Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 10M

Hence Hybrid $\pi$ equivalent circuit becomes,

Av(Voltage Gain) :

The output voltage is,

$V_O= - (ro|| Rc||RL)g_m Vbe$…………………..(1)

The base emitter voltage Vbe, which controls the collector current, Vbe is voltage across Zi, Applying voltage divider rule,

$Vbe=\frac {R1||R2||r\pi}{R_s+R1||R2||r\pi}Vs$

Substituting value of Vbe in equation (1),

$V_o=-(ro||Rc||RL)g_m\frac{R1||R2||r\pi}{Rs+(R1||R2||r\pi)}Vs$

This gives voltage gain Rs,

$Av=\frac{Vo}{Vs}=\frac{-(R1||R2||r\pi)(ro||Rc||RL)g_m}{Rs+(R1|R2||r\pi)}$

$Av=\frac{-(R1||R2||r\pi)\times(ro||Rc||RL)g_m}{Rs+(R1||R2||r\pi)}$

Ai( Current Gain) :

Current Io flows through load resistance RL. Hence applying current divider rule at C,

$Io=\frac{-g_mVbe\times(ro||Rc)}{(ro||Rc)+RL}$----------------------------(2)

From Figure 2,

$Vbe=Ibr\pi$

Apply current divider rule at b,

$Ib=\frac{R1||R2}{r\pi+(R1||R2)}Ii$

$Vbe=\frac{(R1||R2)r\pi}{r\pi+(R1||R2)}Ii$

$Vbe=(R1||R2||r\pi)Ii$

Substitute Vbe in equation (2),

$Ai=\frac{Io}{Ii}=\frac{-g_m(R1||R2||r\pi)\times(ro||Rc)}{(ro||Rc)+RL}$

$Ai=\frac{-g_m(R1||R2||r\pi)\times(ro||Rc)}{(ro||Rc)+RL}$

Zi (Input Resistance):

Zi'=Vb/Ib

But

$Vb = Ib r\pi$

Hence,

$Zi’ = r\pi$

Zi = R1||R2||Zi’

$Zi=R1||R2||r\pi$

Zo (Output Resistance):

For Output Resistance:

Step 1: Set Vs=0V

Step 2: Open the load.

Step 3: Connect an imaginary voltage source Vs at output terminal

Hence an imaginary voltage source delivers current through parallel combination ro and Rc. Where,$ro= V_A/ ICQ$