Peak inverse voltage (PIV) is the maximum voltage a diode can handle in reverse bias condition.
Why does the PIV of the diodes in a full wave rectifier need to be 2×Vm?
(where Vm is the peak voltage of the input AC signal.)
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Consider the circuit diagram of a center-tapped full wave rectifier where D1 is forward biased and D2 is reverse biased.
The maximum reverse voltage appearing across will be 2*Vp. Where Vp is the amplitude of input signal.
So the diode that is used in a center-tapped full wave rectifier should have a PIV of atleast twice the peak voltage of input sine wave. Otherwise diode breakdown will happen and current will flow through the reverse biased diode. And the circuit is not a rectifier anymore.
Similarly, PIV for a full wave bridge rectifier will be Vp.
The minimum required PIV is 2 * Vp where Vp is the peak input voltage if there is a capacitor output filter, because the capacitor holds Vp (minus diode drop) whilst the input voltage goes to -Vp on the negative input cycles. The voltage the diode must stand off is thus 2*Vp.
If the load was a resistor rather than a capacitor, only Vp PIV rating would be required.