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

Marks: 4M

Years: Dec 14

Given:$β_1$ = 50, $β_1$ = 100, $V_C$ = 5V

Calculation with $β_1$ = 50

Step1: Find $I_B$ and $I_c$

Apply KVL to the base loop to write

10 V = $I_EB$ + $I_B R_B$

$I_B$ = $\frac{10-I_EB}{R_B}$ = $\frac{(10-0.7)}{(100 X 10^3 )}$

$I_B$ = 93 μA

$I_C$ = β X $I_B$ = 50 X 93X $10^(-6)$ = 4.65 mA

Step2: Find $R_C$

$V_C$= 5V (Given)

But $V_C$= $I_C$ X $R_C$

5 = 4.65 X $10^(-3)$ X $R_C$

$R_C$ = 1075.3 Ω

ii)Calculation with β_1 = 50

The value of $I_B$ remains unchanged

$I_B$ = 93 μA

New value of collector current is given by,

I_C= β X $I_B$ = 100 X 93X $10^(-6)$ = 9.3 mA

Find $R_C$:

$R_C=\frac{V_C}{I_C} = \frac{(5 V)}{(9.3 mA)}$= 537.6 Ω