ISRO EC 2025 Q.59: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. The op-amp drives the base of an NPN emitter follower biased by a $0.2\text{mA}$ current source, and $V_o$ is the emitter voltage. The follower sits inside the loop, so the whole block acts as a non-inverting amplifier with $V_{in}$ applied to the $+$ input and the divider $10\text{k}\Omega$ / $1\text{k}\Omega$ returned to the $-$ input.

2. Feedback network: the $10\text{k}\Omega$ resistor runs from $V_o$ to the inverting node and the $1\text{k}\Omega$ resistor from that node to ground, so $$V_f=V_o\frac{1}{1+10}=\frac{V_o}{11}.$$

3. Sampling at the output: the feedback network is connected directly across the output node $V_o$, in parallel with the load. If $V_o$ is set to $0$, the feedback signal disappears, so the quantity sampled is the output voltage. This is voltage sampling, which is a shunt connection at the output.

4. Mixing at the input: $V_{in}$ is applied to the $+$ terminal and $V_f$ to the $-$ terminal, so the two voltages are in series around the differential input loop: $$V_d=V_{in}-V_f.$$ A voltage subtracted in series is series mixing, and the feedback signal is a voltage.

5. Series mixing at the input with shunt (voltage) sampling at the output gives the series-shunt topology, a voltage amplifier. Check: $$A_f\approx \frac{1}{\beta }=1+\frac{10}{1}=11,$$ a stable voltage gain set only by the resistor ratio.

6. The other options fail. Series-series would sample the output current through a series resistor. Shunt-series and shunt-shunt need a feedback current summed at the input node, but here the feedback is a voltage applied to a separate input terminal.