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

Marks: 4M

Years: Dec 14

By formula,

$I_D=I_DSS [1- \frac{V_GSQ}{V_p} ]^2$

But, $V_GS$ = $-I_D R_S$

It has been given that$V_RS = I_D R_S$= 1.5 V

$V_GS$ = -1.5 V

Hence,

$I_D=10[1- \frac{-1.5}{-2}]^2 =10[1- 1.5/2]^2$

$I_D$= 0.625 mA

To find $V_DS$:

$V_DD$ = $I_D R_D + V_DS + V_RS$

15 = (0.625 X 2) +$V_DS$ + 1.5

$V_DS$= 12.25 Volts