Good Morning All

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I have a DC link Choke ( Common EI Core), When measuring through LCR meter It gives me reading (see table.). When Adding Inductor in Series ( Pic 3 and Pic 4) it Give the value that is not expected. I am referring formula to calculate inductor in series Leq=L1+L2+2Sqrt(L1*L2). Don't Know why LCR meter is showing different.

Value shown on image is measured form LCR meter. Measured on both 100Hz and 1Khz.

Pic1 :- Measure L1 Inductance using LCR Pic2:- Measure L2 Inductance using LCR Pic 3:- Measure Leq Inductance using LCR ( L1 and L2 in series and in phase ) Pic 4:- Measure Leq Inductance using LCR ( L1 and L2 in series and in out phase )

  • \$\begingroup\$ Are these two coils on the same core? Or really two separate inductances? \$\endgroup\$
    – jcoppens
    Jul 9, 2015 at 5:05
  • \$\begingroup\$ Measure L1 (as in figure 1) but with a short circuit across L2. If Andy's guess is right, you'll see about 300 uH indicating very poor coupling. \$\endgroup\$
    – user16324
    Jul 9, 2015 at 10:07
  • \$\begingroup\$ Yes Inductor 1 and 2 are on same Core. \$\endgroup\$
    – Electra
    Jul 10, 2015 at 5:53

1 Answer 1


Firstly, there is an error in the dot notation - scenario 4 provides the highest inductance yet the dot notation implies that if the two inductors were perfectly coupled, the net inductance would be 0. Because scenario 4 gives the highest value of inductance it can be concluded that it really has the dot notation of scenario 3.

Secondly, you have not considered that the two inductors may not be 100% coupled.

Next is to work out the coupling and a bit of math in my head tells me it's about 70%. Individually each winding has about 600 uH and in series aiding this rises to about 1800uH. If the two windings were 100% coupled they would produce a total inductance of 2400uH when connected in series.

So if 70% of each winding is perfectly coupled then the total inductance is: -

(4 x 0.7 x 600 uH) + (2 x 0.3 x 600 uH) = 2040 uH. OK my head-guess was a little optimistic on coupling. 50% coupling realizes an aiding inductance of 1800 uH.

When put series opposing, 50% of the coupled inductance totally cancels leaving a net inductance of about 2 x 300 uH.

Near enough.

EDIT to explain my math

The standard formula for coupled inductors is: -

\$L_{EQ} = L_1 + L_2 + 2k\sqrt{L_1L_2}\$ and, when both inductors are the same value this results in: -

\$L_{EQ} = L + L + 2kL\$ and, when k=1 (100% coupling), equals 4L

If a fraction (70%) of L1 is 100% coupled to L2, the fraction produces an inductance of 0.7 X 4 L.

The remaining uncoupled parts of L1 and L2 do not interact and are just additive i.e. (1-0.7) X 2 L.

Hope this makes sense.

  • \$\begingroup\$ Thanks Andy for explanation. With 70% Coupling 600uH+600uH+ 0.7*2*sqrt(600uH*600uH)=2040uH. (I used formula L1+L2+K*2*sqrt(L1*L2), K is coupling factor. But the formula you used (4*07.*600uh)+(2*0.3*600uH)=2040uH. What is this formula and why multiply by 4 in fist half and 2 with second half. \$\endgroup\$
    – Electra
    Jul 10, 2015 at 6:31
  • \$\begingroup\$ @electra the 4 is for the coupled fraction of two coils and the 2 is for the uncoupled fraction. I could have used the conventional formula and got the same answer. \$\endgroup\$
    – Andy aka
    Jul 10, 2015 at 7:47
  • \$\begingroup\$ @electra - added a better explanation to my answer. \$\endgroup\$
    – Andy aka
    Jul 10, 2015 at 8:17

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