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.
  • #1
cmb
1,128
128
This is one of those 'oh, I am not sure I know the answer' kind of thoughts.

The strength of magnetic flux in a transformer core is a function of volts per turn, Faraday figured that one out.

So, if I have a light load on a secondary (let's say it is pure ohmic for now, let's not confuse it) and turn up the AC volts on the primary, hey, there it is, voltage 'V' and next to no current going through.

Then I put a heavy ohmic load on the secondary and the volts get pulled down and so I have to crank up the power, more current, to get back to my voltage 'V' across the whole coil.

First case is some lower power with some given voltage across each turn, and second is a higher power at the same voltage across each turn but more current.

Here is my 'duh' moment; the magnetic flux is the same, because that is defined as volts per turn. We have this magnetic core 'transmitting' power from one side to another, but the magnetic domains in the middle would never know there was any difference in the power transmitted.

One might casually assume that if one were to 'look into' the detail of power being transported from one place to another then there would be 'something different' happening within that conduction space, wherein the power is being transported, between different power settings.

Is there? What is it, if magnetic flux is a function of volts only?
 
Physics news on Phys.org
  • #3
anorlunda said:
But Poynting relates to electric charges 'outside' the transformer core, right?

I'm asking what is going on 'within' the magnetic core that is different between power transfer magnitudes?

Maybe the question doesn't make sense. Sure I know what the equations say, I was hoping for a 'note of wisdom' than equations. I guess it is a philosophical POV I was coming to it at; that transfer of EM energy is so 'not like what we know' that there isn't really an analogy (like 'more water in a pipe' or such), and that it's just how it is?
 
  • #4
I think you need to consider the magnetic field created by the current in the secondary, the subsequent induced emf in the primary, and how the circuit driving the primary responds.
 
  • Like
Likes DaveE
  • #5
cmb said:
But Poynting relates to electric charges 'outside' the transformer core, right?
Not right. Fields outside the core caused by charges inside the core. It applies.
 
  • #6
cmb said:
Summary:: Faraday's law says magnetic induction field is down to volt turns, not current, so what transmits 'power' in a magnetic core?

First case is some lower power with some given voltage across each turn, and second is a higher power at the same voltage across each turn but more current.
I don’t think this is correct. Both the current and the voltage should be higher.
 
  • #7
Dale said:
I don’t think this is correct. Both the current and the voltage should be higher.
Intuitively one would think so, but it isn't.

EMF (volt/turn) = - d(flux)/dt

.. look, Ma, no current!

Yet, if the rate of change of flux is zero, then the magnitude of the flux is proportional to the current in the current loop, rather than the voltage. (Biot-Savart).

It's really the dichotomy of Faraday versus Biot-Savart. Hence, my question. Sure, I get all the equations and explanations one at a time, but it just seems odd to me if one stops to think about those two sides of the magnetic core story.

As for charges outside a magnetic core affecting the charges inside, it is an interesting idea I'd like to try to picture that, but how does this apply, in physical terms, to a ferrite with a very high bulk resistance? Why would this not be different for ferrites with low bulk Ohmic resistances versus those with high resistances? If that were true then would the inductive behaviour not be a function of core Ohmic conductivity as well as permeability?
 
  • #8
cmb said:
Intuitively one would think so, but it isn't.

EMF (volt/turn) = - d(flux)/dt

.. look, Ma, no current!
You are forgetting the self inductance. The voltage is proportional to ##di/dt##. So as you increase the voltage you also increase the current.
 
  • #9
Dale said:
You are forgetting the self inductance. The voltage is proportional to ##di/dt##. So as you increase the voltage you also increase the current.
Consider for simplicity the case of a perfectly flux linked transformer with primary and secondary. Then$$N_p\dot{\phi}=V_p\text{ and }N_s\dot{\phi}=V_s$$ So the volts per turn is a characteristic number and that will be unchanging for the (ideal) transformer. Notice core losses and actual resistance will make this slightly untrue but an imperfect flux linkage just changes the numbers.
Of course the loading of the secondary will determine the actual primary and secondary currents flowing in steady state AC configuration. But for an ideal transformer the heat will end up in the secondary load resistor I believe
 
  • #10
Dale said:
You are forgetting the self inductance. The voltage is proportional to ##di/dt##. So as you increase the voltage you also increase the current.
For a given fixed load, yes.

But I am asking what happens to the core for different loads versus currents but a given fixed voltage.
 
