Topic: Analog Circuits
Output response of a diode clipper circuit shown in figure will be
(The circuit is a series diode clipper where $V_i$ connects to the positive terminal of a battery $V_B$, then a resistor $R$, and finally a diode $D_1$ pointing downwards to ground. Output $V_o$ is taken across the diode.)
- A. Output is zero for $V_{i} \le V_{B}$ and $V_{i}-V_{B}$ for $V_{i} > V_{B}$
- B. Output is $V_{B}$ for $V_{i} \le V_{B}$ and $V_{i}$ for $V_{i} > V_{B}$
- C. Output is zero for all $V_{i}$
- D. Output follows $V_{i}$ entirely
Answer: A. Output is zero for $V_{i} \le V_{B}$ and $V_{i}-V_{B}$ for $V_{i} > V_{B}$
Explanation: Case 1 ($D_{1}$ ON): Considering the diode to be ideal, $V_{D}=0$, so $V_{out}=0$. Case 2 ($D_{1}$ OFF): For $D_{1}$ to be off, $I=0$, and applying KVL yields $-V_{in} + V_{B} + V_{out} = 0 \Rightarrow V_{out} = V_{in} - V_{B}$. This occurs when $V_{in} - V_{B} > 0 \Rightarrow V_{in} > V_{B}$. Thus, the output is zero for $V_{in} \le V_{B}$ and follows $(V_{in} - V_{B})$ for $V_{in} > V_{B}$.