Effect of negative feedback on Band width of an Amplifier

Let the Band Width of an amplifier without feedback is = BW. Band width of an amplifier with negative feed back is $BW_{f}= BW (1+A\beta )$ . Negative feedback increases Band width.

Proof:- Consider an amplifier with gain ‘A’

Now the frequency response of the amplifier is as shown in the figure. Frequency response curve means gain (dB) Vs frequency (Hz)

the frequency response of an amplifier consists of three regions

Low frequency region ( $< f_{1}$ -lower cut off frequency).
Mid frequency region ( between $f_{1}$ and $f_{2}$ ).
High frequency region ( the region $> f_{2}$ -upper cutoff frequency)

Gain in low- frequency region is given as $A_{vl} = \frac{A_{v}}{1-j\frac{f_{1}}{f}}---------EQN(I)$ ,

$A_{v}$ -open loop gain,

$f$ – frequency,

$f_{1}$ – lower cut off frequency, where Gain in constant region is $A_{v}$ .

Gain in High-frequency region is $A_{vh} = \frac{A_{v}}{1+j\frac{f}{f_{2}}}$ .

In low-frequency region:-

since open loop gain in low-frequency region is $A_{vl}$ and gain with feedback is $A_{vlf} = \frac{A_{vl}}{1+A_{vl}\beta }$

From EQN(I) $A_{vl} = \frac{A_{v}}{1-j\frac{f_{1}}{f}}$ after substituting $A_{vl}$ in the above equation

$A_{vlf} = \frac{\frac{A_{v}}{1-j\frac{f_{1}}{f}}}{1+\frac{A_{v}}{1-j\frac{f_{1}}{f}}\beta }$

$=\frac{A_{v}}{1-j\frac{f_{1}}{f}+A_{v}\beta }$

$= \frac{A_{v}}{1+A_{v}\beta-j\frac{f_{1}}{f} }$

Now by dividing the whole expression with $(1+A_{v}\beta )$

$A_{vlf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{\frac{1+A_{v}\beta-j\frac{f_{1}}{f}}{(1+A_{v}\beta )} }$

$A_{vlf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{1-j\frac{f_{1}}{f}\frac{1}{(1+A_{v}\beta )} }$

$A_{vlf} = \frac{A_{vf}}{1-j\frac{f_{1}^{'}}{f}}$ , where $A_{vf} = \frac{A_{v}}{1+A_{v}\beta }$ and $f_{1}^{'} = \frac{f_{1}}{1+A_{v}\beta }$

for example lower cut-off frequency $f_{1} =20 Hz$ implies $f_{1}^{'} = \frac{20}{1+A_{v}\beta }$ is decreasing with negative feedback.

In High-frequency region:-

Gain with out feed back in High frequency region is $A_{vh} = \frac{A_{v}}{1+j\frac{f}{f_{2}}}$

Now Gain with negative feed back is $A_{vhf} = \frac{A_{vh}}{1+A_{vh}\beta }$

Substituting $A_{vh}$ in the above equation

$A_{vhf} = \frac{\frac{A_{v}}{1+j\frac{f}{f_{2}}}}{1+\frac{A_{v}}{1+j\frac{f}{f_{2}}}\beta }$

$A_{vhf}=\frac{A_{v}}{1+j\frac{f}{f_{2}}+A_{v}\beta }$

$A_{vhf}= \frac{A_{v}}{1+A_{v}\beta+j\frac{f}{f_{2}} }$

Now by dividing the whole expression with $(1+A_{v}\beta )$

$A_{vhf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{\frac{1+A_{v}\beta+j\frac{f}{f_{2}}}{(1+A_{v}\beta )} }$

$A_{vhf}= \frac{\frac{A_{v}}{(1+A_{v}\beta )}}{1+j\frac{f}{f_{2}}\frac{1}{(1+A_{v}\beta )} }$

$A_{vhf} = \frac{A_{vf}}{1+j\frac{f}{f_{2}^{'}}}$ , where $A_{vf} = \frac{A_{v}}{1+A_{v}\beta }$ and $f_{2}^{'} = f_{2}(1+A_{v}\beta )$

for example lower cut-off frequency $f_{2} =20K Hz$ implies $f_{1}^{'} = 20 K (1+A_{v}\beta )$ is increasing with negative feedback.

In Mid-frequency region:-

Gain with out feed back is $A_{v}$

and the gain with negative feed back is $A_{vf} = \frac{A_{v}}{1+A_{v}\beta }$

With out feedback	With feedback
lower cut-off frequency is $f_{1}$	lower cut-off frequency $f_{1}^{'}= \frac{f_{1}}{1+A_{v}\beta }$ , increases
upper cut-off frequency is $f_{2}$	upper cut-off frequency is $f_{2}^{'} = f_{2}(1+A_{v}\beta )$
BW = $f_{2}-f_{1}$	$BW_{f} = f_{2}^{'}-f_{1}^{'}$ increases