GATE 2020 EE Q.53: solution and answer

Official answer: marks to all candidatesSource: Official answer key, GATE 2020 EE (IIT Delhi)

1. Assumptions. The op-amp is ideal: no input current, a virtual short so $V_{-}=V_{+}$, and it stays out of saturation (no supply rails are given). The nominal coefficient is $\alpha =-4\%{/}^{\circ }\text{C}=-0.04{/}^{\circ }\text{C}$ and its tolerance is $\Delta \alpha =0.25\%{/}^{\circ }\text{C}=0.0025{/}^{\circ }\text{C}$. All resistances are in $\text{k}\Omega$ and all voltages in $\text{V}$.

2. Non-inverting input (voltage divider). No current flows into the input, so $R_{th}$ and $R_1$ form a plain divider across the 3 V source: $$V_{+}=3\cdot \frac{R_1}{R_1+R_{th}}=\frac{3}{1+R_{th}}$$

3. KCL at the inverting node. $$\frac{V_{-}-0.1}{R_2}+\frac{V_{-}-V_{out}}{R_3}=0\Rightarrow V_{out}=\left(1+\frac{R_3}{R_2}\right)V_{+}-\frac{R_3}{R_2}(0.1)=3V_{+}-0.2$$ because $V_{-}=V_{+}$ and $R_3/R_2=2.6/1.3=2$.

4. Output as a function of $\alpha$. With $R_{th}=2(1+\alpha T)$ we get $1+R_{th}=3+2\alpha T$, so $$V_{out}=\frac{9}{3+2\alpha T}-0.2$$

5. The printed data do not describe a physical circuit. With $\alpha =-0.04{/}^{\circ }\text{C}$, $R_{th}=2(1-0.04T)$ is zero at $T=25^{\circ }\text{C}$ and negative at every higher temperature. At $T=150^{\circ }\text{C}$, $\alpha T=-6$ and $R_{th}=2(1-6)=-10\text{k}\Omega$. A thermistor cannot have a negative resistance, and a resistive divider fed from $+3\text{V}$ cannot give $V_{+}=3/(1-10)=-0.333\text{V}$. This is the most likely reason marks were awarded to all candidates. The rest of the solution takes the printed formula and numbers at face value, which is all a candidate in the exam hall could do. The error-propagation method is standard, but the number it gives does not describe any real circuit.

6. Nominal output at $T=150^{\circ }\text{C}$. $$V_{out}=\frac{9}{3+2(-6)}-0.2=\frac{9}{-9}-0.2=-1.2\text{V}$$

7. Sensitivity to $\alpha$. Differentiate the expression from step 4: $$\frac{\partial V_{out}}{\partial \alpha }=-\frac{9(2T)}{(3+2\alpha T{)}^2}=-\frac{9\times 300}{(-9{)}^2}=-33.33\text{V per unit of }\alpha$$

8. Limiting (first-order) error. $$\Delta V_{out}=\left|\frac{\partial V_{out}}{\partial \alpha }\right|\Delta \alpha =33.33\times 0.0025=0.0833\text{V}$$

9. Check at the end points. At $\alpha =-0.0425$: $3+2\alpha T=-9.75$, so $V_{out}=9/(-9.75)-0.2=-1.123\text{V}$. At $\alpha =-0.0375$: $3+2\alpha T=-8.25$, so $V_{out}=9/(-8.25)-0.2=-1.291\text{V}$. Half the spread is $(1.291-1.123)/2=0.084\text{V}$, which agrees with the first-order result. The spread is not symmetric about the nominal value, because $V_{out}$ is not linear in $\alpha$: the deviations are $+0.077\text{V}$ and $-0.091\text{V}$. The conventional limiting error is therefore $\Delta V_{out}\approx \pm 0.08\text{V}$. Quoting the larger one-sided deviation instead would give $0.09\text{V}$.