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_oI_b =\frac{( V_b- V_o)}{r_π}$Substituting value of$ I_bfor V_oV_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))}$