Question

Consider the circuit shown in the figure below:

$\text{The transistor is biased such that}{g}_m=1\text{mA/V. Then,}$

• A. $\text{the 3-dB cut-off frequency due to capacitor}{C}_3\text{is equal to 0.923 rad/sec.}$
• B. $\text{the 3-dB cut-off frequency due to capacitor}{C}_2\text{is equal to 40 Krad/sec.}$
• C. $\text{the 3-dB cut-off frequency due to capacitor}{C}_1\text{is equal to 1 Krad/sec.}$
• D. $\text{the 3-dB cut-off frequency of the amplifier due to capacitor}{C}_3\text{is 923 rad/sec.}$

Answer 1

Answer: (C), (D)

$\text{the 3-dB cut-off frequency of the amplifier due to capacitor}{C}_3\text{is 923 rad/sec.}$

$\text{For the}{C}_1,$

$\therefore \omega =\frac{1}{R_{eq}C}\text{when}{R}_{eq}=5\text{k}\Omega +10\text{k}\Omega ||10\text{k}\Omega$

$\therefore {\omega }_1=1\text{Krad/sec}$

$\text{For capacitor}{C}_2$

$\therefore {\omega }_2=\frac{1}{R_{2}c}=\frac{1}{25\times {10}^3\times 0.1\times {10}^{-6}}=400\text{rad/sec}$

$\text{For capacitor}{C}_C,\text{we have,}$

$\therefore {\omega }_3=\frac{1}{1\text{k}\Omega ||5\text{k}\Omega +10\text{k}\Omega )\times 0.1\times {10}^{-6}}$

$\therefore {\omega }_3=923\text{rad/sec}$