Problem statement

I was wondering whether it's possible to use a coupled inductor (e.g. this) as a signal transformer.

However, these coupled inductors typically have a tradeoff between the ability to store energy in the core (which would then become a vacuum gap, ideally), and the coupling of the coils (which would ideally make the thing an ideal transformer).

In the datasheets, only the inductance in series and parallel configuration are given; is it possible to calculate the Mutual Inductance, and more important, the Coupling Coefficient, from that?


So, if I'm not mistaken, if we unroll the series configuration, we get:


simulate this circuit – Schematic created using CircuitLab

$$L_\text{series} = L_1 + L_2 + 2M = 2(L+M)$$,

\$M\$ being the mutual inductance, which we see once for every side, hence the factor of 2, and assuming \$L_1=L_2=:L\$.

Solving for \$L\$ yields

$$L = \frac{L_\text{series}}2-M\text. \tag S\label S$$

In the parallel configuration:


simulate this circuit

We get opposing EMF; thus, M contributes negatively: \begin{align} L_\text{parallel} &= (L-M)||(L-M)\\ &= \frac1{\frac1{L-M}+\frac1{L-M}}\\ &= \frac1{\frac2{L-M}}\\ &=\frac{L-M}2 \end{align}

Solving for \$L\$ yields

$$L = 2L_\text{parallel}+M\text. \tag P\label P$$

Equating \$\eqref{S}\$ and \$\eqref{P}\$:

\begin{align} \frac{L_\text{series}}2-M &= 2L_\text{parallel}+M\\ \frac{L_\text{series}}2- 2L_\text{parallel}&= 2M\\ M&=\frac{L_\text{series}}4- L_\text{parallel} \end{align}

The coupling coefficient becomes

$$k = \frac{\sqrt{L_1L_2}}M \overset{L_1=L_2=L}= \frac LM$$


A sanity check yields near-zero \$M\$ for above linked datasheet, so that \$k\$ could become larger than one.

  • What am I doing wrong?
  • How to do it right (if possible)?
  • \$\begingroup\$ 3rd para mentions capacitance. Is this an error? I don’t see how you conclude you get opposing emf in the parallel combo given the dots shown. \$\endgroup\$ – Andy aka Jan 4 at 15:45
  • \$\begingroup\$ With two coils in parallel with M=L (as an example), the net end to end inductance is just L i.e. just as it would be for bifilar coils soldered to each other at the ends i.e. it behaves like a single coil of inductance L. And, if M=0 then you get L/2. \$\endgroup\$ – Andy aka Jan 4 at 15:55
  • \$\begingroup\$ I think your mistake is in calling it L minus M and not L+M. \$\endgroup\$ – Andy aka Jan 4 at 15:58
  • \$\begingroup\$ @Andyaka, yeah that capacitance is a mistake. Fixing that. \$\endgroup\$ – Marcus Müller Jan 4 at 17:32
  • \$\begingroup\$ I've redone my answer!! \$\endgroup\$ – Andy aka Jan 5 at 16:40

I'm beginning again with this answer because, like you, I fully expected to be able to derive leakage inductance from the values of parallel and series inductance. I'm answering it by considering that one winding has inductance \$L\$ in series with leakage inductance \$x\$. Inductance \$L\$ 100% couples to its partner inductance hence, M = L

When the two windings are in series and aiding, the total inductance is: -

$$L_{SERIES} = L + L + M + M + x + x = 4L + 2x$$

For the parallel arrangement you can (due to symmettry of the two windings), assume that the leakage inductances (\$x\$) are in parallel and the totally coupled inductances (\$L\$) are also in parallel and both these are connected in series. Like this: -

enter image description here

Hence the total inductance is: -

$$L_{PARALLEL} = (L + M) || (L + M) + x || x = L + \dfrac{x}{2}$$

Now, the ratio of series to parallel is always going to be 4:1 (for identical windings somewhat sharing the same core) hence, any difference between series and parallel inductance values reported in the data sheet are unrelated and therefore cannot be meaningful for understanding the coupling factor or M.


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