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Examination of a two-pulse rectifier
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[QUOTE="Baluncore, post: 6610220, member: 447632"] Lu and Lv are effectively in series with the input current. That will extend the conduction time of the T rectifiers at the end of that phase. The voltage drop across T1+T4 or T2+T3 is higher than the voltage drop across Do. Since negative Lo voltage is less with Do in circuit, the Lo current will fall slower during the rectifier switching crossover; V[SUB]L[/SUB] = L * di/dt ; RMS or peak currents ? I think you must plot the currents and switch voltages against time to understand the effect of Do. [/QUOTE]
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Examination of a two-pulse rectifier
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