In the circuit, $V_{in}=100$ mV. The $10$ nF feedback capacitor initially stores $1$ nC with the shown polarity ($+$ at the op-amp output). The switch $S$ toggles with a $1$ kHz square wave ($1$ ms period): position ‘1’ when High, ‘2’ when Low. Find $|V_0|$ at $t=20$ ms (nearest integer, in mV).
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For an ideal charge amplifier with capacitive input and feedback, output steps by $-\,(C_{in}/C_f)\Delta V_{in}$ at each input edge; with no leakage, the net change over whole cycles is zero.
Initial output from capacitor charge:
\[
V_0(0^+)=\frac{Q}{C_f}=\frac{1~\text{nC}}{10~\text{nF}}=0.1~\text{V}=100~\text{mV}.
\]
Effect of switching (charge amplifier): the inverting node is a virtual ground. Each time the left plate of the $1$ nF input capacitor jumps by $\Delta V$ (between $0$ and $100$ mV), the output steps by
\[
\Delta V_0=-\frac{C_{in}}{C_f}\,\Delta V
=-\frac{1~\text{nF}}{10~\text{nF}}\,(0.1~\text{V})
=-10~\text{mV}.
\]
At the next edge it steps $+10$ mV, and so on—alternating with each transition.
After 20 ms: $1$ kHz $\Rightarrow$ period $1$ ms, so $t=20$ ms is exactly $20$ full periods. Net change over any integer number of periods is zero (equal $+10$ mV and $-10$ mV steps). Therefore,
\[
|V_0(20~\text{ms})|=|V_0(0^+)|=100~\text{mV}.
\]
