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Question Bank (3 marks) · Q3

Q.Relate the bandwidth of an amplifier with and without feedback and comment on the gain-bandwidth product of the amplifier.

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[!TLDR]

Bandwidth increases by (1+Aβ)(1+A\beta): BWf=BW(1+Aβ)BW_f = BW(1+A\beta); and the gain-bandwidth product stays constant, A⋅BW=Af⋅BWfA\cdot BW = A_f\cdot BW_f.

Without feedback the amplifier has a lower 3 dB frequency f1f_1, an upper 3 dB frequency f2f_2, and bandwidth

BW=f2−f1.BW = f_2 - f_1.

When negative feedback is applied, the cut-off frequencies shift outwards:

f1′=f11+Aβ,f2′=f2(1+Aβ).f_1' = \frac{f_1}{1 + A\beta}, \qquad f_2' = f_2(1 + A\beta).

The lower cut-off decreases and the upper cut-off increases, so the new bandwidth is

BWf=BW(1+Aβ),BW_f = BW(1 + A\beta),

which is larger than BWBW by the factor (1+Aβ)(1+A\beta).

Comment on the gain-bandwidth product: negative feedback reduces the gain by the factor (1+Aβ)(1+A\beta) (Af=A/(1+Aβ)A_f = A/(1+A\beta)) and increases the bandwidth by the same factor. The two changes cancel, so the product of gain and bandwidth is the same with or without feedback: …

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