel to the core axis. This is probably why we get induction within that interior closed circuit.
Like this, but without the gap:
Here is the FEMM result where FEMM current into the page models rate of change of flux so that the H field around the conductor models electric field E. Yes there is a difference for a wire along the center where I get 0.3744V compared to one near the core which is 0.4897V. But the tilted E vectors along the top and the bottom also induce voltage that makes the closed circuit voltage nill. But PM gets a current in his closed circuit so maybe the fact that he has series L and C over part of the circuit creates the induction he witnesses. Forget making a rectangular wire circuit with a resistor and concentrate on adding a resistor onto his LC circuit to look for power there.
Smudge
Quote from: Verpies on 2026.07.01, 18:08:48
Like this, but without the gap:
No, it is the E field external to the core from E=-dA/dt, not a magnetic field.
Quote from: Smudge on 2026.07.01, 18:55:15
No, it is the E field external to the core from E=-dA/dt, not a magnetic field.
Do you think that the magnetic field does not leak into the eye/window of the core when the primary/driving winding is narrow (does not span the entire circumference of the the core) ?
Quote from: Verpies on 2026.07.01, 19:15:53
Do you think that the magnetic field does not leak into the eye/window of the core when the primary/driving winding is narrow (does not span the entire circumference of the the core) ?
I know it does but the leakage quantity reduces the greater the pemeability of the core. For high perm cores I tend to think its value is negligible.
I can add more to this theme as we know the current probe clamps a magnetic core around the wire thus adding inductance. This may be the inductance responsible for the resonance PM found, and not the self inductance of the capacitor. If this is the case we have the cap in one side of the rectangular closed circuit and the inductance in a different side. When this is modelled against the induced voltages in each side it opens the door to understanding where the induced current comes from. The text books all tell us that the induced voltage in a closed circuit that does not enclose flux is zero, and simulations do show that. We tend to think that the induced current must therefore be zero. But that is only true if each part of the circuit has impedance that relates current directly to voltage. If one part relates current to the time integral of the induced voltage and another part relates current to the time differential of voltage then clearly it is possible to have a voltage waveform that does induce current in the closed circuit. PM has shown that. The interesting point now is how does the presence of that current feed back to effect the drive into the ring core? I am sure this will eventually be discovered. More on this later as I have to dash off for routine living.
Smudge
Quote from: Verpies on 2026.07.01, 19:15:53
Do you think that the magnetic field does not leak into the eye/window of the core when the primary/driving winding is narrow (does not span the entire circumference of the the core) ?
Looking at the photo in reply#192 PM does have a narrow primary driving coil and the closed rectangular circuit in the core window is orientated to have the leakage flux flow through it so you could be correct here. Easy enough to rotate the closed circuit 90 degrees to see if the effect goes away. Maybe PM has done this and chosen the orientation that maximises the effect. Still thinking about what I said in my previous post and is this really possible.
Smudge
Smudge and Verpies,
After reading your comments, I think it might be best if I were to ask if there is any specific type of test you would like me to perform with/or on this setup that may help answer any questions.
Edit: One thing I might add that I haven't stated before, is the part of the test where the LC assembly is rotated 90 degrees in the core with the result of zero resonant current. For example, if the LC assembly is rotated yet another 90 degrees, resonant current is again seen with positive current flowing from the cap when a positive voltage is applied to the primary. IMO, the H_Field is inducing the foils in the cap. The current level is ~.5A peak and thus lower than the wire induction but this is due to the fact that C is two paralleled 4ufd caps place side by side.
The cancellation of current when the LC network is at 90 degree rotation is due to the induction via the H-Field into both the L and the C when properly positioned in the core. This creates equal currents which cancel each other.
Regards,
Pm
Quote from: partzman on 2026.07.02, 14:25:55
After reading your comments, I think it might be best if I were to ask if there is any specific type of test you would like me to perform with/or on this setup that may help answer any questions.
Yes, axially rotate the following planar metal loop in the eye/window of the core when the narrow primary winding is being pulsed and measure the amplitude of the pulses across the rotating 10K resistor using your scope's voltage probe at various angles.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=56623)
The loop does not have to be in motion when you make your amplitude measurements at various angles.
I have used FEMM to look into leakage flux. This is for a ring core that is infinitely long (an infinite cylinder) but it demonstrates something useful. The first image shows flux in the core from a small winding at the left. The H field normal to the red line is shown and the B field there integrates along the like to 8.26x10-9 webers. In the second image the core has been replaced with air so the H field is what emanates from just the coil. The B field there integrates to 2.83x10-9 webers. Thus the presence of the core has increased the flux from the coil by a factor of nearly 3. There is no doubt that the coil alone would pump some energy into the closed LC circuit, and the core adds to that pumping. Is that knowledge helpful?
Smudge
Quote from: Verpies on 2026.07.02, 16:19:29
Yes, axially rotate the following planar metal loop in the eye/window of the core when the narrow primary winding is being pulsed and measure the amplitude of the pulses across the rotating 10K resistor using your scope's voltage probe at various angles.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=56623)
The loop does not have to be in motion when you make your amplitude measurements at various angles.
OK, I hope this is what you are looking for! The loop is built to just fit into the core window and has a 10K carbon film resistor centered in the top section. I attached a ground wire to the center of the bottom section which is attached to ground. CH2(blu) is then attached to the side of the resistor that is closest to the primary winding and CH3(pnk) is connected to the other side. These are the probe positions for the first pix PR1A.
The assembly is then rotated 90 degrees in PR1B, 180 degrees in PD1C, and 270 degrees in PR1D. It appears there is little to no change in the voltage differences.
I then did the exact same procedural measurements without the ground connected and the voltage differentials were basically the same but the voltage magnitudes changed with each different position as was expected. I have those scope shots if you wish to see them.
I then took it upon myself to do a similar protocol using a shorted loop by replacing the resistor with a piece of wire. Pix PR2A shows the results with the loop aligned with the primary winding and the current probe showing the current leaving the loop lead closest to the primary winding.
The next three pix show the current in the loop after 90,180, and 270 degree rotation respectively. IMO, this indicates H-Field induction into the loop.
Regards,
Pm
My 'guess' would be that the magnetic field traversing across the core window would be weakest in the center of the window, as there is more say stretching of the field there.
Could just measure a wire that was just in different areas of the window. with a 10k voltage most likely be the same anywhere, if efield is what determines the voltage. Maybe make say 20, 30 turns coil that was large in dia to be able to shift the coil portion that is in the window, but using a much lower resistance rather than 10k, to see if current is limited to placement in the window.
mags
Quote from: partzman on 2026.07.02, 22:19:33
I attached a ground wire to the center of the bottom section which is attached to ground.
That is not exactly what I asked for.
I asked:
Quote from: Verpies on 2026.07.02, 16:19:29
...measure the amplitude of the pulses across the rotating 10K resistor using your scope's voltage probe at various angles.
Routing the probe's ground wire/clip randomly can lead to completely different outcomes in these types of experiments.
Quote from: partzman on 2026.07.02, 22:19:33
The assembly is then rotated 90 degrees in PR1B, 180 degrees in PD1C, and 270 degrees in PR1D. It appears there is little to no change in the voltage differences.
These unfortunate choices of measurement angles do not allow to me make a cosine fit because cos(90°)=0 and cos(270°)=0 and cos(180°)=-1.
Making these measurements in 30° degree increments would have been more informative ...and you don't need to post scope shots for each of them (I know how much work that is) - a copy of numeric amplitude readouts is sufficient.
Quote from: partzman on 2026.07.02, 22:19:33
I then did the exact same procedural measurements without the ground connected and the voltage differentials were basically the same but the voltage magnitudes changed with each different position as was expected.
So what were the ground wires/clips doing when they were not connected ? What was scope's chassis ground connected to ?
Quote from: partzman on 2026.07.02, 22:19:33
I then took it upon myself to do a similar protocol using a shorted loop by replacing the resistor with a piece of wire.
That's a different experiment yet because it is measuring current (not voltage). Current is more informative wrt coils than the induced voltage but it is generally harder to measure.
However putting a contactless current sensor in place of the rotating resistor (now wire) is closer to the measurement point I had initially intended.
I have superimposed the two FEMM results showing both the flux in the core and the much smaller magnitude leakage flux. I also show a closed electric circuit in the donut hole depicted as a copper ring to show that the leakage flux lines pass through that ring. PM talks of H field induction and those leakage field lines depict both H field and B field. So the closed loop gets some voltage induction from the rate of change of flux that passes through it just like in any transformer. Flux is B*area and in this closed loop in air it is u0*H*area and d(u0*H*area}/dt is H field induction. Does PM have a different view on what H field induction is?
PM has measured significant current (amps) in his LC loop and I don't know whether this leakage flux pumping is sufficient to do that. It may be that the leakage flux starts the resonant build up process process and some other effect amplifies this, so IMO this still needs consideration to nail it.
Smudge
Quote from: Verpies on 2026.07.02, 16:19:29
Yes, axially rotate the following planar metal loop in the eye/window of the core when the narrow primary winding is being pulsed and measure the amplitude of the pulses across the rotating 10K resistor using your scope's voltage probe at various angles.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=56623)
The loop does not have to be in motion when you make your amplitude measurements at various angles.
Let's try this a different manner so we're on the same page! Specifically state how you wish the voltage probes and grounds to be connected or not connected to the loop plus any other details I may be missing. State what it is you expect to see or what you are looking for. Each sample angle will be taken in 30 degree increments for the reason you've described earlier unless specified otherwise.
On the floating measurements I made, there were no direct ground lead connections from each probe but rather the common scope ground was connected at the CH1(yel) probe which was connected to the power supply ground. IMO, this test showed null results IOW, no appreciable voltage differential across the 1K resistor except for the quiescent channel offset voltage.
Regards,
Pm
The first image below shows the E field inside a ring core across its centre. This is due to a rising flux in the core from a rising current in a coil wound on the ring core (not shown). For the purposes of this lecture, we will assume the E field is uniform.
The next image shows a lump of dielectric material in the E field. The magnetic vector potential derived (MVPD) field penetrates the dielectric to displace the orbital electrons so the dielectric gets charged just as it would if we had voltage applied to electrodes on its upper and lower surfaces. Alongside the dielectric we show a copper wire. The MVPD E field penetrates the copper to drive the mobile conduction electrons to one end. Thus, the wire is also electrically "charged", the potential across the dielectric being the same as the potential across the wire.
The third image shows electrodes on the dielectric connected to the wire, forming a shorted parallel plate capacitor. The MVPD E field has the seemingly impossible property of charging the shorted capacitor. The dielectric stores both charge and energy.
The final image shows the E field suddenly removed by holding the driving coil current (hence also the flux) constant. The energy in the capacitor discharges through the shorting wire. If we measure that current by clipping a current probe onto the top wire we add inductance, so the discharge will be sinusoidal. This is exactly what PM has done. To fully analyse his scheme, we would need to know the inner construction of his capacitor but suffice to say that this little lecture tells us that the E field inside the donut ring core can get a shorted capacitor to both charge and discharge if we drive the input coil in an appropriate manner.
