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Less leakage inductance in full bridge transformer = more loss?

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treez

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Here is a LTspice simulation of two identical full bridge SMPS’s. The only difference between them is that one has more transformer leakage then the other.

The secondary diode snubber resistors snub out the leakage inductance ring. Strangely, the full bridge with more leakage inductance gives rise to less loss in the secondary diode snubber resistors. Why is this?
 

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A bit simplified, the snubber capacitor is charged by a series circuit of resistor and leakage inductance. While R charging involves 0.5CV² losses, L charging is lossless (but oscillating). Varying the L/R ratio gives different fractions 0.5CV².
 
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by making one of the Full bridge txformers with K=1 (no leakage) you get the highest loss....and you can see that the current peak is higher as it wasn't held up by an leakage inductance....so I think the leakage inductor helps to reduce the rms current in the snubber resistor.
This is very remarkable that snubber loss is reduced with increasing leakage inductance.
 

It can be confusing if you associate snubber losses primarly with oscillations that have to be supressed. But the truth is that charging or discharging a capacitor by a resistor or another non-reactive element e.g. a transistor always involves 0.5CV² losses.
 
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The leakage reduces the di/dt the diodes see when they are reversed biased by the next fet turning on, so some is beneficial, too much = too high snubber losses & loss of power transfer due to AC impedance at switching frequency (it takes longer to reverse the power current flow in the Tx)
 
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