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

$I_{csat} = 8 \ mA$

$V_c = 18 \ v$ and $B = 110$

Determine $R_c, R_t$ and $R_B$

Solution:

$\because$ $1_c = \frac{1}{2} 1 \ csat$

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

: $\because$ $\frac{28-vc}{Rc} = 1_c$

$\frac{28-18}{Rc} = HMA$

$R_c = \frac{10}{4MA} = 2.5 MA$

$R_c = 2.5 k$

$1_{csat} = \frac{vcc}{Rc + Re}$

$8 \ MA = \frac{28}{2.5k + RE}$

$\therefore$ RE = 1K

1E = (1+B) 1B

$= 1 + B \times \frac{1_c}{B}$

$= \frac{1+B}{B} 1_c$

$1_c = \frac{110 + 1}{110} \times HMA$

$1_E = 4.036 MA$

From collector loop

28 - Ic Rc - V c e - 1 E RE - 0

From base loop

28 - 1B RB - VBE - 1E RE = 0

$28 - 0.7 = \frac{HMA}{110} RB + 4.036m (1k)$

$27.3 - 4.03 = \frac{4MA}{110} RB$

$23.27 = \frac{4MA}{110} RB$

$RB = \frac{23.27 \times 100}{4MA} = 639.92 K$

RB = 639.92 K