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Author Topic: Transformer Induction  (Read 26502 times)
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img  Centraflow
Re: Transformer Induction « Reply #100 on: 2025-10-06, 08:13:43 »

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Quote from: JimBoot on 2025-10-06, 03:47:17
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.



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img  3D Magnetics
Re: Transformer Induction « Reply #101 on: 2025-10-06, 08:43:07 »
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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 !
   
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Re: Transformer Induction « Reply #102 on: 2025-10-06, 16:12:08 »
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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.
« Last Edit: 2025-10-10, 14:47:06 by partzman »
   
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Re: Transformer Induction « Reply #103 on: 2025-10-08, 21:51:58 »
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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   
   
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Re: Transformer Induction « Reply #104 on: 2025-10-09, 11:58:12 »

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Quote from: partzman on 2025-10-08, 21:51:58
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. 

Quote
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.

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.

Quote
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.

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.

Quote
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?

I hope I have done that.

Smudge
   
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Re: Transformer Induction « Reply #105 on: 2025-10-09, 14:51:06 »
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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 
   
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Re: Transformer Induction « Reply #106 on: 2025-10-09, 15:41:32 »

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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.

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Re: Transformer Induction « Reply #107 on: 2025-10-09, 21:26:57 »
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Quote from: Smudge on 2025-10-09, 15:41:32
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
   
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Re: Transformer Induction « Reply #108 on: 2025-10-10, 00:26:22 »
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Ok Iwill have a go.

if i understood Smudge correctly, this would men that a ceramic cap would also seem to charge.
   
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Re: Transformer Induction « Reply #109 on: 2025-10-10, 00:51:37 »
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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
   
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Re: Transformer Induction « Reply #110 on: 2025-10-10, 08:58:17 »

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Quote from: 3D Magnetics on 2025-10-10, 00:26:22
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.
   
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Re: Transformer Induction « Reply #111 on: 2025-10-10, 15:00:49 »
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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
   
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Re: Transformer Induction « Reply #112 on: 2025-10-10, 15:39:19 »
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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!
   
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Re: Transformer Induction « Reply #113 on: 2025-10-12, 21:48:50 »
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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?


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Re: Transformer Induction « Reply #114 on: 2025-10-13, 16:37:11 »
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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

 
   
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Re: Transformer Induction « Reply #115 on: 2025-10-14, 09:42:32 »

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Quote from: partzman on 2025-10-13, 16:37:11
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.   

Quote
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.

Not charge separation as in charging of C1 but the same induced charge separation you get along a length of wire.

Quote
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!

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.

Quote
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.

That 8V (+-4V) is induced voltage, not induced charge.  What you see there is exactly as expected.

Quote
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,

It is the charge received by C1 during the first phase.

Quote
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 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
   
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Re: Transformer Induction « Reply #116 on: 2025-10-14, 11:03:05 »

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Quote from: 3D Magnetics on 2025-10-10, 15:39:19
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
   
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Re: Transformer Induction « Reply #117 on: 2025-10-14, 11:14:50 »
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Quote from: partzman on 2025-10-13, 16:37:11
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
   
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Re: Transformer Induction « Reply #118 on: 2025-10-14, 15:32:20 »
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Quote from: Smudge on 2025-10-14, 09:42:32
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.
   
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Re: Transformer Induction « Reply #119 on: 2025-10-14, 15:35:17 »
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Quote from: 3D Magnetics on 2025-10-10, 00:26:22
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
   
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Re: Transformer Induction « Reply #120 on: 2025-10-14, 15:44:51 »
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Quote from: web000x on 2025-10-10, 00:51:37
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
   
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Re: Transformer Induction « Reply #121 on: 2025-10-14, 15:50:15 »
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Quote from: Frederik2k1 on 2025-10-14, 11:14:50

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
   
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Re: Transformer Induction « Reply #122 on: 2025-10-14, 23:51:09 »

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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."
   
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Re: Transformer Induction « Reply #123 on: 2025-10-15, 10:44:36 »

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Quote from: partzman on 2025-10-14, 15:32:20
This is not correct as is experimentally shown below.
With respect I say the experiment shown below exactly supports my view.
Quote
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.
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.
Quote
  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
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)
Quote
and 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
   
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Re: Transformer Induction « Reply #124 on: 2025-10-15, 16:23:42 »
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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
   
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