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 $V_G,V_S,I_D$

$V_D=V_{DD}-I_DR_D$

$Thus V_G= V_{DD}-I_DR_D ……………………(1)$

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

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

$I_D=I_{DSS}(1-\frac{-I_D×R_D+V_{DD }}{V_p})^2$

$I_D=8(1-\frac{20-4k×I_D}{-4})^2$

$I_D= \frac{1}{2} m(4+20-4kI_D)2$

$2I_D= 1m(24-4kI_D)2$

$2I_D=1m(576-192k+16MI_D^2)$

$2I_D= 0.576-192I_D+16kI_D^2$

• Solving above equation,

  We get $I_{D1}=5.19 mA$ and $I_{D2}=6.93mA$

  For $I_{D2}=6.93mA; V_{GS}=-7.72V$

  $|V_{GS}|\gt|V_P|$ hence device is in pinch-off region and in this region $I_D=0$

  Hence $I_{DQ}=I_{D1}=5.19mA$

• Substituting value of $I_{DQ}$ in eq.(1)

  $V_G= V_{DD}I_DR_D$

  $V_G= 20-(5.19m×4k)$

  $V_{GSQ}= -0.76V$

  Since $V_G$ and $V_D$ are same.

  $V_{DSQ}= V_{GSQ}= -0.76V$

  $$\boxed{V_{DSQ}= V_{GSQ}= -0.76V , I_{DQ}= 5.19mA}$$