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

Marks: 10M

Years: Dec 15

Emitter follower circuit and their equivalent hybrid-π circuit is as shown below.

For $A_VS: The output voltage $V_o = Voltage across ( r_o || R_E)$ = $I_b ( r_o || R_E) + β I_b ( r_o || R_E)$ = $(1+ β) I_b ( r_o || R_E)$ Since, $ I_b = \frac{V_π}{ I_π} = \frac{( V_b- V_e)}{r_π}$ But $V_e = V_o$ $I_b =\frac{( V_b- V_o)}{r_π}$ Substituting value of $ I_bfor V_o$ $V_o= (1+ β)\frac{( V_b-V_o)}{r_π } ( r_o || R_E)$ = (1+ β) $\frac{V_b}{r_π} ( r_o || R_E) - (1+ β)\frac{V_o}{r_π} ( r_o || R_E)$ $V_o [1+(1+ β)\frac{(r_o ||R_E)}{ r_π} ]= ((1+ β)V_b \frac{( r_o || R_E))}{ r_π }$ $V_o [ r_π+(1+ β)( r_o ||R_E) ]=(1+ β)V_b ( r_o ||R_E)$ But $V_b = \frac{R'_i}{(R'_i+ R_s )} x V_s$ Where $R'_i = R_B || [r_π+(1+ β) R_E]$ By substituting value of $V_b$ we get, $V_o [ r_π+(1+ β)( r_o ||R_E) ]= ( r_o ||R_E)(1+ β) x \frac{R'_i}{(R'_(i )+ R_s )} x V_s$ Hence voltage gain $A_Vs$, $A_Vs =\frac{V_o}{V_s} = \frac{(( r_o ||R_E )(1+ β)xR'_i)}{( r_π+(1+ β)( r_o ||R_E )x (R'_(i )+ R_s))}$