Transformer coupling coefficients

Ah! NOW you have! That's exactly what I was after. .998 is considerably more informative than "close to 1". Thanks!

I'm not exactly just beginning to understand transformers, I was just looking for a "typical number", like John's .998, from an experienced practicioner.

All that mishmash in the appnote is an attempt to boil things down to pat turn-crank formulae. Ugh! It is a LOT easier to deal with and understand transformers by using matrix notation and MathCAD. This is particularly true when dealing with multiply-wound loosely-coupled xfmrs as rod-core ignition transformers where k's of 0.7 and less are not only typical but desirable to produce the desired high-output-Z transfer function. See, e.g., US patents 5,521,444 and 6,191,956.

Fundamental physical terms - ie a mathematical model.

RL

Yup. For every pair of windings i and j, there is a k sub ij and a k sub ji. These are not necessarily equal. This can be seen either with bench tests or in a FEMM simulation.

In the simplified expression for mutual inductance derived from the general matrix equations, M = k * sqrt (L1*L2), that k is the geometric mean of k sub ji and k sub ij for that pair of windings. SPICE uses that k. The geometric mean happens when the system of equations is solved for a particular value.

On Fri, 01 Sep 2006 12:50:52 -0500, Don Foreman

In a linear, passive, bilateral, time-invariant network of several coupled inductances, L1, L2,...,Ln we can write down an impedance matrix for the network, representing the self inductance of each inductor as L11, L22, L33,...,Lnn and the mutual inductances as L12, L21; L13, L31; L23, L32; Lij, Lji; etc. It is an experimental fact discovered by Faraday that Lij = Lji, and this fact can also be demonstrated by energy considerations. I've never seen k defined in such a way as to depend on which inductor is being excited, but if it were then presumably the definitions for a pair of inductors would be kij = Lij/SQRT(Lii*Ljj) and kji = Lji/SQRT(Lii*Ljj). Since Lij = Lji, then kij = kji. This is also evidenced by the application of the reciprocity theorem to the impedance matrix for the network.

I suppose that real-world lab measurements may give slightly different results for kij and kji due to measurement error and n "I believe that the coefficient k is an abstraction introduced to remind the simulator user that something is missing in the model.

It has no physical derivation."

A simple physical interpretation of coupling coefficient would be something like this:

Imagine you have a pair of inductors wound with an equal number of turns in each winding, and with superconducting wire so that IR drops don't figure in; ignore parasitic capacitance, etc., etc. Apply an AC voltage to the first inductor and measure the induced voltage at the terminals of the second inductor. If the inductors were perfectly coupled, the voltage at the second would be equal to the voltage applied to the first. If the inductors are not perfectly coupled the voltage at the second inductor will be some fraction (less than unity) of the voltage applied to the first inductor. That fraction *is* the coupling coefficient; hence its name. It is the ratio of flux linkages. The first inductor links all of its own flux, but the second inductor only links a fraction of the flux produced by the first inductor; that fraction is k.

If the inductors are wound with different numbers of turns, just make the obvious adjustment to the voltage expected at the terminals of the second inductor.

In message , dated Fri, 1 Sep 2006, The Phantom writes

I'm not sure about that. Consider that leakage inductance in a two-winding transformer is a result of flux from one winding not linking with the other winding. Now think of a rectangular core with two 'legs'. On one is a coil which extends over the full width available to it, but the winding depth is just one layer. Very nearly all of the flux is in the core. Now consider the other winding to be very narrow, but with many layers. The pattern of flux generated by this winding is very different from that generated by the first winding. Can we be sure that the non-linking flux is the same in both cases?

For one winding energized, it is certainly true that the two k's are equal, because ether is only a single flux pattern that causes both.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

I should love to ignore IR drops, stray capacty, core loss, leakage, source impedances, fringing, crowding, eddying, self-resonance and the like. It would really be quite interesting to see what was left in transformer action.

It's no wonder I think of 'k' as an abstraction, if all that it identifies is an amplitude error caused by factors that are not being conveniently ignored..

As a Doubting Thomas, I expect that this factor wouldn't repeat, given the simplest turns ratio adjustment, when scaled, unless another factor was identified for removal.

