# 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 $BW_{f}=&space;BW&space;(1+A\beta&space;)$. 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

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

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

$A_{v}$ -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 $A_{vh}&space;=&space;\frac{A_{v}}{1+j\frac{f}{f_{2}}}$  .

In low-frequency region:-

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

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

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

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

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

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

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

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

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

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

In High-frequency region:-

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

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

Substituting $A_{vh}$ in the above equation

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

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

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

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

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

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

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

for example lower cut-off frequency $f_{2}&space;=20K&space;Hz$  implies $f_{1}^{'}&space;=&space;20&space;K&space;(1+A_{v}\beta&space;)$  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 $A_{vf}&space;=&space;\frac{A_{v}}{1+A_{v}\beta&space;}$

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

Thus negative feedback decreases lower cut-off frequency and increases upper cut-off frequency.

$\therefore$ Over all gain decreases with  negative feedback and Band Width increases.

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## Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.