Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 10M

Year: Dec 2015

AC equivalent circuit:

$R_i (Input Resistance):$

$R_i = R_1 || R_2$

$R_i = 2.2M || 1M $

$R_i = 0.6875M\Omega$

$R_O (Output Resistance):$

$R_O = R_D || r_d$

since, $r_d$ is not given $R_O = R_D$

$R_O = 2.2K$

$R_O = 2.2K\Omega$

$A_V(Voltage Gain):$

$A_V = - gm (R_D || r_d)$

Since $r_d$ is not given $A_V = -gm \times R_D$

But, $gm = gmo \sqrt \frac{IDSq}{IDSS}$....(1)

where $gmo = -\frac{2 \times IDSS}{V_P}$

$gmo = - \frac{2 \times 8mA}{-3V} = 5.33mS$

IDSq can be found using DC analysis, For DC analysis all connected capacitor acts as open circuit hence circuit becomes,

We know that,

$IDS = IDSS(1- \frac{VGS}{V_P})^2$

$IDS = 8m(1 + \frac{VGS}{3})^2$

Put different values of VGS and obtain IDS.

Apply KVL from $R_2$ to ground through gate and source,

$VR_2 - VGS - IDS \times R_S = 0$...........(2)

Put $IDS = 0$ in equation (2) we get,

$VR_2 = VGS = \frac{VDD \times R_2}{R_1 + R_2}$

$VR_2 = VGS = \frac{12 \times 1M\Omega}{1M\Omega + 2.2M\Omega}$

$VR_2 = VGS = 3.75V$.......will point on X-axis.

Put $VGS = 0$ in equation (2) we get,

$IDS = \frac{VR_2}{R_S}$

$IDS = \frac{3.75V}{1K\Omega}$

$IDS = 3.75mA$.......will point on Y-axis.

From graph $IDSq = 4.35mA, VGSq = -0.9V$

Substitue value of IDSq in equation (1)

$gm = 5.33m \sqrt \frac{4.35m}{8m}$

$gm = 3.93m\mho$

We get, $A_V = - gm (R_D) = - 3.93m(2.2K\Omega)$

$A_V = - 8.646$

Answers:

1. $A_V = - 8.646$
2. $R_i = 0.6875M\Omega$
3. $R_O = 2.2K\Omega$