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For a network given below determine Zi,Zo,andAV.

Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1

Marks: 10M

Years: May 15

1 Answer
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enter image description here

The circuit is collector DC feedback configuration.

DC: Ib=(VCCVBE)(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

enter image description here

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

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