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I have an EE core with one winding on one outer leg and a second winding on the other outer leg. I am measuring the inductance on one winding by having the other winding open and vice versa.

For the mutual inductance I am measuring the aiding series inductance (L_aiding = L_1 + L_2 + 2M) and opposing series inductance (L_opposing = L_1 + L_2 - 2M) and substitute these two to end up with M = 1/4 * (L_aiding - L_opposing) which gives me the mutual inductance.

Now I can calculate the coupling factor using the following equation: K = M / sqrt(L_1 * L_2).

The problem is with no airgap I get a coupling of around 0.2 and with a 0.4mm airgap I get a coupling of 0.32, which doesn’t make any sense, since an airgap would give me more leakage inductance hence a lower coupling coefficient.

What could I be doing wrong?

ocrdu
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James W.
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  • 1.As air gap increases coefficient of coupling decrease but not the reason you mentioned, leakage inductance in transformer is independent of air gap, coupling factor decrease because K^2= M^2/{(L1leakage +L1) (L2leakage +L2) } since M, L1, L2 all proportional to permeability of core but you can see from formula that decrease in M^2 will be more compare to denominator (due to L1 and L2) hence overall K decreases – user215805 Nov 22 '21 at 18:03

3 Answers3

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What could I be doing wrong?

When the centre limb of the EE core is acting as a "short" route for the driven magnetic field, most of the flux bypasses the undriven coil and hence the coupling appears to be low. As you introduce a gap between EE halves, the "short" route becomes less viable more immediately than the route via the secondary coil. Both routes are of course less viable but, relative to each other, their respective reluctances get closer. This means that coupling will tend to increase as you add a small gap.

What you see is how I would expect it to be.

Andy aka
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  • Now this is on a coupled inductor these measurements, but assume this was for a transformer where coupling should be near 1, how does the transformer then have a good coupling without any airgap? – James W. Nov 22 '21 at 15:49
  • It's quite simple, don't use a core that can short circuit the magnetic path. An E core will do that unless both winding are on the same limb and, you say they aren't so, this makes a bad choice of magnetic component or inappropriate positions of coils – Andy aka Nov 22 '21 at 17:31
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    I see, my measurements makes perfectly sense then. I was designing a coupled inductor and wanted the coupling to be 0.33 to get the desired leakage inductance. Then I became curious about different coupling factors at different airgaps, hence my frustration. But thanks for your answer, really helpfull. – James W. Nov 22 '21 at 17:50
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You can determine the coupling coefficient via leakage inductance. $$ K = {\sqrt{{L_P - L_{lP}}\over {L_P}}}$$ Where:
\$ L_P =\$ primary inductance with secondary open
\$ L_{lP} =\$ primary inductance with secondary shorted (leakage inductance)

This equation holds if the windings are lossless. You can also measure K from the secondary side.

If you introduce loss, then the coupling coefficient becomes: $$ Actual\_K = Apparent\_Measured\_K \sqrt{{1+Q^2}\over {Q^2}} $$ If the winding Q (measured when the secondary is shorted) is higher than 4, you'll have less than 3% error between actual and measured K. This says that you should perform your K measurement at a high enough frequency where the Q will be high enough to reduce error between measured and actual K, but below resonance.

See if this method gives better results.

qrk
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Alternatively connect one winding to the output of a signal generator (sine wave; at around the frequency of interest), and measure the voltage at the terminals of each winding with an oscilloscope.

If the losses are small, both voltages will be in phase, and the coupling coefficient is the ratio of the voltages times the ratio of the number of turns on each winding.

jp314
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