What's different in a transformer core when coupling high & low power?

In summary: For a flux linked transformer, the voltage per turn is unchanging. The currents are determined by the load on the secondary.
  • #36
This is quite a weird thread. After positing in my first post that the flux is the same, irrespective of the current in the coils, we've now come full circle with everyone agreeing with each other but not my original post? (Notwithstanding me creating a misleading aspect by referring to ohmic loads, sorry about that)

It must be me.

So, on to the mystery that I was alluding to from the outset.

Is no-one remotely curious that somehow the primary and secondary coils 'communicate' their currents to each other by influencing and affecting the way each other behaves, yet the behaviour of the thing in between them shows no differences?

I mean, if I send some message to someone, then that message has some tangible effect on the transmission medium in between me and them.

But in the case of transmitting higher reactive currents from one side of a transformer to another, it seems one would never know from analysing the physics in between them that this was happening.
 
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  • #37
cmb said:
This is quite a weird thread. After positing in my first post that the flux is the same, irrespective of the current in the coils, we've now come full circle with everyone agreeing with each other but not my original post? (Notwithstanding me creating a misleading aspect by referring to ohmic loads, sorry about that)

It must be me.

So, on to the mystery that I was alluding to from the outset.

Is no-one remotely curious that somehow the primary and secondary coils 'communicate' their currents to each other by influencing and affecting the way each other behaves, yet the behaviour of the thing in between them shows no differences?

I mean, if I send some message to someone, then that message has some tangible effect on the transmission medium in between me and them.

But in the case of transmitting higher reactive currents from one side of a transformer to another, it seems one would never know from analysing the physics in between them that this was happening.
The "thing" in between them can just as easilly be air as magnetic material. This is no more mysterious than radio waves. Of course it is no less mysterious than radio waves!
I will say it took me a while to show myself analytically that the "deep fields" were rigorously unaffected. But in the final analysis it is one charge in primary wire talking to one charge in secondary wire.
I liked the question very much and it made me think through a few loose ends. Thanks for the question
 
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  • #38
It may help to consider the relation magnetomotive force mmf = flux times reluctance. In your toroid, mmf = ## N_1 i_1 + N_2 i_2 ## with "1" referring to the primary, N = no. of turns and i the current in each winding. Also, mmf = reluctance times flux. Flux is constant thruout the toroid.

With no secondary load ## i_1 = emf/\omega L_1 ## and ## i_2 = 0 ##. Flux times reluctance = ## N_1 i_1 ##.

When a load is applied to the secondary, each winding contributes to mmf the same way: ## \Sigma Ni = mmf ##. But now the current in both windings increases but with the secondary current 180 out of phase with the primary (ignoring the magnetizing current). So we really have 2 sources of flux which cancel each other except for the remanent magnetizing current. Flux therefore does not change.

This is all with zero resistance in both windings and no leakage inductance. If that is not the case then phase shifts between primary and secondary voltage (but not emf) occur but the flux is still unaffected, its magnitude and phase solely determined by the emf (not terminal voltage) applied to the primary coil, i.e. by the magnetizing current.

The magnetic field serves to transfer energy from the generator to the primary to the secondary without changing its own energy level.

Think of two monochromatic laser beams 180 out of phase. Each carries energy but the sum is always zero everywhere in the net beam.
 
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  • #39
cmb said:
But in the case of transmitting higher reactive currents from one side of a transformer to another, it seems one would never know from analysing the physics in between them that this was happening.
I just disagree that ignoring the currents is “analyzing the physics”. I fail to see the point of wondering why there isn’t a difference when there is one that you are aware of but deliberately ignoring.

Also, greater current does increase flux in general.
 
  • #40
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  • #41
Dale said:
Also, greater current does increase flux in general.
But I hadn't really thought through the fact that the induced current in the secondary was coincident in phase with an induced additional countervailing current in the primary even though they produce flux that cancels. It explains why that additional current makes both primary and secondary heat for real rersistive windings, but with no increased flux. Perhaps I should have known all this but apparently I had forgotten it (if ever I knew it).
 
