Question

For the given circuit, \( V_D \) is the threshold voltage of the diode. The graph that best depicts the variation of \( V_o \) with \( V_i \) is: 

• A. A
• B. B
• C. C
• D. D

Hint:

In op-amp limiter circuits, the diode conduction threshold determines the clipping voltage level, beyond which the output voltage becomes constant.

Answer 1

The Correct Option is A

Step 1: Understanding the circuit.
The circuit shown is a precision limiter (clipper) using an operational amplifier and a diode. The non-inverting input of the op-amp is grounded, making the inverting input a virtual ground. The diodes determine the clipping level of the output voltage.
Step 2: Behavior of the circuit for \( V_i<V_D \).
When \( V_i<V_D \), the diode remains reverse-biased, so the op-amp operates in open-loop condition. Therefore, the output \( V_o \) increases linearly and follows: \[ V_o = -V_i \] Hence, there is a negative slope in this region.
Step 3: Behavior of the circuit for \( V_i>V_D \).
When \( V_i>V_D \), the diode becomes forward-biased. The output voltage gets limited to a constant value approximately equal to the diode threshold voltage \( V_D \). Thus, the output stops increasing beyond this point and becomes clipped.
Step 4: Conclusion.
Therefore, the \( V_o \) versus \( V_i \) characteristic shows a slope change (clipping) occurring at \( V_i = V_D \). This matches the graph shown in Option (C).