RL

Circular argument. You redefined k in terms of mutual inductance. I said above that this k will be the geometric mean of kij and kji, so by (your) definition kij = kji =k. If kji is the percentage of flux produced by winding i that links winding j (as originally defined), kij and kji need not and in general are not identical.

The way to measure these experimentally is not by inductance measurements but rather by open-circuit voltage measurements corrected for turns ratio. V2 = (N2/N1) * k * V1 because V2 = N2 dphi2/dt

Depends on how you define k. If it is defined in terms of flux linkages, there can be and in general are differences between kji and kij.

You're having a little problem with consistent definition of k here.

Regarding flux linkages in a transformer as an abstraction is an interesting perspective!

I have a long single layer coil of 180 turns of 26 gauge wire on a porcelain cylinder, .75" in diameter. I wound about 120 additional turns as a fairly compact bundle near the middle. That's fairly close to what you are describing. I posted a picture of it over on abse.

I denote the single layer coil as L1 and call it the primary, and the compact bundle as L2, the secondary. The measurements are as follows:

Self inductance of L1 = 93.6 uH

Self inductance of L2 = 450 uH

Inductance of L1 + L2, connected series aiding = 704 uH

Inductance of L1 + L2, connected series opposing = 385 uH

From this we can determine the mutual inductance to be

(704-385)/4 = 79.75 uH

and the coefficient of coupling to be

k = 79.75/SQRT(93.6 * 450) = .3886

When coils are as loosely coupled as this, it isn't appropriate to use turns ratio to determine the voltage induced in a second winding by the voltage applied to a first winding. The appropriate ratio is the square root of the ratio of the primary self inductance to the secondary self inductance. Note that for coils wound on ungapped ferromagnetic cores (regular old transformers, in other words), this ratio is very nearly the same as the turns ratio. I'm going to call this ratio ~N because it's analogous to a transformer turns ratio. From the measurements above, ~N = .456

So, if the flux generated by applying V1 volts to the primary (L1) were completely linked by the secondary, we would expect to measure V1/~N volts at the secondary. But since not all the primary flux is linked by the secondary, we need to include the coupling coefficient in the expression. Thus, the voltage we should expect to measure at the secondary would be (k / ~N) * V1 volts. If we apply V2 volts to the secondary (L2), we should expect to measure (k * ~N) * V2 volts at the primary.

In the case of the coil assembly whose measurements are given above, we have:

For V1 applied to L1, the voltage V2 at L2 should be V2 = (k / ~N) * V1 = .8522 * V1

For V2 applied to L2, the voltage V1 at L1 should be V1 = (k * ~N) * V2 = .1772 * V2

Now for the measured 25 KHz voltages:

With .4639 volts applied to L1, I measured .3893 volts at L2; I should have expected to measure .8522 * .4639 = .3953 volts.

With 2.19 volts applied to L2, I measured .3859 volts at L1; I should have expected to measure .1772 * 2.19 = .3881 volts.

These measurements don't really justify 4 digits; at 25 KHz with wires draped over the bench, readings are a little jumpy, but those were the numbers I wrote down.

It's pretty clear that k is the same in both directions, even for these two coils where the flux distributions are very different depending on which winding is energized.

This is the rather counterintuitive fact which Faraday discovered.

I should have mentioned in my first post in this thread that these relationships I've been discussing are voltage transfer ratios. Reciprocity only applies as well as it does in this hardware example if the inductances have rather high Q. I implied it when I said that the example would use superconducting wire.

The reciprocity theorem "...kij and kji need not and in general are not identical.

The way to measure these experimentally is not by inductance measurements but rather by open-circuit voltage measurements corrected for turns ratio. V2 = (N2/N1) * k * V1 because V2 = N2 dphi2/dt"

Actually, this is not the way to measure k experimentally. One would not expect the open circuit voltage transfer ratios to be equal; that's not what the reciprocity theorem says. Perhaps this is why Don thinks he has experimental evidence that k is different in the two directions; he measures different voltage transfer ratios in the two directions and believes this means that k is different in the two directions.

"V2 = (N2/N1) * k * V1" is not the fundamental relationship, but rather V2 = SQRT(L2/L1) * k * V1 as I explain above. The N2/N1 ratio is an approximation which is only acceptably accurate for closely coupled inductors such as found in transformers.

