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What I have learned: In a ideal transformer, the power in the primary circuit will equal the power out in the secondary circuit.

Impedance matching: if I take the turns ratio of the transformer in such a way as to allow....

$$Z_{primary}= a^2 Z_{secndary}$$

then maximum power is transferred to the load at the secondary circuit.

My question: if all the power going in is transferred to the secondary circuit, why is the impedance matching required in the first place?

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    \$\begingroup\$ There is a difference between power efficiency of a transformer and the concept of maximum power transfer. If the transformer is ideal and I attach a source to the primary and leave the secondary open circuit, then no power is drawn. If my source was a low impedance, say 10 ohms, 10V, my transformer was 1:1 and I put a load of 1k on the secondary, I will draw 10mA from the secondary. The load will dissipate 100mW (almost). The transformer will draw the same power from the source. No power is lost, it is all transferred but we don't have maximum power transfer. Why not? \$\endgroup\$ Mar 4 '16 at 18:32
  • \$\begingroup\$ ... That is because with a lower impedance load (like 10ohms) we could draw more power from the source. \$\endgroup\$ Mar 4 '16 at 18:33
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First you have to understand why impedance matching is required for maximum power transfer. Consider the circuit below:-

schematic

simulate this circuit – Schematic created using CircuitLab

The source has an output impedance of 50Ω and the load is also 50Ω, so voltage splits 50/50 between source and load. As a result only half the power is transferred - the other half is lost in the source.

This is only 50% efficient, but it is also the maximum power that the source can deliver to the load. Why? If the load impedance is increased then the voltage transferred is higher but current is lower, so the power transfer is less (down to zero Watts at open circuit). If the load impedance is decreased then current is higher but voltage is lower, so power transfer is also less (down to zero Watts with a short circuit).

The purpose of a matching transformer is make both the source and load 'see' the impedance required for maximum power transfer. So if the load is 200Ω then it wants the source to also be 200Ω, while the source wants the load to be 50Ω. A 1:2 transformer achieves this by doubling the input voltage, which is exactly what the 200Ω load needs to draw the same power as a 50Ω load at half the voltage.

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The reason for impedance matching is to efficie tly drive a load.

For example, if you have a soure with source impedance of 200ohms & use it to drive a 50ohm load, then the 50ohm load will pass the full available current from the source, but your source will be heating itself up against its own impedance with 3x as much power as the load receives.

However, if you added a 2:1 turns-ratio, ideal transformer, then the transformer effectively 'trades' half of the source's available voltage in exchange for doubling the available current. The end result here is that the transformer absorbs all of the available power from the source, and pushes all of it through the load, with "just the right amount" of voltage required/supplied to push all of the available current on either side of the transformer.

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