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

Marks: 4M

Years: May 16

$V_D = V_DD - I_D R_D$ 8 = 14 - $I_D$ X 1.6K $1.6kI_D = 6$ $I_D = \frac{6}{1.6k}$ = 3.75 mA Hence $I_D$= 3.75 mA $V_DS = V_DD - I_D R_D$ $V_DS$ = 8 V $I_D = I_DSS (\frac{1- V_GS}{V_p})^2$ $3.75 x 10^(-3) = 8 x 10^(-3) (1+\frac{V_GS}{4})^2$ $\sqrt{(\frac{3.75}{8})} = (1+\frac{V_GS}{4})$ $\frac{V_GS}{4}$ = 1 – 0.6846 = 0.3153 $V_GS$ = -1.2613 For self-bias, $V_GS$ = $-V_GG$ $V_GG$ = 1.2613 V