GATE 2018 EC Q.38: solution and answer

Official answer: 0.4 to 0.6Source: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Assumptions: the op-amp is ideal (infinite open-loop gain, infinite input resistance, so no current flows into either input), but its output cannot swing beyond its supply rails of $\pm 15\text{V}$. First try the linear (virtual-short) model. The (+) input is grounded, so $V_A=0$, and the inverting-amplifier gain gives $$v_o=-\frac{31\text{k}\Omega }{1\text{k}\Omega }\times 1\text{V}=-31\text{V}$$

2. Check this against the supplies. $-31\text{V}$ lies outside $\pm 15\text{V}$, so it cannot happen. The output saturates at the negative rail: $v_o=-15\text{V}$. An ideal op-amp still cannot drive its output past its supplies.

3. In saturation the op-amp is not working in its linear region, so the virtual short no longer holds and $V_A\ne 0$. The input resistance is still infinite, so the (−) input still draws no current.

4. With no current into the (−) input, the same current $I$ flows through the $1\text{k}\Omega$ and $31\text{k}\Omega$ resistors in series, from the $1\text{V}$ source to $v_o=-15\text{V}$: $$I=\frac{1-(-15)}{1\text{k}\Omega +31\text{k}\Omega }=\frac{16\text{V}}{32\text{k}\Omega }=0.5\text{mA}$$

5. The voltage at node A, from the drop across the input resistor: $$V_A=1\text{V}-I(1\text{k}\Omega )=1-0.5=0.5\text{V}$$

6. Check using the voltage-divider (superposition) form: $$V_A=\frac{(1)(31)+(-15)(1)}{31+1}=\frac{16}{32}=0.5\text{V}$$ Now confirm the saturation state. $V_{+}-V_{-}=0-0.5=-0.5\text{V}<0$, so the infinite gain drives the output negative, which matches $v_o=-15\text{V}$. The other rail does not work: with $v_o=+15\text{V}$ we would get $V_A=(31+15)/32\approx 1.44\text{V}$ and $V_{+}-V_{-}<0$, which would drive the output negative, a contradiction. So the only consistent answer is $V_A=0.5\text{V}$.