Transformer coupling coefficients

Aug 29, 2006 75 Replies

I don't dispute that. When working with L, it's correct to use k which is the geometric mean of k ij and k ji, and when working with circuits one uses L. Flux linkages are certainly not an abstraction to a transformer, though they may well be to a circuit designer. When solving the matrix equations for L's, the K ji's and K ij's get cross-multiplied so once in the realm of L it doesn't matter if they're alike or different since they've disappeared as entities.

I didn't find much help in textbooks or in the literature when I was developing my theory some years ago. Seems like it must be somewhere, but I never found it. I've never published it and don't plan to. I can assure you that it works, y'all can decide whether or not you care to believe that. There was a TDK appnote at the time (1995 or so) that stated something like "rod core transformers cannot be designed analytically, they must be developed empirically." Wrong! I wish I still had a copy of that app note! Maybe using FEMM is empirical, but the results from it wouldn't be of any use without the theory.

Let's try a heuristic argument. By reciprocity, the permance of the flux path linking two windings is the same regardless of which winding is excited. But the permeance of the total flux paths for the windings may be quite different, particularly in loosely coupled situations if the windings are geometrically rather different relative to one another and/or in how they're disposed on a core that doesn't comprise a closed path. Therefore, the ratio of flux linking a pair of windings to total flux for a given winding (k ij) can be quite different. FEMM finds these flux linkages as space integrals so flux partially linking a winding (not the whole winding) gets "partial credit".

Let's look at a simple familiar model. This one is limited to 2 windings while the general case is n windings, but it may help clarify. Consider the classic three-terminal model of a transformer:

------ Z1 - M -----+------ Z2-M------- | M |

--------------------------------------------

L1 is primary inductance, L2 is secondary inductance, M is mutual inductance. In conventional mesh notation, L11 = L1, L22 = L2, L21 = L12 = M. Mr. Guillemin smiles. But the ratios of mutual to self, call them k21 and k12, are M/L1 from the left and M/ L2 from the right. Even though L21 = L12 = M, the ratios are not necessarily the same. From any text, M = k * sqrt (L1 * L2). By simple algebra, k21 = k sqrt (L2/L1) and k12 = k sqrt (L1/L2). The geometric mean, sqrt (K21*K12) = k. QED.

If one is working with an existing transformer, it's definitely easiest to measure and think in terms of inductances. If one is designing loosely-coupled transformers, it's easier to use FEMM to see what flux linkages occur with a given geometry. It's vert easy to get k ji and k ij from a FEMM simulation. One can examine variants that way a lot faster than winding, vacuum-potting and testing actual specimens -- and at 30KV in a small xfmr vacuum-potting is not optional though somewhat messy. Once the flux linkage data is converted to inductances in a math model, Lij = Lji in every case as Guillemin and others assert.

Both approaches are correct, and both work. Transformers I've built accurately to a given geometry usually exhibited circuit parameters within 5% or so of what I expected from FEMM modelling and math modelling. The math models were used to predict the effects of distributed capacitances and hence self-resonances, and they did account for winding resistances. Fringing and all that are non-issues because FEMM analyzes and integrates fluxes as they are determined to actually exist. The resulting values of L ij and k were then used in SPICE to predict circuit behavior with associated electronics. Once actual circuits and transformers were made, they performed at the bench as expected.

The application here was very low cost 30KV 30-watt high-frequency oil-ignition transformers operating at resonance with a capacitor in a self-excited power oscillator. Loose coupling is useful here because it enables the desired high output impedance without requiring additional inductances.

The circuit designer could care less about k ij and k ji, since all he can see and all he need care about is inductances. For closely-coupled designs as used in SMPS, the difference between k ji and k ij must be miniscule in any case since k is "very close to 1". I now know, thanks to the experience and helpful post of Mr Woodgate, that values of .998 are quite achievable so I guess I need not worry about k or k ij or such claptrap for my little SMPS project. I'll just wind the suckuh and snub as necessary. I apologize for the diversion.

Funny you should mention that. While I was going through all my texts, I noticed in Skilling's circuit theory book a section about "equality of mutual inductances". In a footnote on that page he gives a brief outline of a proof that if they aren't equal, you would be able to extract energy from an inductor.

I should have said, "...to get the equivalent inductance, divide the imaginary part of the impedance by s.

Asymmetry of things like this is usually a big red flag.

Remembering that you did start this thread with a request for some typical numbers, I found an old EC41 core set in the bin with some wire already on it, and I took some measurements.

The first thing on the bobbin is a single layer, bifilar wound with a pretty red and green pair of wires. The k between these two, calculated from measurements at 1 KHz is .9991.

