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

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

Gain in low- frequency region is given asA_{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

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: vikramarka

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