  • #42
First, I'm too lazy and busy to do the real physical treatment of this, you know, with equations and such. But, I think the way to approach this is to keep in mind that the transformer is a coupled inductor. In the ideal case everything about the core can be modeled as a shunt magnetizing inductance in parallel with the windings.

There will be some magnetizing current which flows that is only due to the excitation from the applied voltage (usually on the primary). The magnetizing current has nothing to do with the secondary current or power delivered (again, ideal model), but there is some reactive "power" associated with it. The value of the core is to redirect the flux in the windings to increase coupling and increases the shunt inductance. It doesn't really change that flux, nor does that flux change the core (again, ideal case).

Then your question about power flow becomes fairly trivial, the ideal transformer part of the model just changes the ac impedance seen by the source. The ideal transformer doesn't deliver, or process, power any more than a superconducting wire does. But it does have a shunt inductive component added, even in an ideal model.

So, your question really revolves around the non-idealities of the transformer, which is necessarily messy. Nevertheless, you can still mostly treat the core as a separate thing primarily related to the voltage excitation of the inductor.
 
  • #43
So, I recalculated the circuit using phasors for a general complex impedance load, ##Z##, and a general mutual inductance ##M=k\sqrt{L_p L_s}##, ##0 \le k \le 1##.

$$i_p=\frac{j V \left(j Z+\omega L_s\right)}{\omega L_p \left(-j Z+k^2 \omega L_s-\omega
L_s\right)}$$ $$ i_s=-\frac{j k V \sqrt{L_p L_s}}{L_p \left(-j Z+k^2 \omega L_s-\omega L_s\right)} $$

So then the flux on the primary side is $$L_p i_p + M i_s = -\frac{j V}{\omega }$$ and the flux on the secondary side is $$L_s i_s + M i_p = -\frac{k V Z L_s}{\omega \sqrt{L_p L_s}
\left(\left(k^2-1\right) \omega L_s-j Z\right)}$$

So interestingly the flux on the primary does not depend on the load ##Z## nor on the transformer's geometry factor ##k##. But the flux on the secondary side does depend on both.
 
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  • #44
Dale said:
So interestingly the flux on the primary does not depend on the load Z nor on the transformer's geometry factor k.
This is a very satisfactory result. In the limit of perfect coupling (k=1) the two fluxes are equal and independent of Z and there is no change in actual internal fields
As soon as there is imperfect coupling and the cancellation of the "out of phase" field no longer obtains. The OP must take care to realize that H field in space will now respond everywhere to changes in the the loading even though the "primary flux" is unchanged. (We sort of cook the books here by our definitions of inductance).
 
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  • #45
For ## k= 1 ## the ## H ## and ## B ## fields are the same for both the primary and secondary windings, but using ## L_p=C N_p^2 ##, and ## L_s=C N_s^2 ##, with ## M=C N_p N_s ##, the result will be that the flux "links", (what @Dale is calling flux) obey ## \Phi_s=\frac{N_s}{N_p} \Phi_p ## , where ## \Phi_p=N_p B A ## and ## \Phi_s=N_s B A ## represents the flux links. This essentially says ## V_s=\frac{N_s}{N_p} V ##.

Note: The magnetomotive force ## MMF=\oint H \cdot dl=NI ##, so that the ##H ## or ## B ## that is generated is proportional to the number of turns ## N ##, but the flux links for the self inductance will then be proportional to ## N^2 ##.

The flux links for mutual inductance will be proportional to ## N_p N_s ##, because the current and the windings in one creates the ## H ## and ## B ## in the other, and the number of links is proportional to the number of windings of the other.

In addition, in this ideal case you should also get ## |N_s i_s| \approx |N_p i_p| ##, to a very good approximation if ## Z ## is small,(a large load), and that is indeed the case. I agree with @Dale 's solution.
 