Here endeth the lesson.
Smudge
Smudge,
The caps used to form the 8uf total capacitance in my resonance test are wound foil types. With the foils placed vertically in the core window, both are charged separated equally with no energy being generated in the cap. This makes your lesson above different as you appear to have horizontal plates in your capacitor which allows for energy to be developed during charge separation.
Therefore, in my experiment as shown in post #192, when the very first positive pulse is applied, the H-Field from the primary induces the amount of current shown in the bare wire connected to the cap at resonance. As seen with each successive pulse to the primary, the resonant current builds to a continuous peak level as determined by the overall circuit losses. The cap receives a small amount of energy during the positive half cycles to the primary with this energy being discharged during the negative half cycles. The average is therefore zero when running in resonance continuously and stabilized.
I do adhere to the notion that the H-Field can also be called leakage flux but I prefer the former.
Now, I will make a statement that will get me classified as a heretic I'm sure! When Faraday did his original induction tests using an iron toroid with a primary and secondary, he created his induction formula from his readings taken with his equipment at the time. His resultant induction law we know is accurate and not in question. However, he had no means at the time to measure the overall voltage differential around the perimeter of the core. IMO therefore, one can apply his induction law to a solenoid coil and the E-Field will be linearly distributed around the perimeter of the coil. However, when we have a folded or closed flux loop as in a toroid core, "U" core, or "E" core, the E-Field is not distributed linearly so we have to be careful if we apply his induction law to such.
My experimentation seems to show that the E-Field (in a rectangular cross sectioned toroid) is nearly all contained between the top and bottom surfaces of the core. I know this is a real point of contention so I am attempting to devise experiments to prove or dispel this apparent anomaly. I already have some to show but I have another test to try before disclosing anything at this time.
Regards,
Pm
Quote from: partzman on 2026.07.03, 20:02:04
Let's try this a different manner so we're on the same page! Specifically state how you wish the voltage probes and grounds to be connected or not connected to the loop plus any other details I may be missing.
Across the resistor.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=56661)
Quote from: Verpies on 2026.07.02, 16:19:29
Yes, axially rotate the following planar metal loop in the eye/window of the core when the narrow primary winding is being pulsed and measure the amplitude of the pulses across the rotating 10K resistor using your scope's voltage probe at various angles.
Quote from: Smudge on 2026.07.04, 14:02:20
The next image shows a lump of dielectric material in the E field. The magnetic vector potential derived (MVPD) field penetrates the dielectric to displace the orbital electrons so the dielectric gets charged ...
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4797.0;attach=56653)
That's right but that dielectric will not stay polarized when the MVPD field is removed (not reversed) - it will relax in picoseconds.
...but it will stay charged indefinitely when free charges are added/removed from its surface ...even when the MVPD field is removed. This is why charged styrofoam peanuts "hold" the charge.
Quote from: Smudge on 2026.07.04, 14:02:20
so the dielectric gets charged just as it would if we had voltage applied to electrodes on its upper and lower surfaces.
There is a great difference in behavior between bound charges and free surface charges on a dielectric.
A charged capacitor with a dielectric between metallic electrodes has electrons permanently added/removed from these metallic electrodes. Additionally, the same agency that added/removed them also performed more work polarizing the dielectric dipoles. Such capacitor "holds" the electrodes' charge imbalance indefinitely because there is no path for it to equilibrize.
The MVPD field will move free charges through space and will deposit them on material objects, but they have to be free to move. Inside the dielectric the charges cannot move linearly but its electric dipoles can rotate and become polarized preferentially in one direction (this takes work) as long as the force that rotated these dipoles persists.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=55323)
The effect of an electric field created by
a moving magnetic field, on free charged particles.Our immediate environment does not have many free moving charges but their number can be increased with e.g.: corona discharge and alpha/beta emitters.
https://youtu.be/ZBHIp967TD8?t=65
You are right to challenge my lecture, I had not thought things through enough.
Quote from: Verpies on 2026.07.04, 21:03:30
That's right but that dielectric will not stay polarized when the MVPD field is removed (not reversed) - it will relax in picoseconds.
Is there proof for your picoseconds claim? I ask because you say "its electric dipoles can rotate and become polarized preferentially in one direction (this takes work)" and I agree it takes work, IOW it stores energy. When the MPVD field is removed what happens to that stored energy? Where does it go in your picosecond time frame?
Quote...but it will stay charged indefinitely when free charges are added/removed from its surface ...even when the MVPD field is removed.
So in the case where we have electrodes on the surfaces that are connected to the short we have to consider what the (temporarily constant along the top of the square wave) MVPD E field does where the free charges in the shorting wire move into a stable distribution and the bound charges in the dielectric displace into a stable condition. I contend that the difference between these does allow the dielectric to become polarised and store energy. I agree the electrodes will not have the necessary free charges added/removed for this polarisation at the instant the MVPD E field is removed. But the stored energy can't just disappear. The E field driving the non-uniform distribution of free electrons in the shorting wire does disappear so IMO the energy stored in the dielectric now takes over control of those free electrons in both the electrodes and the short to discharge that energy via a current pulse. The polarised dielectric rapidly pulls the free electrons in the conductors to the added/removed surface conditions and this movement doesn't involve a significant current flow or capacitance effect with the inductance there (difficullt to come to terms with but the capacitance along conductors is negligible). Thereafter we have the well known discharge of a conventionally charged capacitor. That may be picoseconds if there is no inductance to consider, but PM adds inductance with his current probe. That "taking control" drives the current in the opposite direction to the one I showed in my post so I need to edit that.
Smudge
Quote from: partzman on 2026.07.04, 19:38:14
Smudge,
The caps used to form the 8uf total capacitance in my resonance test are wound foil types. With the foils placed vertically in the core window, both are charged separated equally with no energy being generated in the cap. This makes your lesson above different as you appear to have horizontal plates in your capacitor which allows for energy to be developed during charge separation.
I realize that. I was trying to develop an understanding for your large resonant current pulses from something unexpected. The wound foil plates will have the induced E field along the foil width so putting electron diplacement from top to bottom (your tubular capacitor being vertical). This is polarizing the dielectric along an axis that is not the usual polarisation.
QuoteTherefore, in my experiment as shown in post #192, when the very first positive pulse is applied, the H-Field from the primary induces the amount of current shown in the bare wire connected to the cap at resonance. As seen with each successive pulse to the primary, the resonant current builds to a continuous peak level as determined by the overall circuit losses. The cap receives a small amount of energy during the positive half cycles to the primary with this energy being discharged during the negative half cycles. The average is therefore zero when running in resonance continuously and stabilized.
I do adhere to the notion that the H-Field can also be called leakage flux but I prefer the former.
Then you must accept that the leakage field passes through your shorted capacitor closed loop, or Verpies wire loop, to induce in a conventional way not associated with variation in induced E fields, i.e by V=dPhi/dt where Phi is total flux through the loop.
QuoteNow, I will make a statement that will get me classified as a heretic I'm sure! When Faraday did his original induction tests using an iron toroid with a primary and secondary, he created his induction formula from his readings taken with his equipment at the time. His resultant induction law we know is accurate and not in question. However, he had no means at the time to measure the overall voltage differential around the perimeter of the core. IMO therefore, one can apply his induction law to a solenoid coil and the E-Field will be linearly distributed around the perimeter of the coil. However, when we have a folded or closed flux loop as in a toroid core, "U" core, or "E" core, the E-Field is not distributed linearly so we have to be careful if we apply his induction law to such.
His induction law deals with total flux passing through the loop, independent of whether the magnetic field there is uniform or non-uniform. You have jumped to the E field where its closed integral is Faraday's induced voltage.
QuoteMy experimentation seems to show that the E-Field (in a rectangular cross sectioned toroid) is nearly all contained between the top and bottom surfaces of the core. I know this is a real point of contention so I am attempting to devise experiments to prove or dispel this apparent anomaly. I already have some to show but I have another test to try before disclosing anything at this time.
But simulations show this not to be the case, the E field there supplying only about 0.4 of the volts per turn. The problem comes in trying to measure the voltage on things in that part of the core where the routing of ground connections become important. It is very easy to be misled.
Smudge
Quote from: Verpies on 2026.07.04, 20:49:13
Across the resistor.
OK. Using CH3(pnk) with the ground connected across the 10K resistor as you show, the loop is then rotated in 45 degree increments as 30 degrees was difficult to do accurately. The CH3 offset was measured at -300uv with the vertical deflection set to 20mv/division. The loop was then rotated cw as viewed from the top and the device was run continuously.
This is the data with the first column being the degrees, the second being the average voltage measured across the 10K during the positive pulse application to the primary, and the third being the average voltage measured during the negative pulse application to the primary.
0, -5.20mv, +4.42mv
45, -3.70mv, +3.32mv
90, +57uv, +560uv
135, +2.87mv, -1.62mv
180, +5.20mv, -4.27mv
225, +2.09mv, -960uv
270, +380uv, +160uv
315, -991uv, +2.37mv
These measurements were taken with hand positioning of the loop and are therefore subject to variations as if compared to a more accurate mechanical positioning method. Hopefully they will be helpful as is!
Regards,
Pm
Smudge,
I lost my detailed response to your last post so here is my overall response showing one proof IMO of my position.
Here is a test where the scope probe is charge separated in the toroid core window and appears to indicate that the ~V/T to the primary exists only in the core window.
First pix is the actual test set, second pix is the circuit schematic, and the third is the scope results.
What is seen here is a wire that covers the outsides of the toroid with a scope probe CH3(pnk) placed in the core window and attached to one end. The other end is connected to circuit ground with the scope probe ground wire also connected to this same circuit ground. So, we have zero volts at the probe tip due to the grounded outer wire on the toroid and zero volts at the coax shield of the probe cable.
We now see that there is a voltage across the probe as indicated on the scope pix that is the opposite polarity from the applied voltage on the primary. This is due to the fact that the internal 10M resistor and remaining wire within the probe tip are charge separated by an amount very close to the V/T of the primary. Notice this occurs within the upper and lower surfaces if you will of the toroid's core window. Little to no voltage appears across the open circuited outer wire on the toroid.
Regards,
Pm
Quote from: partzman on 2026.07.05, 14:49:04
What is with the timeout of a user??? After spending some time preparing a response to a user and then trying to post, I was forced to again sign-in and after doing so, I lost all my responding post! Very $@#*^&%* irritating >:(
Indefinite until the server's resources are exhausted if you check the "Always stay logged in:" checkbox ...and they are getting exhausted with thousands of Guests hammering.
Quote from: partzman on 2026.07.05, 14:29:13
This is the data with the first column being the degrees, the second being the average voltage ...
The average of a sine waveform over 360° is zero.
The average of a square waveform over 360° is ½V
P-P.
Average peak-to-peak voltage or average amplitude would have been better.
Quote from: partzman on 2026.07.05, 14:29:13
...measured across the 10K during the positive pulse application to the primary, and the third being the average voltage measured during the negative pulse application to the primary.
