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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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  • \$\begingroup\$ What is your question? \$\endgroup\$
    – Andy aka
    Mar 9, 2015 at 11:27
  • \$\begingroup\$ And what is a "Vm"? perhaps a "mV"? \$\endgroup\$ Mar 9, 2015 at 11:44
  • \$\begingroup\$ Vm is the amplitude of the input signal \$\endgroup\$ Mar 9, 2015 at 12:02
  • \$\begingroup\$ what is the need of calculating PIV and how can we say that piv is 2Vm \$\endgroup\$ Mar 9, 2015 at 12:04

2 Answers 2

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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.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.

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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.

schematic

simulate this circuit – Schematic created using CircuitLab

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

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  • \$\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, 2015 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\$ Mar 9, 2015 at 17:42

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