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

Marks: 10M

Year: Dec 2015

To find Icq, Vceq all connected capacitors acts as open circuit, hence circuit becomes,

Apply thevenin's theorem at base terminal to determine Rth and Vth,

$Vth = \frac{Vcc \times R_2}{R_1 + R_2}$

$Vth = \frac{10 \times 10K\Omega}{10K\Omega + 33K\Omega}$

$Vth = 2.32V$

$Rth = R_1 || R_2$

$Rth = 10K\Omega || 33K\Omega = 7.67K\Omega$

Apply KVL from Vth to ground through base emitter,

$Vth - Rth \times I_B - Vbe - I_E \times R_E = 0$

But, $I_E = I_C + I_B = (1+\beta)I_B$

$Vth - Rth \times I_B - Vbe - (1+\beta)I_B \times R_E = 0$

$I_B = \frac{Vth - Vbe}{Rth + (1+\beta)R_E}$

$I_B = \frac{2.32 - 0.7}{7.67K\Omega + (1+150) 500\Omega}$

$I_B = 19.47\mu A$

We know that, $I_Cq = \beta I_B$

$I_Cq = 150 \times 19.47\mu A$

$I_Cq = 2.92mA$

Apply KVL from Vcc to ground collector emitter,

$Vcc - Icq R_C - Vceq - I_E R_E = 0$

But $I_E \approx I_C$

$Vcc - