$I_{cq} = \frac{1}{2}$ $I_{csat}$

$I_{csat} = 8mA$

Vc = 18 v and B = 110

Determine $R_c, R_t$ and $R_B$

Solution: Drawing DC load line for above ckt

$\because$ $I_c = \frac{1}{2} Icsat$

$= \frac{1}{2} \times (8MA) = 4 \ mA $

$I_{cq} = 4 \ MA$

$V_{cc} = 28$

$\therefore$ $I_c = \frac{v_{cc} - V_c}{R_c}$

$4 \ mA = \frac{28-18}{R_c}$

$\therefore$ $R_c = \frac{10}{4 \ mA} = 2.5 K$

$R_c = 2.5 K$

Selecting $V_{CEQ} = \frac{1}{2} V_{cc} = \frac{1}{2} (28)$

$V_{CEQ} = 14 \ V$

From collector loop

$28 - I_c \ R_c \ - VCE - I_E \ R_E \ - 0$

$28 - 4 \ mA \times 2.5 k - 14 = I_E \ R_E $

$4 = 4.036 \ mA \times \ R_E$

$ R_E = 990.99$

From base loop

$28 - I_B R_B - VBE - 1_E R_E = 0$

$\therefore$ $R_B = \frac{28 - 0.7 - 4.036 MA \times 990.99}{36.36 \mu A} = 640.82$

$R_B = 640.82$