Question

Consider the OP AMP based circuit shown in the figure. Ignore the conduction drops of diodes \(D_1\) and \(D_2\). All the components are ideal and the breakdown voltage of the Zener is 5 V. Which of the following statements is true? \begin{center} \includegraphics[width=0.55\textwidth]{16.jpeg} \end{center}

• A. The maximum and minimum values of the output voltage \(V_O\) are +15 V and -10 V, respectively.
• B. The maximum and minimum values of the output voltage \(V_O\) are +5 V and -15 V, respectively.
• C. The maximum and minimum values of the output voltage \(V_O\) are +10 V and -5 V, respectively.
• D. The maximum and minimum values of the output voltage \(V_O\) are +5 V and -10 V, respectively.

Hint:

Zener-based clippers limit voltage at asymmetric levels depending on Zener breakdown and diode conduction. Always analyze positive and negative cycles separately.

Answer 1

The Correct Option is C

Step 1: Understand the circuit.
The given circuit is an op-amp with diodes and a Zener diode arrangement, forming a limiter. - For positive half cycles, \(D_1\) conducts and \(D_Z\) (Zener) enters breakdown after 5 V. - For negative half cycles, \(D_2\) conducts.

Step 2: Positive cycle.
When input \(V_{in}\) is positive and large: - The op-amp drives its output positive. - Once the Zener diode conducts in reverse at \(5 \, V\), additional drop across the 1k\(\Omega\) resistor allows the output to rise up to approximately \(+10 \, V\). Thus, the positive maximum output is \(\; +10 \, V\).

Step 3: Negative cycle.
When input \(V_{in}\) is negative and large: - The diode \(D_2\) clamps the negative excursion. - This clamps the voltage to approximately \(-5 \, V\). Thus, the minimum output is \(\; -5 \, V\).

Step 4: Final range.
Therefore, \[ V_O \in [-5, +10] \]

Final Answer:
\[ \boxed{+10 \, V \; \text{and} \; -5 \, V} \]