I have an Eicor 600 ohm to 15K center tap matching transformer could the 600 ohm primary as
the output and use just one side of the 15K center tap as the input. Wouldn't that be 7.5K to
600 ohm and give me a voltage reduction of about 2.5 ?
the output and use just one side of the 15K center tap as the input. Wouldn't that be 7.5K to
600 ohm and give me a voltage reduction of about 2.5 ?
7500/600 = 1/Vs^2 (taking Vp as 1 volt)
12.5 = 1/Vs^2
Vs^2 = 1/12.5 = 0.08
Vs = 0.28
So, for 1 V in you would get 0.28 V out
Check my maths - it's late here!
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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
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
I am still in my morning coffee, so I am a little confused. Isn't the impedance ratio (7500/600) which results in a voltage ratio of 5/sqrt(2) as the fellow calculated above?
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.
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! 🤓
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.
Mark is right if the center tap is indeed at half the number of turns. The inductance of a coil is proprtional to N^2. If we half the number of turns in a true center tap the inductance is 1/4 of the full impedance and the math works out to be 5/2 voltage ratio. If the center tap is done by making half the inductance rather than the number of turns your math is correct. Confusing.
Except a transformer is not typically inductive, it simply multiplies or divides the load impedance as is, and this should swamp the inductive component if designed right.
Just go from the fundamental equations of an ideal transformer if stuck, Vp/Np = Vs/Ns, Ip*Np = -Is*Ns
There are components where the inductance starts to become important, such as resonant transformers or gapped transformers - but a standard transformer has so much inductance that you can assume Ip*Np + Is*Ns = 0 (no net current) to a good approximation, which gives the current ratio equation directly.
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.
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.
Think about a transformer with a ratio of 1:1
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.
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.