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

Marks: 10M

Year: May 2016

AC equivalent circuit

$A_V(Voltage Gain):$

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

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

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

We know that,

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

$IDS = 12m(1 + \frac{VGS}{3})^2$

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,

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

$VR_2 = VGS = \frac{20 \times 11M\Omega}{11M\Omega + 82M\Omega}$

$VR_2 = VGS = 2.36V$.......will point on X-axis.

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

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

$IDS = \frac{2.36V}{610\Omega}$

$IDS = 3.87mA$.......will point on Y-axis.

From graph $IDSq = 5.33mA, VGSq = -1V$

Substitue value of IDSq in equation (1)

$gm = 8m \sqrt \frac{5.33m}{12m}$

$gm = 5.33m\mho$

We get, $A_V = - gm (R_D || r_d) = - 5.33m(2K\Omega || 100K\Omega)$

$A_V = - 10.45$

$Z_i (Input Resistance):$

$Z_i = R_1 || R_2$

$Z_i = 82M || 11M $

$Z_i = 9.69M\Omega$

$Z_O (Output Resistance):$

$Z_O = R_D || r_d$

$Z_O = 2K || 100K$

$Z_O = 1.96K\Omega$

Answers:

1. $A_V = - 10.45$
2. $Z_i = 9.69M\Omega$
3. $Z_O = 1.96K\Omega$