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

Marks: 4M

Years: Dec 15, May 16

Apply KVL at base loop,

10 - $I_E R_E$-270k $I_B- V_BE$ = 0

10 = $I_E R_E$+270k $I_B+ V_BE$

10 = (1+β) $I_B$ x 1.5k +270k $I_B+ V_BE$

10-0.7 = 101 x $I_B$ x 1.5k +270k $I_B$

9.3 = $I_B x 421.5 x 10^3$

$I_B = 0.022 x 10^(-3) $A

$I_E = (1+β)I_B$= 101 x 0.022 x $10^(-3)$ A = 2.228 mA

$I_E$ = 2.228 mA

Voltage drop across $R_E = I_Ex R_E$= 2.228 mA x 1.5k = 3.342 V

$V_E =V_EE$ – Voltage drop across $R_E$

= -5 + 3.342 = -1.658 V

$V_E$= -1.658 V