(i) Find thevenin’s equivalent voltage $V_th$ and resistance $R_th$ for base circuit.

(ii) Determine ICQ and VCEQ

Given β = 100

To find $V_th$ for the base circuit:

Assume that the base is acting as load. So in order to calculate $V_th$, we will assume the base terminal to be disconnected from $R_1$, $R_2$ and $R_3$.

Assume the branch currents to be $I_1$, $I_2$ and $I_3$ as shown in fig (a).

Apply KCL at point X in fig 3.8 (a) to write,

$I_1= I_2 +I_3$

$\frac{(5- V_th)}{R_1} = \frac{(V_th+5)}{R_2} + \frac{(V_th-3)}{R_3}$

$\frac{(5- V_th)}{500} = \frac{(V_th+5)}{70} + \frac{(V_th-3)}{500}$

$\frac{(5- V_th)}{500} = \frac{(500 (V_th+5)+70 〖(V〗_th-3))}{(70 X 500)}$

By solving,

$V_th= \frac{(5- V_th)}{500}$= -3 Volts

$V_th$= -3 Volts

To calculate $R_th$ Refer fig 3.8 (b), Figure is drawn by connecting all three DC sources to ground.

$R_th$ = $R_1$ || $R_2$ || $R_3$

= 500K || 500K|| 70K = 54.7 kΩ

$R_th$= 54.7 kΩ

To find $I_B$ and $I_CQ$:

Refer fig 3.8 (c), which shows the Thevenin’s equivalent circuit.

Apply KVL to the base loop,

$V_th +I_B R_th + V_BE + I_E R_E = V_EE$

3 + $I_B R_th$ + 0.7 + (1 + β)$I_B R_E$ = 5

By solving, $I_B$ = 2.32 µA

$I_CQ$ = $βI_B$ = 100 X 2.32 µA = 0.232 mA

$I_CQ$ = 0.232 mA

$I_E$ = (1 + β) $I_B$ = 101 X 2.32 µA = 0.234 mA

Calculate $V_CEQ$:

Apply KVL to the collector loop of fig(c),

15 V + 5 V = $I_CQ R_C$ + $V_CEQ$ + $I_E R_E$

$V_CEQ$ = 20 – (0.232 X 50) – (0.234 X 5) = 7.23 Volts

$V_CEQ$ = 7.23 Volts