GATE 2016 EC (EC2) – Question 50
The Zener diodes in the op-amp circuit clamp $V_O$ to +5 V or −5 V. Switch S is initially closed and opens at t=0. The time in seconds when $V_O$ changes state is ________.
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Correct answer: 0.7 to 0.9
Explanation
With S closed, the capacitor is uncharged and the inverting input is at +10 V. The output is consequently −5 V. Its positive-feedback divider puts the noninverting input at $-5(1)/(4+1)=-1$ V.
After opening, the inverting node falls toward −10 V through 10 kΩ with $\tau=10^4(100\ \mu\mathrm F)=1$ s. Its voltage is $-10+20e^{-t}$.
Switching occurs at −1 V: $e^{-t}=9/20$, so **$t=\ln(20/9)=0.79851$ s**.