Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 5M

Year: May 2016

Given: $\beta = 80, I_c = 2mA, Vcc = 12V, V_c = 7.6V, V_E = 2.4V$

Apply KVL through Rc,

$Vcc - I_cR_c - V_c = 0$ ................(1)

Since $V_c = 7.6V, Vcc = 12V, I_c = 2mA$, substitute in equation (1).

$12 - 2m \times R_c - 7.6 = 0$

$R_c = \frac{12-7.6}{2\times 10^-3}$

$R_c = 2.2 K \Omega$

Apply KVL through base emitter,

$Vcc - I_B R_B - Vbe - V_E = 0$.............(2)

Since $I_C = \beta I_B $

$I_B = \frac{2mA}{80} = 25 \mu A$

Substitute in equation (2),

$12 - 25 \mu A \times R_B - 0.7 - 2.4 = 0$

$R_B = \frac{12-0.7-2.4}{25 \mu A}$

$R_B = 356K \Omega$

Answers:

• $R_B = 356K\Omega$
• $R_C = 2.2 K \Omega$