Answer to Question #91491 in Electric Circuits for Sahil

Question

Product of open loop gain & its bandwidth =10^6 Hz
Calculate close loop bandwidth of amplifier .given R(I)=10k ohm
&R(f)=150k ohm

Answer 1

For an amplifier the product of gain and bandwidth is constant.

For a non-inverting amplifier, the closed loop gain ($A_{CL}$) is given by

$$A_{CL}=1+\frac{R_f}{R_i}$$

$$A_{CL}=1+\frac{150}{10}$$

$$A_{CL}=16$$

If closed loop bandwidth is given by $B_{CL}$ then

                                             $A_{CL}B_{CL}=$  Product of open loop gain & its bandwidth

$$A_{CL}B_{CL}=10^6$$

$$16B_{CL}=10^6$$

$$B_{CL}=6.25\times 10^4 \ Hz$$

Answer: The closed loop bandwidth of the amplifier is  $6.25\times 10^4\ Hz$