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For the network of figure 6b with RD=2K?, RG=10M?, and VDD=12volts, Determine the following(i) IDQ(ii) VDSQ
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The base Voltage of Voltage Divider Configuration is given by

$V_B=V_{th}=V_{CC}\dfrac{R_2}{R_1+R_2}$

The Collector current is given by the formula

$I_c=\beta I_b$

Hence the base current is given by

$I_b=0.4mA$

Applying KVL in clockwise direction in Collector Emitter Junction we have

$V_{CE}=V_{CC}-I_cR_c-I_ER_E$

$12=24-9.4-(2.4 \times10^{-3}R_E)$

The Emitter Resistance is given by

$R_E=\dfrac{-2.6 \times10^3}{-2.4}=1.08K\Omega$

The Stability Factor for a Voltage Divider BJT is given by

$S=(\beta+1)\dfrac {R_{th}+R_E}{R_E(\beta+1)+R_{th}}=51\dfrac {R_{th}+1080}{1080(51)+R_{th}}=5.1$

$280908+5.1R_{th}=51R_{th}+55080$

$R_{th}=\dfrac{225828}{45.9}=4920\Omega=4.920K\Omega$

Since Resistors R1 and R2 are in parallel we have

$R_{th}=\dfrac{R_1R_2}{R_1+R_2}$

The value of R2 can be computed as

$R_2=0.1\beta R_E=24600=2.460K\Omega$

Hence the value of R1 can be calculated from above equation as

$4.920R_1+12.1032=2.460R_1$

$2.46R_1=12.1032$

$R_1=4.92K\Omega$

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