Center-tapped transformer and inductance calculation for transformer driver

Question

Using a push-pull transformer driver, for example the LT3439 like below with a center-tapped transformer, more precisely this one here

So we there is a center tapped transformer (with 1:1.3 ratio between two "full" windings, without the tap) and a half-bridge rectifier. At each half-cycle of the switching frequency, a current flows through half the primary and half the secondary at all time.

I wonder now: The linked transformer datasheet says the primary inductor (the whole primary winding if I understand correctly) is 475uH, and the ratio is \$a = V_p/V_s = 1/1.3\$.

So I wonder what is the equivalent inductance at the secondary. Is it \$L_p \cdot 0.5^2 \cdot (1/a^2) = 200uH\$ or I'm I wrong.

Given \$|Z_L| \propto L\omega\$ and the general formula \$|Z_{sec}| = Z_{Lprim}/a^2\$ (load seen at the secondary), and the fact \$L_{coil} \propto N^2\$ (N=# of turns), I would think it's a correct assumption but I'm far from being sure. The 0.5^2 factor thus would be because current flows only in half the transformer coils at all instant.

Can anyone confirm? The reason I'd like to know the inductance at the secondary is to calculate some RLC filter with it (to match the output filtering to my desire), taking directly the L of the transformer into my calculations. I'm also interested at the answer for the sake of theory.

Thanks!

Answer 1

The inductance of the whole primary winding (pins 1 to 3) is given as >475uH. The inductance of a half winding is therefore (going as N^2) >119uH.

As the turns ratio is 1.3:1, and again inductance going as N^2, the inductance of half of a secondary winding is >200uH.

This is the inductance of the winding, not the inductance that the transformer presents to the outside world when it's being used as a transformer. Assuming that one of the half-primaries is being driven all the time, the impedance at the primary side is very low, and this is the impedance (times the turns ratio squared, so times about 2) that the transformer secondary presents.

If you want to filter the output with an RLC, you will need to provide each of the R, L, and C externally.

The transformer output does look a bit inductive, but this is the leakage inductance, not the winding inductance. In a good transformer, the leakage inductance is designed to be as small as possible, and is usually an order of magnitude or two smaller than the winding inductances. It depends on the geometry of the windings, and the spaces between them. You can think of it this way, anything that's good for low leakage inductance is bad for low inter-winding capacitance, you can optimise one or the other, but not both.

Some transformers enhance leakage inductance deliberately as a circuit component, for instance microwave oven transformers. Some have high leakage inductance as an unwanted byproduct of having high inter-winding isolation. But most general purpose transformers will have striven for reasonably low leakage inductance.

Answer 2

The reason I'd like to know the inductance at the secondary is to calculate some RLC filter with it (to match the output filtering to my desire),

You are misunderstanding how this forward converter works; it appears you might be thinking it operates like a flyback converter, which it doesn’t.

The inductance that can be used in the secondary in terms of usefulness as a filter is leakage inductance and not the primary to secondary projected magnetization inductance. Leakage inductance is not given in the data sheet but, can be assumed to be between 2% and 5% of the magnetization inductance as projected via the turns ratio squared.