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Homework Help: Zin(s) of magnetically coupled inductors

  1. Nov 3, 2005 #1
    I am not quite getting how this works:
    i have a serially connected two inductors L1 and La (primary) and a secondary circuit with also two inductors (L2 and Lb), the problem asks to find input impedance in s-domain.
    The answer given is
    s[La + L1 - M2/(L2 + Lb)]
    so, could someone explain how i am supposed to get negative term?
    thanks in advance...
  2. jcsd
  3. Nov 3, 2005 #2


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    Staff: Mentor

    From the answer it looks like the secondary windings are shorted. That is, there is no explicit load impedance being driven by the secondary windings, and the far end of the winding L2 and the far end of Lb are just connected together with a wire.

    The secondary current causes a flux in the transformer magnetics that goes in the opposite direction of the flux that generated the secondary voltage. The lower the load impedance connected to the secondary, the higher the secondary current, so the higher the reverse flux is. In the extreme case of a shorted secondary (like in your homework problem), the reverse flux gets close to the forward flux, and the input impedance of the primary winding goes down to zero. The main mitigating effect is the quality of the magnetic coupling coefficient M. If M is high (near 1.0), you will see a short at the input to the primary. If M is lousy (like lots of leackage inductance), then that will keep the shorted secondary from making the input impedance of the primary look like a short.

    Remember, the input voltage on the primary generates a primary current, that current generates the forward magnetic flux in the core, the changing of that flux with respect to time is what generates a secondary voltage, and the current flowing through the output load of the secondary is what generates the reverse flux. The changing of the reverse flux with respect to time is what generates a reverse voltage at the primary, which increases the primary current. All these voltages, currents and fluxes have stable AC values where Faraday's Law (changing flux generates a voltage) and the Biot-Savart Law (a current generates a magnetic field and flux) are satisfied.

    Hope that helps. The negative sign comes from the direction of the reverse flux, and the resulting reverse voltage that increases the input current based on the output loading.
  4. Nov 3, 2005 #3
    thanks much for reply....and yes, you understood the circuit correctly.

    dumb question #1: if Zin(s) = Z(primary) + Z(secondary) why does Z(secondary) part in the answer look like L2||Lb? (yeah... there's just two inductors connected together through a wire)

    dumb question #2: why is it true that L2Lb = M2?

    thanks again
  5. Nov 4, 2005 #4


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    Staff: Mentor

    Those are good questions, not dumb questions. A big part of technical college work is learning to ask the right questions (of yourself, your profs, and your tutors).

    -1- The secondary load impedance transforms into a parallel impedance, in parallel with the input impedance of the transformer. In the simple equations, the input impedance of the transformer is the input magnetizing inductance in parallel with the load impedance (scaled with the turns ratio squared). The next better model is to include the primary leakage inductance before the input magnetizing inductance and transformed secondary impedance. The next better model includes the coupling ratio M and the output secondary leakage inductance before the parallel transformed load impedance on the secondary side.

    -2- To be honest, the squared stuff versus linear terms stuff -- I always have to check my cribsheet or re-derive it. It's not intuitive to me, unfortunately. If you set up the simultaneous equations that I mentioned in my earlier post and play with them some, you might get a more intuitive feel than I have so far.

    This is good stuff, Evler. Transformer modeling is something that you may use every day as a working engineer, depending on what you choose as your specialty coming out of school. Every 10Base-T and 100Base-T connection in the world uses a comm transformer phy....
  6. Nov 4, 2005 #5
    Thanks a lot, berkeman!
    actually in class they told us to "take it on faith" that M2 = LbL2, i.e. it is covered in Electromagnetics class which we are not...
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