SOLUTION

Peak inverse voltage:- Peak inverse voltage (P.I.V) is the maximum voltage across the diode when it is not conducting. This happens during the negative half of the applied voltage when the diode does not conduct. At this stage, the voltage across the diode is equal to the applied voltage.

Peak Inverse Voltage of the bridge Rectifier:-

 Bridge Rectifier:- A bridge circuit acts as a full-wave rectifier without the use of a center-tapped transformer. It consists of four diodes in two pairs D₁,  D₃, and D₂,  D₄, connected to form a bridge as shown in Fig. The A.C supply to be rectified is applied to the diagonally opposite ends of the bridge through a transformer. The load resistance R_L is connected between the other Iwo ends of the bridge.

Working:- During the positive half cycle of A.C when the end A of the secondary of the transformer is positive and the end B is negative the diodes D₁,  D₃ is forward biased while the diodes D₂,  D₄ are reverse biased. Therefore, only the diodes D₁,  D₃ conduct and the direction of flow of current are shown by full line arrows. The current flows from A to the first diode D, to C, through the load resistance R_L lo D, to the second diode D_3, and then to B. thus completing the circuit.

During the negative half-cycle when the end B of the secondary of the transformer is positive and the end A is negative the diodes D₂,  D₄ are forward biased, whereas the diodes D₁,  D₃ are reverse biased. Therefore. only the diodes D₂,  D₄ conduct, and the direction of flow of current is shown by dotted line arrow heads. The current flows from B to the first diode D2 to C through the load resistance R_L to D, the second diode D_4, and then to A  thus completing the circuit. During both the half cycles the current through the load resistance R_L, flows in the same direction l.e., C to D. The peak inverse voltage of each diode is equal to the maximum secondary voltage, E_o of the transformer as proved below.

Peak inverse voltage:- If during the half-cycle of a c. input the end A of the secondary of the transformer is positive and end B is negative, diodes D₁ and D₃ are forward biased, while the diodes D₂ and D₄ are reverse biased. Since the diodes are considered ideal. the diodes D₁ and D₃ have negligible forward resistance and can be taken to be simple connecting wires. In such a case the two reverse-biased diodes D₂ and D₄ and the secondary of the transformer are in parallel. Hence the peak inverse voltage of each diode D₂ or D₄ is equal to the maximum E_o across the secondary. Similarly, during the next half-cycle, D2 and D4 are forward biased and can be taken to be simple connecting wires. In such a case diodes D₁ and D₃ and the secondary of the transformer are in parallel. Hence again the peak inverse voltage of D₁ or D₃. is equal to the maximum voltage E_o.

Hence each diode has only transformer voltage across it on the inverse cycle. In other words, the peak inverse voltage is equal to the peak transformer secondary voltage E_m or E_o