12v to 340V Push Pull Transformer Design with TURN RATIO PROBLEM

[arsalanarahim]

Can anyone help me with Push Pull Transformer Design....12V on the Low voltage side....Switching Frequency is 100Khz and Demand on Secondary side is 340V....so the Primary No of turns by formula N = (V x ton) / (B x Ac) *verified by many books


where ton is the switching on time for one MOSFET and B is the change in flux density which i have kept as 0.18T which is very reasonable. and Ac iis cross sectional Area of the core........The value of Np comes out to be 1 (one).........Is this one turn Transformer possible...my First question.......Anyone? Then i have many questions....after this?

Secondly can a Push PUll transformer work at Open Secondary? If not why? and thirdly.....this one turn primary transformer will not give a high current at no-load....i.e. no load current

 

[u04f061]

Yes this one turn is possible. Infact fractional turns are also possible. focus.ti.com/lit/ml/slup200/slup200.pdf

At 100khz 0.18T is not reasonable. It should be 0.12T as the core will go into saturation during transients. As for my opinion, a push pull transformer should work with open secondary.

 

[arsalanarahim]

With open secondary i dont think it can work....! as i saw one thread on edaboard.com https://www.edaboard.com/threads/163501/

Exact words there are these "if you open the second wind, the transfomer be change to two connect inductor , when on of FET is off , the magnetize current must charge thecapacitor of FET Cds, if your transformer have no leakage inductor between two primary winds, the Voltage of one FET Cds must be clamped up to 2*Vin by body diode of another fet , if your transformer leakage inductor is too much , and your magnetize current is bigger enough , the energy of leakage inductor can charge the voltage of fet's Cds too high maybe broken your fet .
you can decrease your leakage inductor and increase your snubber to decrease the peak voltage of fet."

So is this authentic...? What u say?