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Exercise Problems · Q2

Q.An amplifier has a mid-frequency gain of 400 and lower and upper 3dB frequencies are 100Hz and 10 kHz. A negative feedback network with β\beta=0.01 is incorporated into the amplifier circuit. Calculate a) gain with feedback b) new bandwidth

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[!TLDR]

With A=400A=400, band 100 Hz–10 kHz and β=0.01\beta=0.01: the gain falls to 80 while the bandwidth widens to 49.5 kHz.

The loop gain is Aβ=400×0.01=4A\beta = 400\times0.01 = 4, so 1+Aβ=51+A\beta = 5.

  1. Gain with feedback:

    Af=A1+Aβ=4005=80A_f = \dfrac{A}{1+A\beta} = \dfrac{400}{5} = 80

  2. The open-loop bandwidth is BW=f2−f1=10 000−100=9900 Hz=9.9 kHzBW = f_2-f_1 = 10\,000 - 100 = 9900\text{ Hz} = 9.9\text{ kHz}. With feedback,

    BWf=BW(1+Aβ)=9.9×5=49.5 kHzBW_f = BW(1+A\beta) = 9.9\times5 = 49.5\text{ kHz}

    [!ANSWER]

(a) Af=80A_f=80 (b) BWf=49.5BW_f=49.5 kHz

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