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

Question

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.

Answer 1

... 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 +V_m (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 \$.

Answer 2

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.