Question

A negative feedback amplifier with open-loop gain $\frac{A_0}{1+j\frac{\omega }{{\omega }_0}}$ is connected with a negative feedback circuit with $A_0>0$ and feedback factor $\beta (>0)$. The 3 dB cut-off frequency at which the gain of the resultant feedback circuit reduces by a factor of 0.707 is

• A. ${\omega }_0/(1+A_0\beta )$
• B. ${\omega }_0{A}_0\beta$
• C. ${\omega }_0(1-A_0\beta )$
• D. ${\omega }_0(1+A_0\beta )$

Answer 1

The correct option is D ${\omega }_0(1+A_0\beta )$

$A_{CL}=\frac{A_{OL}}{1+A_{OL}\beta }=\frac{\frac{A_0}{(1+j\omega /{\omega }_0)}}{1+\frac{A_0}{(1+\frac{j\omega }{{\omega }_0})}\beta }=\frac{A_0}{1+j\frac{\omega }{{\omega }_0}+A_0\beta }$

$=\frac{\frac{A_0}{1+A_0\beta }}{1+j\frac{\omega }{{\omega }_0(1+A_{0\beta })}}=\frac{A_{CL}}{1+j\frac{\omega }{{\omega }_0}}$

$\therefore {\omega }_0={\omega }_0(1+A_0\beta )$