Author Topic: Find primary mutual inductance given secondaries with different coupling  (Read 1267 times)

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Offline 741Topic starter

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Hi

For a single secondary, this is my understanding:
   L1=Primary
   L2=secondary
   Coupling = K <= 1

To model this explicitly I'd do this:
   L1.Mutual = K.L1
   L1.Leakage = (1-K)L1
   L2.Mutual = K.L2
   L2.Leakage = (1-K)L2
See also https://www.eevblog.com/forum/projects/what-is-a-good-approach-for-estimating-far-side-impedances-across-a-transfomer/   

Now, for simplicity let all inductors have 1uH self-inductance.
   .param K_12          = 0.4
   .param K_13          = 0.3
   .param K_23          = 0.1

Question: What is the mutual inductance at L1? It wouod seem to be a function of both L2 and L3.
Taking them separately, then 'at L1' I'd "hazard a guess" of (0.4uH | 0.3uH) but then I never used K_23 here.
 

I can enter this into LTSpice and the simulation runs. I might eventually infer the rule this way I suppose.
I'd like to understand this better, both intuitively/physically and analytically.

See also
   https://groups.io/g/LTspice/topic/three_winding_transformer/50189668
   
BTW, on one site I saw the rule "-1<= K <=1". If this -ve coupling valid? Never seen it anywhere else.

Offline mtwieg

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Re: Find primary mutual inductance given secondaries with different coupling
« Reply #1 on: January 29, 2026, 01:26:30 pm »
Hi

For a single secondary, this is my understanding:
   L1=Primary
   L2=secondary
   Coupling = K <= 1

To model this explicitly I'd do this:
   L1.Mutual = K.L1
   L1.Leakage = (1-K)L1
   L2.Mutual = K.L2
   L2.Leakage = (1-K)L2
In a coupled inductor model, there's also an ideal transformer with some turns ratio. Without knowing that turns ratio, and how these other inductances are arranged around it, can't say whether they're correct or not.

Except I can say with certainty that including the mutual/magnetizing inductance twice in the circuit model is incorrect.

Quote
Now, for simplicity let all inductors have 1uH self-inductance.
   .param K_12          = 0.4
   .param K_13          = 0.3
   .param K_23          = 0.1

Question: What is the mutual inductance at L1? It wouod seem to be a function of both L2 and L3.
There is no "mutual inductance of L1". L1 has some mutual inductance to L2 and some mutual inductance to L3.

Quote
I can enter this into LTSpice and the simulation runs. I might eventually infer the rule this way I suppose.
I'd like to understand this better, both intuitively/physically and analytically.

See also
   https://groups.io/g/LTspice/topic/three_winding_transformer/50189668
When dealing with more than two windings, making equivalent circuit models out of individual inductances and ideal transformers becomes extremely messy, often impractical. A more tidy way to express the model is with an inductance matrix. See attached images (can't get it to display correct in the post):
2741217-0

L is the inductance matrix. The diagonal terms L11, L22, etc are the self-inductances of each coil (their inductances measured with all other coil open-circuit). The off-diagonal terms are mutual inductances. By nature, the matrix is symmetric (i.e. L12 = L21, L13 = L31, etc). The L matrix contains everything necessary to define the behavior of the coupled coils (so long as they only consist of inductances, that is).

Often, the inductance matrix is represented a little differently, with self-inductances represented as L1, L2, etc and mutual inductances represented as M12, M13, etc. But the values are the same:
2741247-1

You could also represent all the coupling factors with a K matrix, defined as:
2741209-2

Again, K is always symmetric (i.e. K12 = K21, K13 = K31, etc).

These matrix representation are extensible to any arbitrary number of windings.

In LTspice, you can implement these circuits like this (here, one side of each winding is connected to global ground):
Quote
.param L11 = 1u
.param L22 = 1u
.param L33 = 1u
.param K12 = 0.4
.param K13 = 0.3
.param K23 = 0.1
L1 n001 0 {L11}
L2 n002 0 {L22}
L3 n003 0 {L33}
K12 L1 L2 {K12}
K13 L1 L3 {K13}
K23 L2 L3 {K23}
Quote
BTW, on one site I saw the rule "-1<= K <=1". If this -ve coupling valid? Never seen it anywhere else.
Yes, K values cannot have magnitude greater than unity. This is basically saying that mutual induction can't be greater than self-induction. But mutual inductance can have negative sign, so -1 <= K <= 1 is a valid range.
« Last Edit: January 29, 2026, 01:52:27 pm by mtwieg »
 
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Offline 741Topic starter

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Re: Find primary mutual inductance given secondaries with different coupling
« Reply #2 on: January 30, 2026, 11:52:01 am »
Thanks for your repy, I will try to understand it better in due course.  :)

Quote
There is no "mutual inductance of L1". L1 has some mutual inductance to L2
Yeah, I was aware something was wrong with that statement, but could not think of the right way to say it at the time.

For the -ve coupling: Is this related to the sense the coils are wound WRT one another? I am still unsure about it.
« Last Edit: January 30, 2026, 11:56:46 am by 741 »
 

Offline mtwieg

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Re: Find primary mutual inductance given secondaries with different coupling
« Reply #3 on: January 30, 2026, 01:30:24 pm »
For the -ve coupling: Is this related to the sense the coils are wound WRT one another? I am still unsure about it.
Not sure what you mean by "-ve coupling".

When coils are coupled, you need to define a polarity for each winding (these polarities can be chosen arbitrarily). If K12 is positive, then putting positive voltage on L1 creates positive voltage on L2. If K12 is negative, you get negative voltage on L2. But you could also choose to define L2 with opposite polarity and also flip the sign of K12, and the model would still be correct.

 


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