Semiconductor Electronic: Material, Devices And Simple Circuits — Physics STD 12 Science — Question

4. $220\sqrt{2}\text{V}.$

Solution:

Key concept: Half wave rectifier: When the P-N junction diode rectifies half of the ac wave, it is called half wave rectifier.

1. During positive half cycle,

Diode → Forward biased

Output signal → obtained

2. During negative half cycle,

Diode → reverse biased

Output signal → not obtained

3. Output voltage is obtained across the load resistance RL. It is not constant but pulsating (mixture of ac and dc) in nature.
4. Average output in one cycle

$\text{I}_\text{dc}=\frac{\text{I}_0}{\pi}\text{ and }\text{V}_\text{dc}=\frac{\text{V}_0}{\pi};\text{I}_0=\frac{\text{V}_0}{\text{r}_\text{f}+\text{R}_\text{I}}$

(rf = forward biased resistance)

5. r.m.s. output: $\text{I}_\text{rms}=\frac{\text{I}_0}{2},\text{V}_\text{rms}=\frac{\text{V}_0}{2}$

As p-n junction diode will consuct during positive half cycle only, during negative half cycle diode is reverse biased. During this diode will not give any output. So, potential difference across capacitor C = peak voltage of the given AC voltage

$=\text{V}_0=\text{V}_\text{rms}\sqrt{2}=220\sqrt{2}\text{V}$