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

$I_{csat} = 8mA$

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} …