Question

A Piezoelectric transducer with a sensitivity of 2.0 pC/N having a capacitance of 1600 pF and a leakage resistance of ${10}^{12}\Omega$ is connected to a charge amplifier as shown in figure. If a force of 0.1 sin 10 t N is applied to the transducer, the output-amplitude of the charge amplifier is

• A. 0.141 V
• B. 0.232 mV
• C. 1, 414 mV
• D. 1,732 mV

Answer 1

The correct option is A 0.141 V

Charge, $Q=dF=2\times {10}^{-12}\times 0.1$ sin 10t
$=0.2\times {10}^{-12}sin10t$
$=0.2sin10t\left(pC\right)$
Voltage developed (V) =
$\frac{Q}{C_p}=\frac{0.2}{1600}sin\left(10t\right)V$
$=0.125sin10t\left(mV\right)$
$R_F\Rightarrow$Resistance of feedback circuit
$R_F={10}^8||X_c$
$X_c=\frac{1}{\omega C}=\frac{1}{10\times {10}^{-9}}={10}^8\Omega$
$\therefore R_F={10}^8||{10}^8$
$=0.5\times {10}^8\Omega$
$R_1\Rightarrow Resistance$ of the inverting terminal of the op-amp
$C_p=1600pF=1600\times {10}^{-12}F$
$X_{C_p}=\frac{1}{\omega C_p}=\frac{1}{10\times 1600\times {10}^{-12}}$
$\frac{{10}^{12}}{16000}$
$R_L={10}^{12}$
$R_1=R_L||X_{C_p}={10}^{12}||\frac{{10}^{12}}{16000}$
$\frac{{10}^{12}\frac{{10}^{12}}{16000}}{{10}^{12}+\frac{{10}^{12}}{16000}}\simeq \frac{{10}^{12}}{16000}$
$V_{out}=-\frac{R_F}{R_1}\left(V\right)$
$|V_{out}|=\frac{R_F}{R_1}\left(V\right)$
$V_{out}=\frac{0.5\times {10}^8}{{10}^{12}}\times 16000\times 0.125\times {10}^{-3}$
$=0.1\times {10}^{-3}=0.1mV$
$\therefore V_{out}isnearestto0.141mv$