current gain, input resistance, output resistance using hybrid-π model which include early effects.

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

Marks: M

Years: Dec 14

Common emitter amplifier with voltage divider bias with bypassed emitter resistance with ac equivalent circuit is as shown in fig 4.7(b).

The hybrid π equivalent circuit is as shown below

For voltage gain $A_vs$:

$A_vs = \frac{V_o}{Vs}$

$V_o$ is voltage developed across ($r_o ||R_C ||R_L $) due to current $g_m V_π.$

$V_o= -g_m V_π(r_o ||R_C ||R_L )$

$V_π$ = Voltage across $R'_i$ where $R'_i$ = $R_B ||r_π$

$V_π= \frac{(R_B ||r_π)}{(R_S+R_B ||r_π )} x V_S$ …… ($R_B = R_1 ||R_2$)

$V_S= \frac{(R_S+R_B ||r_π)}{(R_B ||r_π )} x V_π$

Substituting value of $V_o$ and $V_S$ into $A_vs$ formula,

$A_vs = \frac{(-g_m V_π (r_o ||R_C ||R_L )}{([ R_S+R_B |├|r_π ] V_π )} x (R_B ||r_π) $A_vs = -g_m (r_o ||R_C || R_L ) x \frac{(R_B ||r_π)}{([ R_S+R_B |├|r_π ] V_π )}$ For input resistance $R'_i$: $R'_i= R_B ||r_π= R_1 ||R_2 ||r_π$ For output resistance $R'_o$: The output resistance is obtained by setting $V_S$ to 0. As $V_S$ = 0, $V_π$ = 0. Hence $g_m V_π$= 0 and the hybrid circuit modified as shown in fig 4.9 below. ![enter image description here][3] $R'_o = r_o ||R_C || R_L$ For current gain ($A_IS $): $A_IS = \frac{I_o}{I_i}$ Refer fig above, ![enter image description here][4] Consider current division at node A. The current $g_(m ) V_π$ is divided between $r_(o ) ||R_(C )and R_L.$ By using current division rule, $I_o = \frac{(-g_m V_π (r_o ) ||R_C ))}{(r_o |├|R_C )+ R_L )}$ $I_b$ is obtained by current division rule at node B. $I_b = \frac{R_B}{(R_B+ r_π )} x I_i$ $V_π = \frac{(R_B r_π)}{(R_B+ r_π )} x I_i = (R_B || r_π) I_i$ Substitute value of $V_π$ for getting $I_o$, $I_o = (-g_(m ) [(R_B || r_π) I_i ] \frac{(r_(o ) ||R_(C )) )}{( (r_(o ) |├|R_(C ) )+ R_L )}$ Hence current gain is $A_IS = \frac{I_o}{I_i} = (-g_m r_o ||R_C) \frac{(R_B || r_π))}{(R_L+ (r_(o ) |├|R_(C ) ) )}$