GATE 2020 IN Q.44: solution and answer

GATE 2020 · INQ.44 · 2 marks · NAT

The circuit shown uses ideal opamp powered from a supply $V_{C C} = 5 V$ . If the charge $q_{p}$ generated by the piezoelectric sensor is of the form $q_{p} = 0.1 sin (10000 π t) μ C$ , the peak detector output after 10 cycles of $q_{p}$ (in volts, rounded off to one decimal place) is ____

GATE 2020 IN question 44: Charge-amplifier circuit. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

The peak detector output is 3.5 V. The charge amplifier gives $v_{o} = 2.5 - q_{p} / C_{f} = 2.5 - sin (10000 π t)$ V, which swings from 1.5 V to 3.5 V without clipping, so the peak is 3.5 V.

Official answer: 3.4 to 3.6Source: Official answer key, GATE 2020 IN (IIT Delhi)

Read the circuit correctly. The non-inverting input is wired to the bottom rail, so $v_{+} = \frac{1}{2} V_{C C} = 2.5 V$ . The ground under the op-amp is its negative supply pin, so the output can swing only between $0$ and $5 V$ . The sensor, the $10 pF$ and the $100 pF$ all sit between the inverting input and this same $2.5 V$ rail. The feedback element is $C_{f} = 100 nF$ .
Virtual short (ideal op-amp, linear operation, zero input current). Negative feedback through $C_{f}$ gives $v_{-} = v_{+} = 2.5 V$ .
The input capacitors carry no current. The $10 pF$ and $100 pF$ capacitors are connected between $v_{-} = 2.5 V$ and the $2.5 V$ rail, so the voltage across them is always $0$ : $i_{C} = (10 + 100) pF \cdot \frac{d ( v _{-} - 2.5 )}{d t} = 0$
KCL at the inverting node. Take $i_{p} = \frac{d q _{p}}{d t}$ as the sensor current entering the node. No current flows into the op-amp or the input capacitors, so all of it flows through $C_{f}$ towards the output: $\frac{d q _{p}}{d t} = C_{f} \frac{d ( v _{-} - v _{o} )}{d t} = - C_{f} \frac{d v _{o}}{d t} (v_{-} is constant)$
Integrate. Assume $C_{f}$ starts uncharged, so $v_{o} (0) = v_{-} = 2.5 V$ . Also $q_{p} (0) = 0$ . Then $v_{o} (t) = 2.5 V - \frac{q _{p} ( t )}{C _{f}}$
Decisive computation. $\frac{q _{p, m a x}}{C _{f}} = \frac{0.1 \times 1 0 ^{- 6} C}{100 \times 1 0 ^{- 9} F} = 1 V \Rightarrow v_{o} (t) = 2.5 - sin (10000 π t) V$ The sensor's polarity is not marked. Reversing it only flips the sign of the sine term, so the output range is the same either way.
Check that the op-amp stays linear. $v_{o}$ swings between $2.5 - 1 = 1.5 V$ and $2.5 + 1 = 3.5 V$ . That is well inside the $0$ to $5 V$ supply range, so the output never clips and the virtual-short assumption holds for the whole time.
Peak detector after 10 cycles. $ω = 10000 π rad/s$ , so $f = 5 kHz$ and $T = 0.2 ms$ ; 10 cycles take $2 ms$ . Since $q_{p}$ is periodic with zero average, $v_{o}$ repeats the same $1.5$ to $3.5 V$ swing every cycle, and nothing builds up. An ideal peak detector holds the largest value reached: $V_{P D} = 2.5 + 1 = 3.5 V$

KEY IDEA

This is a charge amplifier. With the + input biased at $V_{C C} /2 = 2.5 V$ , the virtual short puts the sensor's own capacitance and the cable capacitance ( $10 + 100 pF$ ) at zero volts across, so they carry no current. All of the sensor charge goes onto $C_{f}$ , and $v_{o} = V_{C C} /2 - q_{p} / C_{f}$ . Biasing at mid-supply lets a single-supply op-amp handle a bipolar charge signal. The peak is the bias plus $q_{p, m a x} / C_{f} = 1 V$ .

WATCH OUT FOR

Reading the ground symbol under the op-amp as the + input's reference. That ground is the op-amp's negative supply pin; the + input is wired to the 2.5 V rail. With the + input taken as 0 V you get $v_{o} = \mp sin (10000 π t)$ , which clips at the 0 V rail and gives a wrong 1.0 V. Another mistake is dividing by $C_{f} + 110 pF$ : the input capacitors have zero voltage across them under the virtual short, so they take no charge. The phrase '10 cycles' is a distractor. With zero-mean charge nothing accumulates, and the peak is the same in every cycle.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2020 IN Q.44: solution and answer

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