written 8.1 years ago by | • modified 8.1 years ago |
Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 10M
Years: May 15
written 8.1 years ago by | • modified 8.1 years ago |
Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 10M
Years: May 15
written 8.1 years ago by |
The circuit is collector DC feedback configuration.
DC: Ib=(VCC−VBE)(RF+βRC)
= \frac{(12-0.7 V)}{((120kΩ+68 kΩ)+(140)3 kΩ)} = \frac{(11.3 V)}{(608 kΩ)} = 18.6 µA
〖 I〗_E = (1+ β) 〖 I〗_B = (141) (18.6 µA)
= 2.62 mA
r_e =\frac{ (26 mA)}{I_E} = \frac{(26 mA)}{(2.62 mA)} = 9.92 Ω
Now, β r_e = (140) (9.92 Ω) = 1.39 kΩ
The ac equivalent circuit appears in fig below.
Z_i = R_F1 ||β r_e = 120 kΩ || 1.39 kΩ = 1.37 kΩ
Z_i = 1.37 kΩ
c. Testing the condition r_o \gt= 10 R_C, we find
30 kΩ >= 10 (3 kΩ) = 30 kΩ
Which is satisfied through the equals sign in the condition. Therefore,
Z_o= R_C || R_F2 = 3 kΩ || 68 kΩ
= 2.87 kΩ
Z_o= 2.87 kΩ
d. r_o\gt= 10 R_C ; therefore,
A_V = \frac{(- R_C || R_F2)}{ r_e} = \frac{(- 3 kΩ || 68 kΩ)}{(9.92 Ω)}
Approximately, \frac{(- 3 kΩ || 68 kΩ)}{(9.92 Ω)} = \frac{(- 2.87 kΩ)}{(9.92 Ω)} = -289.3
A_V = -289.3