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No phase shift why?
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Cschmidt 8 years ago
I'm expecting a voltage phase shift on the Vout here. Inductors of inverse polarity (L3) should cause a phase shift from the Vin. [No phase shift why?][1] [1]: https://easyeda.com/editor#id=WExrlDVdv
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andyfierman 8 years ago
With only `k12` and `k13`, your inductor coupling is unrealisable. You need to add `k23 L2 L3 1`. Please also see https://easyeda.com/example/Transformers_and_coupled_inductors-LWewOI0ic
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andyfierman 8 years ago
In the primary side of your circuit, you have an inductor of 1mH. In the secondary side of your circuit you have a 4mH inductor in phase with the primary in series with a 2mH inductor in antiphase with the primary. This produces a total effective secondary inductance of 2mH in phase with the primary. Where you have probed the circuit, you are looking directly across the primary with one probe and across the secondary with the other. The inductors are ideal: they have no parasitic capacitances, resistances, leakage inductances and there are no core losses associated with your transformer. Therefore the phase difference between the two probes is always zero. If you reversed the phasing of both of the secondary side inductors then the resultant secondary would be 2mH in anti-phase with the primary so the phase difference would always be 180 degrees. However, if you probe across the voltage source on the primary side and across either the primary or the secondary, then you will see then a phase shift and an attenuation will be observed because the inductances of the transformer creates a high pass filter with the 1k in series with the voltage source together with the transformed 1k resistance on the secondary side. To help visualise this, the transformer circuit can be simplified by reflecting the secondary side back to the primary. Since inductance is proportional to the square of the turns ratio, the turns ratio is 1/sqrt(2):1 i.e. 0.7071:1. The resistances are also transformed by the square of the turns ratio. Therefore, if the total 1k resistance in the secondary is transformed (or 'reflected') back into the primary side, it becomes 0.5k to ground. Hence the total resistance of the LR high pass filter is 1k in parallel with 0.5k and the voltage source is attenuated by 0.5k/(1k+0.5K). This is illustrated here: https://easyeda.com/andyfierman/Reflecting_a_secondary_impedance_to_the_primary-cUgMGXR9r For more on this, see `Eliminating ideal transformers` in: https://en.wikipedia.org/wiki/Equivalent_impedance_transforms
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