Physics - Semiconductors Question with Solution | TestHub

The input RMS voltage is $V_{in,rms}=220\text{ V}$. The peak input voltage is $V_{in,peak}=V_{in,rms}\sqrt{2}=220\sqrt{2}\text{ V}$. The transformer turns ratio from primary to half-secondary is $N_p:N_s/2=10:1$. Therefore, the RMS voltage across the half-secondary winding is $V_{s/2,rms}=\frac{V_{in,rms}}{10}=\frac{220}{10}=22\text{ V}$.

The peak voltage across the half-secondary winding, which is the peak output voltage $V_m$, is $V_m=V_{s/2,rms}\sqrt{2}=22\sqrt{2}\text{ V}$. For a full-wave center-tapped rectifier, the Peak Inverse Voltage (PIV) across each diode is $2V_m$. So, PIV $=2\times 22\sqrt{2}=44\sqrt{2}\approx 44\times 1.414=62.216\text{ V}$. Rounding to the nearest integer, the PIV is 62 V.