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UCC28950: Question about calculation magnetizing inductance Lmag

Other Parts Discussed in Thread: UCC28950

Hi,

I want to design a phase shifter via full bridge and secondary rectifier diodes. Now I'm studying the >>UCC28950 600-W, Phase-Shifted, Full-Bridge
Application Report <<, which is also available as a excel and mathcad file.

Unfortunately I don't understand formula (9), that describes a calc. for the minimum magnetizing inductance. Until now, I only read about the advantage of small Lmag for supporting @ light loads. The text above the formula is drescribing the negative effect of current sensing, but I don't get it with full understanding.

My first try was to play with that formula and to find out the original thought. So:

(1-D)/fs = free wheeling time (in conventional topologies "toff")

DeltaIout/(a1*2)= half output ripple reflected to the primary

When I want to have a Lmag ripple, less than half of the output ripple, why do I use the "off time" and Vin. Shouldn't be taken the range of "on time", so Vin*ton=Vin*D/f? And why the value 0.5*DeltaIout' ?

Maybe someone can describe this section with more details and can help me?

Thanks in advance,

Christian