0, -5.20mv, +4.42mv
45, -3.70mv, +3.32mv
90, +57uv, +560uv
135, +2.87mv, -1.62mv
180, +5.20mv, -4.27mv
225, +2.09mv, -960uv
270, +380uv, +160uv
315, -991uv, +2.37mv
Could be a cos and -cos:
Quote from: Verpies on 2026.07.05, 18:51:33
The average of a sine waveform over 360° is zero.
The average of a square waveform over 360° is ½VP-P.
Average peak-to-peak voltage or average amplitude would have been better.
Could be a cos and -cos:
I'm sorry I wasn't clear on the probe measurements and calling it the "average" was incorrect. The first voltage measurements were the peaks of the waveforms taken with the "A" cursor and the second voltage measurements were also the peaks taken with the "B" cursor.
Pm
Here is another test of a piece of wire inserted into the hole of the toroid where it's charge separated voltage produces a current in a separate voltage supply.
The first pix is of the test circuit showing the wire, diode, ground, and the CH3(pnk) probe connection. The current probe is just to the right out of the pix, but attached to the red lead coming from the power supply. Now one could argue that the top and bottom surfaces are contributing to the over all E-Field but this will be addressed in a following post.
The second pix shows the schematic of the circuit. We see a Schottky diode connecting the wire to a 3.40v stabilized and bypassed power supply. The idea being that when the charge separated voltage across the wire was greater than one Schottky diode drop above 3.40v, we should see a current flow from the wire to the power supply.
The third pix show the scope results of this test. We see an average current flow of ~43ma in CH4(grn) when the wire CS voltage reaches 3.844v average as seen on CH3. This means the Schottky diode has a conduction voltage of ~.444v which is within reason.
Pm
Here is yet another test showing no E-Field along the upper, lower, and outer surfaces using a resonance circuit made up of a formed shorting wire and an 8uf capacitor.
The first pix if a photo of the test setup.
The second pix is the circuit schematic showing how the loop is formed around the outside perimeter of the toroid with the cap positioned in the near center of the outside leg. The other vertical leg of the wire is placed close to the primary 12T winding for maximum H-Field induction. The leakage current on the outside of the toroid should be less than that on the inside of the core, but we see resonant current near equal to that of the inside LC test. However, some of this is attributable to the lower resonance frequency.
The third pix is the scope shot showing the measurement results. With the bottom of the cap grounded and the top connected to Ch3(pnk). we see little to no overall average E-Field present. What we do see is the charging and discharging of the cap due to the resonant current flowing thru the loop. The resonant frequency here is 138kHz which is lower than the 164kHz with the internal LC because of the longer length of the shorting wire.
Pm
Quote from: partzman on 2026.07.05, 15:46:01
Smudge,
I lost my detailed response to your last post so here is my overall response showing one proof IMO of my position.
Here is a test where the scope probe is charge separated in the toroid core window and appears to indicate that the ~V/T to the primary exists only in the core window.
First pix is the actual test set, second pix is the circuit schematic, and the third is the scope results.
What is seen here is a wire that covers the outsides of the toroid with a scope probe CH3(pnk) placed in the core window and attached to one end. The other end is connected to circuit ground with the scope probe ground wire also connected to this same circuit ground. So, we have zero volts at the probe tip due to the grounded outer wire on the toroid and zero volts at the coax shield of the probe cable.
We now see that there is a voltage across the probe as indicated on the scope pix that is the opposite polarity from the applied voltage on the primary. This is due to the fact that the internal 10M resistor and remaining wire within the probe tip are charge separated by an amount very close to the V/T of the primary. Notice this occurs within the upper and lower surfaces if you will of the toroid's core window. Little to no voltage appears across the open circuited outer wire on the toroid.
It is clear to me that this configuration measures the full volts per turn and doesn't measure just the volts across the core window. My first image below shows inducement into the entire probe connection circuit. The next image shows that entire connection circuit rotated through succesive 90 degrees and in each case it measures the full volts per turn.
Smudge
Quote from: partzman on 2026.07.05, 20:29:03
Here is yet another test showing no E-Field along the upper, lower, and outer surfaces using a resonance circuit made up of a formed shorting wire and an 8uf capacitor.
The first pix if a photo of the test setup.
The second pix is the circuit schematic showing how the loop is formed around the outside perimeter of the toroid with the cap positioned in the near center of the outside leg. The other vertical leg of the wire is placed close to the primary 12T winding for maximum H-Field induction.
You accept that your H field induction is leakage flux so what you measure in this circuit is divorced from flux in the core. You can make this outside circuit any shape you like as long as it encloses some leakage flux. The induced voltage from your wire close to the core is completely cancelled by the induced voltage around the rest of your circuit, so has no effect. The A field from the core is everywhere in your circuit and induces voltage all round it that integrates to zero as the circuit does not enclose the core flux.
QuoteThe leakage current on the outside of the toroid should be less than that on the inside of the core, but we see resonant current near equal to that of the inside LC test.
Because that inside circuit also did not enclose core flux so it is the same situation. What you have done here is clarify why that circuit get its high resonant current, it is all due to leakage flux (not leakage current which I am sure you did not mean).
QuoteHowever, some of this is attributable to the lower resonance frequency.
The third pix is the scope shot showing the measurement results. With the bottom of the cap grounded and the top connected to Ch3(pnk). we see little to no overall average E-Field present. What we do see is the charging and discharging of the cap due to the resonant current flowing thru the loop. The resonant frequency here is 138kHz which is lower than the 164kHz with the internal LC because of the longer length of the shorting wire.
Ignoring the HF bursts we do see Ch3 voltage rising and falling on alternate 4uS periods by an amount that agrees with the 8uF carrying the measured current. It would not show the inducing E field from the core flux for reasons already stated.
Smudge
Quote from: Smudge on 2026.07.06, 06:45:55
It is clear to me that this configuration measures the full volts per turn and doesn't measure just the volts across the core window. My first image below shows inducement into the entire probe connection circuit. The next image shows that entire connection circuit rotated through succesive 90 degrees and in each case it measures the full volts per turn.
Smudge
You are correct that the configuration is measuring the full volts per turn. Where we differ is in the location of exactly where this V/T resides. IMO based on all my tests, is there is little to no voltage in the loop around the upper, outer, and lower sides of the toroid! If this is correct, this places the very tip of the probe at zero volts. We already know the coax of the probe cable is at zero volts so this means the inner conductor/resistor of the probe is charge separated. The polarity of the charge separation would force the tip positive but it can't go positive due to the fact (IMO) that the tip is at virtual ground. So, the inner wire feeding the input amplifier to the scope is forced negatively. This is exactly what we see in the scope pix when a positive voltage is applied to the primary.
If your interpretation was correct, the tip again will be charge separated in the positive direction and it's value would be displayed by a positive voltage on the scope. However, this is not what we see!
Pm
Quote from: Smudge on 2026.07.06, 07:36:46
You accept that your H field induction is leakage flux so what you measure in this circuit is divorced from flux in the core. You can make this outside circuit any shape you like as long as it encloses some leakage flux. The induced voltage from your wire close to the core is completely cancelled by the induced voltage around the rest of your circuit, so has no effect. The A field from the core is everywhere in your circuit and induces voltage all round it that integrates to zero as the circuit does not enclose the core flux. Because that inside circuit also did not enclose core flux so it is the same situation. What you have done here is clarify why that circuit get its high resonant current, it is all due to leakage flux (not leakage current which I am sure you did not mean).Ignoring the HF bursts we do see Ch3 voltage rising and falling on alternate 4uS periods by an amount that agrees with the 8uF carrying the measured current. It would not show the inducing E field from the core flux for reasons already stated.
Smudge
Let me state my position on the H-Field. The primary wire that is on the inside (or outside) of the core has two H-Fields. One is on the 'inside' of the wire closest to the core and one on the 'outside' of the wire. The 'inside' H-Field creates the flux in the core as well as the leakage flux.
The 'outside' is what creates the resonant current via induction. I respectively disagree on the rest of your comments above for reasons I will show in a later post.
Pm
Smudge,
This is a test of the outside loop placed on the core but now open circuited. The first pix is the test setup and the second pix is the schematic.
The third pix is the scope measurements of this loop. No appreciable voltage is seen using the A and B cursors on CH3(pnk). The scope probe and ground lead form a loop that is basically in parallel with the open loop so if there were any voltage present in either produced by the E-Field or the A_Field, it would be displayed as such on the screen.
Pm
hey pm
what if the open wire on just the outside of the core were on the other end of the core. just saying not right on the primary. if its not any trouble.
mags
Quote from: Magluvin on 2026.07.08, 02:46:15
hey pm
what if the open wire on just the outside of the core were on the other end of the core. just saying not right on the primary. if its not any trouble.
mags
Mags,
The voltage readings will be the same.
Pm
im just thinking....
mags
I would like to point out the following which I thot I had already posted but couldn't find it on the forum so if I did, here it is again.
The observation with this experiment is that the total voltage drops across each primary turn in the hole of a closed flux core (in this case our same toroid) sum to nearly the total V/T applied to the primary. IOW, there is little to no voltage across the wire on the outside of the core. IMO, any voltage drop in the outside wire is due to product of the current in the primary times the resistance of that portion of the primary winding.
The attached pix shows the test setup with a four turn primary that has little loops formed at the ends of each primary wire in the center or hole of the toroid. Two probes are then used to take differential measurements of the average voltages present at the ends of each turn of the primary. Measurements are taken when the primary has a positive voltage applied and also when the 3/4 bridge collapses with the voltage to the primary will then be negative. With the intrinsic nature of a 3/4 bridge, the collapse or negative application to the primary will always be slightly higher than the positive application due to a diode in the flyback or collapse path back to the power supply.
This is a table of the voltage measurements. The first voltage differential listed is taken during the application of a positive voltage to the primary and the second differential is taken during the primary collapse and is therefore negative.
#1 = 3.608v, -4.080v
#2 = 3.640v, -4.202v
#3 = 3.704v, -4.241v
#4 = 3.600v, -4.131v
__________________
Tot =14.552v, -16.654v
Vap=14.56, -16.670v
VaP is the differential voltage measured across the leads attached to the primary for an accurate V/T for both polarities. The results seem to indicate that nearly all the voltage drop across the excited primary exists in the wire in the hole of the toroid between the upper and lower surfaces,
Regards,
Pm
Quote from: Magluvin on 2026.07.08, 20:35:56
im just thinking....
mags
Could you maybe explain what you're thinking?
Pm
Quote from: partzman on 2026.07.08, 21:01:49
Could you maybe explain what you're thinking?
