GATE 2022 EC Q.41: solution and answer

GATE 2022 · ECQ.41 · 2 marks · MCQ

For the following circuit with an ideal OPAMP, the difference between the maximum and the minimum values of the capacitor voltage ( $V_{c}$ ) is ________.

GATE 2022 EC question 41: An ideal op-amp with supply rails +15 V (top) and −12 V (bottom). — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
15 V
(B)
27 V
(C)
13 V
(D)
14 V

Answer and solution

ANSWER

The capacitor voltage swings between the two Schmitt-trigger thresholds, $+ 5 V$ and $- 8 V$ , so the difference between its maximum and minimum is $13 V$ : option (C).

Official answer: CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Model. The op-amp is ideal: its inputs draw no current, and with positive feedback to its $+$ input it acts as a Schmitt trigger. Its output therefore sits at a rail, $V_{o} = + 15 V$ or $V_{o} = - 12 V$ . The capacitor charges through the feedback $R$ towards $V_{o}$ . Its $+$ terminal is the inverting input and its $-$ terminal is grounded, so $V_{C} = v_{-}$ . The $I_{D}$ – $V_{D}$ plot says each diode is an ideal switch: $I_{D} = 0$ for $V_{D} < 0$ (an open circuit), and it conducts at $V_{D} = 0$ (a short).
Label the ladder. From ground: $R$ to node $A$ , $R$ to node $P$ (which is $v_{+}$ ), $R$ to node $B$ , then $R$ to $V_{o}$ . $D_{1}$ sits across $A$ – $P$ with its anode at $P$ and its cathode at $A$ . $D_{2}$ sits across $B$ – $V_{o}$ with its anode at $B$ and its cathode at $V_{o}$ . No current enters the $+$ input, so the ladder is a plain voltage divider.
Output at $+ 15 V$ . Current flows down the ladder from $V_{o}$ to ground. $D_{2}$ : current flows from $V_{o}$ to $B$ through its resistor, so $V_{B} < V_{o}$ and $V_{D 2} = V_{B} - V_{o} < 0$ . $D_{2}$ is OFF. $D_{1}$ : current reaches $P$ and must continue towards $A$ , so $V_{P} > V_{A}$ and $D_{1}$ is forward biased. It turns ON and shorts the $A$ – $P$ resistor, carrying current from $P$ to $A$ (forward, which is consistent). The path is $V_{o} \to R \to B \to R \to P \to D_{1} \to A \to R \to$ ground: a total of $3 R$ , with $R$ below $P$ .
Upper threshold, by the voltage divider rule: $V_{U T} = (+ 15) \cdot \frac{R}{3 R} = + 5 V$
Output at $- 12 V$ . Current now flows up the ladder from ground to $V_{o}$ . $D_{1}$ : current flows from $A$ to $P$ through its resistor, so $V_{A} > V_{P}$ and $V_{D 1} = V_{P} - V_{A} < 0$ . $D_{1}$ is OFF. $D_{2}$ : current reaching $B$ must continue towards $V_{o}$ , so $V_{B} > V_{o}$ and $D_{2}$ is forward biased. It turns ON and shorts the $B$ – $V_{o}$ resistor, carrying current from $B$ to $V_{o}$ (forward, which is consistent). The path is ground $\to R \to A \to R \to P \to R \to B \to D_{2} \to V_{o}$ : a total of $3 R$ , with $2 R$ below $P$ .
Lower threshold, by the voltage divider rule: $V_{L T} = (- 12) \cdot \frac{2 R}{3 R} = - 8 V$
Capacitor swing. With $V_{o} = + 15 V$ , the capacitor charges towards $+ 15 V$ . When $v_{-} = V_{C}$ passes $v_{+} = + 5 V$ , the output flips to $- 12 V$ . The capacitor then discharges towards $- 12 V$ . When $V_{C}$ falls below $v_{+} = - 8 V$ , the output flips back to $+ 15 V$ . Both thresholds lie strictly between the rails, so each is always reached and the circuit keeps oscillating. In steady state $V_{C, m a x} = + 5 V$ and $V_{C, m i n} = - 8 V$ .
Result: $V_{C, m a x} - V_{C, m i n} = 5 - (- 8) = 13 V$ This is option (C).

KEY IDEA

This is an RC relaxation oscillator: the capacitor on the inverting input charges towards whichever rail the output is at, and the output flips when $V_{C}$ crosses $v_{+}$ . So $V_{C}$ swings between the two thresholds, not between the rails. The diodes make the thresholds asymmetric. At $+ 15 V$ output, $D_{1}$ shorts the $A$ – $P$ resistor, so $v_{+}$ is $\frac{1}{3}$ of the output. At $- 12 V$ output, $D_{2}$ shorts the $B$ – $V_{o}$ resistor, so $v_{+}$ is $\frac{2}{3}$ of the output. The thresholds are therefore $+ 5 V$ and $- 8 V$ .

WATCH OUT FOR

(1) Taking the rails as the capacitor's extremes gives $15 - (- 12) = 27 V$ , option (B). The capacitor stops at the thresholds, because the output flips before $V_{C}$ gets anywhere near a rail. (2) Swapping the diode states, for example by misreading which end is the anode, puts $D_{2}$ ON at $+ 15 V$ (giving $+ 10 V$ ) and $D_{1}$ ON at $- 12 V$ (giving $- 4 V$ ). That yields $14 V$ , option (D). Always check each diode's state against the direction the ladder current actually flows. (3) Ignoring the diodes gives a $4 R$ ladder with $2 R$ below $P$ : thresholds of $+ 7.5 V$ and $- 6 V$ , and a swing of $13.5 V$ , which is not an option.

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GATE 2022 EC Q.41: solution and answer

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