GATE 2024 EC Q.49: solution and answer

Official answer: DSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Assumptions. Both op-amps are ideal (infinite gain, no input current) except that their outputs saturate at $\pm 10\text{V}$. Resistances below are in $\text{k}\Omega$.

2. OPA1 is a non-inverting Schmitt trigger. Its inverting input is grounded and the $100\text{k}\Omega$ resistor gives positive feedback, so the output always sits at $V_{01}=\pm 10\text{V}$. Superposition at the $+$ input gives $$V_{+}=\frac{100V_i+10V_{01}}{110}$$

3. Thresholds. The output switches when $V_{+}$ crosses $0$, i.e. when $V_i=-\frac{10}{100}{V}_{01}=-0.1V_{01}$. With $V_{01}=-10\text{V}$ the output flips to $+10\text{V}$ as $V_i$ rises through $V_{UT}=+1\text{V}$. With $V_{01}=+10\text{V}$ it flips to $-10\text{V}$ as $V_i$ falls through $V_{LT}=-1\text{V}$.

4. Delay of $V_{01}$. The triangle swings $4\text{V}$ (from $-2$ to $+2\text{V}$) in each $4\mu \text{s}$ half-period, so its slope is $1\text{V}/\mu \text{s}$. After each zero crossing, $V_i$ needs $1\mu \text{s}$ to reach $\pm 1\text{V}$. So $V_{01}$ is a $\pm 10\text{V}$ square wave with period $8\mu \text{s}$, in phase with $V_i$ but delayed by $1\mu \text{s}$. That rules out (A), (B) and (C).

5. OPA2 in its linear range is an inverting integrator built with an inductor. The $-$ input is a virtual ground, so $V_{01}=L\frac{d{i}_L}{dt}$, i.e. $i_L=\frac{1}{L}\int V_{01}dt$. No current enters the op-amp, so all of $i_L$ flows through the $1\text{k}\Omega$ resistor to the output: $$V_{02}=-R{i}_L=-\frac{R}{L}\int V_{01}dt,\frac{R}{L}=\frac{10^3}{10^{-3}}=10^6{\text{s}}^{-1}$$

6. Ramp rate. $\left|\frac{d{V}_{02}}{dt}\right|=10^6\times 10=10^7\text{V/s}=10\text{V}/\mu \text{s}$. An unlimited integrator would ramp $40\text{V}$ in each $4\mu \text{s}$ half-period, but the output can only move between $-10\text{V}$ and $+10\text{V}$.

7. Saturation clips the ramp. The full $20\text{V}$ swing at $10\text{V}/\mu \text{s}$ takes $2\mu \text{s}$. For the rest of each half-period $V_{02}$ stays flat at $\pm 10\text{V}$. (While OPA2 is saturated its $-$ input is no longer at $0\text{V}$, so the inductor current relaxes with time constant $L/R=1\mu \text{s}$ instead of ramping. This only moves the moment the next ramp starts: linear operation resumes once $i_L$ is back at $\pm 10\text{mA}$, and each ramp from $+10\text{mA}$ to $-10\text{mA}$ (or back) at $10\text{mA}/\mu \text{s}$ still lasts $2\mu \text{s}$. In steady state each $8\mu \text{s}$ period is a $2\mu \text{s}$ ramp, $2\mu \text{s}$ flat at one rail, a $2\mu \text{s}$ ramp back, and $2\mu \text{s}$ flat at the other rail.)

8. Result. $V_{02}$ is made of linear ramps joined by flat $\pm 10\text{V}$ levels, so it is a trapezoidal waveform. Together with the $1\mu \text{s}$ delay of $V_{01}$, the correct option is (D).