# How is the peak inverse voltage of diodes in a center tapped rectifier twice than the bridge?

I am not able to understand how the PIV becomes twice in case of centre-tapped rectifier. PIV is the maximum inverse voltage that a diode can handle before having a breakdown. For a half wave rectifier, in the negative half of the cycle the diode acts as a open circuit so the applied voltage signal with Vm as peak value appears on the terminal of that diode, and so the PIV of that diode becomes Vm. I am not able to understand this concept on a full wave rectifier.

... in the negative half of cycle the diode acts as a open circuit, so the applied voltage signal having Vm as peak value appears on the terminal of that diode, and so the PIV of that diode becomes Vm.

simulate this circuit – Schematic created using CircuitLab

Figure 1. The situation when the secondary dot end is at +Vm (the peak voltage).

• In the situation shown in Figure 1, D1's anode is at +V and it's cathode is one diode drop less.
• D2's anode is at -V while it's cathode is at +V.
• The voltage across D1, $$\ V_{D1} = +V_m - (+V_m - 0.7) = 0.7 \ \text V \$$.
• The voltage across D2, $$\ V_{D2} = -V_m - (+V_m - 0.7) = -2V_m + 0.7 \ \text V \$$.

Ignoring the diode voltage drop, the peak reverse voltage across D2 is $$\ 2V_m \$$.

In half wave rectifier inverse cycle, output is zero volt (no capacitor connected), so inverse peak is Vm peak. In a full rectifier, in the inverse cycle of one diode, the other polarity diode will supply +Vm peak, so the difference is 2xVm peak.