GATE EE 2014 Set 3 — Question 32

Stage 1 - is it a linear amplifier?The input $v_i$ is applied to the inverting ($-$) terminal of the first op-amp, while the feedback resistor $R_2$ is returned from its output to the non-inverting ($+$) terminal, with $R_1$ from that terminal to ground. Feeding the output back to the $+$ terminal is positive feedback, so the amplifier has no stable linear operating point: it behaves as a comparator with hysteresis (Schmitt trigger). Its output can only sit at one of the supply rails,$$v_{o1} = +V_{sat} \quad \text{or} \quad v_{o1} = -V_{sat}$$depending on $v_i$ and the previous state. The virtual-short assumption used to obtain gains such as $-R_2/R_1$ or $1 + R_2/R_1$ is therefore not valid here.Stage 2The second op-amp receives $v_{o1}$ at its non-inverting terminal; its inverting terminal is grounded through $R$ with the other $R$ as feedback. Being driven by a signal that is already at a supply rail, this stage is also driven into saturation, so its output again rests at a rail:$$v_o = +V_{sat} \quad \text{or} \quad -V_{sat}$$Why the other options failOptions (A), (B) and (C) are the closed-loop gains of ordinary negative-feedback inverting and non-inverting amplifiers. They would require the first stage to operate linearly, which the positive feedback through $R_2$ prevents. Hence the output is never proportional to $v_i$; it only toggles between the saturation levels.Answer: (D) $+V_{sat}$ or $-V_{sat}$