Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1

Marks: 10M

Years: May 16

It is also called as common drain circuit.

The input is applied to the gate and output is obtained at the source. The source follower amplifier is shown in fig 5.1. Ac equivalent circuit is drawn by connecting drain to ground.

Fig 5.3 shows the small signal equivalent circuit.

For voltage gain A_V: $A_V = \frac{V_o}{V_i} $

Output voltage is measured across $r_o|| R_S$

$V_o = g_m V_gs (r_o|| R_S)$

$V_in = V_gs+ V_o$

$V_in = V_gs+g_m V_gs (r_o|| R_S) = V_gs [1+g_m (r_o|| R_S)]$

$V_gs = \frac{V_in}{(1+g_m (r_o || R_S))}$

But $V_in = \frac{R_i}{(R_si+ R_i )}$ x $V_i$ where $R_i$ = $R_1 ||R_1$

$V_gs = \frac{((\frac{R_i}{(R_si+ R_i ))}V_i)}{(1+g_m (r_o || R_S))}$

Substitute this into expression for $V_o$ to get,

$V_o= \frac{((\frac{R_i}{(R_si+ R_i }))g_m (r_o || R_S)])}{(1+〖g_m (r〗_o || R_S))}$ = $\frac{(g_m (r_o || R_S)])}{(1+〖g_m (r〗_o || R_S))(\frac{R_i}{(R_si+ R_i }))}$

Input resistance ($R_i$):

$R_i= R_1 ||R_2$

Output resistance ($R_o$): The equivalent circuit to calculate $R_o$

$R_o = \frac{V_o}{I_o}$

Apply KCL at node A to get,

$I_o + g_m V_gs = I_r+I_RS$

But $I_r= \frac{V_o}{r_o}$ and $I_RS = \frac{V_o}{R_S}$

$I_o + g_m V_gs = V_o( \frac{1}{r_o} +\frac{1}{R_S} )$

$V_gs = -V_o$

$I_o = V_o( \frac{1}{r_o} +\frac{1}{R_S} ) + g_m V_o = V_o [g_m+\frac{1}{r_o} +\frac{1}{R_S} ]$

$R_o = \frac{V_o}{I_o}$ = = $\frac{1}{([g_m+\frac{1}{r_o} +\frac{1}{R_S} ])}$

$R_o = g_m ||\frac{1}{r_o} ||\frac{1}{R_S}$