Note that for the (loosely coupled) transformer experiment described above, we would expect the voltage V2 to be (N2/N1) * k * V1 = 120/180 * .3886 * .4639 = .12 volts if it were true that V2 = (N2/N1) * k * V1. But the measured voltage V2 was .3893 volts, considerably different. The expression V2 = SQRT(L2/L1) * k * V1 gives the correct result.

The DC resistance and the skin and proximity effect (equivalent) resistances in the windings will generally cause unequal (adjusted for turns ratio if closely coupled, otherwise adjusted for square root of self-inductance ratio) voltage transfer ratios.

You can see why the voltage transfer ratio isn't necessarily the same (adjusted for turns ratio) in both directions by a simple thought experiment. Imagine you have a two winding transformer on a low loss ferromagnetic core, with the two (equal turns) windings closely coupled (bifilar wound). Since the self-inductances will be finite, there will be some magnetizing current; let's say that for an applied voltage of 100 volts, the magnetizing current is .1 amps (a nearly pure reactance of 1000 ohms), for either winding. If we apply the 100 volt excitation to the primary of this transformer, we will measure nearly 100 volts at the secondary. And if we apply 100 volts to the secondary, we will measure nearly 100 volts at the primary; the same voltage transfer ratio in both directions. Now put a 1000 ohm resistor in series with the primary (and consider it part of the primary), and apply 141 volts to the combination. We will measure about 100 volts at the secondary, so we might think the coupling coefficient is about .707. But if we apply 100 volts to the secondary, we will measure nearly 100 volts at the primary (with the 1000 ohm resistor in series), because our measurement is made open circuit and no current passes through the 1000 ohm resistor, and we would think the coupling coefficient is very near 1.

The resistor has caused the voltage transfer ratio to be different in the two directions. If the external 1000 ohm resistor bothers you, just imagine the primary to be wound with resistance wire of 1000 ohms total DC resistance, and you get the same result. This is how (probably different) losses in the different windings can cause the voltage transfer ratio to be different in the two directions.

But if we apply 100 volts to the primary and short the secondary, the secondary current will be .1 amp. If we apply 100 volts to the secondary, the short-circuit primary current will be .1 amp. The transfer

*impedances* are equal as the reciprocity theorem requires.

What is left is most of what a transformer does. All of those things are generally not necessary to consider for a first cut at a transformer design. It's up to the designer to know when to take them into account.

When you analyze a circuit made up of R, L and C, don't you ignore the L and C in the R, the R and L in the C, and the R and C in the L? Even though you know that each element is, in the real world, contaminated by the other two, don't you do a lot of analysis of your schematic assuming most of them are pure?

In the example I gave, the coupling coefficient would behave just as I described. And it is a necessary component of the expression (formulated without mutual inductance, as may be convenient) for the behavior of coupled inductors, as I explain in my long post in reply to John Woodgate.

Can you give us a citation of a book (with page number) or other authority that explains how it is that kij # kji?

Dear Phantom, My questions below pertain to two windings, loosely coupled, K

"Electric Machinery" Fitzgerald & Kingsley, 1961, McGraw Hill, page

26.

Kingsley... what a horse's ass... just about the worst professor I ever had.

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

| 1962 | I love to cook with wine. Sometimes I even put it in the food.

I won't go into all the details; any circuit theory text should give them. What's going on is that when the two windings are connected series aiding, the measured inductance is L1 + L2 + 2*m, and when they're connected series opposing, the measured inductance is L1 + L2 - 2*m. So, when we subtract, we get (L1 + L2 + 2*m)-(L1 + L2 - 2*m) = 4*m. Notice that when the coils are closely coupled, m will be small, and using this method will require us to subtract two nearly identical numbers (because the L's are much bigger than m), causing severe numerical cancellation. This will happen when k is some number like .998, very close to 1. For coils not so tightly coupled, the method works ok.

Yes. When you measure the inductance of one winding of a two winding transformer with the other winding shorted, you are measuring the leakage inductance of the primary plus the leakage inductance of the secondary referred to the primary. One caveat is that the frequency used to measure the inductance should be high enough that the measured reactance dominates the value. In other words, the imaginary part of the impedance should be much greater than the real part.

