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

enter image description 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!

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  • \$\begingroup\$ ”The 0.5^2 factor thus would be because current flows only in half the transformer coils at all instant.” Are you looking for one winding on the secondary? No square on the 0.5 term then. I get 140.5 uH on each secondary winding. \$\endgroup\$ – winny Feb 15 at 9:16
  • \$\begingroup\$ I'm looking for the inductance of the "half-winding" at the secondary in which current flows, since it's this inductance that would be next to C1 on the picture, thus necessary to do LC filter calculations. \$\endgroup\$ – Yannick Feb 15 at 10:18
  • \$\begingroup\$ Updated my image to reflect better what I'm seeking. \$\endgroup\$ – Yannick Feb 15 at 10:27
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    \$\begingroup\$ As Obi Wan Kenobe said, 'this is not the inductance you are looking for'. The inductance of the winding itself does not 'appear' at the terminals, it's substantially shorted out by transformer action and the primary drive. The inductance you can 'see' at the output terminals and is available to build a filter with is the leakage inductance. In a good transformer, this is an order of magnitude or two below the winding inductances, though in some transformers it's enhanced, sometimes as a circuit element, sometimes as a by product of inter-winding isolation. \$\endgroup\$ – Neil_UK Feb 15 at 10:57
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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.

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  • \$\begingroup\$ I indeed thought I could somewhat use the transformer inductance as a filter (flyback style). Wurth says the value of 475uH is "L1" on their site. Are you implying the corresponding reactance is "Xm" (central branch of the T model), and not L2,s as I assumed (right reactance of the T model)? So my calculations would be right if that was L1,p which is not? Sorry for double confirming but I want to be sure! And how come flyback transformers can do LC filtering directly as a side question. \$\endgroup\$ – Yannick Feb 15 at 11:17
  • \$\begingroup\$ Flyback transformers have current flowing in primary, or secondary, but not both. This means the primary is open circuit when secondary current is flowing, so that's the impedance that appears across the secondary L. You have a 'forward' transformer, where current flows in both windings. The primary impedance is a short circuit, which 'shorts out' the secondary L. \$\endgroup\$ – Neil_UK Feb 15 at 11:42
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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.

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  • \$\begingroup\$ Yes, the distinction I was looking for, and it answers mot of the misunderstandings people have, is that in a forward transformer, load current flows in primary and secondary at the same time, flux is due to the difference which can (should) be very small so high ur, no need for energy storage, winding L is shorted out by the primary drive. In a flyback, current flows in primary or secondary but not both, flux is due to the full winding current, high energy storage is required, primary is open so Ls is not shorted. Only non-intutive thing now is how low ur gets good energy storage. \$\endgroup\$ – Neil_UK Feb 15 at 11:50

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