## Explain the working of rectifier circuits, Electrical Engineering

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Q. Explain the working of Rectifier Circuits?

A simple half-wave rectifier using an ideal diode is shown in Figure(a). The sinusoidal source voltage vS is shown in Figure (b). During the positive half-cycle of the source, the ideal diode is forward-biased and closed so that the source voltage is directly connected across the load. During the negative half-cycle of the source, the ideal diode is reverse-biased so that the source voltage is disconnected from the load and the load voltage as well as the load current are zero. The load voltage and current are of one polarity and hence said to be rectified. The output current through the load resistance is shown in Figure (c). In order to smooth out the pulsations (i.e., to eliminate the higher frequency harmonics) of the rectified current, a filter capacitor may be placed across the load resistor, as shown in Figure (a). As the source voltage initially increases positively, the diode is forward-biased since the load voltage is zero and the source is directly connected across the load. Once the source reaches its maximum value VS and begins to decrease, while the load voltage and the capacitor voltage are momentarily maintained at VS, the diode becomes reverse-biased and hence open- circuited. The capacitor then discharges over time interval t2 through RL until the source voltage vS(t) has increased to a value equal to the load voltage. Since the source voltage at this point in time exceeds the capacitor voltage, the diode becomes once again forward-biased and hence closed. The capacitor once again gets charged to VS. The output current of the rectifier with the filter capacitor is shown in Figure(b), and the circuit configurations while the capacitor gets charged and discharged are shown in Figure(c). The smoothing effect of the filter can be improved by increasing the time constant CRL so that the discharge rate is slowed and the output current more closely resembles a true dc current.

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