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

Marks: 10M

Year: May 2015

AC equivalent circuit:

$R_i (Input Resistance):$

$R_i = R_1 || R_2$

$R_i = 10M || 6.8M $

$R_i = 4.04M\Omega$

$A_V(Voltage Gain):$

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

Since, $r_d$ is not given $A_V = - gm(R_D)$

$gm = 2K_n [VGSq - V_T]$......(1)

$IDS = k_n(VGS - V_T)^2$

$k_n = \frac{IDS(ON)}{[VGS(ON) - V_T]^2} = \frac{5mA}{[6-3]^2}$

$k_n = 0.55mA/V^2$

Hence, $IDS = 0.55mA/V^2 \times (VGS - 3)^2$ .........(2)

But voltage divider biasing equation is,

$VR_2 - VGS - IDS \times R_S = 0$

Hence, $VGS = VR_2 - IDS \times 0.75K\Omega$

$VR_2 = \frac{VDD \times R_2}{R_1 + R_2} = \frac{24 \times 6.8M\Omega}{10M\Omega + 6.8M\Omega}$

$VR_2 = 9.71V$

Hence, $VGS = 9.71 - IDS \times 0.75K\Omega$ ........(3)

Put equation (3) in (2),

$IDS = 0.55mA/V^2 \times (9.71 - IDS \times 0.75K\Omega - 3)^2$

$IDS = 0.55m \times (6.71 - IDS \times 0.75K\Omega)^2$

$IDS = 4.94mA or 16.15mA$

Select IDS with which VGS is greater than or equal to $VGS(T)$

Hence, $IDSq = 4.94mA$

with $IDSq = 4.94mA$ in equation (3),

$VGSq = 6.005V$

Substitute value of VGSq in equation (1)

$gm = 2K_n [VGSq - V_T] = 2 \times 0.55m [6.005 - 3]$

$gm = 3.305mA$

$A_V = -gm(R_D) = -3.305m \times 2.2K = -7.27$

$R_O (Output Resistance):$

$R_O = R_D || r_d$

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

$R_O = 2.2K$

$R_O = 2.2K\Omega$

Answers:

1. $R_O = 2.2K\Omega$
2. $R_i = 4.04M\Omega$
3. $A_V = -7.27$