Given $V_CC$ = +20V, $V_EE$= -20V, $R_i$ = 8.2KΩ, $R_2$ = 2.2KΩ, $R_C$ = 2.7KΩ, $R_E$ = 1.81KΩ, $C_1= C_2$ = 10µF and β = 120.

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

Marks: 10M

Years: May 15

The Thevenin resistance and voltage are determined for the network to the left of the base terminal as shown in fig (a) and (b).

$R_th = R_1 || R_2$ = 8.2K || 2.2K = 1.73 kΩ

I = $\frac{(V_CC+ V_EE)}{(R_1 + R_2 )} = \frac{(20+ 20)}{(8.2k +2.2k)} = \frac{(40 V)}{(10.4 KΩ)}$ = 3.85 mA

$V_th = I R_2 - V_EE$= 3.85 mA x 2.2 kΩ - 20 V

= -11.53 V

By redrawing network as shown in fig and apply KVL,

Substituting $I_E$ = (1+β) $I_B$ gives,

$V_EE-V_th - V_BE- (1+β) I_B R_E - I_B R_th$= 0

$I_B = \frac{(V_EE -V_th - V_BE)}{( R_th+ (1+β) R_E )}$

= $\frac{(20-11.53-0.7)}{(1.73 k+121x1.8k)} = \frac{(7.77 V)}{(219.53 k)}$ = 35.39 µA

$I_C= βI_B$ = 120 x 35.39 µA = 4.25 mA

$V_C = V_CC - I_C R_C$

= 20 V – (4.25 mA)(2.7k)

= 8.53 V

$V_B = -V_th - I_B R_th$

= -11.53 V – 35.39µA x 1.73k

= -11.59 V