Transformer Induction

Author Topic: Transformer Induction (Read 26539 times)

Topic Options

partzman

Transformer Induction « on: 2025-09-05, 21:58:57 »

Group: Experimentalist
Hero Member

Posts: 2232

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

Dielectric Charging Via the E_Field.pdf (634.97 kB - downloaded 71 times.)

Transformer Induction

Transformer Induction.png (24.06 kB, 1188x639 - viewed 520 times.)

Power flow in transformers via the poynting vector.pdf (295.41 kB - downloaded 87 times.)

Smudge

Re: Transformer Induction « Reply #1 on: 2025-09-06, 09:24:55 »

Group: Professor
Hero Member

Posts: 2332

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

Transformer Induction

Dielectric1.png (130.33 kB, 1187x639 - viewed 473 times.)

Transformer Induction

Dielectric2.png (122.33 kB, 1187x639 - viewed 461 times.)

Smudge

Re: Transformer Induction « Reply #2 on: 2025-09-06, 09:48:29 »

Group: Professor
Hero Member

Posts: 2332

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

Verpies

Re: Transformer Induction « Reply #3 on: 2025-09-06, 13:40:16 »

Group: Administrator
Hero Member

Posts: 4513

Quote from: Smudge on 2025-09-06, 09:48:29

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..

Centraflow

Re: Transformer Induction « Reply #4 on: 2025-09-06, 13:46:37 »

Group: Experimentalist
Hero Member

Posts: 3914

Quote from: verpies on 2025-09-06, 13:40:16

Neither will spirally wound capacitors...

Bifilar coils will.

---------------------------

"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.

Verpies	Re: Transformer Induction « Reply #5 on: 2025-09-06, 14:21:24 »
Group: Administrator Hero Member Posts: 4513	Quote from: Centraflow on 2025-09-06, 13:46:37 Bifilar coils will. ...but we are not discussing coils in this subtopic. We are discussing capacitors.

Centraflow

Re: Transformer Induction « Reply #6 on: 2025-09-07, 10:24:02 »

Group: Experimentalist
Hero Member

Posts: 3914

Quote from: verpies on 2025-09-06, 14:21:24

...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 »

« Last Edit: 2025-09-07, 11:25:33 by Centraflow »

---------------------------

"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.

Hakasays

Re: Transformer Induction « Reply #7 on: 2025-09-07, 14:33:32 »

Hero Member

Posts: 603

Quote from: partzman on 2025-09-05, 21:58:57

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.

---------------------------

"An overly-skeptical scientist might hastily conclude by scooping-up and analyzing a thousand buckets of seawater that the ocean has no fish in it."

partzman

Re: Transformer Induction « Reply #8 on: 2025-09-07, 15:15:14 »

Group: Experimentalist
Hero Member

Posts: 2232

Quote from: Hakasays on 2025-09-07, 14:33:32

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

Smudge	Re: Transformer Induction « Reply #9 on: 2025-09-07, 16:31:56 »
Group: Professor Hero Member Posts: 2332	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 Energy Gain using Dielectric Displacement.pdf (135.9 kB - downloaded 90 times.)

Hakasays

Re: Transformer Induction « Reply #10 on: 2025-09-08, 03:35:53 »

Hero Member

Posts: 603

Quote from: Smudge on 2025-09-07, 16:31:56

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, 15:15:14

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?

---------------------------

"An overly-skeptical scientist might hastily conclude by scooping-up and analyzing a thousand buckets of seawater that the ocean has no fish in it."

Smudge

Re: Transformer Induction « Reply #11 on: 2025-09-08, 10:56:43 »

Group: Professor
Hero Member

Posts: 2332

Quote from: Hakasays on 2025-09-08, 03:35:53

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.

Quote

I 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.

Quote

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?

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.

Quote

Is 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

Hakasays	Re: Transformer Induction « Reply #12 on: 2025-09-08, 15:46:47 »
Hero Member Posts: 603	Thanks for all the clarifications, Smudge --------------------------- "An overly-skeptical scientist might hastily conclude by scooping-up and analyzing a thousand buckets of seawater that the ocean has no fish in it."

Allcanadian

Re: Transformer Induction « Reply #13 on: 2025-09-08, 17:16:20 »

Hero Member

Posts: 3064

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

---------------------------

Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman

Verpies

Re: Transformer Induction « Reply #14 on: 2025-09-08, 20:52:15 »

Group: Administrator
Hero Member

Posts: 4513

Quote from: Smudge on 2025-09-07, 16:31:56

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.

Also, there is a problem with the following statement:

Quote from: Energy_Gain_using_Dielectric_Displacement_by_Smudge on 2025-09-07, 16:31:56

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.

partzman

Re: Transformer Induction « Reply #15 on: 2025-09-08, 20:57:15 »

Group: Experimentalist
Hero Member

Posts: 2232

Quote from: Smudge on 2025-09-07, 16:31:56

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

Smudge

Re: Transformer Induction « Reply #16 on: 2025-09-09, 07:54:16 »

Group: Professor
Hero Member

Posts: 2332

Quote from: verpies on 2025-09-08, 20:52:15

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.

Quote

Also, 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.

Verpies

Re: Transformer Induction « Reply #17 on: 2025-09-09, 13:38:36 »

Group: Administrator
Hero Member

Posts: 4513

Quote from: Smudge on 2025-09-09, 07:54:16

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: 20cm² and ~10µm of fine barium titanate powder.

Quote from: Smudge on 2025-09-09, 07:54:16

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

Smudge

Re: Transformer Induction « Reply #18 on: 2025-09-09, 14:38:04 »

Group: Professor
Hero Member

Posts: 2332

Quote from: partzman on 2025-09-08, 20:57:15

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.

Quote

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.

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.

Quote

This 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.

Quote

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.

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.

Quote

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.

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.

Quote

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.

Agreed, in which case it is not a secondary loop.

Quote

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.

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.

Quote

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.

No, the primary current is at zero so the return to the supply has already happened.

Smudge

Pm

[/quote]

Smudge

Re: Transformer Induction « Reply #19 on: 2025-09-09, 14:56:32 »

Group: Professor
Hero Member

Posts: 2332

Quote from: verpies on 2025-09-09, 13:38:36

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?

Quote

OK 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.

F6FLT

Re: Transformer Induction « Reply #20 on: 2025-09-09, 18:00:11 »

Group: Professor
Hero Member

Posts: 2435

Quote from: verpies on 2025-09-06, 13:40:16

...
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.

---------------------------

"Open your mind, but not like a trash bin"

Verpies

Re: Transformer Induction « Reply #21 on: 2025-09-09, 18:09:42 »

Group: Administrator
Hero Member

Posts: 4513

Quote from: Smudge on 2025-09-09, 14:56:32

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, 14:56:32

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.

F6FLT

Re: Transformer Induction « Reply #22 on: 2025-09-09, 18:23:55 »

Group: Professor
Hero Member

Posts: 2435

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)

---------------------------

"Open your mind, but not like a trash bin"

Verpies	Re: Transformer Induction « Reply #23 on: 2025-09-09, 18:34:09 »
Group: Administrator Hero Member Posts: 4513	Quote from: F6FLT on 2025-09-09, 18:23:55 KNO₃ Are you sure? The stuff is hygroscopic and conductive when even slightly damp.

Smudge

Re: Transformer Induction « Reply #24 on: 2025-09-10, 15:36:42 »

Group: Professor
Hero Member

Posts: 2332

Quote from: verpies on 2025-09-09, 18:09:42

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.