Half-wave rectifier circuit with filter capacitor

2. The DC voltage output by the half-transition rectifier power supply. It is actually a pulsating DC voltage, which has a large ripple and is limited in practical applications. In order to reduce the ripple, a filter capacitor
C needs to be connected to the output end of the voltage , as shown in Figure 7-2 (a). The diode VD in the positive half of the AC is turned on , and while supplying power to the load, it also charges the capacitor c. The charging time constant r cd = [Rn ● RL / (Rn + RL)]. Ca; Ru.C (when ruler n < RL) o Since the value of R n is small, the time constant f is also small , so the voltage Uc on the capacitor C is quickly charged to close to the maximum AC voltage E2. When the AC value ez reaches the maximum value and then begins to decrease, when Uc>ez, the diode VD is cut off due to reverse bias. After that, the capacitor C begins to discharge to RL, and the discharge time constant is: rfd-RL. C o Since RL》R n, the discharge time constant rfd is larger, so soon after the discharge starts , the next charging cycle has arrived, and the voltage on the capacitor C drops very little. If this is repeated, the load will get as shown in Figure 7- The voltage waveform shown in 2(b). When the power supply is no-load, the output voltage Vd=√2 E2 0