Consider a two coil transformer having a primary inductance of L1, with resistance R1, and a secondary having inductance of L2, with resistance R2. If we set R1 and R2 equal to zero and calculate the impedance seen at the primary with the secondary shorted, we get (s is the complex frequency variable, jw):

Zin = s*(L1 - m^2/L2) = s*[L1 * (1-k^2)]

If R1 and R2 are not zero, the impedance is more complicated:

s^2*(L1*L2 - m^2) + s*(R1*L2 + R2*L1) + R1*R2 Zin = --------------------------------------------- s*L2 + R2

In each case, to get the equivalent inductance, divide by s. In the first case, the calculated inductance is L1*(1-k^2). In the second case, it's more complicated. But, the thing to note is that the presence of non-zero resistances changes the apparent (leakage) inductance measured with the secondary shorted. For the two coil inductor for which I gave all the measurements in the earlier post, if we calculate (with the frequency f set to 1000 Hz) the inductance using the complicated formula with R1 =

1.858 ohms and R2 = 2.256 ohms (these are the measured DC values), we get 84.96 uH. But, if R1 and R2 are set to zero, we get 79.46 uH, the same as the simple formula.

Now set the frequency to 10,000 Hz and re-evaluate the formulas. With R1 and R2 set to zero, the result is the same, 79.46 uH. But with their non-zero values, the calculated apparent inductance is 79.55 uH. If you increase the frequency to 100,000 Hz, the value with non-zero resistance gets even closer to the value with zero resistance.

I measured the inductance of L1 with L2 shorted on a GenRad Digibridge using a test frequency of 1000 Hz and got 85.0 uH, mighty close to the calculated value of 84.96 from the paragraph above. Theory triumphs again!, as my former EE professor Don Reynolds used to say.

The lesson here is to measure the inductance of a transformer's primary with the secondary shorted at a frequency near the operating frequency. This is especially important for transformers with ferrite cores, because ferrite can have a fairly low permeability, giving a winding reactance not enough higher than the resistance to avoid error. Most low cost meters use a 1000 Hz (120 Hz is also sometimes available; do not use this frequency with a ferrite cored transformer!), and may not give good results. This may be why you aren't getting good results with shorted winding measurements.

(Try this: Get a 60 Hz (or 50 Hz) power transformer and connect a 25 ohm rheostat across the secondary. Measure the primary inductance with a meter using a test frequency of 1000 Hz. Start with the rheostat set to zero ohms (thereby providing a short on the secondary), and mote the indicated inductance. Now increase the resistance of the rheostat and watch the indicated inductance change. How can just adding resistance to the secondary cause the *inductance* to change so much? The answer is in the complicated formula above. If you can, change the test frequency to a higher one, and you will see that the indicated inductance doesn't change as much as you increase the rheostat's resistance.)

It should be possible to manipulate the network algebra to get an expression for the true leakage inductance, corrected for the error caused by the non-zero winding resistances when the measurement is made with a low test frequency. I'll look at this possibility later.

And, of course, as shown in step 4 of the AN1679 procedure, you can compute k from a good short-circuit measurement.

In the App note, the author uses a symbol, Lps, apparently standing for open circuit inductance, primary side. I will use Lss for the secondary side open circuit inductance.

In step 1, you must change the computation for N. Instead of computing N = Np/Ns, compute N (in my long post I called this value ~N, because it's different from the turns ratio when k is substantially less than 1) as N = SQRT(Lps/Lss). Use this value for N in later calculations. This is the change that will give you good results when the coupling is loose. You will have to *measure* Lss because you can't calculate it from measurements made only on the primary side.

In step 8, he uses the DC values for Rp and Rs. I suspect that at the high frequencies typically used for flyback operation, these values will be too low.

Finally, he made a major boo-boo in Figure 8. The resistance Rp should be in series with LI1, not Lm. There should be another resistance in parallel with Lm to represent the core loss.

Did you ever take a class from Guillemin? In his book, "Introductory Circuit Theory" on page 380-381, he discusses coupling coefficients, and explicitly says L12=L21 in an example. Nowhere does he suggest that k is different in different directions. I am unable to find any reference to that effect.

Nope.

I was in course 6B... honors EE.

We had Harry B. Lee for network theory.

He was fabulous... taught from his own hand-outs... didn't use Guillemin's book.

That's why I'm still so good at nodal and loop analysis after all these years.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

If they were different, wouldn't you have invented a free energy machine?