written 7.8 years ago by | • modified 7.8 years ago |
Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 10M
Years: May 15
written 7.8 years ago by | • modified 7.8 years ago |
Mumbai University > Electronics and telecommunication engineering > Sem 3 > Analog electronics 1
Marks: 10M
Years: May 15
written 7.8 years ago by |
The input impedance at the base emitter by applying KVL is given by:
$Z_b = \frac{V_i}{ I_b}$ = β r_e + (1+ β) $R_E$
The input impedance can also be written as
$Z_b = \frac{V_i}{ I_b}$ =β ( $r_e+ R_E$)
Since $R_E$ is often much greater than $r_e$
$Z_b$ = β $R_E$
= 120 X 0.56 X $10^3$
= 67.2
Calculate the output impedance $Z_o$:
$Z_o = R_C$ = 2.2k
Calculate the voltage gain $A_V$:
$A_V = \frac{V_o}{Vi}$
Refer fig to write,
$V_o = -I_C R_C$
But $I_C = β I_b$
$V_o = -β I_b R_C$
And $V_i= Z_i X I_b= β R_E I_b$
$A_V= \frac{V_o}{Vi} = \frac{(-β I_b R_C)}{(β R_E I_b )} = \frac{-R_C}{ R_E} = \frac{(-2.2 X 10^3)}{(0.56 X 10^3 )}$= 3.92V