Question:
The diode used in the circuit has a fixed voltage drop of 0.6 V when forward biased. A signal \( v_s \) is given to the ideal OpAmp as shown. When \( v_s \) is at its positive peak, the output \( (v_{OA}) \) of the OpAmp in volts is \(\underline{\hspace{2cm}}\).
The diode used in the circuit has a fixed voltage drop of 0.6 V when forward biased. A signal \( v_s \) is given to the ideal OpAmp as shown. When \( v_s \) is at its positive peak, the output \( (v_{OA}) \) of the OpAmp in volts is \(\underline{\hspace{2cm}}\).
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When a diode is forward biased, it will limit the output voltage of the OpAmp by its voltage drop.
Updated On: Jan 8, 2026
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Correct Answer: 1
Solution and Explanation
The given signal is \( v_s = 0.4 \sin(100\pi t) \). The diode has a forward voltage drop of 0.6 V. The OpAmp in the circuit will produce an output \( v_{OA} \) which follows the input signal, but it will be limited by the diode's voltage drop. Since the peak of \( v_s \) is 0.4 V, the diode will not conduct as the signal is lower than the diode's threshold voltage. Therefore, the output voltage \( v_{OA} \) will be limited to the peak value of the input signal minus the diode drop, which gives:
\[ v_{OA} = 0.4 - 0.6 = -0.2 \, \text{V} \] Thus, the value of \( v_{OA} \) is \( -0.2 \, \text{V} \).
\[ v_{OA} = 0.4 - 0.6 = -0.2 \, \text{V} \] Thus, the value of \( v_{OA} \) is \( -0.2 \, \text{V} \).
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