ISRO EC 2023 Q.53: solution and answer

Official answer: DSource: ISRO's revised and final answer key of 16 February 2024

1. The input $V_{in}$ is applied to the inverting (-) terminal. The non-inverting (+) terminal is connected to the junction of the $10\text{k}\Omega$ and $5\text{k}\Omega$ resistors, which form a divider from $V_{out}$ to ground. The feedback is therefore positive and the circuit is an inverting Schmidt trigger.

2. The fraction of the output fed back to the + input is $\beta =\frac{5\text{k}\Omega }{10\text{k}\Omega +5\text{k}\Omega }=\frac{1}{3}$.

3. With supplies of $\pm 5\text{V}$ and an ideal op-amp, the output saturates at $V_{sat}=+5\text{V}$ or $-5\text{V}$.

4. The output changes state when $V_{in}$ equals the voltage at the + input, $V_T=\beta V_{out}$.

5. With $V_{out}=+5\text{V}$ the threshold is $V_{UTP}=\frac{5}{3}=1.667\text{V}$. As $V_{in}$ rises above this value the output switches to $-5\text{V}$.

6. With $V_{out}=-5\text{V}$ the threshold is $V_{LTP}=\frac{-5}{3}=-1.667\text{V}$. As $V_{in}$ falls below this value the output switches back to $+5\text{V}$.

7. Hence the lower threshold point is $-1.667\text{V}$, option (d). Option (c) is the upper threshold, and options (a) and (b) are the saturation levels, not thresholds.