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.)

  • \$\begingroup\$ What is your question? \$\endgroup\$ – Andy aka Mar 9 '15 at 11:27
  • \$\begingroup\$ And what is a "Vm"? perhaps a "mV"? \$\endgroup\$ – Vladimir Cravero Mar 9 '15 at 11:44
  • \$\begingroup\$ Vm is the amplitude of the input signal \$\endgroup\$ – harika naidu Mar 9 '15 at 12:02
  • \$\begingroup\$ what is the need of calculating PIV and how can we say that piv is 2Vm \$\endgroup\$ – harika naidu Mar 9 '15 at 12:04

Consider the circuit diagram of a center-tapped full wave rectifier where D1 is forward biased and D2 is reverse biased.enter image description here

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.


simulate this circuit – Schematic created using CircuitLab

If the load was a resistor rather than a capacitor, only Vp PIV rating would be required.

  • \$\begingroup\$ Your and Nidhin's answers have complementary points. Full wave can be achieved by several means and his example does not need a capacitor to deliver 2Vp to a diode. \$\endgroup\$ – Russell McMahon Mar 9 '15 at 14:28
  • \$\begingroup\$ @RussellMcMahon As the question refers specifically to a FW rectifier, Nidhin's answer is more applicable to the actual question, +1 for him. \$\endgroup\$ – Spehro Pefhany Mar 9 '15 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.