Perfectly coupled transformer with different coil diameters

[741]

When coupling is perfect, the transformer voltage ratio formula is
   "primary voltage" / "secondary voltage" = Np / Ns

with Np =  turns on primary and Ns =  turns on secondary.

If the coils are of differing diameters:
   Can we also have k = 1?
   For example, maybe we can make the core of larger diameter for the larger coil.

[moffy]

Since the core material acts as a low impedance path for the flux you could have a stepped core with two different winding sizes, but if the change in diameter is abrupt I would expect fringing fields, but if tappered I guess the fringing would be lower. Something like water flow between two different diameter pipes, if the change is abrupt you get turbulence, if tappered you get a more laminar flow.You will not however achieve 100% coupling as there will always be stray fields, not fully coupled, even if small.

[741]

Thanks. The "perfect" attribute, was only for simplicity of reasoning.

My question is leading to another:

Suppose K= 0.99 for both situations, suppose the turns ratio is unity.

For the case where the two core diameters differ, is the voltage ratio still 1:1?

[RoGeorge]

No, the voltage is smaller.

If your secondary turn only sees a fraction from the total magnetic lines produced by the primary turn, then that secondary turn will produce only a fraction of the voltage.  Induced voltage in a coil is proportional with the area (and the field is uniform inside the primary turn).

[T3sl4co1l]

What permeability?

How is core area varying -- or was that a typo for coil?

Note that it matters precisely which question you're asking, and which one you really need.  An ideal core material has µ→∞, which gives k = 1.  But an otherwise-real transformer still has leakage inductance between the windings, i.e. you short the secondary and you read the same LLp as you would with an average µ.  It's not that leakage has been eliminated, it's that magnetizing inductance is infinite.

The voltage ratio in such a case will also be Ns/Np, but only when unloaded.

The voltage under load, will differ, because leakage remains present.

Tim

[RoGeorge]

the two core diameters differ, is the voltage ratio still 1:1

Only now I've noticed you are talking about two cores.  Well, if you have, for example an E+I transformer, and you put one primary turn on the middle of E core, and one secondary turn on a lateral leg of the E shaped core, then you will have half the voltage, no matter the diameter, because the magnetic lines split equally through the 2 lateral legs of the E core.

If you have an E core where both prim/sec are in the midlle of the E core, yet you file away half of the middle leg, where you place the 1 turn secondary, the the voltage ratio will be the same 1:1, because the magnetic lines like to travel through the core rather than through air, so all the magnetic lines produced by the primary will be squeezed through the interior of the smaller secondary turn, thus the same voltage.

Magnetic lines like to travel through the magnetic materials, just like current like to travel through wires, and the magnetic lines split and merge together through the paths of the lowest magnetic resistance, similar with how the electric current likes to go through the paths with the smallest electric resistance.

Depends a lot of the setup you have in mind.  Post a drawing with the arrangement of the 2 turns and their core(s).

[mag_therm]

Yes, a diagram would help!

The two windings each have inductance Lp and Ls which is related to the absolute permeability of the core.
The maximum possible permeability of body centred cubic iron/steel is at absolute zero but varies with the magnetic field strength.
Considering Curie-Weiss (variation with temperature) at 30 C, the relative permeability of iron/steel can be estimated by an empirical approximation
Ur = 734000 / (H^0.92)  Easy enough with a calculator, and H [A/m] is rms value

When the flux produced by the primary does not completely link the secondary, the coefficient of coupling reduces below 1.
For analysis of this , the Mutual Inductance is used.
M = k * sqrt( Lp*Ls)
Then the ideal transformer model ( neglecting magnetizing inductance) becomes a two loop Kirchoff where the Lp, Ls are the diagonals,
and M is the two off diagonals.
By that method the all important leakage reactance can be calculated and compared with measurement. (coefficient of coupling k is abstract)

In the transient simulators (I use qucs, presume others too) The k value is entered by the user and the leakage reactance is internally calculated.
The difficulty with that k entry is that it can't be visualized and is crammed to nearly = 1 for ordinary transformers.
Also k is not an output in FEM models of inductors ( I used QuickField which outputs L & M and don't know if that is same in others)

I would recommend that those in transformer design /specification should have familiarity with the 2 loop transformer model.

Edit
Added a S Parameter simulation  in qucs comparing qucs transformer model  against 2 loop Kirchoff, showing inductances.

https://app.box.com/s/yx8s3wxb73xarhtzo198cmz06pvnhhaa

[741]

I'm not at all experienced with transformer design, and I'm having to wade through someone else's calculations. The actual setup is hard to describe succinctly, it relates to the need to induce current across an air gap.

At this stage at least, I'm not asking about "how can we improve k", "how can we change the coil" or anything like that. At this stage, I'm simply wanting to understand/agree with the reasoning.


The calculations have brief notes, one says they find mutual inductance by considering the action of the larger coil on the smaller.

Then the thing I am wondering about is the formula they use they find the ratio Vp/Vs using mutual inductance: Vp/Vs = Lp/M

I can make this formual "work" if I
(1) Recall that L is proportional to N^2
(2) Use the transformer voltage ratio formula

Now it's part (2) that worries me becuase the coil areas are different - and so I worry "does the transformer rule that Vp/Vs = Np/Ns" apply here?

[T3sl4co1l]

You don't use the transformer voltage ratio formula; this is the ratio formula, for nonideal transformers.  M encodes everything you need to know about it.

M is a bit hard to measure in practice, but it does have a straightforward definition (a mutual source EMF_2 = M dI_1/dt in series with winding 2, and vice versa), and all the variations on that (like all the formulas with k, open/short windings, etc.).

Tim

[moffy]

Now it's part (2) that worries me becuase the coil areas are different - and so I worry "does the transformer rule that Vp/Vs = Np/Ns" apply here?

Vp = Np*dPhi_p/dt = Lp*dI_p/dt and Phi = B*A
Vs = Ns*dPi_s/dt

So the voltage across the secondary depends upon the flux (Phi) coupled from the primary coil into the secondary coil. If the secondary has a reduced area its B field increases with relation to the field in the primary, assuming reasonably good coupling, as long as it doesn't saturate.

[mag_therm]

Old reference here, Dwight "Electrical Coils and Conductors" Chapters 26 to 32
give some bibliography along with derivations of M for different sized spaced circular solenoids with concentric and parallel axis.
The methods involving calculation and summation of long analyticaL series, are superseded by FEM these days

For applications of both small signal radio and vehicular inductive power transfer, both primary and secondary side can be resonated with capacitors to 
increase circuit Q factor and power transfer efficiency.

These days the way to do it is by FEM. For the case with capacitors, a circuit consisting of  lumped capacitors, load & source resistors etc connected to the FEM modeled inductive blocks (Lp and Ls) is prepared and pre-coupled to the solver.
Then the results are presented for post processing in engineering units.

[741]

As I understand the responses, "Vp/Vs = Lp/M" is correct.

Is there a simple derivation and/or how do I look this up?