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  • #46
hutchphd said:
The "thing" in between them can just as easilly be air as magnetic material. This is no more mysterious than radio waves. Of course it is no less mysterious than radio waves!
I will say it took me a while to show myself analytically that the "deep fields" were rigorously unaffected. But in the final analysis it is one charge in primary wire talking to one charge in secondary wire.
I liked the question very much and it made me think through a few loose ends. Thanks for the question
You're sort of simultaneously concluding but also re-opening my original dilemma.

I accept, of course, that photons mediate the electromagnetic field. As counter intuitive as it seems, in the absence of apparent photons, magnetics must also use photons as their mediators, including permanent magnets. Where one would look to see such photons, between two moving permanent magnets, I struggle to imagine.

This is probably the same question, actually. I don't really know the answer to that and if I did I'd probably know the answer to this.

The thing is that I tended to have assumed that the flux in the core of an inductor/transformer is the manifestation of the mediating photons between the two coils, just like measuring a flux of photons between two radio transmitting/receiving antenna. It seems there must be more photons being mediated between the coils than 'just' those associated with the magnetic flux?

[I think it is reasonable to assume that in a high Q (>400), low conductivity (>10^8 ohm.cm), ferrite toroid that k is as close to 1 as can be measured and there are no electrical currents present as can be measured within it.]
 
  • #47
I think I see what @cmb is puzzled by, and that is basically the "magic" of the transformer, that it can adjust its primary current when the load increases, and the magnetic flux in the core basically stays the same, with the primary mmf ## N_p I_p ## balancing the secondary mmf ##N_s I_s ##.

With the "links" of post 40, the upper limit to the secondary and primary currents seems to be governed by how much current the copper wire can handle, rather than by any changes that may be happening in the iron core due to the increased loading.

Perhaps there is something that is present in the calculations, in addition to this, but with the balance ## |N_p I_p|=|N_s I_s |##, it seems that higher powers are readily accommodated by the transformer core, at least when ## k \approx 1 ##. I think this is what the OP has been asking several times throughout this thread. See also https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/ post 17, where Jim Hardy talks about this self-balancing. @cmb Perhaps you would find this entire thread good reading, where the operation of the transformer is discussed in detail.

Perhaps it would be useful to take the limit in the equations of post 43 for ## Z \rightarrow 0 ##, when ## k \neq 1##, but I haven't figured out the implications yet.
 
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  • #48
cmb said:
You're sort of simultaneously concluding but also re-opening my original dilemma.
I understand your dilemma but the answer is simple.
The magnetic field is a calculational fiction. We have shown that for ana idealized transformer with perfect coupling (and no relativistic effects incidentally) the internal field does not change because it is instantly compensated. But the observable quantities (the currents) do change with more power and just happen to cancel the field changes. Charges interact with other charges. The fact that our fiction behaves a certain way is not fundamental: we set it up this way.
This is, to quote @etotheipi, an argument about the picture in your head and whether the baby panda occasionally becomes invisible...
 
  • #49
cmb said:
The thing is that I tended to have assumed that the flux in the core of an inductor/transformer is the manifestation of the mediating photons between the two coils, just like measuring a flux of photons between two radio transmitting/receiving antenna. It seems there must be more photons being mediated between the coils than 'just' those associated with the magnetic flux?

So there is a very small energy associated with the turning motion of the microscopic magnets in the core. And at each moment that energy must be increased by a large amount by the primary coil and decreased by equally large amount by the secondary coil.

So it must be so that the two coils exert opposing torques on the microscopic magnets. The coil that is winning the wrestling match is losing energy, the loser is gaining energy.

Now I try to get into photons, hmm, somehow virtual photons are able to transfer energy, like when I spin a ball, the ball and my hand exchange virtual photons, and the ball gains energy. So I guess now we could ask what is different in the empty space between my hand and a ball spun by my hand, when the ball is a one with small rotational inertia , compared to a case when the ball has a large rotational inertia.

In the latter case more angular momentum is exchanged, so I guess in the latter case the group of virtual photons has more angular momentum. Right?

