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I have an inductor which has three coils, L1, L2, L3. In normal use L2 and L3 are connected in series, and resonated with a capacitor. L1 would then drive the tank circuit.

In Spice, I need the coupling factor "K" which is found by measuring L1 with a second winding open and shorted, K=sqrt(1-(L2s/L2o))

Is it more accurate to take the K measurement for each winding independently against the others, or should I just series connect L2 and L3 and measure a single K for L1 vs the series combination of L2 and L3?

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  • \$\begingroup\$ What has SPICE got to do with this question? \$\endgroup\$
    – Andy aka
    Sep 23 '20 at 18:13
  • \$\begingroup\$ I need to create a spice model of the inductor, including the proper K value to work out the rest of the design. \$\endgroup\$
    – user103218
    Sep 23 '20 at 18:23
  • \$\begingroup\$ Does "K" imply LTspice? If so, the K statement applies this (one) coupling factor to all elements of the statement. Here is some info about it. \$\endgroup\$
    – rdtsc
    Sep 23 '20 at 18:55
  • \$\begingroup\$ For many cases, the simplifying assumption that all windings are equally coupled may be true. In this case it won't be. L1 will be less coupled to L2 and/or L3 than L2 is coupled to L3. \$\endgroup\$
    – user103218
    Sep 23 '20 at 18:59
  • \$\begingroup\$ I can probably assume that the K between L1 and either L2 or L3 will be the same, and that L2 to L3 will have a high K value. \$\endgroup\$
    – user103218
    Sep 23 '20 at 19:00
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Is it more accurate to take the K measurement for each winding independently against the others, or should I just series connect L2 and L3 and measure a single K for L1 vs the series combination of L2 and L3?

Choose one inductor as the reference and measure the others individually with respect to the chosen reference inductor. I say this because when there are more than two coils you cannot have true independence of coupling factors i.e. you can have L1 and L2 coupled at 0.9 AND L2 and L3 coupled at 0.8 but you can't then couple L3 to L1 with an independent coupling value because it has an inherent coupling (via L2) of 0.72 to L1.

So, choose your path a little carefully or it might unravel.


Having just read a bit on-line it seems that some simulators will not like the situation if there are not 3 coupling factors for 3 mutual coils and, additionally, those mutual coupling factors must make numerical sense. So, in the example above with L1, L2 and L3 you might need to add a coupling factor between L1 and L3.

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  • \$\begingroup\$ So from measurement, I have K values of .48 between the two windings of the split primary, and .70 between the secondary and either of the two primary windings, so it looks like my LTSpice coupling statements need to be: K1 L1 L3 .7, K2 L2 L3 .7 and K3 L1 L2 .48 Does that make sense? \$\endgroup\$
    – user103218
    Sep 30 '20 at 17:32
  • \$\begingroup\$ The windings are laid down as AAAAAA__CC__BBBBBB where A and B are the split primary and C is the secondary \$\endgroup\$
    – user103218
    Sep 30 '20 at 17:32
  • \$\begingroup\$ I have no idea because I don't know what inductors you are referring to. I would need a picture. I also have no idea what AAAAAA__CC__BBBBBB means. \$\endgroup\$
    – Andy aka
    Sep 30 '20 at 17:35
  • \$\begingroup\$ Physically, AAAAAA represents L2, CC is L1, BBBBBB is L3. There is a couple mm between the windings. These are all on a common ferrite core. \$\endgroup\$
    – user103218
    Sep 30 '20 at 21:09
  • \$\begingroup\$ Picture and schematic please. \$\endgroup\$
    – Andy aka
    Sep 30 '20 at 21:57
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LTSpice requires all possible coupling coefficients to be specified, and I strongly suspect most other flavours of Spice would as well.

As L2 and L3 are connected in series in normal use, you could justifiably regard them as being perfectly coupled. The K12 and K13 would therefore be equal.

It would simply then be a case of measure it as you use it. Measure the coupling from L1 to the series combination of L2 and L3.

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  • \$\begingroup\$ Measuring the actual K, L2 and L3 are coupled at about 0.4951, and L1 to L2 or L3 is about 0.635 in the actual inductor. \$\endgroup\$
    – user103218
    Oct 7 '20 at 17:06

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