transformer coupling question

[benet]

When using a transformer is it the number of turns that determines the coupling or the inductance? For instance, if you had two of the exact same Coils and one was inducing current to another coil that had twice the terms but the same inductance. And the other was inducing current to a coil that had twice the inductance but the same number of terms which would double the voltage? Thank you.

[klugesmith]

A picture might help some of us understand your word question.

In most transformers, all turns of all coils link a shared magnetic core.
Then coils with same number of turns automatically have same inductance.
Different coils have different inductance in proportion to square of the turns ratio.

In circuit simulators like SPICE, it's common to represent a transformer as two coupled inductors.
For example, a 120:1200 volt plate transformer might be modeled with inductors of 1 henry and 100 henries, and coupling factor of 0.99.

You don't need a magnetic core to get that N-squared effect.   Consider a pair of air-core solenoids filling identical bobbins, so they have the same length, diameter, and thickness.  If one is wound with thinner wire, to have 10 times more turns, it will have 100 times more inductance.

[davekni]

Kludgesmith described turns ratios and inductance well.  Concerning coupling factor, consider his example:

For example, a 120:1200 volt plate transformer might be modeled with inductors of 1 henry and 100 henries, and coupling factor of 0.99.

That 0.99 coupling factor means that unloaded output voltage will be 0.99 * 1200 = 1188V.  Or, to compensate for imperfect (non-unity) coupling, the transformer manufacturer might make the turns ratio 10.1:1, for 102 henries output inductance.  Then the ideal output would be 1212V, for a real unloaded output of 1200V.  Of course, there will be additional output voltage drop with load.  A transformer rated for 120V input and 1200V output will output more than 1200V unloaded.  It may really be a 10.5:1 transformer.

In other words, coupling factor is the actual unloaded output voltage divided by the theoretical output voltage, where "theoretical" is defined by sqrt(inductance ratio) as Kludgesmith described.

[klugesmith]

Yes, what Dave said. (I was trying to keep it simple).
With a transformer in hand, you can get the turns ratio and coupling factor by measuring unloaded voltage ratio forward and reverse.  There's a worked example for the little red transformer in this post:

https://highvoltageforum.net/index.php?topic=1225.msg9075#msg9075
"Voltage ratio 8:1 in normal direction and 1:7.5 driven backwards, indicating turns ratio of 7.75 with 97% coupling."

[klugesmith]

From the reported magnetizing current, the 120-V winding of that itty bitty transformer has inductance of 28 henries. 

The inductance of same coil at half voltage is 50 henries.  Explanation: the steel core is only half as strongly magnetized, thus operating lower on its B-H curve, so its effective permeability is higher.

Nonlinear magnetization is also important in ferrite cores.   For example, fixed inductors for SMPS have their inductance specified at a stated DC current.

[benet]

Thank you very much for your responses but I am still a little confused. It sounds like it's not possible to have two coils with the same number of turns and different inductances but in the equation for inductance (or at least the one that I have been using) it has more factors contributing than only the number of turns. L = (n^2*r^2)/(9r+10h) Where n = umbers of terns, r = Radius, H = height. Given this equation, a coil with h = 6, n = 10, d = 6 would have and inductance of 10.244 uH, while a coil with h = 6, n = 10, d = 60 would have and inductance of 272.727 uH. As well attached should be an image illustrating the initial question. 

 

[davekni]

Yes, it is certainly possible to have different inductances for the same turn counts, as you have described.  The preceding posts referred to typical transformer construction where both coils are wound around a common high-permeability core.  In that case, inductance is very close to proportional to turns squared.  If two inductors do not share the same core (or have only air-core), or core permeability is not much higher than air, inductance ratio is not so dominated by turns squared.

Even with such low-coupling situations, unloaded voltage ratio will be:
coupling_factor * sqrt(secondary_inductance / primary_inductance).
That holds for almost-ideal transformers and for far-separated inductors where coupling is almost zero.  This low-coupling case can be relevant when considering high-sensitivity electronics being disturbed by adjacent power electronics (through weak coupling of a high-power inductor with a small-signal inductor).

[klugesmith]

Thanks for the picture, Benet.  And sharing your formula, which is valid & gives good insights.

In top half of your picture, the right coil can't have more turns and same inductance.  The formula shows that it has more inductance, because of the n^2 term. 
Suppose those two coils were in the same place, with just enough size difference to fit together. Or wound on same tube, alternating 1 thick wire turn with several thin wire turns.  Then their magnetic field space (as shown in your sketch) would be common to both coils. Coupling could be almost 100%.  You would have a real transformer, with voltage ratio same as turns ratio (and square root of inductance ratio).

It would not be a practical transformer, at low frequency, because of the air core. Inductances would be small, so the no-load primary current (magnetizing current) would be very high.  With that geometry, magnetic core would help by greatly increasing the inductances. It would not change the coupling factor or voltage ratio.

For practical power transformers, more inductance is always better.  Limited by the need to use real magnetic materials, with non-infinite permeability.  The solid "conductor" of magnetic flux must be narrow enough to fit through middle of coils, and long enough to connect from top to bottom outside the coil.  To increase inductance by using more turns, would need thinner wire (with more resistance & less current capacity) or more physical volume for coils and core.

[AbrahamCross]

When using a transformer is it the number of turns that determines the coupling or the inductance? For instance, if you had two of the exact same Coils and one was inducing current to another coil that had twice the terms but the same inductance. And the other was inducing current to a coil that had twice the inductance but the same number of terms which would double the voltage? Thank you.

In a transformer, when alternating current passes through the primary coil, it creates a changing magnetic field. This changing magnetic field induces a voltage in the secondary coil. The magnitude of this induced voltage depends on the rate of change of the magnetic field and the inductance of the coils involved.

If you have two identical coils, and one coil has twice the number of turns but the same inductance as the other, the induced voltage in the coil with twice the number of turns would be twice that of the coil with fewer turns. This is because the rate of change of the magnetic field is directly proportional to the number of turns.

On the other hand, if you have two identical coils, and one coil has twice the inductance but the same number of turns as the other, the induced voltage would not be doubled. The inductance affects the rate at which the magnetic field builds up in the coil when the current changes, but it does not directly affect the coupling between the coils.