Question

Please find:

1) High frequency cutoff

2) Mid Band gain

Version: EECE_3120 2018 33k VSS C4 -12V C1 4 R3 2.2kΩ 1uF-POL LF353N C5 12V 0.1uF士 VDD

Answer 1

R_3 C_1 > R_4 C_2 Therefore Given Band filter f_L = 1/2 pi R_3 c_1 F_H = 1/2 pi R_4 c_2 rightarrow f High cut off frequency (f_H) = 1/2 pi times 33 times 10^3 times 1000 times 10^- 12 f_H = 10^6/2 pi times 33 f_H = 4.82 K H Z mid Brand A_n = -R_4/R_3 = -33/2.2 A_M = -15 At mid band frequency c_1 acts as short circuit and �c_2� act as open circuit than it acts as inverting amplifier For inverting amplifier gain A_V = -R_f/R_1