Dual Active Bridge leakage inductance

[abhishek.2138]

In bidirectional dual active bridge topology, how to split the leakage inductance between primary & secondary windings.
If my total leakage = 32uH & want to split it in both primary & secondary windings, then how it can be done.

Can anybody suggest formula for this.

Reference schematic below, how Lk can be split in two windings -

 

[Easy peasy]

For a 1:1 transformer, you can assume the measured leakage is split evenly to both sides - assuming a relatively symmetric transformer design and build.

For other turns ratios you can assume each side has its proportion proportional to the turns ratio squared

However - it really doesn't matter - as the driving side see the leakage of "both" sides - no matter which side is driving, referred to that side, i.e. a LV driving side will see a low leakage ( proportional to N^2 ) and the high side will see a higher leakage. In each event the leakage seen can be measured by shorting out the other side.

 

[FvM]

Physically, a transformer can have asymmetrically distributed leakage inductance. As a result, the voltage ratio is slightly (e.g. by several percent) different from winding ratio. If you consider the transformer as black box formed by coupled inductors, it is characterized just by L1, L2 and k. It relates to the winding ratio of a physical transformer under the assumption of a specific leakage inductance distribution.

 

[Easy peasy]

Ah .... can you please provide a basic sketch of a transformer suitable for power electronics use, ideally a 1:1 one, that has asymmetric leakage ? I for one would be very interested to see this ... ?

( without cheating by using shaped or longer lead out wires )

 

[FvM]

The winding arrangement is prim-sec-sec-prim (top - down). The winding position to airgap causes asymmetrical leakage inductance, about 5:1. See Unitrode/TI application about transformer calculation.

 

[Easy peasy]

Assymetric, usually means non proportional to turns ratio squared, i.e. different dependent on which side is driven .... even for 1:1 turns ....

i.e. there is significantly more flux that doesn't couple when driven from one wdg, than you get when driven from the other.

Is there a case of this ? consider a winding on a smaller cylinder, concentrically inside a much larger cylinder with a wdg on it, same turns on each .... ???