Question

Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one : 

• A. A
• B. B
• C. C
• D. D

Hint:

In clipper circuits with a series battery \( V_B \), the output across the load starts appearing only when the input exceeds the bias voltage \( V_B \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The circuit contains an AC source \( V_i = 10 \sin \omega t \), an ideal diode \( D \), a resistor \( R \), and a \( 3\text{V} \) DC battery.
The diode is ideal, acting as a short circuit when forward-biased and an open circuit when reverse-biased.
The diode will conduct only when the potential at its anode is greater than the potential at its cathode.
Step 2: Key Formula or Approach:
Using Kirchhoff's Voltage Law (KVL), the condition for the diode to conduct is:
\[ V_i>3\text{V} \]
When the diode conducts (\( V_i>3\text{V} \)), the voltage across the resistor \( V_R \) is:
\[ V_R = V_i - 3 \]
When the diode does not conduct (\( V_i \leq 3\text{V} \)), the current in the circuit is zero:
\[ V_R = 0 \]
Step 3: Detailed Explanation:
1. During the positive half cycle, the input voltage \( V_i \) increases from \( 0\text{V} \) to \( 10\text{V} \).
2. As long as \( V_i<3\text{V} \), the diode is reverse-biased, and \( V_R = 0 \).
3. Once \( V_i \) exceeds \( 3\text{V} \), the diode becomes forward-biased (ON).
4. The voltage \( V_R \) follows the shape of the input wave but is shifted down by \( 3\text{V} \).
5. The peak value of \( V_R \) will be \( 10\text{V} - 3\text{V} = 7\text{V} \).
6. During the negative half cycle, \( V_i \) is negative, making the anode lower than the cathode, so the diode remains OFF and \( V_R = 0 \).
Step 4: Final Answer:
The resulting waveform should be zero until the input reaches \( 3\text{V} \), then rise to a peak and return to zero when the input falls below \( 3\text{V} \).
This matches the waveform shown in option (A).