Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1

Marks: 10M

Years: Dec 15

Step1: Calculate the value of k

k = $\frac{(I_D (on))}{[V_GS (on)- V_TH ]^2} $

$∴k = \frac{(6 X 10^(-3))}{(8-3)^2}$

$∴ k = 0.24 X 10^(-3) \frac{A}{V^2}$

Step2: Obtain I_DQ

$V_GS = V_DD - I_D R_D = 12 - I_D R_D$

We have, $I_DQ$ =$ k [V_GS- V_TH ]^2$

Substituting value of $V_GS$ we get,

$I_DQ = 0.24 X 10^(-3) [12 - I_D R_D- V_TH ]^2$

=$ 0.24 X 10^(-3) [12 - I_D x2x10^3-3 ]^2 = 0.24 X 10^(-3) X [9 - 2000I_D ]^2$

= $0.24 X 10^(-3) X [81-36000I_D + 4000000I_D^2]$

$I_DQ= 0.01944 – 8.64 I_DQ + 960 I_D^2$

960 $I_D^2-9.64I_DQ$+0.01944 = 0

By solving above quadratic equation we get,

$I_DQ$= 2.794 mA or$ I_DQ$= 7.2477 mA

Step3: Obtain $V_DS$

If we calculate $V_DS$ taking $I_D$= 7.2477 mA we get,

$V_DS = V_DD - I_D R_D = 12-7.2477 x 10^(-3) x 2 x 10^3 $= 12 – 14.495 = -2.495

Practically, the value of $V_DS$ must be positive, hence$ I_DQ$= 7.2477 mA is invalid.

Now, calculate value of $V_DS$ taking $I_DQ$= 2.794 mA,

$V_DS = 12-2.794 x 10^(-3) x 2 x 10^3$ = 12- 5.588 = 6.412 V

$V_DS$= 6.412 V