# Input impedance of an air core transformer?

Discussion in 'Homework Help' started by Insertname, Jul 15, 2012.

1. ### Insertname Thread Starter New Member

Mar 15, 2011
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Hi all,

just a quickie here on transformer theory. I want to know how input impedance varies with load impedance for (a) an iron core transformer and (b) an air core transformer. I'm trying to understand these devices for building a Tesla coil.

For a simple iron core transformer, the input impedance is given by the turns ratio squared times the load impedance. This means that as the load impedance tends to infinity, so does the input impedance. In contrast, for an air core transformer we have this formula:

V_S is the source voltage, Z_1 is the impedance on the primary side (e.g. a resistor wired in series), L_1 is the primary leakage inductance, L_2 is the secondary leakage inductance, M is the mutual inductance, and Z_L is the load impedance. I_1 is the current flowing in the primary

The above formula implies that if input impedance DOES NOT go up consistently with load impedance. If Z_L is zero, we get input impedance is Z_1. If load impedance is infinity, input impedance is Z_1 + jwL_1. Can anyone explain why input impedance doesn't go up consistently with load impedance in the air core case, but it does in the iron core case

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
In reality there is no difference for either case in terms of the relevant equation. The issue is about the relative magnitude of the various terms. Even the simple transformer model doesn't apply under careful analysis. It just isn't as obvious in the closely coupled magnetic core case.

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,255
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I believe your expression has an error; the sign after the term jωL1 should be -, not +.

The common assumption that the impedance seen on one winding of a transformer is related to the impedance seen at the other winding by the square of the turns ratio is an approximation that is only valid if the coefficient of coupling is nearly unity, and the winding inductances are very large, preferable infinite.

To see this with your formula, correct the error I mentioned, let the inductance of the primary be a multiple of the secondary inductance, namely L1 = (N^2)*L2 (N is the turns ratio). Now calculate the limit of your formula as M -> L2*N; this is equivalent to a coupling coefficient of unity. Next, calculate the limit as L2 -> ∞. You should get Zin = Z1 + (N^2)*ZL.

Typically, "air core" transformers have a coefficient of coupling considerably less than unity, so the impedances are not related as the square of the turns ratio.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Hi Electrician,

I'm confident the OP's original equation is correct. It should be a +ve term.

5. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,255
311
You're quite right. i overlooked the fact that what would have been a (j ω)^2 term in the numerator became -(ω)^2 and that minus sign accounts for the + sign where I said there should be a minus sign; I missed the missing j^2.

The rest of my comments are still relevant.

6. ### Insertname Thread Starter New Member

Mar 15, 2011
12
0
I realise the iron core case is an approximation but it gives such different answers as load impedance is set to 0, then to ∞. Hence the confusion

I'll try this as soon as I can. Many thanks for the feedback. It's much appreciated