Pm
was thinking, if you see the pic of the toroid with the winding on the left, it shows that there can be some flux outside the toroid where the winding is, but none on the outside of the toriod on the right. unless ofcourse there is over saturation of the core from driving the primary.
but, if you have tried my earlier suggestion and found it will be the same having the test wire outside the core where the primary is and on the other side of the toroid where the primary isnt, then it is what it is.
mags
Quote from: Magluvin on 2026.07.09, 02:17:10
was thinking, if you see the pic of the toroid with the winding on the left, it shows that there can be some flux outside the toroid where the winding is, but none on the outside of the toriod on the right. unless ofcourse there is over saturation of the core from driving the primary.
but, if you have tried my earlier suggestion and found it will be the same having the test wire outside the core where the primary is and on the other side of the toroid where the primary isnt, then it is what it is.
mags
Yes, with an open loop, the voltage remains at or near 0v anywhere around the perimeter of the core. If you meant measuring the resonant current in the L/C loop on the outside of the core, then as you move away from the influence of the H-Field of the primary. the current drops off considerably but there is some still measurable due to the small amount of leakage flux outside the core and it varies.
Pm
I have been busy setting myself up to do my own measurements on things I find interesting. I expected my Rigol scope to be of a size to fit a standard 19 inch rack, and that tells you how long ago I last did any experiments. But I digress, back to the topic in hand.
Quote from: partzman on 2026.07.06, 16:16:33
Smudge,
This is a test of the outside loop placed on the core but now open circuited. The first pix is the test setup and the second pix is the schematic.
The third pix is the scope measurements of this loop. No appreciable voltage is seen using the A and B cursors on CH3(pnk). The scope probe and ground lead form a loop that is basically in parallel with the open loop so if there were any voltage present in either produced by the E-Field or the A_Field, it would be displayed as such on the screen.
Please do that test again with the scope probe connection as shown in the attached image, and not the alignment you used. This allows the leakage H field from the primary coil to pass through the area I have shaded green. The scope should record the voltage induced around the periphery of that area.
Smudge
Quote from: Smudge on 2026.07.10, 09:20:36
I expected my Rigol scope to be of a size to fit a standard 19 inch rack, and that tells you how long ago I last did any experiments.
You bought yourself a Rogol scope !? What vertical resolution ? Which model ?
Quote from: Smudge on 2026.07.10, 09:20:36
I have been busy setting myself up to do my own measurements on things I find interesting. I expected my Rigol scope to be of a size to fit a standard 19 inch rack, and that tells you how long ago I last did any experiments. But I digress, back to the topic in hand.
Please do that test again with the scope probe connection as shown in the attached image, and not the alignment you used. This allows the leakage H field from the primary coil to pass through the area I have shaded green. The scope should record the voltage induced around the periphery of that area.
Smudge
OK this is the test you requested with the first scope pix below showing the differential of the peaks between the positive and negative pulses using the A and B cursors on CH3(pnk). This does show that the H-Field of the primary is inducing this loop.
However, when using my differential test where the grounds are the scope probe lead's outer coax, the loop is much larger and not affected that greatly by the H-Field induction. IMO, this would result in a more accurate measurement of any voltage developed across this loop by the E or A field only.
Although you didn't ask for it, the second scope pix shows your test with the loop rotated 90 degrees on the core.
Regards,
Pm
Although
Quote from: Verpies on 2026.07.10, 13:53:05
You bought yourself a Rogol scope !? What vertical resolution ? Which model ?
It is Rigol DS 1102 that has an 8 bit resolution. That is enough for the initial work I have in mind where my theory tells me the effect I am looking for is quite significant.
Quote from: Smudge on 2026.07.10, 16:21:51
It is Rigol DS 1102 that has an 8 bit resolution.
That model has not been manufactured in years. I hope you paid only a few pounds for it.
Quote from: Verpies on 2026.07.10, 18:49:04
That model has not been manufactured in years. I hope you paid only a few pounds for it.
£224 via Amazon UK.
Quote from: partzman on 2026.07.10, 14:25:16
OK this is the test you requested with the first scope pix below showing the differential of the peaks between the positive and negative pulses using the A and B cursors on CH3(pnk). This does show that the H-Field of the primary is inducing this loop.
The
leakage H field from the primary, being in in air is producing B=u
0H tesla, which integrated over the area of the loop produces flux Phi whose rate of change gives the votages you measure. Ignoring the HF ringing, the square wave voltage you measure and the time period tells you that rate of change, so you could establish the magnitude of that loop leakage flux. And it will tell you that it is responsible for the current you observed in your closed loop LC circuit in post #253; that has a smaller area loop but since the leakage field is greatest near the primary coil will intercept almost the same value of flux. There is of course the other non-leakage flux within the ring core that does not pass through this loop and therefore has no effect on the measurement.
QuoteHowever, when using my differential test where the grounds are the scope probe lead's outer coax, the loop is much larger and not affected that greatly by the H-Field induction. IMO, this would result in a more accurate measurement of any voltage developed across this loop by the E or A field only.
Without a view of your differential measurement set up to see how the leakage flux passes through the loops of the probe connections I can't comment on what you are actually measuring. But I must emphasize that your judgement concerning the E field from the non-leakage flux in the core is wrong and you have not persuaded me otherwise. Your attempt to obtain evidence that the U shaped piece of wire placed close to the outer surfaces of the ring core is in a zero E=-dA/dt electric field ignores the effect of that field further away from that surface on the remainder of the external closed loop where its effect is completely nulled. Perhaps the image below showing the E field from the core flux is present all around the external loop will help convince you. At any point in that external loop the induced current is zero, and zero current into the scope's internal impedance is zero voltage measured.
Regards
Smudge
Smudge,
Here is the schematic of the differential test performed on the open loop outside the core. The ground connection for each probe is made at the scope and transferred thru each probe's outer shield. The CH1(yel) probe has a probe ground wire connected to the circuit supply ground and therefore is the overall ground for the CH2 and CH3 probes.
The probes appear to be able to be positioned at any location without any appreciable differential voltage being measured.
Pm
Quote from: partzman on 2026.07.11, 20:13:19
Smudge,
Here is the schematic of the differential test performed on the open loop outside the core. The ground connection for each probe is made at the scope and transferred thru each probe's outer shield. The CH1(yel) probe has a probe ground wire connected to the circuit supply ground and therefore is the overall ground for the CH2 and CH3 probes.
The probes appear to be able to be positioned at any location without any appreciable differential voltage being measured.
Pm
There is an induced voltage that varies with position of the single probe as you described in your post #268. There is a reasonable explanation for voltage being induced into the scope ground connection so it is different from the usual scope measurement. A differential measurement of a voltage using two probes requires the common ground connection to be at an intermediate potential, usually half the value so that each one yields half the actual voltage. How do you achieve that when the ground leads themselves have induced voltage? For your differential measurement with two probes the system is much more complicated and your 2D schematic is not sufficient for an analysis. Photos showing the layout in 3D would be better. My inital reaction is that the potential of the common ground connection of the two probes is undetermined in relation to your post #268 results and will always result in a zero measurement.
Smudge
Quote from: Smudge on 2026.07.12, 09:35:07
There is an induced voltage that varies with position of the single probe as you described in your post #268. There is a reasonable explanation for voltage being induced into the scope ground connection so it is different from the usual scope measurement. A differential measurement of a voltage using two probes requires the common ground connection to be at an intermediate potential, usually half the value so that each one yields half the actual voltage. How do you achieve that when the ground leads themselves have induced voltage? For your differential measurement with two probes the system is much more complicated and your 2D schematic is not sufficient for an analysis. Photos showing the layout in 3D would be better. My inital reaction is that the potential of the common ground connection of the two probes is undetermined in relation to your post #268 results and will always result in a zero measurement.
Smudge
The connections to the outside wire with the differential probes was very similar to the pix of the probes measuring the 4 turn primary in my post #262. With that test, it appears that the total voltage applied to the primary is divided between the vertical wires in the hole of the toroid!
I have to ask myself, since this also appears to be the conditions on the various secondary tests I have run, how is this possible? I mean if the E-Field is a function of the A-Field and this field is thought to be circulating the toroid in an azimuth direction, how is this possible. Or, is there some other mechanism at work here? This is the focus of my work at this time but I have to admit, I don't have a clue!!!
Pm
Quote from: partzman on 2026.07.13, 20:52:56
The connections to the outside wire with the differential probes was very similar to the pix of the probes measuring the 4 turn primary in my post #262. With that test, it appears that the total voltage applied to the primary is divided between the vertical wires in the hole of the toroid!
I have to ask myself, since this also appears to be the conditions on the various secondary tests I have run, how is this possible? I mean if the E-Field is a function of the A-Field and this field is thought to be circulating the toroid in an azimuth direction, how is this possible.
The E field coming from flux change in the core via E=-dA/dt where A lines form closed loops around that flux has a unique property. An electrical circuit that encloses that core flux will have an induced voltage. An electrical circuit that does not enclose that core flux will have zero induced voltage. If such a latter cicuit does produce a voltage there has to be an explanation and it appears from the work you have done the explanation is leakage flux that passes outside the core material. So you have your probes and ground connections within a magnetic field that is changing with time. so your scope will give you the voltage induced into your circuit. In some of your measurements it is a piece of wire shorting out your probe tip to the probe ground connection. That will give you a voltage from the total leakage flux passing through that loop. It is not the voltage that you think is across the piece of wire, it is from the flux passing through the loop. You can vary the size of the loop and get varying voltage just by squeezing the ground connection so that it runs close to the probe or by altering their angle as you have discovered after Verpies' suggestion.
QuoteOr, is there some other mechanism at work here? This is the focus of my work at this time but I have to admit, I don't have a clue!!!
I think you are right to follow a new path but not by trying to measure the E field at different positions around the core flux using your oscilloscope. Your suggestion to PhysicsProf that a cell placed in the core window will react to the E field there is good and that could be an indirect method for establishing what that field value is. I think my suggestion that dielectric placed there will react can also provide an indirect measurement.
Smudge
Smudge,
Thanks for your comments and suggestions.
Regards,
Pm
so the efield, it is only present when there is a flux change?
mags
Quote from: Magluvin on 2026.07.14, 19:32:40
so the efield, it is only present when there is a flux change?
mags
Yes.
Pm
and that flux change, if it happens in thin air, is there an efield developed in the air?
mags
Quote from: Magluvin on 2026.07.14, 22:29:34
and that flux change, if it happens in thin air, is there an efield developed in the air?
All free ions will me moved by this electric field.
ok. so how do we measure that? can we ionize a gas in a sealed glass tube, like neon?
mags
The reason I started this thread is because I began to see (via experimentation) that perhaps transformer induction, as we now know it, was either incorrect of incomplete! Let me try to explain.
Take the experiment in my post #262 which IMO is quite revealing! Here we have a toroid (2) with a 4 turn primary and we apply 16v to this primary resulting in ~4v/turn using mosfet switching in a 3/4 bridge configuration. A full bridge would be better but this is what I'm presently working with!
So, we have a relatively low source impedance power supply and a primary winding with an inductance of ~260uH that yields the voltages in the table as shown. My interpretation of this table is that the voltages measured across the portion of the primary wire that is between the upper and lower surfaces of the toroid, sum to nearly equal the total applied voltage to the primary. I see no way that the differential probes taking these voltage measurements could or would have any effect on each wire segment's measurements!
