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My goal is to program a FDM simulation of a transformer. So, we have the coupled inductor model to start with ... mutual inductor transformer model

However, if L1 = 1, L2 = 9, and coupling is perfect then M = sqrt(1*9) = 3 and L1 - M = -2.

So, this seems problematical to me. Also, if I try to simulate this model using SPICE, just for fun you understand, my goal is to program the sim myself, SPICE chokes on the negative inductor value and no output signal is generated.

I note that the model above does produce the right differential equations, that is, v1 = L1*i1' + M*i2' v2 = L2*i2' + M*i1'

However, if you try to simulate these differential equations directly, and the coupling is perfect, the the ratio L1 / M = M / L2 = sqrt(L1) / sqrt(L2) and you cannot solve them for i1' and i2'. So, I'm stuck starting with these differential equations too.

One more thing - running the SPICE transformer model, i.e. specifying L1 = 1, L2 = 9, and k = 1, with a 1 v. 60 Hz input produces the expected 3 v. output. However, changing k to .9 reduces the output to a few mv. Even .99 and its still a few mv. That seems very peculiar to me (I'm not an EE).

Any assistance would be appreciated.

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If the coupling is perfect then M = L1 = L2 i.e. the model you are using has coupling inbuilt into the formulas so you can't mathematically do what you are wanting to do.

L1 is the total inductance looking into the primary and this equals magnetization inductance + leakage inductance so M cannot be bigger than L1 and therefore L1-M cannot be negative.

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  • \$\begingroup\$ L1 = primary self-inductance, L2 = secondary self-inductance, M = k * sqrt(L1 * L2) where k is the coupling coefficient. \$\endgroup\$ – DrWill Mar 21 '16 at 16:04
  • \$\begingroup\$ When coupling is perfect on a 1:1 transformer, M=L1=L2 because of the formula you stated. You cannot suddenly break the rules due to having a non-unity turns ratio. L2 is only valid when transcribed to the primary side via the turns ratio squared. \$\endgroup\$ – Andy aka Mar 21 '16 at 17:31
  • \$\begingroup\$ With the parameters given, it's not a 1:1 transformer, it should be a 1:3 voltage transformer with perfect coupling. And that's what I see using SPICE. However, when k = .9 instead of 1, the behavior completely changes, which completely baffles me. \$\endgroup\$ – DrWill Mar 21 '16 at 17:33
  • \$\begingroup\$ You have to take the 9 henry inductance and divide it by the turns ratio squared (9) in order to make sense of the model you use. The "T" model is for a 1:1 transformer and cannot model any other ratio. \$\endgroup\$ – Andy aka Mar 21 '16 at 17:37

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