Hmm actually that last paragraph is messed up in many ways. Like for example it's the ball with the smaller rotational inertia that gains more energy per some amount of angular momentum.
 
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  • #50
cmb said:
If the only thing that changes is the current in the windings, then nothing 'different' happens in the core. Right?

Yes, the magnetic field inside the transformer core will not change, but when the power flow changes, in addition to the inevitably changing current, there is indeed another thing that will change. I think you should be able to find the answer from the Poynting vector.

https://en.wikipedia.org/wiki/Poynting_vector
 
  • #51
cmb said:
So long as I keep good volt/turns that works well for the core's loss characteristics, I can put in and draw out low impedance power and make the core as small as I like, it seems!? I'm now picturing putting 10kW through a diddy 35mm toroid! The only question on size seems to be if one can wind wire of sufficient ampacity around the core

This is indeed a very interesting question.

First of all, of course, under ideal conditions, the magnetic flux inside the magnetic core will not change when the output current increases. Since the number of turns of primary winding and the number of turns of secondary winding of the transformer have not changed, if the magnetic flux changes, it means that the input and output voltages have changed, but this will not happen.

Since the magnetic flux inside the magnetic core will not change due to the increase of current, as long as the diameter of the winding wire is continuously increased to keep the ohmic loss constant, then we can continue to increase the power of the transformer while maintaining the size of the magnetic core. This is really great.

In theory, this idea seems to be feasible, but for engineers, as long as they think about it carefully, they will know that it is actually not feasible.

This is because, in addition to the limited winding window area of all types of magnetic cores that can be purchased on the market, when the wire diameter or the number of winding layers continue to increase, the leakage inductances will inevitably continue to increase, so the efficiency of the transformer will continue to decrease until it can no longer be used. At this time, the transformer will be equivalent to a transformer without a magnetic core.

Although in principle we can make a power transformer without a magnetic core, the size of the transformer will be large, the leakage of magnetic flux may cause a lot of interference, and it does not comply with economic and environmental principles. :smile:
 
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  • #52
alan123hk said:
Since the magnetic flux inside the magnetic core will not change due to the increase of current, as long as the diameter of the winding wire is continuously increased to keep the ohmic loss constant, then we can continue to increase the power of the transformer while maintaining the size of the magnetic core. This is really great.
I believe @Dale's excellent analysis shows this only to be absolutely true for perfectly coupled (k=1) transformers. The salient point in your analysis is still approximately correct: in fact it gets worse than you say because as the coupling degrades and the secondary load increases the core flux directly
 
  • #53
hutchphd said:
I believe @Dale's excellent analysis shows this only to be absolutely true for perfectly coupled (k=1) transformers. The salient point in your analysis is still approximately correct: in fact it gets worse than you say because as the coupling degrades and the secondary load increases the core flux directly

Of course, the coupling factor must be equal to 1. If this condition is met, the core flux will not be affected by the load current, so I said that under ideal conditions (It may also include the condition that the winding resistance is zero, but there seems to be a contradiction. If the winding resistance is zero, all problems have been resolved). :smile:

When leakage inductance appears, the coupling factor will decrease with the increase of leakage inductance, that is, the coupling factor is no longer equal to 1, so as I mentioned earlier, the working efficiency of the transformer will decrease.

At this time, there may be different situations, for example, the increase of the wire diameter should have reduced the voltage drop of the internal winding wire of the transformer, but the appearance of leakage inductances may reduce the actual output voltage.

Another situation is the user may choose to increase the input voltage to force the output voltage to return to the original value, so the overall situation may become a bit complicated.

As for whether the core flux increases or decreases, we can first set all the parameters of the model, and then determine the final analysis result.
 
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  • #54
Dale said:
So interestingly the flux on the primary does not depend on the load Z nor on the transformer's geometry factor k. But the flux on the secondary side does depend on both.

