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# Reflecting a secondary impedance to the primary

5 years ago 1707
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## Description

In the primary side of this circuit, there is an inductor of 1mH.

In the secondary side of there is inductance of 2mH in phase with the primary.

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 `Vin` and `Vout` probes is always zero. If the reversed the phasing of the secondary side inductor was reversed then the resultant secondary would be 2mH in anti-phase with the primary so the phase difference would always be 180 degrees.

However, if probes are placed across the voltage source on the primary side and across either the primary or the secondary, 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 parallel 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).

For more on this, see `Eliminating ideal transformers` in:

https://en.wikipedia.org/wiki/Equivalent_impedance_transforms

## BOM

ID Name Designator Footprint Quantity
1 SIN(0 1 100k) AC 1 0 V1 2P-5.0 1
2 1k R1,R2,R3 R3 3
3 1m L1,L3,L4 INDUCTOR-1206 3
4 2m L2 INDUCTOR-1206 1
5 0.5k R4 R3 1
6 V=V(sig)*0.5k/(1k+0.5k) B1 NONE 1
7 {(1k*0.5k)/(1k+0.5k)} R5 R3 1

None

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