So, if this all true and correct, then consider the following- the instant we apply 16v to the primary winding, it will initially be evenly distributed around each turn of the primary. As current then begins to flow thru the primary winding and continues to increase, flux will begin to flow in the core and also continue to increase. Now here is my question- at what point in time and by what means, does the equally distributed voltage in the primary only appear in the primary wire that is between the upper and lower surfaces of the toroid? This is the elephant in the room!
Regards,
Pm
Quote from: Magluvin on 2026.07.15, 14:13:16
ok. so how do we measure that? can we ionize a gas in a sealed glass tube, like neon?
Definitely. And when these charged particles slam into something with sufficient speed, they glow.
There are free ions in atmospheric air, too, but they don't live very long before they get neutralized. These ions also get moved by such fields, but to get to decent speeds they need a strong field that can accelerate them meaningfully in the short time before they get neutralized. Corona discharge makes more of them. In partial-pressure systems the lifetime of these ions is much longer because the gas molecules are not packed together as densely as in the atmospheric air.
Quote from: partzman on 2026.07.15, 15:38:48
...the instant we apply 16v to the primary winding, it will initially be evenly distributed around each turn of the primary.
No. Read this (http://www.classictesla.com/download/corum_lumped_failure.pdf).
Quote from: partzman on 2026.07.15, 15:38:48
The reason I started this thread is because I began to see (via experimentation) that perhaps transformer induction, as we now know it, was either incorrect of incomplete! Let me try to explain.
Take the experiment in my post #262 which IMO is quite revealing! Here we have a toroid (2) with a 4 turn primary and we apply 16v to this primary resulting in ~4v/turn using mosfet switching in a 3/4 bridge configuration. A full bridge would be better but this is what I'm presently working with!
So, we have a relatively low source impedance power supply and a primary winding with an inductance of ~260uH that yields the voltages in the table as shown. My interpretation of this table is that the voltages measured across the portion of the primary wire that is between the upper and lower surfaces of the toroid, sum to nearly equal the total applied voltage to the primary. I see no way that the differential probes taking these voltage measurements could or would have any effect on each wire segment's measurements!
So, if this all true and correct, then consider the following- the instant we apply 16v to the primary winding, it will initially be evenly distributed around each turn of the primary. As current then begins to flow thru the primary winding and continues to increase, flux will begin to flow in the core and also continue to increase. Now here is my question- at what point in time and by what means, does the equally distributed voltage in the primary only appear in the primary wire that is between the upper and lower surfaces of the toroid? This is the elephant in the room!
Regards,
Pm
ok. so i have a rotor on a motor, 1/4 in disk mags alternating n and s up and down on the outer face of the rotor. give me a suggestion as to how i can do this to see. probably not just a simple neon bulb with the 2 electrodes because of the electrodes.
mags
how can we detect the efield of a changing magnetic field without conductors being affected by the magnetic field???
OR, is it that it has to be a conductor that IS affected by the changing magnetic field presence, in order for that feield to show itself?????
mags
Quote from: Magluvin on 2026.07.15, 18:07:07
ok. so i have a rotor on a motor, 1/4 in disk mags alternating n and s up and down on the outer face of the rotor. give me a suggestion as to how i can do this to see. probably not just a simple neon bulb with the 2 electrodes because of the electrodes.
mags
Mags,
I wasn't responding to your question on ionization but rather pointing out my view on transformer induction.
Forgive me if I'm missing your point but I don't have an answer for your question above.
Regards,
Pm
Quote from: Magluvin on 2026.07.15, 20:31:32
how can we detect the efield of a changing magnetic field without conductors being affected by the magnetic field???
OR, is it that it has to be a conductor that IS affected by the changing magnetic field presence, in order for that feield to show itself?????
mags
I think basically this is the question I'm asking. If we were to take these same measurements on a straight solenoid coil, we would find the E-Field and the H-Field to be evenly distributed around the outside of the coil assembly.
Again, I hope I understand your questions!
Pm
Quote from: Verpies on 2026.07.15, 17:33:16
No. Read this (http://www.classictesla.com/download/corum_lumped_failure.pdf).
Yes, I agree with this paper. With a careful single cycle measurement of the setup, I do not see evidence of the applied voltage being evenly distributed across the primary, so my assumption is incorrect!
Pm
GREAT STUFF Partsman.
It is hard to see how the secondary loop is not complete.
I somewhat agree that differential measurement should deal with this .
Is there a difference between a tightly wrapped secondary and a very loosely wrapped one ?
Quote from: partzman on 2026.07.15, 21:46:33
Mags,
I wasn't responding to your question on ionization but rather pointing out my view on transformer induction.
Forgive me if I'm missing your point but I don't have an answer for your question above.
Regards,
Pm
no problem. let me put it this way...
if there is no conceivable way to detect, let alone use, the efield in the presence of a changing magnetic field unless there is a conductor present in that changing magnetic field, then i have to assume that it only appears in a conductor when that conductor is in the presence of a changing magnetic field.... sooo...
why is it so inconceivable that what you call an efield, isnt simply produced within the magnetically influenced, induced, conductor, and the magnetic field is the key to triggering that efield within the conductor?
ive asked for years and years for an example of how we can determine the presence of the 'efield' in the presence of a magnetic field, like in thin air. in thin air or even say a vacuum, and never get a definitive answer.
i have my own theories of what is going on, but i dont think anyone has gotten what i suggested. so i leave it to you all to prove that there is an efield associated with a changing magnetic field at ALL TIMES, even in the absence of a conductor being influenced by that changing magnetic field.. makes no sense that the efield can only coexist with a magnetic field when that magnetic field is changing.
mags
The electric field or changing magnetic field can be detected without a conductor because these fields accelerate free charges outside of conductors, too. For example like this:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=55323) The effect of a moving magnetic field,
on free charged particles.
Like deflecting electrons in a television's CRT but that deflection has always been described as the yoke's 'magnetic field' causing the deflection. OR, in a CRT like an oscilloscope, the deflection is done by electrostatic charge of deflection plates. It's one, or the other. I've never seen it described that the yoke's changing magnetic field produces, or is accompanied by an e-Field to cause the deflection.
Thats a nice 'animation' you have there, but its not an example of what i can do on the bench to prove that a changing magnetic field is accompanied by an e-Field. In fact, if in the animation it were the particles that were moving and the magnet stationary, would the particles still deflect?? If the magnets field is no longer changing, well where is that e-Field now??? Same deflection, no changing magnetic field.
Now with all that said, in a TV's CRT with 'magnetic' deflection yokes, with no input to the yokes we get a bright dot in the middle of the screen because there is no deflection. Well what will happen to that bright dot in the middle of the screen if we apply some DC to one of the, say horizontal yokes? Will the dot move from the center at first and then back to center because the dc input to that yolk is no longer changing??? or will it stay deflected?
If it stays deflected with DC applied to that yoke, and from what you guys describe that the e-Field is only present when the mag field is changing, then what is it that keeps the electron beam deflected if the e-Field is no longer present once the yoke's magnetic field is no longer changing?
mags
I'm trying to stay on topic because of the claim of the e-Field here in the experiments. If it is way off or not related, we can move this to a new topic but if I'm correct, then what I have presented shows a large issue to the claims of the e-Field here and how and when it works, if it is even factual by description here. In that case, things will have to be thought about differently when it comes to the theories in this thread, and should remain here.
mags
Quote from: Verpies on 2026.07.16, 06:43:21
I've never seen it described that the yoke's changing magnetic field produces, or is accompanied by an e-Field to cause the deflection.
Actually, television CRTs accelerate the electrons electrically (thus creating motion between the magnetic field, created by the deflection yoke, and the electrons) while the magnetic field deflects the moving electrons. So
both the electric and magnetic fields are at play in television CRTs.
Quote from: Magluvin on 2026.07.16, 05:34:12
... if in the animation it were the particles that were moving and the magnet stationary, would the particles still deflect??
Yes. Only the relative motion matters for this effect ...and all motion is relative.
Quote from: Magluvin on 2026.07.16, 05:34:12
Will the dot move from the center at first and then back to center because the dc input to that yolk is no longer changing??? or will it stay deflected?
The dot will stay deflected.
What deflected it was the relative motion between the electron and the constant (not changing) magnetic field generated by the yoke.
It doesn't matter whether the electron is moving wrt the magnetic field or the magnetic field is moving wrt the electron.
Once the moving electron exits the magnetic field region, its path does not revert the previous angle but continues in a straight line (deflected).
Of course a moving magnetic field is a changing magnetic field ...and vice versa.
* wrt = with respect to
Quote from: 3D Magnetics on 2026.07.15, 22:15:04
GREAT STUFF Partsman.
It is hard to see how the secondary loop is not complete.
I somewhat agree that differential measurement should deal with this .
Is there a difference between a tightly wrapped secondary and a very loosely wrapped one ?
Thanks! No, there appears to be no difference is how tight or loose the secondary is wound.
Pm
Quote from: Verpies on 2026.07.16, 06:43:21
Actually, television CRTs accelerate the electrons electrically (thus creating motion between the magnetic field, created by the deflection yoke, and the electrons) while the magnetic field deflects the moving electrons. So both the electric and magnetic fields are at play in television CRTs.
Yes. Only the relative motion matters for this effect ...and all motion is relative.
The dot will stay deflected.
What deflected it was the relative motion between the electron and the constant (not changing) magnetic field generated by the yoke.
It doesn't matter whether the electron is moving wrt the magnetic field or the magnetic field is moving wrt the electron.
Once the moving electron exits the magnetic field region, its path does not revert the previous angle but continues in a straight line (deflected).
Of course a moving magnetic field is a changing magnetic field ...and vice versa.
* wrt = with respect to
ok. but im talking about the efield specifically and its definition of it only being there when the magnetic field is changing. not sure why, but im feeling im either being misunderstood, or replies to my question are ignoring the core of the question.
ok. ill make it simple one more time..
1. is there an efield associated with a non changing, non moving magnetic field?
2. if not, then why?
3. when a magnetic field does move or change, why does this efield as described and use of its term in this thread and many others, happen to 'appear' and exist when that magnetic field moves or changes??
4. and finally, does this efield only present itself when a wire or any conductor is present in the changing magnetic field??
i see that it is used in theory as to calculating 'voltage' per turn when considering self induction and mutual induction, but i find that nobody really wants to answer the questions or claim that they dont understand the questions. my questions ar not that complicated.
mags
"What deflected it was the relative motion between the electron and the constant (not changing) magnetic field generated by the yoke.
It doesn't matter whether the electron is moving wrt the magnetic field or the magnetic field is moving wrt the electron.
Once the moving electron exits the magnetic field region, its path does not revert the previous angle but continues in a straight line (deflected)."
ok. so the deflection is due to a relationship between the 'magnetic' field and the 'electron', nothing more. correct?
mags
Quote from: Magluvin on 2026.07.16, 17:53:28
ok. so the deflection is due to a relationship between the 'magnetic' field and the 'electron'
The motion between the magnetic field and the electron.
Quote from: Magluvin on 2026.07.16, 17:53:28
...nothing more. correct?
Not only - the motion of the observer comes into play, too.