  • #11
hutchphd said:
Consider for simplicity the case of a perfectly flux linked transformer with primary and secondary. Then$$N_p\dot{\phi}=V_p\text{ and }N_s\dot{\phi}=V_s$$ So the volts per turn is a characteristic number and that will be unchanging for the (ideal) transformer. Notice core losses and actual resistance will make this slightly untrue but an imperfect flux linkage just changes the numbers.
Of course the loading of the secondary will determine the actual primary and secondary currents flowing in steady state AC configuration. But for an ideal transformer the heat will end up in the secondary load resistor I believe
Thanks, yes, exactly.

The thought has come up because of the following scenario; I am working with the latest LDMOS transistors, that can withstand >65:1 VSWR, and a tuneable load that can 'set' such a mismatch if I actually wanted to.

There is a 'legacy' impedance matching 9:1 transformer in there between the two (I don't think it is necessary but it is there!), and as I can basically set what I like either side of it, the question is what is best load to set for minimum losses in that transformer for a given power?

It seems to me that if I set the load so that it pulls the two turn transformer to (say) 50V/T and 1A then I get 100W through but a flux that is a function of the 50V/T. Just for the sake of numbers let's say it is 13.56MHz in a 1cm^2 xs core, so the flux will be 92 Gauss, which for 61 material would be a loss of about 400mW/cc according to data sheet.

But if I set the load so that it is 10V/T but drawing 20A, then I get through 400W but a flux of 46 Gauss. This is a power loss of about 50mW/cc.

I end up transmitting 4 times MORE power 'through the core' for LESS core loss, about an 8th in fact.

If one can set the load arbitrarily, a low impedance load will always give a much lower power loss in the core, it seems?

What is the 61 material core doing differently in those two scenarios?
 
  • #12
cmb said:
For a given fixed load, yes.

But I am asking what happens to the core for different loads versus currents but a given fixed voltage.
The load doesn’t change the self inductance nor the mutual inductance. You still wind up with the same relationship between current and voltage at the transformer.
 
  • #13
Dale said:
The load doesn’t change the self inductance nor the mutual inductance. You still wind up with the same relationship between current and voltage at the transformer.
Could you please help me understand this.

If the load on the secondary is one ohm the relationship would be one amp in the coil for each volt across it.

If it is 10 ohms then are you suggesting it is still one amp for each volt across the coil?

Is this something to do without of phase currents, or something else?
 
  • #14
Dale said:
The load doesn’t change the self inductance nor the mutual inductance. You still wind up with the same relationship between current and voltage at the transformer.
This is true but the question had to do with the magnetic flux deep in the core. After some thought I believe it does not change for a fixed input . The phase between current and voltage will be driven by output loading
 
  • Like
Likes Dale
  • #15
hutchphd said:
This is true but the question had to do with the magnetic flux deep in the core. After some thought I believe it does not change for a fixed input . The phase between current and voltage will be driven by output loading
Yes, thanks, I'm coming to the same sort of conclusion.

Analogies aren't the greatest things, but I think the best analogy is like a tug-o-war rope, you can pull on either end lightly or really strongly. The rope itself can conduct small amounts of power (work x distance) or large amounts of power, the rope itself doesn't know but is put under tension. I think magnetic domains themselves don't have 'tension' as such so just exhibit the net balance.

The consequence is that the phases of current become increasingly displaced in phase for increasing loads.

As for the losses in the ferrite, if the rope doesn't stretch and lose work that way (i.e. magnetic domains don't 'stretch' like rope, that's where the analogy ends) the actual tension doesn't matter.

The actual power loss in the material we can measure is like a damper attached to the middle of the rope and the other end fixed to a stationary point on the ground. As voltage is like displacement and current like the tension, the damper will absorb the displacement not the power (displacement x tension).

Does this sound like a reasonable stab at the scenario we're thinking about?
 
  • #16
I will not comment on your anology (we all have our own pictures in our heads you can deal with yours) but the big losses in cores are due to real induced currents in laminated steel. And there are other problems when saturation impends because the domains all get aligned. And of course the hysteresis comes from the domain structure and mechanical pinning thereof. One need to be careful about where the energy actually goes, if that matters.
 
  • #17
hutchphd said:
I will not comment on your anology (we all have our own pictures in our heads you can deal with yours) but the big losses in cores are due to real induced currents in laminated steel. And there are other problems when saturation impends because the domains all get aligned. And of course the hysteresis comes from the domain structure and mechanical pinning thereof. One need to be careful about where the energy actually goes, if that matters.
Of course, saturation and steel excepted. These carry with them some very specific physics related to their material type.

I'm thinking primarily here of 'more ideal' transformers, like RF frequencies in ferrites, so no currents (albeit not the tiny currents in high resistance ferrite grains) and fields are milliTesla and far from saturation.
 
