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So I am trying to simulate a wireless power transfer via coupled coils. I am using the cantilever model (source: http://denverpels.org/Downloads/Denver_PELS_20070410_Hesterman_Magnetic_Coupling.pdf).

I am just having doubts about the current. From what I understand, the two inductances, L1 and L2 in the cantilever model are the leakage and magnetizing inductances which aren't physical inductances. In the figure I uploaded, I am simulating two coils that have: k = 0.3 (coupling) L1 = L2 = 7.64 uH (self-inductances)

Now, I just wanted to know, what is the actual current that goes through the transmitter coil (primary winding in this case)? Is it the current going through the red junction or blue junction?

In other words, what is the simulated current that I should use in order to properly choose the wire gauge if I wanted to coil (based on current)?

I believe that it should be the current going through the red junction since the said figure is the equivalent model of the coupled coil with some mutual inductance M, but I just wanted to make sure.

Cantilever Model

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  • \$\begingroup\$ Your lack of knowledge about leakage vs. magnetizing inductance vs. wire gauge is not going to draw much attention, +50 or not. Please re-word your question to be more precise and to the point. \$\endgroup\$
    – user105652
    Apr 25, 2018 at 5:58
  • \$\begingroup\$ what is SRF aand ESR? \$\endgroup\$ May 1, 2018 at 18:52
  • \$\begingroup\$ What is the max power that will force the cantilever into saturation? bang-bang break.... etc poorly defined problem (-1) insufficient research \$\endgroup\$ May 1, 2018 at 19:06

1 Answer 1

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From what I understand, the two inductances, L1 and L2 in the cantilever model are the leakage and magnetizing inductances which aren't physical inductances. In the figure I uploaded, I am simulating two coils that have: k = 0.3 (coupling) L1 = L2 = 7.64 uH (self-inductances)

Leakage is leakage and it doesn't couple with magnetizing inductance. That's the whole point about the leakage inductance - it is regarded as that fraction of the turns that don't produce a useful magnetic field that couples to the secondary.

In other words you don't use a "k" to couple them.

Next: you have leakage and magnetizing inductances in the wrong positions. Mag inductance should be in position L1 and leakage should be in position L2 with reference to your schematic.

So, for the remainder of my answer I'm calling L2 the leakage and L1 the magnetizing inductance.

With the secondary unloaded, the measured primary inductance is L2 + L1

With the secondary shorted (assuming you are not modelling secondary leakage seperately), the measured primary inductance will be L2.

Try and correct your understanding of this before asking the question about choosing the wire gauge based on the current.

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  • \$\begingroup\$ Sorry for the confusion. L1 is the magnetizing while L2 is the leakage... Just a faulty typing.. As for this: "In other words you don't use a "k" to couple them" What I am doing is that I am simulating the circuit for different values of k (different coupling), which corresponds to different values of L1 and L2 in the corresponding cantilever model (as well as 1:N, see the link). \$\endgroup\$
    – user139731
    Apr 16, 2018 at 9:53
  • \$\begingroup\$ L1 and L2 (if they are what you say they are do not couple). They are independant. You are misreading what it says in the linked document. \$\endgroup\$
    – Andy aka
    Apr 16, 2018 at 9:59
  • \$\begingroup\$ To clear things up, let's start calling L1 and L2 (in the figure above) as Lmag and Lleak respectively. They are independent with respect to each other, but the values of both are dependent on k. That's why when I am simulating a different value of k, I change both Lmag and Lleak. Both self-inductances are given and equal (7.64uH). Thus, whenever I vary k, I change Lmag and Lleak (as well as turns ratio, since the turns ratio in cantilever model is not the real turns ratio).. I am not saying Lmag and Lleak couples with each other.. \$\endgroup\$
    – user139731
    Apr 16, 2018 at 10:03
  • \$\begingroup\$ OK, what you are doing is altering k and this changes the relative values of L1 and L2. When k = 1 there is zero leakage and when k = 0 there is zero magnetization inductance. \$\endgroup\$
    – Andy aka
    Apr 16, 2018 at 10:15
  • \$\begingroup\$ yes.. If k = 1, then Lleak = 0, and Lmag = self-inductance of coil1 (or coil2 since coil1 = coil2).. If k = 0, Lmag = 0 and Lleak is maximum and equal to self-inductance of coil 1). \$\endgroup\$
    – user139731
    Apr 16, 2018 at 10:32

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