For the network of figure 6b with RD=2K?, RG=10M?, and VDD=12volts, Determine the following(i) IDQ(ii) VDSQ
1 Answer

The base Voltage of Voltage Divider Configuration is given by


The Collector current is given by the formula

$I_c=\beta I_b$

Hence the base current is given by


Applying KVL in clockwise direction in Collector Emitter Junction we have


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

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

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



Since Resistors R1 and R2 are in parallel we have


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




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