According to the last two equations, the flux on the primary side and on the secondary side are as follows,

$$ \theta_p = - \frac {jV} \omega ~~~~~~~~~ \theta_s = \frac {jV} \omega \left( \frac {N_s} {N_p} \right) $$
But when ## ~k=1 ~##, shouldn't ## ~\theta_p = \theta_s ~ ## ?

In addition, it seems that ## ~ \theta_p ~ ## should be more like to equal to ##~ -\frac {jV} {N_p ~\omega} ~##, so that ##~ V = j~Np ~\omega ~ \theta_p ~##.

In this way, the voltage on the primary side can be proportional to the product of the three parameters, the number of turns of the primary winding, the operating frequency and the magnetic flux on the primary side.
 
  • #55
alan123hk said:
But when ## ~k=1 ~##, shouldn't ## ~\theta_p = \theta_s ~ ## ?
No. When ##k=1## they share their field, in other words every field line through one also goes through the other. But flux is more than just the field, it also involves the number of turns. Those are different for the two. So the flux will not be the same even in the ideal case, simply because the same field lines are going through more turns on the high voltage side.
 
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  • #56
Thank for your explanation, then I think the flux you mentioned refer to ## ~ \lambda = N\theta ~##, or can be called flux linkage.
 
  • #57
alan123hk said:
Thank for your explanation, then I think the flux you mentioned refer to ## ~ \lambda = N\theta ~##, or can be called flux linkage.
With all due respect @Dale carefully defined the flux as ##\Phi=Li##
So it required no particular pedantry.

You can call it whatever you want but the maths are where the physics lives.
 
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  • #58
alan123hk said:
Thank for your explanation, then I think the flux you mentioned refer to ## ~ \lambda = N\theta ~##, or can be called flux linkage.
See post 45. I think it is perfectly ok though for @Dale to call it flux, even though he is referring to flux linkage.
 
  • #59
hutchphd said:
With all due respect @Dale carefully defined the flux as Φ=Li
So it required no particular pedantry.
You can call it whatever you want but the maths are where the physics lives.

Of course I absolutely agree with what you said.
Because I did not noticed that the flux has been defined as Φ=Li before, nor have I correctly understood what those equations express, I am sorry for that.

Charles Link said:
See post 45. I think it is perfectly ok though for @Dale to call it flux, even though he is referring to flux linkage.

Of course, how to call it is not a problem. I just misunderstood

Now, I understand why the flux (Li) on the primary side is not affected by K and Z, while the flux (Li) on the secondary side changes with the changes in K and Z. This is because it equivalently represents the input voltage on the primary side and the output voltage on the secondary side.
 
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  • #60
I am curious how to know the core flux ##~ \theta_c ~##(magnetic flux density multiplied by the surface area) changes when the K (coupling factor) decreases, so I try to find an equation to express it. For simplicity, I have to assume that the resistance of the winding wire inside the transformer is zero.

First, I need to obtain the primary current ##~ I_p ~## and secondary current ##~ I_s ~##. Then, I use the relationship between mutual inductance and flux linkage to derive the core flux as shown below.

$$ ~ \theta_c ~= \theta_{sp} + \theta_{ps} = \frac {M I_p} {N_s} + \frac {M I_s} {N_p} $$

After a long period of tedious algebraic operations, I got the following equation.

$$ ~ \theta_c ~= ~N_p \frac {V_p} {R_c} \left[ \frac { \frac {N_s^2} {N_p} (1-k) + \frac Z {L_p} } { j \omega Z - \omega^2 L_s (1-K^2) }\right] ~=~
V_p \left[ \frac { K \frac {N_s} {N_p} \sqrt {Lp L_s} (1-k) + K \frac Z {N_p} } { j \omega Z - \omega^2 L_s (1-K^2) } \right] $$
where
Vp = Voltage on primary side
Lp = Inductance on the primary side
Ls = Inductance on the secondary side
M = Mutual Inductance
Rc = Mutual Reluctance
Np = Number of turns on the primary side
Ns = Number of turns on the secondary side
K = coupling factor
Z = loading impedance

Although I omitted the resistance of the internal winding wire, the derived equation is still very complicated. It can be imagined that in the actual engineering design, it is inevitable to use computer simulation for analysis and design.
 
