In the given circuit the input voltage Vᵢₙ is shown in figure. The cut-in voltage of p-n junction diode (D₁ or D₂) is 0.6 V. Which of the following output voltage (V₀) waveform across the diode is correct?

Solution

Physics sir · 4 steps

Two diodes face opposite ways, so one clips the top at 0.6 V and the other clips the bottom at −0.6 V.

1. Read the diodes. Both diodes are in parallel with the output, and V₀ is the voltage across them. D₁ points downward, so it turns on when the top wire is positive. D₂ points upward, so it turns on when the top wire is negative.
2. Positive half. While Vᵢₙ < 0.6 V, D₁ stays off, no current flows through R, and V₀ = Vᵢₙ. Once Vᵢₙ crosses 0.6 V, D₁ conducts and holds V₀ at 0.6 V. The extra voltage drops across R, which gives a flat top.
3. Negative half. Now D₁ is reverse biased. D₂ turns on once Vᵢₙ falls below -0.6 V and holds V₀ at -0.6 V, which gives a flat bottom.
4. So. The output follows the sine wave near zero and is flattened at +0.6 V and -0.6 V. This double-clipped trapezoidal wave is option (d).

The idea to keep. A diode across the output clips the voltage at its cut-in value in the direction it conducts. Two diodes facing opposite ways clip both halves, and the series resistor takes up the extra voltage.

Answer from the NTA final answer key. Solution by PrepWiser.