When the observer is not moving wrt the electron then the electron appears as the source of an electric field only.
However, the same electron appears as the source of magnetic field when the observer is moving wrt this electron.
Quote from: Magluvin on 2026.07.16, 15:09:32
1. is there an efield associated with a non changing, non moving magnetic field?
No, but there is an electric field associated with non-moving charges (e.g. ions).
Quote from: Magluvin on 2026.07.16, 15:09:32
2. if not, then why?
This is a very basic question. "Basic" does not mean easy. Understanding it requires the prerequisite understanding what a charge is and what a field is and what is the root cause of the difference between electric, magnetic and gravitation fields. All of that requires understanding what space and time are. Mainstream science treats these as axioms. Geometric algebra (https://youtu.be/60z_hpEAtD8?t=2137) and projective geometry help a lot. I don't think you will appreciate them.
Quote from: Magluvin on 2026.07.16, 15:09:32
3. when a magnetic field does move or change, why does this efield as described and use of its term in this thread and many others, happen to 'appear' and exist when that magnetic field moves or changes??
I could just reply "because of relativity" or "because Maxwel's equations say so", but without understanding the axioms mentioned in pt.2, you will not be satisfied. These tools will tell you how fields behave and how they depend on each other but they will not tell you why and what they are, e.g. Coulomb's law will tell you how much charges attract or repel as a function of distance but it will not tell you what a charge is or what space is.
Quote from: Magluvin on 2026.07.16, 15:09:32
4. and finally, does this efield only present itself when a wire or any conductor is present in the changing magnetic field??
No.
"Not only - the motion of the observer comes into play, too."
nothing else? efield??
mags
Quote from: Magluvin on 2026.07.16, 19:21:56
nothing else? efield??
I thought we were discussing only the deflection in the TV CRT.
Yes, the path of the electron is also affected by electric and gravitational fields in addition to the magnetic field and the relative motion of the observer.
Quote from: Magluvin on 2026.07.16, 15:09:32
ok. but im talking about the efield specifically and its definition of it only being there when the magnetic field is changing. not sure why, but im feeling im either being misunderstood, or replies to my question are ignoring the core of the question.
ok. ill make it simple one more time..
1. is there an efield associated with a non changing, non moving magnetic field?
2. if not, then why?
3. when a magnetic field does move or change, why does this efield as described and use of its term in this thread and many others, happen to 'appear' and exist when that magnetic field moves or changes??
4. and finally, does this efield only present itself when a wire or any conductor is present in the changing magnetic field??
i see that it is used in theory as to calculating 'voltage' per turn when considering self induction and mutual induction, but i find that nobody really wants to answer the questions or claim that they dont understand the questions. my questions ar not that complicated.
mags
I will attempt to answer your question-
1) Yes, electrostatics provide us with a scalar E-Field.
3) According to classical electrodynamics, the E-Field appears due to E=-dA/dt . IOW, the E-Field is generated by a change in the A-Field or the magnetic vector potential.
4) The E-Field is always present and will manifest itself in a wire located in the field via charge separation.
Pm
Quote from: partzman on 2026.07.16, 19:46:45
I will attempt to answer your question-
4) The E-Field is always present and will manifest itself in a wire located in the field via charge separation.
pm. your answer number 4. are we saying the magnet nor the wire needs to be moving and the efield is there?
mags
Quote from: Magluvin on 2026.07.16, 21:13:45
pm. your answer number 4. are we saying the magnet nor the wire needs to be moving and the efield is there?
mags
My answer for 4) was poorly constructed. It should have read-
4) When conditions exist for the electromagnetic production of an E-Field, the field will always be present and will manifest itself in a wire located in the field via charge separation.
Pm
Please remove if too off topic
Here a poster at Mooker made a new thread with a claim
Screen shot below
Be advised link he attached went to second screen shot ( notified Jim already)
Claims strange gain mechanism 650 times with no grounding ( capacitor experiment)
Respectfully submitted
Chet
Ps for additional clarity!! link at above post is connected to malware !!
Quote from: partzman on 2026.07.17, 13:52:13
My answer for 4) was poorly constructed. It should have read-
4) When conditions exist for the electromagnetic production of an E-Field, the field will always be present and will manifest itself in a wire located in the field via charge separation.
Pm
well my theory is that it is all a relationship between the magnetic field and the electron. and that gives the illusion or the impression that this efield is attached or comes along with the magnetic field, as you say, at all times.
and if you think more on what you have said twice in answer 4 and the newly constructed answer 4, "the field will always be present and will 'manifest itself' in a wire located in the field via charge separation", then it shouldnt be hard to think that what i stated as my theory, that it is all just a relationship between the magnetic field and the electron, and nothing more. ;) except for one thing.. and i would like anyone to demonstrate a stationary magnet and a stationary wire showing a manifestation of an efield. you didnt specify the 'when conditions exist' part....
as you say 'manifest itself', then really it either doesnt exist, and from what i get from the the idea of the efield that you guys present, that the efield cant be detected unless the magnetic field is moving/changing, or the wire moving or changing position, then i believe that the efield that is said to 'coexist' with the magnetic field at all times is malarky by definition. and, i believe that if we can detect the efield, of which indicates potential difference, of which you choose to use the words charge separation, then i can say that yes, the efield can and will 'only appear' as potential difference from one end of that wire to the other due to the magnetic fields interaction with the electrons in that wire. nothing more.
mags
Quote from: Magluvin on 2026.07.17, 15:55:39
well my theory is that it is all a relationship between the magnetic field and the electron. and that gives the illusion or the impression that this efield is attached or comes along with the magnetic field, as you say, at all times.
Referring to my toroid experiments, preset theory says there is no magnetic field in the hole of the toroid except for a small leakage flux, however, CE (classical electrodynamics) says there is an A-Field (magnetic vector potential) in the hole that varies with the current in the primary and this A-Field produces the E-Field.
Quote
and if you think more on what you have said twice in answer 4 and the newly constructed answer 4, "the field will always be present and will 'manifest itself' in a wire located in the field via charge separation", then it shouldnt be hard to think that what i stated as my theory, that it is all just a relationship between the magnetic field and the electron, and nothing more. ;) except for one thing.. and i would like anyone to demonstrate a stationary magnet and a stationary wire showing a manifestation of an efield.
Edit: If at any time one would add a load to a charge separated wire, one would have electron flow as a result but of course you already know that.
I don't know of any such experiment but perhaps someone else here may!
Quote
you didnt specify the 'when conditions exist' part....
AFIK, a voltage across an inductor can only be produced by one of two ways and that is, transformer induction or by passing a PM past the coil. I was referring to the former.
Quote
as you say 'manifest itself', then really it either doesnt exist, and from what i get from the the idea of the efield that you guys present, that the efield cant be detected unless the magnetic field is moving/changing, or the wire moving or changing position, then i believe that the efield that is said to 'coexist' with the magnetic field at all times is malarky by definition. and, i believe that if we can detect the efield, of which indicates potential difference, of which you choose to use the words charge separation, then i can say that yes, the efield can and will 'only appear' as potential difference from one end of that wire to the other due to the magnetic fields interaction with the electrons in that wire. nothing more.
With my toroid experiments, I refer to charge separation because the E-Field that is present separates any free charges in the wire or any object that consists of free charges. I have charge separated semiconductors, ferrites, electrolytes, etc. The claim of all this work is that the charge separation, or the evident E-Field, exists only between the upper and lower surfaces or a rectangular toroid. This shouldn't be according to CE. Also as I stated above, there is no magnetic field in the center of the toroid except for a small leakage flux but most is contained in the core.
Pm
Quote
mags
Quote from: Chet K on 2026.07.17, 14:05:35
Please remove if too off topic
Here a poster at Mooker made a new thread with a claim
Screen shot below
Be advised link he attached went to second screen shot ( notified Jim already)
Claims strange gain mechanism 650 times with no grounding ( capacitor experiment)
Respectfully submitted
Chet
Ps for additional clarity!! link at above post is connected to malware !!
That link at Mooker's site no longer seems to work!
Pm
"With my toroid experiments, I refer to charge separation because the E-Field that is present separates any free charges in the wire or any object that consists of free charges."
free charges.. so in a wire, just a wire alone, there are free charges? im used to hearing it as free electrons.. has something changed more recently that the word electrons has been replaced by charges?
mags
Quote from: Magluvin on 2026.07.17, 21:19:40
...has something changed more recently that the word electrons has been replaced by charges?
No but other charge carriers are mentioned less frequently.
How should we define 'charge carriers'?
mags
I could look it up... I just want your reference for the sake of our discussion... ;)
mags
Quote from: Magluvin on 2026.07.17, 22:19:00
How should we define 'charge carriers'?
Any particle that carries an electric charge: electrons, positrons, protons, muons, ions, alphas, Cooper pairs, polarons, solitons, pions, kaons.
Larger structural charge carriers are also possible, e.g. Lycopodium spores and holes in semiconductors.
Quote from: Verpies on 2026.07.18, 09:17:02
Any particle that carries an electric charge: electrons, positrons, protons, muons, ions, alphas, Cooper pairs, polarons, solitons, pions, kaons.
Larger structural charge carriers are also possible, e.g. Lycopodium spores and holes in semiconductors.
so in pm's case when he talks about free charge carriers in wire it is electrons, correct? also when the wire is not induced or not connected to input electrically, the electron are not free at these times, correct?
mags
Quote from: Magluvin on 2026.07.18, 16:57:57
so in pm's case when he talks about free charge carriers in wire it is electrons, correct? also when the wire is not induced or not connected to input electrically, the electron are not free at these times, correct?
mags
When the piece of wire is charge separated, we will find electrons bunched up, so to speak, at the negative terminal and protons at the positive terminal. If we connect an external resistive load to the wire outside the core, we will then see a current flow that will be called one of two types depending on one's view.
One type is conventional current flow where the reference is the flow of positive charge from the positive terminal to the negative. In semiconductors this is called 'hole flow'.
The other type is just called 'current flow' or 'electron flow' which is referencing a flow from negative to positive.
If the load however is placed in the hole of the toroid along with the wire, there is equal charge separation in both the wire and the load with the result being no current flow no matter which type of reference you choose.
Pm
Quote from: partzman on 2026.07.18, 20:45:32
When the piece of wire is charge separated, we will find electrons bunched up, so to speak, at the negative terminal and protons at the positive terminal. If we connect an external resistive load to the wire outside the core, we will then see a current flow that will be called one of two types depending on one's view.
absolutely! im with you there.. years ago messing with AV plug and charging a cap via 1 wire from 1 lead of the sec of a small hv 1200v neon transformer is where i came to that conclusion. so i ask this...
do you believe that on the neg end of that sec(at the time that the ac out is at peak neg), 1200v, that the bunched up neg charge would release some into that av plug to charge up that cap, 'as if it were just an extension of that sec winding'? lets say even further, that sec hv lead wire coming out of that hv transformer, would it also contain bunched up electrons? further more, extend that wire to a greater length, would there be bunched up electrons in that extended wire also?
