flyback converter and related transformer issues

Transformer issues

I got two questions that need your help,

(1) For an ideal transformer, which does not consist of leakage inductance, magnetizing inductance, core loss resistance, etc., its primary winding is connected to an AC source while the secondary winding is unloaded, theoretically, the voltage across the primary winding is equal to the source voltage while the current flowing through the primary winding is zero. But In my understanding, since there is no current flowing through the primary winding, there should be no varying magnetic field and thus no voltage is induced across the primary winding, am I missing anything here?

(2) A real transformer can be modeled as a magnetizing inductance in parallel to an ideal transformer, what is the magnetizing inductance? I checked some documents, some define it as the mutual inductance, is that right?

Thanks in advances!

1) An ideal transformer doesn't deal with magnetic fields, it just transforms voltage. The problem is that you can't calculate with infinity, e.g. infinite inductance. Assume a high finite nductance and a small magnetizing and the problems falls apart.

2) With perfect coupling, mutual and magnetizing inductance are the same. In a real transformer, coupling is below unity and both quantities become different.

Thanks, FvM, it helps!
Based on your answer, I got a few more questions:

(1) If an ideal transformer does't deal with magnetic fields, can I just see the two windings as a voltage gain? like a dependent voltage source with a gain equal to the turns ratio. And the self-inductance of the two windings in an idea transformer are infinity?


(3) The reason why a real transformer has a coupling coefficient below unity is that it has a leakage inductance in addition to the magnetizing inductance? it this right?
The mutual inductance = magnetizing inductance + leakage inductance? is this equation right?

Hi FvM,

I have been waiting for your further explanation to my two additional questions indicated in my last reply, thanks a lot!

1) It's voltage gain, if you are driving the primary side with a voltage source. Otherwise it's a bidirectional translator.
2) Depends on your transformer equivalent circuit, how you define e.g. magnetizing inductance. In the popular T circuit, mutual inductance and magnetizing inductance are the same, the total (terminal) inductance is then mutual/magnetizing inductance + 1/2*leakage inductance

What is the terminal inductance you were referring to ? why we do not need to care about the self-inductance of each winding?

I meaned the total inductance measured at a winding with all other windings open. You can also consider it as self inductance Lxx in contrast to mutual inductance Lxy.

ux = ∑Lxy * diy/dt ; y = 1..N

FvM! I got it better, but still not fully get it. Can you recommend to me some transformer related documents that might be helpful? I checked some documents the last few days but they sometimes contradict each other in terms of the contents, I have been struggling with the transformer issues for a while, so I want to get a thorough understanding of it.
Thanks for your help!

2) Yes, the magnetizing inductance is also the mutual inductance.

Thanks, wky!

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FvM said:

I meaned the total inductance measured at a winding with all other windings open. You can also consider it as self inductance Lxx in contrast to mutual inductance Lxy.

ux = ∑Lxy * diy/dt ; y = 1..N

Click to expand...

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FvM, I am assuming ux in your equation is the voltage across the xth winding, this equation does make sense. But you also said in your previous reply that self-inductance of a winding = magnetizing inductance/mutual inductance + 1/2 * leakage inductance, in my understanding, the self-inductance should be independent of mutual inductance like in the coupled inductor, there is no quantitative relation between self inductance and the mutual inductance, anything wrong?

there is no quantitative relation between self inductance and the mutual inductance, anything wrong

Click to expand...

It's true for the Lxy matrix model, but not for a physical transformer. Again, with ideal coupling k=1, self and mutual inductance are identical (at least for 1:1 winding ratio). A current variation dI/dt in any of the windings will induce the same voltage in the primary (or any other) winding, in other words L11 = L12 = L13 ...

FvM, you said self-inductance of a winding = magnetizing inductance/mutual inductance + 1/2 * leakage inductance? Why 1/2 ?

You understand that "/" means *or*, not divided by in my post #5?

The 1/2 is originated from the standard "T" equivalent circuit which has a "primary" and "secondary" leakage inductance component as series elements. Each must be half the measured leakage inductance for consistency with the real component behaviour.

FvM, thanks for the reply.
If the measured leakage inductance Lk is measured at the primary with secondary shorted,
we have Lk = Lkp + (Np/Ns)^2 *Lks , Lpk and Lks being respectively the primary and secondary leakage inductances,
The primary self inductance Lp = Lkp + Lmag
Per your saying, Lkp = 1/2 *Lk = 1/2*Lkp +1/2*(Np/Ns)^2*Lks,
Giving Lkp = (Np/Ns)^2*Lks
For a 1:1 transformer, it means Lkp always = Lks, is this true for a real component ?

For a 1:1 transformer, it means Lkp always = Lks, is this true for a real component ?

Click to expand...

Yes and no. It totally depends on the involved equivalent circuit.

You can identify a real transformer with a set of measurements. (For simplicity, I'm restricting the discussion to inductance, a real transformer also exposes a frequency dependent real impedance component.)

The previously mentioned Lxy inductance matrix is one possible representation of the real component. Another would be a open and short circuit inductance measured at both terminals.

It's your decision how you translate the empirical measurement into an equivalent circuit. You'll restrict the degree of freedom if you demand a physically plausible model.

E.g. if you measure different open circuit and short circuit inductances at both terminals of a transformer known to have 1:1 windings ratio. A physically plausible model would refer to asymmetrical leakage inductance, but you can also adjust the windings ratio of the model. In other words, there are different models that can represent the same real component.

Another possible model is the SPICE transformer model comprised of primary and secondary inductance and coupling factor. You'll find that a set of three parameters fully describes a transformer without loss elements, and the different models can be translated into each other.

A previous discussion related to the same problem: https://www.edaboard.com/threads/264462/
Or: https://www.edaboard.com/threads/235835/

FvM, my own answer to my question is a NO, and may be YES in very special cases.
Also I don't think this has something to do with equivalent circuits.
(I understand that there can be different equivalent circuits to describe the behaviour of a real component, they being 'equivalent' to each others)
For a 1:1 transformer, if Lkp = Lks , it requires that the magnetic reluctances (including core, gap, leakage paths etc) seen by the primary and secondary windings are the same.
For a conventional transformer in which one winding wrapping around the other, the two magnetic reluctances are definitely not equal, and therefore Lkp=Lks is not true.

The problem is that Lkp and Lks in your calculation aren't directly observable. They are assigned to a set of measurements based on an equivalent circuit or a "model".

The reluctance based model derived by Dixon is a "pi" rather than a "tee" model with only one element representing leakage inductance. **broken link removed**

Lkp and Lks are parameters of the "tee" model, but you can of course doubt their physical plausibility. I just notice that they are often used in literature and circuit descriptions, so I try to calculate with it.

I noticed that TI has reorganized literature archives once more so that the above link became invalid. Attached the original file.

Lkp and Lks are not diectly measurable but deducible from other measuremnts (OK based on a T-model).
Without revisiting Dixon's paper, if we take the self inductance as (leakage inductance + magnetizing inductance),
then for primary , Lp = Lkp + Lm = N^2/Rp, secondary, Ls = Lks + Lm = N^2/Rs (for a 1:1 xformer)
where the R's are reluctances seen by the windings,
As Rp≠Rs, Lkp≠Lks

Yes, sounds reasonable, based on the T-model.