GATE 2024 IN Q.52: solution and answer

GATE 2024 · INQ.52 · 2 marks · NAT

Consider the capacitive sensor circuit and its output voltage shown in the figure. The circuit is switched ON at $t = 0$ . Assuming the opamp to be ideal, the frequency of the output voltage $V_{o}$ is ______ kHz (rounded off to two decimal places).

GATE 2024 IN question 52: Left: an op-amp astable (relaxation oscillator) with supplies +V_CC and -V_EE. — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

ANSWER

The output frequency is $f \approx 6.17 kHz$ ( $T = 2 R C ln 2.25 = 162.2 μ s$ ).

Official answer: 6.00 to 6.30Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Identify the circuit. This is a symmetric op-amp astable (relaxation oscillator). $R$ and $C$ feed the inverting input; the $10 k Ω$ / $16 k Ω$ divider feeds a fraction of the output back to the non-inverting input (positive feedback). Assumption: the op-amp is ideal (no input current, instant switching) and its output sits at $\pm V_{s a t} = \pm 13 V$ , as the figure shows.
Feedback fraction (voltage divider). No current enters the $+$ input, so $β = \frac{10}{10 + 16} = \frac{10}{26}, V_{P} = β V_{s a t} = \frac{10}{26} (13) = 5 V$ The comparator therefore switches when $V_{C}$ reaches $+ 5 V$ (while $V_{o} = + 13 V$ ) or $- 5 V$ (while $V_{o} = - 13 V$ ).
Start-up (first half-cycle only). At $t = 0$ , $V_{C} = 0 < V_{P}$ , so $V_{o} = + 13 V$ and $C$ charges from 0 toward $+ 13 V$ . This first interval, from 0 to $+ 5 V$ , is shorter than a normal half-period ( $R C ln \frac{13}{8}$ ), which is why the figure marks $T$ later on. It does not set the frequency.
Steady-state half-period (RC charging law). With $V_{o} = + 13 V$ , $C$ charges through $R$ from $- 5 V$ toward $+ 13 V$ : $V_{C} (t) = V_{s a t} + (- V_{P} - V_{s a t}) e^{- t / R C}$ Setting $V_{C} (t_{1}) = + V_{P}$ gives $t_{1} = R C ln \frac{V _{s a t} + V _{P}}{V _{s a t} - V _{P}} = R C ln \frac{13 + 5}{13 - 5} = R C ln 2.25$ The discharge from $+ 5 V$ toward $- 13 V$ , ending at $- 5 V$ , is the mirror image, so it takes the same time.
Period. $T = 2 R C ln \frac{1 + β}{1 - β} = 2 R C ln \frac{36/26}{16/26} = 2 R C ln 2.25$ With $R C = (1 k Ω) (100 nF) = 1 0^{- 4} s$ : $T = 2 (1 0^{- 4}) (0.81093) = 1.6219 \times 1 0^{- 4} s$
Frequency. $f = \frac{1}{T} = \frac{1}{1.6219 \times 1 0 ^{- 4} s} \approx 6165.8 Hz \approx 6.17 kHz$

KEY IDEA

In a symmetric op-amp astable, the capacitor swings between $\pm β V_{s a t}$ , charging each time toward $\pm V_{s a t}$ . Each half-period is $R C ln \frac{1 + β}{1 - β}$ , so $T = 2 R C ln \frac{1 + β}{1 - β} = 2 R C ln (1 + \frac{2 R _{1}}{R _{2}})$ . Here $β = \frac{R _{1}}{R _{1} + R _{2}}$ , $R_{1}$ is the resistor from the $+$ input to ground (10 kΩ) and $R_{2}$ is the one from the $+$ input to the output (16 kΩ). The value of $V_{s a t}$ cancels, so the ±13 V only confirms that the thresholds are ±5 V.

WATCH OUT FOR

Swapping the divider resistors. $β = \frac{16}{26}$ , or $T = 2 R C ln (1 + \frac{2 \cdot 16}{10}) = 2 R C ln 4.2$ , gives about 3.48 kHz. Taking $β = \frac{10}{16}$ (the resistor ratio instead of the divider fraction) gives about 3.41 kHz. A third mistake is timing the start-up interval (0 to $+ V_{P}$ , which is $R C ln \frac{13}{8}$ ) as if it were a half-period; the figure marks $T$ in steady state, where $V_{C}$ swings the full $- V_{P}$ to $+ V_{P}$ .

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GATE 2024 IN Q.52: solution and answer

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