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For the circuit shown below, calculate IDQ & VDSQ. It is given that ID(ON)=6 mA, VGS(ON)=8V. Vth=3V.
1 Answer
written 3.0 years ago by |
Answer: Step 1 : Calculate the value of k :
$k=\dfrac{I_D(on)}{[V_{GS}(on)-V_{TH}]^2}$
$\therefore k=\dfrac{6\times 10^{-3}}{(8-3)^2}$
$\therefore k=0.24 \times 10^{-3}A/V^2$...................(1)
Step 2 : obtain $I_{DQ}$
$I_D=k[V_{GS}-V_{TH}]^2$..........................(2)
Now , $V_{GS}=V_{DD}-I_DR_D$
$V_{GS}=12-I_DR_D$......................(3)
Putting the value of (1) and (3) in (2),
But $R_D=2k=2000 \Omega$
$\therefore I_D=0.24 \times 10^{-3}[9-2000I_D]^2$
$\therefore I_D=0.24 \times 10^{-3}[81-36 \times 10^3I_D+4 …