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

Marks: 10M

Year: Dec 2016

Figure 1: Circuit Diagram of Common Source amplifier.

At mid frequency all connected capacitors acts as short circuit. Hence AC equivalent using JFET becomes,

Figure 2: Small signal equivalent circuit

Voltage Gain ($A_V$):

$V_O = I_O * R_D$

But $I_O = - Ids$

$V_O = - Ids * R_D$ ....(1)

Apply KCL at drain terminal,

$Ids = I _1 + g_m Vgs$

$I_1 = \frac{V_O}{r_d}$

$Ids = \frac{V_O}{r_d} + g_m Vgs$ .....(2)

substitute equation (2) in (1),

$V_O = - (\frac{V_O}{r_d} + g_m Vgs) R_D$

$V_O + \frac{V_O}{r_d} R_D = - g_m Vgs R_D$

$V_O ( 1+ \frac{ R_D}{r_d} ) = - g_m Vgs R_D$

$V_O ( \frac {r_d + R_D}{r_d} ) =- g_m Vgs R_D$

$V_O = - g_m Vgs (\frac{R_D r_d}{r_d + R_D}) $

$V_O = - g_m Vgs (R_D || r_d)$

From input side we get vin = vgs,

hence, $V_O = - g_m Vin (R_D || r_d)$

$A_V = \frac{V_O}{Vin} = -g_m(R_D || r_d)$

Input Resistance (Rin):

$Rin = \frac{Vin}{Iin}$

Since Iin passes only through $R_1||R_2$,

$Rin = \frac{Vin}{Iin} = R_1||R_2$

Output Resistance ($R_O$):

$R_O$ can be obtained using following steps:

i. Set Vin=0.

ii. Connect and imaginary voltage source $V_O$ that delivers current $I_O$.

$R_O = \frac{V_O}{I_O}$

Figure 3 : AC equivalent for $R_O$

$R_O = \frac{V_O}{I_O} = R_D || r_d$