Quote from: partzman on 2026.07.18, 20:45:32
One type is conventional current flow where the reference is the flow of positive charge from the positive terminal to the negative. In semiconductors this is called 'hole flow'.
"hole flow" definition of a 'hole' is a valence band electron missing from an atom, of which makes that atom positively charged. are you saying that the hole itself flows or that the atom itself flows?
when an atom is normal, equal amount of electrons and protons give that atom a neutral charge. remove an electron and the atom becomes positively charged. thus in my thinking, the pos end of that wire is just filled with atoms in the wire structure that have a pos charge and nothing more. just missing electrons on that pos end.
if you are talking about 'ions', atoms with a non neutral charge, moving through an electrolyte, then we are not talking about wires.
Quote from: partzman on 2026.07.18, 20:45:32
If the load however is placed in the hole of the toroid along with the wire, there is equal charge separation in both the wire and the load with the result being no current flow no matter which type of reference you choose.
thats understandable. as both the load and the wire of the loop are induced equally. but, as i believe you said earlier, i think, you can measure the top of the loop and the bottom, each out of the window of the core, and see voltage, similarly like just 2 wires connected at top and connected at bottom, where in our case the load is like one of the wires. so there should be say bunched up electrons at one connection of the loop out one side of the window, and pos charge(atoms with holes) at the connection on the other side of the window. if so, then you did have current flow to get those electrons to one end of that loop that is outside of that end of the window. maybe not a lo but current flow did happen both in the wire and the load of the loop.
mags
missed one part of what i wanted to say as i was interrupted here at work... here is the total part of my previous post...
Quote from: partzman on 2026.07.18, 20:45:32
When the piece of wire is charge separated, we will find electrons bunched up, so to speak, at the negative terminal and 'protons at the positive termina'l. If we connect an external resistive load to the wire outside the core, we will then see a current flow that will be called one of two types depending on one's view.
absolutely! im with you there..
here is what i didnt get into...
so, do protons themselves leave the nucleus and flow through the wire in your understanding?
mags
Quote from: Magluvin on 2026.07.18, 16:57:57
so in pm's case when he talks about free charge carriers in wire it is electrons, correct?
Correct.
Quote from: Magluvin on 2026.07.18, 16:57:57
also when the wire is not induced or not connected to input electrically, the electrons are not free at these times, correct?
Incorrect.
They are free to move but only within the confines of the space occupied by the metal.
They can't get out of that space unless other conditions are satisfied, e.g.: thermionic emission (https://en.wikipedia.org/wiki/Thermionic_emission), photoelectric emission (photoemission), field emission (cold emission), secondary emission (when energetic particles strike the electrode), Schottky (field-enhanced thermionic) hybrid emission, photoionization, etc...
In my opinion the electrons confined to matter are fundamentally different particles than electrons ejected outside of matter ...but that is a heresy in mainstream science.
Quote from: Verpies on 2026.07.19, 01:07:14
Incorrect.
They are free to move but only within the confines of the space occupied by the metal.
They can't get out of that space unless other conditions are satisfied, e.g.: thermionic emission (https://en.wikipedia.org/wiki/Thermionic_emission), photoelectric emission (photoemission), field emission (cold emission), secondary emission (when energetic particles strike the electrode), Schottky (field-enhanced thermionic) hybrid emission, photoionization, etc...
In my opinion the electrons confined to matter are fundamentally different particles than electrons ejected outside of matter ...but that is a heresy in mainstream science.
if you are saying electrons are free to move around the nucleus of the say copper atom, then i agree. but you seem to imply that there are more electrons free to go where they will throughout the wire, with no bond to the copper element atoms.
are these free electrons you speak of broken from their bonds of the copper atoms and do as they please without external influence, or are you saying that all the copper atoms of the wire have all their electrons, making those atoms of neutral electrical charge, but there are other free electrons in the wire? extra electrons in the wire than what is required to make all the copper atoms whole?
and your opinion of different kinds of electrons.. what has given you the idea of that opinion?
i have some opinions myself about electrons. an alternate theory to put it bluntly. my idea has more to do with how the electrons do what they do when they interact with a magnetic field rather than an efield related directly to the magnetic field that supposedly only 'manifests' itself when the mag field moves or changes only if there is a say wire for example being affected by the moving mag field, and or oppositely when a conductor moves within a stationary mag field. id say that the efield that manifests is brought about by the magnetically affected electron, rather than directly from the moving magnetic field itself. if the magnetic field moves on its own without any, lets just say, wires in the vicinity, and the efield cannot be detected there, then i find that harder to believe than my theory. im not doubting there is an efield. i believe it is associated or say comes directly from the wire itself rather than from the magnetic field itself. it seems to be a more likely mechanism. id say the idea that there is an efield associated with the magnetic field comes from the fact that the electron moves at 90deg of the moving magnetic field. but i believe that the electrons have a magnetic field themselves. and their reaction to magnetic fields has to do with that rather than an efield that exists with with the magnetic field itself. my theory makes more sense that the efield only manifests 'from' in the wire when reacting to the magnetic field.
but id like to here your opinion of more than one kind of electron.
mags
Quote from: Magluvin on 2026.07.19, 03:03:26
...but you seem to imply that there are more electrons free to go where they will throughout the wire, with no bond to the copper element atoms.
If there were more electrons than protons then the copper lump would not be electrically neutral and would have a net electric charge, i.e. it would make your hair stand up if you touched it.
Also, if the electrons inside a lump of metal have negative charge and can move freely then they should all repel themselves to the surface of that metal, shouldn't they ?
Unless the free electrons inside metals are not electrically charged...
Quote from: Magluvin on 2026.07.19, 03:03:26
...what has given you the idea of that opinion?
I have developed my own physical paradigm which is at odds with mainstream science and unacceptable to most people because it upends their cherished ideas about space and time as a 3D+1D container and everything follows from that. I don't want to convince anyone about its merits. Most are not ready to abandon the comfort of the ST aquarium anyway.
The significant thing about electrons in conductors is that they do become detached from the atoms and whiz about randomly at great speed (Fermi velocitty) but they do not move very far unless there is an externally applied electric field. The random movement gives rise to electrical signals measured as voltage across the length of the conductor known as thermal noise. It is these detached electrons that then obtain the small drift velocity when an E field is present.
Smudge
Quote from: Magluvin on 2026.07.18, 22:52:48
missed one part of what i wanted to say as i was interrupted here at work... here is the total part of my previous post...
absolutely! im with you there..
here is what i didnt get into...
so, do protons themselves leave the nucleus and flow through the wire in your understanding?
mags
Let's say our wire has no outside influence from any type of field or heat source so that the atoms in the wire are basically charge balanced. We would therefore have a net zero potential measured across the ends. Now, let's place the wire in a charge separation environment. This will result in a separation of the electrons from the atoms and they will collect in one end of the wire that becomes negative. This electron deficiency around the nucleus leaves the remaining protons and/or neutrons on the other end of the wire that produces the positive end. Yes, I do believe that these electron deficient protons will move in the wire.
Pm
Quote from: partzman on 2026.07.19, 14:18:08
Let's say our wire has no outside influence from any type of field or heat source so that the atoms in the wire are basically charge balanced. We would therefore have a net zero potential measured across the ends. Now, let's place the wire in a charge separation environment. This will result in a separation of the electrons from the atoms and they will collect in one end of the wire that becomes negative. This electron deficiency around the nucleus leaves the remaining protons and/or neutrons on the other end of the wire that produces the positive end. Yes, I do believe that these electron deficient protons will move in the wire.
Pm
PM,
IMO, in a metal, there are only the fixed metal atoms (neutral), free electrons (negative) and fixed metal ions (positive) all electrically in balance (neutral).
Is it not so that putting this metal in a "charge separation environment" (what is that?), will attract / repel the free electrons to say one end of the metal wire which then becomes negative, but as the atoms and ions are fixed they cannot being attracted / repelled (moved) to the other end causing it to be positive.
But as the negative electrons are moved to one end causing it to become negative, that means the other end will become positive only due to lack of negative electrons.
Itsu
Sorry, but I'm with Itsu. Actually, I learned this in school in the 80s.
I only bring up the possibility that the efield as we speak of is not a magnetic field component. my ideas on it is that the efield is a product of the wire that is induced, or to say influenced by a moving magnetic field.
I see the influence on an electron by a moving magnetic field similar to refraction of light through glass at an angle. if we can agree that the refraction 'influenced' by the glass can happen and is real, then it should not be hard to imagine a similar effect between an electron moving at right angles to a magnetic field. not saying it happens exactly the same way, but in a way that makes it a possibility.
Hey, maybe I'm wrong. Don't know for sure yet but my mind keeps working on it. Working on it as if there is something wrong with the conventional thinking on the subject. Like the cap-to-cap deal. Something just didn't feel right about losing 50% of the energy by way of resistance. The logic just wasn't there for me. but when Milehigh said that in an ideal situation that if we started with 10V in a 10uf cap and connected it to a 10uf cap that was 0V, we would end up with 7.07V in each cap in order to have no loss. Then it clicked... Ampere's law proves that there is no way to end up wit 7.07V in each cap starting with 10V in the source cap. We still would have 5V in each cap and a 50% loss. Simple as counting the electrons transferred from pos and neg plates.
So this has me going the same way.
mags
So do you think you would be able to permanently push (or pull) electrons from metal spheres this way ?:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=56763)
...or more practically, like this:
(https://www.overunityresearch.com/index.php?action=dlattach;topic=3926.0;attach=41031) Spheres are connected to coils with several microwave-oven diodes in series.
...and would a spark occur if you disconnected these spheres and brought them close together ?
https://www.youtube.com/watch?v=gPXv063O5B8
mags
Quote from: Magluvin on 2026.07.20, 14:23:14
https://www.youtube.com/watch?v=gPXv063O5B8
This was a well-made video illustrating moving capacitor plates apart.
However it was charged with an electrostatic machine - a very different source than the half-dipole antenna or balanced TC source, because one generates surface charges and the other conducted current.
It also answers a different question - what happens to surface charges on the dielectric in a capacitor as its plates move apart rather than the pumping of electrons by conducted current and moving the electrodes closer together.
Itsu, Mags, Verpies,
I have concluded that in my recent research and experimentation, I have been both correct and incorrect in my analysis of "charge separation" in these experiments. Correct in the fact that certain entities such as electrolytes and dielectrics have been charge separated but not so with a piece of wire! A piece of wire must have a source of induction to allow it to exhibit a voltage potential across it's ends. Charge separation can happen in a piece of wire but it will have a short life due to the conductivity of the wire. So, in my experiments when I measure the V/T across a piece of wire in the center hole of a toroid assembly, this is a result of a flux change that is inducing that wire. This realization came while studying the work of J. Edwards and T.K. Saha and in particular their paper titled "Establishment of Flux in Magnetic Cores" that is attached below.
What I've been seeing in my work is what they describe as their interpretation of what happens in transformer induction. I now realize the cause of the E-Field generation from the primary I, the H-Field and the B field in the hole of the core. Essentially, we have a magnetic transmission line that uses the inside of the core hole as a wave guide. Instead of me try to explain it in my words, please read the attached paper for the details.