  • #61
alan123hk said:
Now, I understand why the flux (Li) on the primary side is not affected by K and Z, while the flux (Li) on the secondary side changes with the changes in K and Z. This is because it equivalently represents the input voltage on the primary side and the output voltage on the secondary side.
I prefer the following explanation (the maths tell the exact tale so this is superfluous)
For k=1 the flux is the same on both "sides". There is additional flux from the primary current that exactly cancels (is ##\pi## out of phase and exactly equal in magnitude to) the flux from the current in the secondary and so nothing changes with Z.
When k<1 that cancelation is no longer exact then the flux depends upon Z
I will also point out that this entire result was previously obtained
Dale said:
So, I recalculated the circuit using phasors for a general complex impedance load...
 
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  • #62
For a slightly different take on transformers, here is the lumped element model that EEs use for circuit analysis. The same material, different jargon. EEs that work with this will speak of "coupling" and "flux" in conversation (because we did actually take some physics classes); but, in my experience, they really work with "leakage inductance", "magnetizing inductance", voltages, and currents. The ideal transformer with imperfect coupling is modeled as a "T-section" shown below:
img001.jpg

Then, applying KVL, we get this solution:
$$ \begin{align}
v_1 &=~ L_1 \dot {i_1} + L_m ( \dot {i_1} + \dot {i_2 ^ \prime} ) ~~\Rightarrow ~~
v_1 =~ (L_1 + L_m) \dot {i_1} + n L_m \dot {i_2}
\\
v_2^ \prime &=~ L_2 ^ \prime \dot {i_2 ^ \prime} + L_m ( \dot {i_1} + \dot {i_2 ^ \prime} ) ~~\Rightarrow ~~
v_2 =~ n L_m \dot {i_1} + (L_2 + n^2 L_m) \dot {i_2}
\end{align}
$$
Which gives us this solution:
$$
\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}
~=~
\begin{bmatrix} (L_1 + L_m) & n L_m \\ n L_m & (L_2 + n^2 L_m) \end{bmatrix}
\dot {\begin{bmatrix} i_1 \\ i_2 \end{bmatrix}}
~\equiv ~
\begin{bmatrix} L_{11} & L_{12} \\ L_{21} & L_{22} \end{bmatrix}
\dot {\begin{bmatrix} i_1 \\ i_2 \end{bmatrix}}
$$
Where ##L_{11}## and ##L_{22}## are the "self inductances" and ##L_{12} = L_{21}## is the "mutual inductance" that physicists speak of.

The coupling constant ## k = \frac {M}{\sqrt{\rm {L_1L_2}}} = \frac {L_{12}}{\sqrt{L_{11}L_{22}}} ## Note that L1 & L2 here are the ones you find in physics books, different than the ##L_1## & ##L_2## component values in the solution above.

I'll (probably) post tomorrow on how this is used (or not, actually). In practice this model is either too complex, or too simple.

edit: Oops! Typo - The dependent current source ni2 is drawn backwards.
 
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  • #63
hutchphd said:
For k=1 the flux is the same on both "sides". There is additional flux from the primary current that exactly cancels (is π out of phase and exactly equal in magnitude to) the flux from the current in the secondary and so nothing changes with Z.
When k<1 that cancelation is no longer exact then the flux depends upon Z
I will also point out that this entire result was previously obtained

Of course this is a concise, correct and commonly used explanation

I just understand it from another angle, sometimes, it can be less dull to describe thing in another way.

Because it is assumed that the primary side is connected to an ideal constant voltage source, and the resistance of the winding wire is not included in the equations, the primary side coil flux will not change due to the change of K and the load current. Otherwise, it will be contradicted with the constant voltage source, so the only place where the flux can change is the secondary coil.

On the other hand, if the secondary side is not connected to the load but connected to a constant voltage source too, neither the primary coil flux nor the secondary coil flux will change due to changes in K factor.
 
