Examination of a two-pulse rectifier

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SUMMARY

The discussion focuses on the operational characteristics of a two-pulse rectifier, specifically examining the series connection of Lu and Lv with the input current, which increases the conduction time of the T rectifiers. It highlights that the voltage drop across T1+T4 or T2+T3 exceeds that across the diode (Do), resulting in a slower decline of the Lo current during the rectifier switching crossover. The relationship between voltage (VL) and the rate of change of current (di/dt) is emphasized, indicating the importance of plotting currents and switch voltages over time to fully grasp the impact of Do.

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  • Understanding of two-pulse rectifier circuits
  • Familiarity with voltage drop concepts in semiconductor devices
  • Knowledge of inductance and its effects on current (VL = L * di/dt)
  • Experience with plotting electrical waveforms over time
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  • Study inductance and its role in current behavior during switching
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Electrical engineers, power electronics specialists, and students studying rectifier circuits will benefit from this discussion, particularly those interested in optimizing rectifier performance and understanding current dynamics.

polibuda
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Homework Statement
I'm testing a two-pulse rectifier. I am considering a system with an inductive load. I don't understand why the current I1 is slightly greater than the current I2, and when we add the diode D0, the current I2 is slightly greater than the current I1. Does it have to do with charging and discharging the coil? I don't understand why, with a resistive load, the current I1 is much greater than the current I2, and adding the D0 diode doesn't change the parameter values. I put the results in the tables. Could somebody me help with these problems?
Relevant Equations
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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; VL = 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.
 

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