Question 8.9: A negative feedback amplifier has an open-loop gain of 50,00...
A negative feedback amplifier has an open-loop gain of 50,000. The open-loop upper cut-off frequency (i.e. BW) is 100 KHz. Calculate the closed-loop upper cut-off frequency (i.e. BW) if negative feedback of 2 per cent is introduced. Also, calculate the total harmonic distortion with feedback if there is a 15 per cent distortion without feedback.
Given,
BW_{(OL)}= 100 KHz
A = 50,000.
β = 0.02
= 49.95.
Using the relation, gain–bandwidth product as constant quantity,
Open-loop gain × Bw_{(OL)} = Closed-loop gain × Bw_{(CL)}
or, 50,000 × 100 ×10^{3} = 49.95 × BW_{(OL)}
or, BW_{(CL)}=\frac{50,000 × 100 ×10^{3}}{49.95 }
= 100,100 ×10^{3} Hz
= 100.1 Hz
Since the value of the lower cut-off frequency is low, the upper cut-off frequency is approximately taken as bandwidth.