Then there is some Nomex, about 3 wraps, and another winding. The k between one of the first layer windings, and the isolated winding on top of the Nomex is .9962.

I guess I've had my head up my butt on this issue, having spent too much time in conventional power conversion. The only time k ever showed signifigant effect was with pluggable, rfid or transcutaneous coupling (and even this was mostly on paper).

At 3 to 5KW, I tended to take the heating of adjacent hardware as the effect of leakage, rather than as the product of accidental coupling. Even there, amplitude errors in the main coupler wasn't a measurable or signifigant symptom.

RL

Wow! When Woodgate said "close to 1", he wasn't kidding! Thank you for that data, Phantom. I have some of those EC cores also -- somewhere. I think I'll hunt for them today.

Thanks Phantom, great stuff. Now let's say that the two windings are on two different cores, but different sizes and material and in a fixed position. Does any of the above change? Regards, Harry

Phantom, one problem! I measure my DUT for: Lpo, Lps, Lso, Lss, Rs, and Rp at the operating freq (130KHz) on a precision LCR meter. Then do the indicated calculations and make my SPICE model and run it. Simulations are right on. Then, being a doubting Thomas, I interchange secondary and primary readings and do indicated calculations, which are all different. Lm winds up on secondary instead of primary so I do the obvious T^2 trick to transform it to other side. Everything gets transformed into a perfect mirror image but the K factor (0.111) has not changed! Just kidding, this looks great. I will call it "Phantom's Reversible Coupling Factor". Highest Regards, Harry

If the core material had really low permeability, like molypermalloy with a perm of 30, or something like that, you might have watch out for that error that occurs when you measure the leakage inductance with one winding shorted.

Other than that, it should all work. As a general rule, you can *always* use the SQRT(L1/L2) formulation rather than N1/N2. You can even use it when the coils are tightly coupled, because it's the fundamental relation for coupled coils. The N1/N2 ratio is only an approximation, although a quite good approximation when L1 and L2 are tightly coupled.

There is another technique which I'll describe later tonight which I think may avoid the error caused by winding resistance..

Also, I want to do further analysis and see if it isn't possible to compensate for the error caused by the winding resistance.

As a matter of curiosity, what make and model is your LCR meter?

Are you telling me that you have a transformer with a coupling coefficient of only .111? No wonder you couldn't get the model in the App note to work!

Have you ever noticed that if you attempt to measure the turns ratio of a 60 Hz power transformer by applying a voltage to the primary and then measuring the open circuit voltage at the secondary, you don't get the reciprocal of the value you get when you do the measurement in the other direction? It's close for a 60 Hz power transformer, but if you try it with a loosely coupled transformer, it isn't even close. That's because the expression, V2 = V1 * N2 / N1 is only an approximation. The correct expression is V2 = V1 * k * SQRT(L2 / L1); the simpler expression is as good as it is because for a power transformer, k is usually very nearly 1, and N2/N1 is very nearly SQRT(L2 / L1). You could think of SQRT(L2 / L1) as a pseudo turns ratio.

Let ~N be SQRT(L1 / L2). The notation, V1' and V2' will mean that V1 or V2 is the excitation voltage, respectively. Then V2 = V1' * k / ~N, or k / ~N = V2/V1' = V12 (note that L1 was excited for this measurement). Also, V1 = V2' * k * ~N, or k * ~N = V1/V2' = V21 (L2 was excited for this measurement). The two quantities V1/V2' and V2/V1' are the result of open circuit measurements, and

*note well*, V2/V1' is *not* equal to 1/(V1/V2').

Let V12 be the result of applying excitation to L1 and dividing it into the open circuit voltage at L2 (that is, V2/V1'). Let V21 be the result of applying excitation to L2 and dividing it into the open circuit voltage at L1 (that is, V1/V2').

Thus, we have V12 = k / ~N, and V21 = k * ~N.

Finally, V21 * V12 = (k * ~N) * (k / ~N) = k^2

And, V21 / V12 = (k * ~N) / (k / ~N) = ~N^2

So here is a way to determine k and ~N with only open ciruit measurements, and the winding resistances shouldn't cause any error. We don't even need to be able to measure inductance to get k and ~N.

Let's see how it works out with the loosely coupled inductors I have been using as an example.

I applied .4639 VAC to L1, and measured .3893 VAC at L2; this gives V12 = .8392. I also applied 2.19 VAC to L2 and measured .3859 VAC at L1; this gives V21 = .1762.

Then V21 * V12 = k^2 = .14787, and SQRT(.14787) = .3845 = k.

And V21 / V12 = ~N^2 = .20997, and SQRT(.20997) = .458 = ~N.

Looking back and comparing to the results I got with the other method (measuring short circuit inductance), this method did quite well.