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  • #64
Just a comment on post 60 for @alan123hk : I think you might try ##\theta_c=L_p I_p+M I_s ##. I'm not sure it will give the same result as what you have.

You also might try ## \theta_c=L_s I_s+M I_p ##. One problem is for the case of incomplete coupling, I don't think the flux is the same everywhere.

See also @Dale 's posts 22 and 43. (One thing that may be worth noting depending on the way the transformer is coupled, it is possible for ##M ## and ## k ## to be negative).
 
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  • #65
alan123hk said:
I just understand it from another angle, sometimes, it can be less dull to describe thing in another way.
But your description is, to me, misleading. You have defined the various fluxes to make it true, yet only in the k=1 (or k=0) case do such "fluxes" correspond to directly accessible quantities. The interesting fact is that current in the primary current does vary according to the secondary impedance even though the primary voltage is fixed, and et seq.
As @etotheipi expertly pointed out it is less "dull" to include a bear and a bicycle in the description, but not necessarily more useful.

No Mas.
 
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  • #66
Charles Link said:
Just a comment on post 60 for @alan123hk : I think you might try θc=LpIp+MIs. I'm not sure it will give the same result as what you have. You also might try θc=LsIs+MIp. One problem is for the case of incomplete coupling, I don't think the flux is the same everywhere.

First of all, I want to explain that my definition of flux is only the surface integral of the normal component of the magnetic field B, not including the number of turns.

What I call core flux ## \theta_c ## only means the flux linkage (common flux) between the two inductors (Lp and Ls), that does not include their respective leakage flux. That leakage flux usually enters the surrounding air space, which will not lead to the saturation of the magnetic core, so it is not the key consideration for the time being.

$$ \text { Core Flux,} ~~\theta_c ~= \theta_{sp} + \theta_{ps} = \frac {M I_p} {N_s} + \frac {M I_s} {N_p} $$
$$ \text { Lp Flux,}~~\theta_p = \theta_{p(Leakage)} + \theta_c = \frac {L_p I_p} {N_p} + \frac {M I_s} {Np} $$
$$ \text { Ls Flux,}~~\theta_s = \theta_{s(Leakage)} + \theta_c = \frac {L_s I_s} {N_s} + \frac {M I_p} {Ns} $$

So of course ## ~~\theta_p \neq \theta_c \neq \theta_s ~~##, is there a little misunderstanding in communication here ?
 
  • #67
See also post 45: ##M=C N_p N_s ##. I'm more than a little puzzled. With a proportionality constant ## C ##, (perhaps along with a ## k ## for the non-ideal case), you essentially have ## \theta_c=C (N_p I_p+N_s I_s) ##, and since ## |N_p I_p|=|N_s I_s |##, neglecting the zero load (open circuit) ## I_p ##, I'm not sure what your ## \theta_c ## represents.

It would make sense to talk about ## \theta_p ## or ## \theta_s ##, but I don't know that it makes sense to define a third quantity.
 
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  • #68
Charles Link said:
M=CNpNs. I'm more than a little puzzled. With a proportionality constant C, (perhaps along with a k for the non-ideal case), you essentially have θc=C(NpIp+NsIs), and since |NpIp|=|NsIs|, neglecting the zero load (open circuit) Ip, I'm not sure what your θc represents

Please note that ## ~|N_p I_p|=|N_s I_s |~ ## only happens when the magnetization inductance is infinite, in this case ## ~\theta_c=C (N_p I_p+N_s I_s) = 0~ ##. When the magnetization inductance is finite, ## ~ I_p \neq \frac {N_s} {Np} I_s ~ ##, and ## ~\theta_c=C (N_p I_p+N_s I_s) ~## represents the non-zero common flux between ##N_p## and ##N_s##.
 
