For the circuit shown in the figure below, assume that D₁, D₂ and D₃ are ideal. v(t) = π sin (100πt) V



The DC components of voltages v₁ and v₂, respectively are

Concept:



Forward Bias: When the p-type side of the diode is connected to a higher potential than the n-type side, the diode is said to be forward-biased i.e., ON

Reverse Bias: When the case is opposite and n-type side is kept at a higher potential than the p-type side i.e., OFF

Analysis:

At positive pulse, D₁ is ON, D₂is OFF and D₃ is OFF.

and in the negative half cycle, D₁ is OFF and D₂ and D₃ are ON. The circuit acts as a half-wave rectifier.

The output of the half-wave rectifier circuit will be:

$=\frac{V_m}{\pi }$

For positive pulse:

$V_1=\frac{V_m}{2\pi },V_2=\frac{V_m}{2\pi }$

$V_1=\frac{\pi }{2\pi },V_2=\frac{\pi }{2\pi }$

For negative pulse, D₁ is OFF, D₂ is ON and D₃ is ON. Hence resister across D₃ gets short-circuited. total voltage appears across the first resistor.

$V_1=\frac{-\pi }{\pi },V_2=0$

$\Rightarrow V_1=\frac{\pi }{2\pi }+\left(\frac{-\pi }{\pi }\right)=-0.5V$

${\mathrm{V}}_2=\frac{\pi }{2\pi }+0=0.5\mathrm{V}$