Mumbai University > EXTC > Sem 3 > Analog Electronics 1

Marks: 5 M

Year: Nov 2013

• This is an feedback arrangement. $I_G=0mA$ and $V_{1MΩ}=0V$ the dc equivalent circuit is

Fig1 DC equivalent circuit

• Now from diagram we can clearly notice that $V_S=0$ and since there is direct connection between gate and drain $V_G=V_D(V_{GS}=V_{DS})$

1. Calculation for $V_G,V_S,I_D$

$V_D=V_{DD}-I_DR_D$

$Thus V_G= V_{DD}-I_DR_D ……………………(1)$

$I_D=I_{DSS}(1-\frac{V_{GS}} {V_p})^2$

$I_D=I_{DSS}(1-\frac{V_G}{V_p})^2$

$I_D=I_{DSS}(1-\frac{-I_D×R_D+V_{DD …