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

[alan123hk]

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.

 

[hutchphd]

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

 

[alan123hk]

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).

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.

 

[alan123hk]

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.

 

[Dale]

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.

 

[alan123hk]

Thank for your explanation, then I think the flux you mentioned refer to $~ \lambda = N\theta ~$, or can be called flux linkage.

 

[hutchphd]

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.

 

[Charles Link]

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.

 

[alan123hk]

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.

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.

 

[alan123hk]

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.

 

[hutchphd]

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

So, I recalculated the circuit using phasors for a general complex impedance load...

 

[DaveE]

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:

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.

 

[alan123hk]

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.

 

[Charles Link]

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).

 

[hutchphd]

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.

 

[alan123hk]

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 ?

 

[Charles Link]

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.

 

[alan123hk]

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 $N_s$.

 

[artis]

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.

 

[cmb]

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.

 

[artis]

@cmb well I thought the size matter was most relevant only in lower frequencies because as you get to Mhz your core size goes down anyway , in fact most modern few hundred watt SMPS supplies have match box sized transformers at Khz frequencies

 

[cmb]

@cmb well I thought the size matter was most relevant only in lower frequencies because as you get to Mhz your core size goes down anyway , in fact most modern few hundred watt SMPS supplies have match box sized transformers at Khz frequencies

Indeed so. It just struck me how such a small core could transmit what seemed to be 'unlimited' amounts of power when I got down to designing it.

I think there are other 'real' reasons why there is a finite limit, I think as you push the core harder (more coil current but same voltage, thus same flux) the reactance of the primary keeps increasing and it becomes more a problem for the power driver to keep pumping power in, rather than the core reaching some limit.

I'm not actually sure I can see whether that becomes a bigger or lesser problem as the transformer frequency is lowered.

 

[hutchphd]

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.

What are you talking about??
The whole point of this exercise was to show that the flux is independent of the secondary load.

 

[artis]

What are you talking about??
The whole point of this exercise was to show that the flux is independent of the secondary load.

Yes I am aware of that and that indeed is the reason why one can in theory use a small core but push through high power.
But I was referring to the OP talking about how he during his calculations started to think one could use a small core and still transfer high power. Then members started pointing out that it would not be practical due to various limitations.

 

[cmb]

I think there is a crossed purpose in the last couple posts, which is OK in the context of the discussion.

The flux is (theoretically, at least, Faraday's law and all that) a function of volts/turn. I took what @artis said to [ultimately] mean that if we try to increase the current capacity (to increase power) then we end up reducing the number of turns (thicker gauge), and if one reduces the number of turns due to thicker gauge (for higher current) then that would increase flux.

Not sure if he did mean that, but I took it that way as this makes sense to discuss in that context.

Ultimately, in the case of a 'perfect' core material, it implies total maximum power for a transformer will be a function of the wire, rather than the core. Of course, cores are always imperfect, so the optimum is the balance between the total number of turns of a given wire capable of a given current, and the physical space to wind them.

For RF purposes, one never wants to get anywhere near saturation of the material, this would be incredibly lossy. Typical RF ferrites should operate, where possible, at no more than a few units of mT. Reason is that if you are pumping the core several million times a second, even small microJoule amounts of energy to polarise the core leads to several watts of power to dissipate. For laminated steel mains cores at 10's of Hz, they can lose several Joules per polarisation and the heating can be managed.

Horses for courses/YMMV, etc...

Thanks for the discussion points.

 

[hutchphd]

For RF purposes, one never wants to get anywhere near saturation of the material, this would be incredibly lossy.

We can all agree with that. Nice thread.

 

[artis]

@cmb well just to clear up what I meant, the reason I said that a thicker wire with less turns should be employed for this "small core high power test" is simply because on a small core there is very little space and making many turns will not be possible so one can simply push the current instead of the voltage the end result will be the same in terms of core flux whether you have more turns less current or less turns more current.

A nice thread indeed, I personally never before had thought about this that the core size is not critical for maximum power as long as flux doesn't saturate it.

The mechanical analogies with ropes and pushing also fail to fully show the beauty of this because mechanically even if you apply perfectly even but opposite forces the total force is still a sum of both and eventually something will break or tear , most likely at the midpoint.
Yet in the transformer, as load current increases so does primary supply current (assuming a limitless power supply) yet the flux stays the same because as primary tries to increase the flux the secondary opposes this increase even more, net result balance.

As far as I can think of mechanical analogies fail here. One could think of power transmitted through a shaft, like a motor driving a generator (primary - secondary) but it fails because the more power you transmit through the shaft the higher the torque until the shaft snaps violently at some point.

This "magnetic stuff" truly remarkable, and not that much intuitive.

 

[Charles Link]

@artis See
https://www.physicsforums.com/threa...to-this-transformer-homework-problem.1002895/
and
https://www.physicsforums.com/threa...rmers-how-to-apply-them.1002399/#post-6485075
for a couple of other threads where we discussed some of the features of transformers in detail.