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

Marks: 5M

Years: Dec 15

Applying KVL to the base loop we get,

$V_BB - I_B R_B - V_BE$ = 0

Substituting the given values,

4 - 220k $I_B$–0.7 = 0

∴$I_B = \frac{3.3}{220}$ = 0.015mA

$I_B$= 0.015mA

Now the collector current can easily calculated by the following formula

$I_C$ = $I_B$ β = (0.015x10-3x200) = 3mA

$I_C $= 3mA

Applying KVL to the collector loop we get,

$V_CC - I_c R_c - V_CE $= 0

$V_CC - I_c R_c = V_CE$

$V_CE$ = 10 – 3 mA x 2k = 10 – 6

= 5 V

$V_CE$ = 5 V