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Author Topic: Transformer Induction  (Read 41664 times)
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Re: Transformer Induction « Reply #175 on: 2025-12-19, 15:21:46 »
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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
« Last Edit: 2025-12-19, 16:22:23 by partzman »
   
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Re: Transformer Induction « Reply #176 on: 2025-12-20, 07:16:24 »

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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
   
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Re: Transformer Induction « Reply #177 on: 2026-01-22, 10:50:07 »
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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.
« Last Edit: 2026-01-22, 11:56:52 by F6FLT »


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Re: Transformer Induction « Reply #178 on: 2026-01-23, 05:01:31 »

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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.
   
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Re: Transformer Induction « Reply #179 on: 2026-01-23, 10:08:49 »

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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
Quote
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
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).
Quote
and 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. 
Quote
  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.
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. 
Quote
Therefore 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
   
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Re: Transformer Induction « Reply #180 on: 2026-01-24, 03:58:42 »

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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
« Last Edit: 2026-01-24, 05:15:48 by Magluvin »
   
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Re: Transformer Induction « Reply #181 on: 2026-01-24, 05:05:02 »

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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
   
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Re: Transformer Induction « Reply #182 on: 2026-01-24, 07:54:40 »

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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
   
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Re: Transformer Induction « Reply #183 on: 2026-01-24, 08:12:54 »

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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
   
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Re: Transformer Induction « Reply #184 on: 2026-01-24, 15:25:58 »
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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 
« Last Edit: 2026-01-24, 16:26:43 by partzman »
   
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Re: Transformer Induction « Reply #185 on: 2026-01-24, 17:41:15 »

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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).
Quote
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.
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 (dA/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.
Quote
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.
Which it does through this build-up phase, producing the E field that drives the electrons in the secondary wire to create the voltage.
Quote
If 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
« Last Edit: 2026-01-24, 18:43:14 by Smudge »
   
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Re: Transformer Induction « Reply #186 on: 2026-01-25, 06:01:07 »

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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.
   
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Re: Transformer Induction « Reply #187 on: 2026-01-25, 08:33:15 »

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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. 
   
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Re: Transformer Induction « Reply #188 on: 2026-01-25, 09:00:25 »

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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.
   
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Re: Transformer Induction « Reply #189 on: 2026-02-19, 23:03:18 »

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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
   
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Re: Transformer Induction « Reply #190 on: 2026-02-20, 01:10:18 »

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Are you powering the primaries in a phased sequence ?
   
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Re: Transformer Induction « Reply #191 on: 2026-02-20, 02:21:38 »

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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
   
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Re: Transformer Induction « Reply #192 on: 2026-02-26, 16:13:09 »
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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.

« Last Edit: 2026-02-26, 21:54:20 by partzman »
   
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Re: Transformer Induction « Reply #193 on: 2026-02-26, 19:21:31 »

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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
   
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Re: Transformer Induction « Reply #194 on: 2026-02-26, 20:53:24 »
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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
« Last Edit: 2026-02-26, 21:54:49 by partzman »
   
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Re: Transformer Induction « Reply #195 on: 2026-02-27, 10:41:41 »

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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).
   
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Re: Transformer Induction « Reply #196 on: 2026-02-28, 15:05:55 »
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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
   
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Re: Transformer Induction « Reply #197 on: 2026-02-28, 16:05:47 »

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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.

  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):

  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.
   
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Re: Transformer Induction « Reply #198 on: 2026-03-01, 15:15:06 »
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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
« Last Edit: 2026-03-01, 16:16:06 by partzman »
   
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Re: Transformer Induction « Reply #199 on: 2026-03-01, 16:16:57 »

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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:


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:

  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:


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.
   
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