Pm
just thonking...
i wonder if we suspended a wire through the core, positioned in the middle straight up and down, but loose, and loaded it, even short, would it move if we have input to a primary winding on the core just on one side, not all the way around? maybe...
mags
i think i have what i need to do that test.
mags
will put up some pics tonight. have toroid core that is in 2 pieces. the mating surfaces are very clean cut. got them from an induction fluorescent ring light.
have 1 side wound for input. making a stand to mount it and hang loose wire through it.
lets see if it moves....
mags
mags
just a survey... will the wire will move or not?
pulse the input winding in the pic. hang wire verrtically through the core and short the wire ends.
mags
Quote from: Magluvin on 2026.07.28, 18:59:37
just a survey... will the wire will move or not?
pulse the input winding in the pic. hang wire verrtically through the core and short the wire ends.
mags
With the wire ends shorted around the outside of the core then yes, a thin wire should move with a pulse to the primary due to the Primary's H-Field or the leakage flux whichever one wishes to use.
Pm
which way do you think the wire will move in reference to primary position?
mags
Quote from: partzman on 2026.07.28, 20:05:31
With the wire ends shorted around the outside of the core then yes, a thin wire should move with a pulse to the primary due to the Primary's H-Field or the leakage flux whichever one wishes to use.
Pm
The leakage flux (which is the field outside the core and of greatest magnitude in the core hole) does not have the same pattern as the Primary's H-Field (the field from the primary coil if no core were present). So you shouldn't use the latter, you should only use the leakage flux. The current induced into the shorted turn will influence the flux within the core hence also the leakage flux so it is a complicated procedure to establish the forces on it. But the experimental movement observation will obey Fleming's LH rule and give you an indication.
Smudge
Quote from: Smudge on 2026.07.29, 07:21:31
The leakage flux (which is the field outside the core and of greatest magnitude in the core hole) does not have the same pattern as the Primary's H-Field (the field from the primary coil if no core were present). So you shouldn't use the latter, you should only use the leakage flux. The current induced into the shorted turn will influence the flux within the core hence also the leakage flux so it is a complicated procedure to establish the forces on it. But the experimental movement observation will obey Fleming's LH rule and give you an indication.
Smudge
I maintain that the H-Field on the inside (in the hole) of the primary wire on the toroid is the same as (or creates) the so-called leakage flux. Therefore, if one sets up an experiment (which I have not done) that is similar or equal to mine in respect to polarities and phase, the deflection of the thin wire will be away from the primary.
Edit:
My last statement is ambiguous! It should have read, "Therefore, if one sets up an experiment (which I have not done) that is similar or equal to mine in respect to polarities and phase, the deflection of the thin wire will be away from the primary with a positive pulse applied to the primary and deflected towards the primary with a negative pulse."Edit2: Had it right the first time.
Pm
if the test wire through the toroid is shorted, it will move in an opposite direction depending on the input pulse polarity??
mags
Quote from: Magluvin on 2026.07.30, 13:15:34
if the test wire through the toroid is shorted, it will move in an opposite direction depending on the input pulse polarity??
mags
Actually after more consideration on your question above, the deflection of the wire will depend on the current in the primary and not the voltage. So, the current in the primary will not change polarity but will simply rise to a peak and then return to zero. Therefore, the deflection will be in the same direction for the entire cycle.
Thanks for the heads up!
Pm
Quote from: partzman on 2026.07.30, 14:34:19
Actually after more consideration on your question above, the deflection of the wire will depend on the current in the primary and not the voltage.
The deflection will also depend upon the direction of current induced into the wire.
QuoteSo, the current in the primary will not change polarity but will simply rise to a peak and then return to zero.
And the magnetic fields from that current will rise and fall. The risng field wil induce one polarity of voltage (hence current) in the wire while the falling field will induce the opposite polarity.
QuoteTherefore, the deflection will be in the same direction for the entire cycle.
I disagree for the reason just stated.
Smudge
Quote from: partzman on 2026.07.30, 14:34:19
So, the current in the primary will not change polarity but will simply rise to a peak and then return to zero.
That's right but only below some L/R ratio vs. risetime and always when the resistance of the circuit is zero.
For an electric current to flow along the wire, the wire must be a part of a closed circuit / loop.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=56816)
Current induced in a conductive circuit by magnetic flux rising and falling in time (X axis)
in the same manner to the left and right of the Y axis, for different circuit resistances.
Quote from: Smudge on 2026.07.31, 06:39:20
The deflection will also depend upon the direction of current induced into the wire.And the magnetic fields from that current will rise and fall. The risng field wil induce one polarity of voltage (hence current) in the wire while the falling field will induce the opposite polarity. I disagree for the reason just stated.
Smudge
This is what I thought originally but after more consideration, I came to the last conclusion that the force would be unidirectional. I plan to do this simple experiment but the resulting deflection may be hard to see when the operating frequency is ~100kHz.
Pm
Quote from: Verpies on 2026.07.31, 15:50:45
That's right but only below some L/R ratio vs. risetime and always when the resistance of the circuit is zero.
For an electric current to flow along the wire, the wire must be a part of a closed circuit / loop.
(https://www.overunityresearch.com/index.php?action=dlattach;topic=4525.0;attach=56816) Current induced in a conductive circuit by magnetic flux rising and falling in time (X axis)
in the same manner to the left and right of the Y axis, for different circuit resistances.
So, in my experimental device, the resistance of the primary is low so this will allow a unidirectional induction!?! I will run some tests and we shall see!
Pm
OK, I ran a test with the two toroids stacked with a 12T primary which is pulsed at 100kHz when using a 48v DC supply. The fine wire is 36ga (.005" dia) single urethane coated and is ~4"OAL. The shorted outside part of the turn is 22ga for support and to complete the circuit.
I first tried single pulses using manual triggering with a Rigol DG4162 generator. Using a white piece of paper for backing in the toroid hole, I could not detect any movement with these old eyes!
Next I tried a series of 10 pulses again using manual triggering, and no detectable movement was seen!
Lastly, I tried a short burst of continuous pulses whereby the wire turned red from the current (~11.5A rms). There was noticeable movement toward the primary which I consider to be the result of the wire expansion along with the wire's curvature.
Edit: Mags, have you seen any deflection?
Pm
havnt had time yet. been working 60-70hrs wk. that there needs to end..
was going to do this weekend. i get sun off
mags
how many primary turns did you do?
mags
Quote from: partzman on 2026.07.31, 19:27:31
So, in my experimental device, the resistance of the primary is low so this will allow a unidirectional induction!?!
Primary ?! No, we are talking about "induced" current in a wire so this implies "secondary". There is a fundamental difference between the "inducing" and the "induced" circuit.
The resistance of the primary (the winding that generates the main magnetic flux in the core and leakage flux in its hole) is not relevant to this discussion.
oops.. 12turn. i see it now
so it pulled toward the primary....
firstly, to understand more is, lets say that maybe the wire 'was' pulled in to the primary by magnetic interaction.
if we hung 2 wires losely and horizontally instead. quarter in between them. if we power both wires, same polarity, if there is movement of the wires in reference to each other, the movement should be opposite of just powering 1 of those wires and loop shorting the other. the loop shorting example should be the same as your example. by the way, thanks for doing that.
simple stuff. like i said, will do some of these simple things this weekend. trying to figure out why i cant log in to my magluvin youtube channel.
mags
Quote from: Verpies on 2026.07.31, 21:44:59
Primary ?! No, we are talking about "induced" current in a wire so this implies "secondary". There is a fundamental difference between the "inducing" and the "induced" circuit.
The resistance of the primary (the winding that generates the main magnetic flux in the core and leakage flux in its hole) is not relevant to this discussion.
PM stated- "So, the current in the primary will not change polarity but will simply rise to a peak and then return to zero."Your reply-
"That's right but only below some L/R ratio vs. risetime and always when the resistance of the circuit is zero.
For an electric current to flow along the wire, the wire must be a part of a closed circuit / loop."
Context is everything and here you respond to my comment about the primary's current waveform. Now you say you were referring to a secondary??!??
Also, the primary induces the single wire via the H-Field.
Pm
Quote from: Magluvin on 2026.08.01, 03:40:27
oops.. 12turn. i see it now
so it pulled toward the primary....
firstly, to understand more is, lets say that maybe the wire 'was' pulled in to the primary by magnetic interaction.
if we hung 2 wires losely and horizontally instead. quarter in between them. if we power both wires, same polarity, if there is movement of the wires in reference to each other, the movement should be opposite of just powering 1 of those wires and loop shorting the other. the loop shorting example should be the same as your example. by the way, thanks for doing that.
simple stuff. like i said, will do some of these simple things this weekend. trying to figure out why i cant log in to my magluvin youtube channel.
mags
I am not saying the loose wire was pulled to the primary by induction force but rather heat expansion gave it the ability of moving towards the primary.
Pm
how can you be sure? other than that, do you conclude that the wire will not move in he core when the primary is active, because it was heat flexing the wire?
mags
Quote from: Magluvin on 2026.08.01, 18:41:37
how can you be sure? other than that, do you conclude that the wire will not move in he core when the primary is active, because it was heat flexing the wire?
mags
I find no visible movement in the wire that should be caused by forces from the H-Field in the many tests I've run. Either the wire is too stiff, or the forces are too weak, or occur at too high a speed to be visible, or.....? I only see movement when the wire is heated and the direction is easily influenced by the slight bend in the wire. The core can be rotated so the primary is in a different position relative to the wire and the deflection of the wire still remains the same direction.
My conclusion is that I see no deflection in the wire resulting from H-Field forces.
Pm
Quote from: partzman on 2026.07.30, 14:34:19
So, the current in the primary will not change polarity but will simply rise to a peak and then return to zero. Therefore, the deflection will be in the same direction for the entire cycle.
Quote from: partzman on 2026.08.01, 14:51:20
Context is everything and here you respond to my comment about the primary's current waveform. Now you say you were referring to a secondary??!??
I was certain that your statement referred to the secondary because it mentioned a "deflection". I answered about the current induced in the secondary because I thought that is where the "deflection" took place.
My answer is not applicable to the primary current.
I will not provide a reply about the current induced in the primary because the current in the primary can be anything that the power supply (i.e. bridge) pushes into it.
Quote from: Verpies on 2026.08.02, 05:22:29
I was certain that your statement referred to the secondary because it mentioned a "deflection". I answered about the current induced in the secondary because I thought that is where the "deflection" took place.
My answer is not applicable to the primary current.
I will not provide a reply about the current induced in the primary because the current in the primary can be anything that the power supply (i.e. bridge) pushes into it.
OK, I understand your explanation above.
Thank you.
Pm
thats fair. if it didnt move, but it seems like it should have, what do you think is happening here vs 2 hanging wires close to one another as described earlier. have a vid.
https://m.youtube.com/watch?v=1JZLKvWO0ks
im at home. its raining cats and electrons. see if i can get to my shop to do some things later.
mags