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

Marks: 5M

Years: Dec 15

To find $I_B$,

Applying KVL to the base emitter loop, we get

$V_BE +I_E R_E$ = 5

Substitute $I_E$ = (1+β) $I_B$

∴ 5 – 0.7 = (1+β)$ I_B R_E$

∴ 5 – 0.7 = (1+100) $I_B$ x 1 x $10^3$

∴I_B = 0.0425 mA

The emitter current is given by

$I_C$ = $I_Bβ$ = (0.0425x10-3x100) = 4.25 mA

$I_C$= 4.25 mA

$I_E$ = (1+β)$ I_B$ = 101 x 0.0425 mA = 4.2925 mA

Applying KVL to the collector loop we get,

$V_CC - I_c R_c - V_CE $= 0

10 =$ I_c R_c + I_E R_E+ V_CE$

10 = 4.25 mA x 0.5 k + 4.2925 mA x 1k + $V_CE$

$V_CE$= 3.5825 V