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  • #69
In the best traditions of this thread let me just say that " look Ma, I'm late to the thread"

I think @cmb was actually correct in his original thought that it seems indeed the size of the magnetic core doesn't matter as long as you have good coupling , an increased load current in the secondary will result in an increased current in primary and in theory as long as the wires can keep up and the primary power supply too it seems the power through the core can just keep on increasing because the core flux never enters saturation because for every increase in primary current and core flux there is an opposed increase in secondary current creating a increased back EMF and cancelling the increase in flux , so net result flux stays the same yet power transferred through the core increases. Very interesting.

One problem though why I think this might not be practical , because having a small core but enough wire turns and thickness to allow for a high power transfer will result in heavy core saturation during light load or no load condition. Say this transformer is attached to mains voltage. Under no load or light load the core will be driven into saturation on each half cycle of the AC, so the impedance will be highly non linear and the coil will draw excessive current on each cycle right after saturation of core begins.
Result is this transformer will heat up excessively etc.

But if one can control the voltage and current that enters into the primary as to decrease the volts under light load and increase volts as necessary during heavy loads then in theory I think one can indeed "get away" with a small core that is capable of high power.
I guess one way to test this scenario would be to use an autotransformer and attach it to a small core that has a low turn heavy gauge wire as it's primary and secondary. Then have a low resistance resistive type of load in the secondary.
 
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  • #70
artis said:
In the best traditions of this thread let me just say that " look Ma, I'm late to the thread"

I think @cmb was actually correct in his original thought that it seems indeed the size of the magnetic core doesn't matter as long as you have good coupling , an increased load current in the secondary will result in an increased current in primary and in theory as long as the wires can keep up and the primary power supply too it seems the power through the core can just keep on increasing because the core flux never enters saturation because for every increase in primary current and core flux there is an opposed increase in secondary current creating a increased back EMF and cancelling the increase in flux , so net result flux stays the same yet power transferred through the core increases. Very interesting.

One problem though why I think this might not be practical , because having a small core but enough wire turns and thickness to allow for a high power transfer will result in heavy core saturation during light load or no load condition. Say this transformer is attached to mains voltage. Under no load or light load the core will be driven into saturation on each half cycle of the AC, so the impedance will be highly non linear and the coil will draw excessive current on each cycle right after saturation of core begins.
Result is this transformer will heat up excessively etc.

But if one can control the voltage and current that enters into the primary as to decrease the volts under light load and increase volts as necessary during heavy loads then in theory I think once can indeed "get away" with a small core that is capable of high power.
I guess one way to test this scenario would be to use an autotransformer and attach it to a small core that has a low turn heavy gauge wire as it's primary and secondary. Then have a low resistance resistive type of load in the secondary.
Thanks for the thoughts.

In general my thoughts were in regards RF transformers than perhaps mains frequency/AF transformers, but the principle should still apply. In that case, as you say, it would be down to whether you can physically get the coil wrap around the core for the necessary wire gauge to carry that power.

Reason I came up with the thought in the first place was because of exactly what you are saying there, so long as the wires and such can cope, etc., and they could [in the case I was working on] because I designed a 6.78MHz (ISM band) power coupler and the maths was saying all I needed was a teeny little core to transmit kW power. (Only needed 4~9 turns on this particular core at 6.78MHz and could still stay well under 100mT flux.)

Seemed odd at the time.

In the end I used an air core. It makes for a bigger device but I had the space to build that and avoided the fraught complexities of ferrite selection.

In some future life if I ever need to squash a kW class RF power transformer into a matchbox then I think it is possible and will give it a go. Another benefit would also be that the ferrite, being always slightly lossy (obviously, always need to design these things so bulk temperature of the core is not excessive) will supress harmonics whereas I did find a few harmonics coming through with the air core solution.

There are also some ferrites I have been looking at that seem to have very good performance specifications when operating around 2MHz, this also could work out very favourably. Sort of mid-range between current SMPS frequencies and what I would regard as 'RF'. I suspect SMPS will probably move towards this sort of 2MHz frequency, if what I am seeing in material performance (and SiC power modules to supply the switching power) is something considered further by that industry.
 

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