  • #18
hutchphd said:
we all have our own pictures in our heads

true, I have a picture of a baby panda riding a bicycle through a field full of daisys in my head right now
 
  • Haha
  • Like
Likes hutchphd and Dale
  • #19
etotheipi said:
true, I have a picture of a baby panda riding a bicycle through a field full of daisys in my head right now
Hey! Me too (now)
 
  • Love
Likes etotheipi
  • #20
My panda always rides a unicycle. He's talented.
 
  • Wow
Likes etotheipi
  • #21
cmb said:
I'm thinking primarily here of 'more ideal' transformers, like RF frequencies in ferrites, so no currents (albeit not the tiny currents in high resistance ferrite grains) and fields are milliTesla and far from saturation.
Wouldn't your basic question about currents and voltages apply equally to a transformer with no core, i.e. just two coils? It seems to me like focusing on what's going on in the core is just an unnecessary complication.
 
  • Like
Likes DaveE and hutchphd
  • #22
OK, so consider a simple circuit with two loops. The primary loop has a voltage source ##v_p(t)= V \cos(\omega t)## and an inductor ##L_p##, and the current through the primary loop is ##i_p(t)##. The secondary loop has an inductor ##L_s##, and a resistive load ##R##, and the current through the secondary loop is ##i_s(t)## with the voltage across the secondary loop being ##v_s(t)##. The primary and secondary are perfectly coupled with mutual inductance ##M=\sqrt{L_p L_s}##.

$$v_p(t) = L_p i_p'(t) + M i_s'(t)$$ $$v_s(t)=L_s i_s'(t) + M i_p'(t)$$ $$v_s(t)=R i_s(t)$$

Solving we get

$$v_p(t) = V \cos(\omega t)$$ $$i_p(t)= -\frac{V L_s}{R L_p} \cos(\omega t) + \frac{V}{L_p \omega} \sin(\omega t)$$ $$v_s(t)= V \sqrt{\frac{L_s}{L_p}} \cos(\omega t)$$ $$i_s(t)=\frac{V}{R}\sqrt{\frac{L_s}{L_p}}\cos(\omega t)$$

cmb said:
Intuitively one would think so, but it isn't.

EMF (volt/turn) = - d(flux)/dt

.. look, Ma, no current!
As you can see, changing the load does change the current in both the primary and the secondary. It does not change the voltage in the primary since that is fixed by the supply, and it also does not change the voltage in the secondary, which surprised me.

So with a fixed voltage supply, the additional power is delivered as additional current.
 
  • Like
Likes artis, hutchphd and vela
  • #23
Dale said:
So with a fixed voltage supply, the additional power is delivered as additional current.
Of course it is. That was never in question.

So, then, what is the answer to the question?

There is no physical manifestation I am aware of in the magnetic core other than to the voltage, because EMF (volt/turns) is rate of change of flux.

If the only thing that changes is the current in the windings, then nothing 'different' happens in the core. Right?

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.
 
  • #24
etotheipi said:
true, I have a picture of a baby panda riding a bicycle through a field full of daisys in my head right now
I'm trying to perform the serious business of making the most efficient kW class RF amplifier I can make into the smallest space. I'm unclear on the origin of your comment. Could we keep 'off-topic and humour' to at least until after there is an answer to the question?
 
  • Sad
Likes etotheipi
  • #25
cmb said:
There is no physical manifestation I am aware of in the magnetic core other than to the voltage, because EMF (volt/turns) is rate of change of flux.
There is the current too.

cmb said:
If the only thing that changes is the current in the windings, then nothing 'different' happens in the core. Right?
I am not sure why you think the current being different is not something different.

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.
And what would happen if you put a large current through that 35 mm toroid?

You seem to know that the current is different and yet want to ignore the current and be confused as a result. Current is rather important in general.

cmb said:
Intuitively one would think so, but it isn't.

EMF (volt/turn) = - d(flux)/dt

.. look, Ma, no current!
Don’t forget ##L = \Phi/I##. You cannot ignore current simply because it doesn’t show up in one single equation.
 
Last edited:
  • #26
I'm not following sorry.

What is the physical manifestation of more or less coil current in the core? That is my question. I don't think there is any.

The current is 'coil current' and only coil current, in the windings, not in the core.

Does the core do anything differently with more or less current in the windings? This has been the only question.

There isn't any 'current' in the core, only magnetic flux which is a function of voltage only.

Is that right, or does the 'core' behave in some different way with more or less current in the windings?
 
  • #27
More current gives a larger self induced flux. As I said above ##L=\Phi/I##. Your “look ma no current” is only one equation out of many. You cannot separate the core from the windings they go together. Without the windings there isn’t a core.
 
  • Like
Likes etotheipi
  • #28
I think I understand your confusion.

I mentioned nothing about having a fixed inductance.

