GATE 2024 IN Q.29: solution and answer

GATE 2024 · INQ.29 · 1 mark · NAT

Consider the readout circuit of a piezoelectric sensor shown in the figure.

When the piezoelectric sensor generates a charge $q_{p}$ , the resulting change in voltage $V_{x}$ is $- 2$ V. Then the corresponding change in the voltage $V_{o u t}$ is ______ V (rounded off to nearest integer).

Note: Assume all components are ideal.

GATE 2024 IN question 29: A charge-amplifier readout circuit. — Figure from the GATE 2024 IN paper. It is part of the question.

Answer and solution

ANSWER

The change in $V_{o u t}$ is $- 3 V$ .

Official answer: -3Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Assumptions: the op-amp is ideal (infinite open-loop gain, zero input current), and nothing loads the node $V_{x}$ , so no current leaves it. All voltages and charges below are changes caused by the sensor charge $q_{p}$ .
Virtual short at the op-amp inputs: the $+$ input is grounded, so $V_{+} = 0$ and negative feedback forces $V_{-} = 0$ . The $2 μ F$ capacitor from the inverting node to ground therefore has $0 V$ across it and takes no charge. It does not affect the result.
Charge conservation at the inverting node: no current enters the op-amp and none goes into the $2 μ F$ capacitor, so all of $q_{p}$ goes into the feedback branch. The $100 pF$ and $200 pF$ capacitors are in series and nothing is drawn from $V_{x}$ , so each one carries the same charge $q_{p}$ .
Charge on the 100 pF capacitor ( $C_{1}$ ): the voltage across it is $V_{-} - V_{x} = 0 - (- 2) = 2 V$ , so $q_{p} = C_{1} (V_{-} - V_{x}) = (100 pF) (2 V) = 200 pC$
Same charge on the 200 pF capacitor ( $C_{2}$ ): $V_{x} - V_{o u t} = \frac{q _{p}}{C _{2}} = \frac{200 pC}{200 pF} = 1 V$
Output: $V_{o u t} = V_{x} - 1 = - 2 - 1 = - 3 V$ .
Check using the series capacitance: $C_{e q} = \frac{100 \times 200}{100 + 200} = \frac{200}{3} pF$ , so $V_{o u t} = - \frac{q _{p}}{C _{e q}} = - \frac{200 pC}{( 200/3 ) pF} = - 3 V$ The same result in one line: $V_{o u t} = V_{x} (1 + \frac{C _{1}}{C _{2}}) = - 2 \times 1.5 = - 3 V$ .

KEY IDEA

This is a charge amplifier. The virtual ground holds the inverting node at 0 V, so the sensor's shunt capacitance (2 µF) takes no charge and all of $q_{p}$ goes onto the feedback capacitors. Since nothing loads $V_{x}$ , the two series capacitors carry the same charge. So $V_{x}$ gives you $q_{p}$ through the 100 pF capacitor, and $q_{p}$ then gives you the extra drop across the 200 pF capacitor.

WATCH OUT FOR

Taking $V_{x}$ as the output, or splitting the voltage the wrong way. With equal charge, the drop across each series capacitor goes as $1/ C$ , so the 200 pF capacitor drops half as much as the 100 pF one (1 V, not 4 V). Another mistake is letting the 2 µF sensor capacitance share the charge. It sits at virtual ground and takes none.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2024 IN Q.29: solution and answer

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