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

Marks: 10M

Years: Dec 15

$Y_os$ = 40 us

$I_DSS$ = 8 mA

$V_(GS (off))$ = -4 V

$g_mo = \frac{(2I_DSS)}{V_P} = 2x\frac{8}{4}$ = 4 mS

$V_(GS )= - I_D R_S$

$I_D = I_DSS [1- \frac{V_GS}{V_P} ]^2$

Therefore, $V_(GS )= - I_DSS R_S [1- \frac{V_GS}{V_P} ]^2$

$V_(GS )= - 8 x 0.75 [1- \frac{V_GS}{4} ]^2$

$V_(GS )$ = 10.4721 and $V_(GS )$= 1.527

$g_m= g_mo(1- \frac{V_GSQ}{V_P} ) = 4 mS (1-\frac{V_GS}{V_P} )$

The voltage gain is given by the formula

$A_V =\frac{V_o}{V_i} $

The drain current is given by the formula

$I_D = \frac{I_DSS}{2}$

Substitution give us

$I_D = \frac{(8 x 10^(-3))}{2}$

Hence the value of $I_D$ is found as

$I_D = 4 x 10^(-3) A or I_D $= 4mA

Hence the drain current is$ I_D$ = 4mA

$V_o = -I_D X R_D = -4 X 10^(-3) X 2.2X 10^3$ = - 9.8 V

$V_i = V_(GS (off)) + I_D X R_s$

= -4 + 4X 10^(-3)X 750 = -1 V

$A_V =\frac{V_o}{V_i} =\frac{(-9.8)}{(-1)}$= 9.8