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

Marks: 10M

Years: May 15

AC equivalent circuit:

Let us draw the AC equivalent circuit of voltage divider bias configuration with $R_E$ unbiased

Approximate h-parameter equivalent circuit:

Now replace the transistor in the AC equivalent circuit fig by its hybrid equivalent circuit.

Obtain the expression for $R_i$ and $R'_i$:

$R_i = \frac{V_b}{ I_b}$ … (1)

But $ V_b = I_b r_π+ I_e R_E = I_b r_π+ (1+β)I_b R_E$

$ V_b = I_b [ r_π+ (1+β)R_E]$ ….. (2)

Substituing eq 2 in 1

$R_i = \frac{(I_b [ r_π+ (1+β)R_E] )}{ I_b} = r_π+ (1+β)R_E$

$R_i = r_π+ (1+β)R_E$

$R'_i=R_1||R_2|| r_π+ (1+β)R_E$

Obtain the expression of voltage gain $A_VS$:

$A_VS = \frac{V_o}{ V_s}$

$V_o$= Voltage across ( $R_C || R_L$)

$V_o= -β I_b ( R_C || R_L)$ …. (3)

Now refer circuit Fig4.2,

$V_b= I_b R_i$

Substitute this in eq 3

$V_o=\frac{(-β I_b ( R_C || R_L))}{R_i }$

But $V_b= \frac{R'_i}{( R'_i+ R_s )}$ X $V_S$

$V_o=\frac{(-β R'_i V_S)}{( R'_i+ R_S )}$ X $\frac{(( R_C || R_L))}{ R_i}$

$A_VS$ = $\frac{V_o}{ V_s}$ = $\frac{(-β R'_i)}{( R'_i+ R_S )}$ X $\frac{(( R_C || R_L))}{ R_i}$

But $R_i = r_π+ (1+β)R_E$

$A_VS = \frac{(-β ( R_C || R_L))}{( r_π+ (1+β)R_E )}$ X $\frac{R'_i}{( R'_i+ R_S )}$

If $R'_i\gt\gt R_S and (1+β)\gt\gt r_π$ then

$A_VS = \frac{(-( R_C || R_L))}{R_E }$

This is the expression for voltage gain.

Output resistance:

Let $V_S$ = 0, and $V_x$ is connected between the output terminals and current supplied by this source be $ I_x$.

As $V_S$ = 0, $I_b$ will reduce to zero. Hence $βI_b$ = 0 so the dependent current source is replaced by an open circuit.

Therefore, $ R'_o$ = $\frac{V_x}{ I_x}$ = ( $R_C || R_L$)