If one puts on more or less turns, one can set the impedance to anything one wants.

I am discussing what the difference in the core material is between one condition where there is (say) 10V/turn and 1A, versus 10V/turn and 100A in the windings? Put on whatever loads and numbers of turns you want to, to set those conditions, the inductance and impedance are not fixed here in my question.
 
  • #29
I never said they were fixed. That is why I put them as variables.

Regardless of whether the inductance is fixed or not there is inductance and the equation ##L=\Phi /I## holds
 
  • Like
Likes etotheipi
  • #30
Dale said:
I never said they were fixed. That is why I put them as variables.

Regardless of whether the inductance is fixed or not there is inductance and the equation L=Φ/I holds
In that case, my bad for wholly confusing the picture by referring to ohmic loads.

If I add a capacitance to the load on the secondary, the circuit impedance changes. The current in the coils goes up while the volt/turn can remain the same.

More current in the coils but the same volts/turn. What happens to the flux? Can we funnel endless VAr via the core, ever increasing coil current, with no effect on it so long as the volt/turns is constant?
 
  • #31
cmb said:
If I add a capacitance to the load on the secondary, the circuit impedance changes.
This is starting to feel like a moving target.

cmb said:
More current in the coils but the same volts/turn. What happens to the flux?
##\Phi=LI## for self inductance. ##\Phi_1=L_1 I_1 \pm M I_2## for mutual inductance with ##M=k\sqrt{L_1 L_2}##
 
  • #32
Dale said:
This is starting to feel like a moving target.
It's totally stationary from post #1.

I have only asked what difference is there in the core of an inductor when the current increases but the volt/turn stays the same.

Only one question.

Still wondering.
 
  • #33
cmb said:
It's totally stationary from post #1.

I have only asked what difference is there in the core of an inductor when the current increases but the volt/turn stays the same.

Only one question.

Still wondering.
Only one question and only two answers:
1) The short answer: Nothing changes in the core.
2) The rather longer answer is that when the secondary current increases (which would tend to cancel out the flux in the core), the current in the primary increases (thus keeping the flux the same). This is illustrated in the attachment for the very simple case of a DC source connected to the primary (obviously a situation that cannot be allowed to continue).
 

Attachments

  • Part 4.pdf
    238.7 KB · Views: 114
  • Like
Likes Merlin3189 and DaveE
  • #34
cmb said:
It's totally stationary from post #1.
Your last post was the first one that mentioned a capacitative load. You also didn’t mention a variable inductance until post 28. Even if those things were in your mind and intended, I am telling you that to me it feels like a moving target.

cmb said:
I have only asked what difference is there in the core of an inductor when the current increases but the volt/turn stays the same.

Only one question.

Still wondering.
Not sure why you are still wondering. It is easy enough to calculate ##\Phi=LI## from what I posted earlier for a resistive load and ##k=1##. Or you can change the load to a complex impedance and use phasors to simplify the calculations.

The full differential eq results for ##k\ne 1## are particularly interesting, but too big to post. That might be a good one to use phasors on.
 
Last edited:
  • #35
agrumpyoldphysicstec said:
Only one question and only two answers:
1) The short answer: Nothing changes in the core.
2) The rather longer answer is that when the secondary current increases (which would tend to cancel out the flux in the core), the current in the primary increases (thus keeping the flux the same). This is illustrated in the attachment for the very simple case of a DC source connected to the primary (obviously a situation that cannot be allowed to continue).
I agree wholeheartedly.

cmb said:
If I add a capacitance to the load on the secondary, the circuit impedance changes. The current in the coils goes up while the volt/turn can remain the same.

If we redo @Dale solution for abitrary (complex) impedance ##Z ## on the secondary, the primary complex current is $$I_p=\frac {V_p} { Z_{in}}$$
$$=\frac {V_pL_s} {ZL_p}\left[ 1+\frac Z {j\omega L_s}\right]$$
Where J is for the EE imaginary.
So there are two terms in quadrature. The second term is the response of the primary coil with no secondary current. The first term is the added current supplied by the source to counteract the "Lenz Law" response of the secondary if current flows (at 90deg phase to primary current).
So the the magnetic flux is unchanged for any secondary impedance, reactive or not.
 

Similar threads

  • Electrical Engineering
Replies
8
Views
1K
Replies
6
Views
2K
  • Electrical Engineering
Replies
8
Views
1K
Replies
3
Views
913
  • Mechanics
Replies
7
Views
2K
Replies
8
Views
822
  • Mechanics
Replies
2
Views
3K
  • Electromagnetism
Replies
3
Views
845
  • Electrical Engineering
2
Replies
36
Views
3K
Replies
21
Views
1K
Back
Top