The winding resistances shouldn't have much effect on the results, but the flux distribution in the windings will be slightly different for different measurement frequencies, giving rise to slightly different values for k and ~N.

Harry, it would be most instructive if you have time to make these measurements on your transformer and compare results to those made with the other method.

I proof read all this several times, but Murphy has probably inserted several dumb mistakes.

Ph, It's a Quad Tech Model 1920 and gives AC resistance and inductance at the measuring frequency range of 20Hz to 1.0MHz. Nice feature when Rac/Rdc>>1.0 My windings are on different cores, separated by up to 1.00", hence 0.100" < K< 0.60" even if measured from either direction. Just to answer Don question, I have wound many transformers for power electronics and K

Hey Ph. will do it tomorrow, Got to go get my G+T, mostly gin and cook pizza on the Barbie. Harry

Phantom, By inspection the above will be true:

If: V2 = V1*K*SQRT(L2/L1) And: K= SQRT(V21*V12) And: N= SQRT(V21/V12) And: N= SQRT(L2/L1) Then: V2=V1*V21 which is obviously true and tell us to measure V21 and use it in place of N because K, L2 and L1 are not known. Here are my new measurements at 130KHz: V12= 0.0678/9.720 = 6.97mV V21= 4.920/3.870 = 1.27V and K= SQRT(1.27*6.97m)= .094 compared to 0.111 using inductor measurement method. N= SQRT(1.27/6.97m)= 13.5 compared to 9.4 using inductor measurement method. Close but the inductor method yields all parasitic values and can be used in a SPICE model. Also, Lpo, Lps, Lso, Lss, Rp and Rs can be easily measured at the operating frequency by a transformer fab house on one setup, a LCR meter. What are your thoughts? Harry

Hi Don

I just found this site so in this case am late in replying but can help if you want me to.

I have been designing switch mode power supplies since 1980. Am currently employed and have held top engineering positions including Senior Staff Scientist for Emerson Electric in 1980 thru 1989. In my earlier years, designed pulse transformers, now designing switch mode transformers, so am very experienced in calculating leakage inductance, ac conductor losses both skin and proximity effect and effective capacity of stepup HV transformers.

I can send you the appropriate formulas which are quite easy to use that will allow you to calculate leakage inductance referred to any winding (usually the driven primary winding). The only rules for the equations to valid for are: (1) all windings must be of the same traverse or length-of-winding. (2) the length of the windings must be at leat 30 times the thickness of the insulation between the windings. These conditions are what you want anyway to reduce the leakage figure.

If you wish I can phone you to discuss how its done and send the documentation to you. If you are in the USA I can call you for nothing since I use Vonage as my phone service. My email is snipped-for-privacy@comcast.net

Email me and I will call you back.

By the way the coupling coefficient for real practical switch mode transformers is typically about 1% of the primary open circuit inductance (OCL) or a figure of 0.99498 It has nothing to do with the OCL but is often close to the figure above. It is in fact the square root of the value (1-LLP/LP) where LLP is the leakage inductance referred to the primary winding.

Also a good choice for your forward converter, subject to the specification details, would be a forward diagonal topology which clamps the two switches to the DC line voltage and returns any stored leakage energy to the DC source capacitor.

Best Regards

Wendell Boucher

Hi Wendell, I think we all would be interested in your formulas, please publish to this group. Above you state that coupling coefficient (K) has nothing to do with OCL, then you state: K=SQRT(1-LLP/LP). But is not OCL the same as LP? Please explain. Harry

Harry, can you provide the following:

Inductance of L1 and L2 at both 1 KHz and 130 KHz

Resistance of L1 and L2 at both 1 KHz and 130 KHz

Inductance of L1 at 1 KHz and 130 KHz with L2 shorted.

Resistance of L1 at 1 KHz and 130 KHz with L2 shorted.

Inductance with L1 and L2 series aiding at both 1 KHz and 130 KHz.

Inductance with L1 and L2 series opposing at both 1 KHz and 130 KHz.

With all these numbers I should be able to give a full account of what's going on.

I'm guessing you have most of them already, so a few additional measurements should provide the rest.

In my work above, I defined ~N to be SQRT(L1/L2), not SQRT(L2/L1) which won't be consistent with the rest of the expressions.

I'm surprised you didn't get better agreement. Maybe it's the presence of a core. My little coupled inductor arrangement was air core. With a winding shorted, the core effect is different than for open-circuit measurements.

This last method will be mainly useful with 60 Hz power transformers where the series aiding and series opposing method to determine m (and subsequently, k) may suffer from numerical cancellation. The method is only for determining ~N and k anyway. If you need anything else, it's back to the LCR meter.

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