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- Thread starter woody
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The reduction ratio will be 2.5, but not because half a secondary is 7.5K: since the two halves are ~perfectly coupled, one half is 3.75K.

The impedance ratio for the whole winding is 15000/600=25, resulting in a voltage ratio of √25=5.

Thus, the ratio for one half is effectively 5/2=2.5

However, Elvee tells us that using only one half of the 15K winding gives an impedance of 3.75K.

May I ask him if this has this something to do with leakage inductance caused by imperfect flux coupling?

Says me, sounding erudite after a quick google!

No, its because impedance ratio is always the square of turns ratio. Its a transformer because it transforms impedance.I assumed, like woody, that using only one half of the 15K winding would give an impedance of 7.5K.

However, Elvee tells us that using only one half of the 15K winding gives an impedance of 3.75K.

If we half the number of turns in a true center tap the inductance is 1/4 of the full impedance

Thanks, I get it: 15K/4 = 3.75K!

Except a transformer is

Just go from the fundamental equations of an ideal transformer if stuck, Vp/Np = Vs/Ns, Ip*Np = -Is*Ns

The voltage ratio equation is just a restating of the fact that the magnetic flux is the same for every turn.

Combine that with an assumption of each winding having zero resistance and you have an ideal transformer.

voltage ratio is 1:1 secondary voltage is equal to primary voltage

current ratio is 1:1 secondary current is equal to primary current

and impedance ratio is 1:1

Now imagin we double the number of turn in the secondary winding

You have now a tranformer with a ratio 1:2

voltage ratio is 1:2 secondary voltage is double of primary voltage

current ratio is 2:1 secondary current is half of primary current

now if at the secondary voltage is double and current is only half of what we had with the first transformer. ( no change in primary parameters )

the impedance ratio is now 1:4

Remember turn ratio control voltage ratio AND current ratio in the opposite way.

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