Question about transformers and inductors...

[zachomega]

Lets suppose that I measure a center tapped primary of a transformer for inductance and get 3 henries...Would the center tap to either leg give me 1.5 henries or would the inductance be a different value?

-Zach

[George Giblet]

The tapped inductance will the 1/4 * the inductance across the outside.

The transfomer core is assumed to have an AL factor such that,

   L = AL * N^2 where N is the number of turns on the core

The taps only use half the windings but the core has the same AL factor, so

   Ltap = AL * (N/2)^2 = AL * N^2 / 4  = (AL * N^2)/4 =  L / 4

  ie 1/4 the total.

You may be puzzled that the two half have an inductance of L/4 but the sum of the inductances (L) doesn't agree with series inductance formula, which would give a total of L/4 + L/4 = L/2.   The reason for this is the inductors are not independent the two halves are on the same core and the magnetic fields of each are coupled.   The coupling is modelled using the mutual inductance concept, in short the combination of the two coupled inductances is given by,

L  = L1  + L2  + 2*M  ; assuming the field of L1 and L2 add
where M = k * sqrt(L1 * L2); k is called the coupling factor.

For a transformer the k value is very nearly 1 (if the coils were far away it would be near zero, uncoupled), so M = sqrt(L1 * L2).  Also in our case the two halves are equal so Ltap = L1 = L2.  So M = Ltap  and,

L  = Ltap + Ltap + 2*Ltap = 4*Ltap

ie.  Ltap = L/4

Which is the same answer we got using the AL factor method.

[zachomega]

Not sure I got all of that...but I definitely picked up some of it...

Thank you for the very details explaination...I'm going to have to digest that a bit more.

-Zach