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

Marks: 10M

Years: May 15

The circuit is collector DC feedback configuration.

DC: $I_b = \frac{( V_CC- V_BE)}{( R_F+ β R_C )}$

= $\frac{(12-0.7 V)}{((120kΩ+68 kΩ)+(140)3 kΩ)} = \frac{(11.3 V)}{(608 kΩ)}$ = 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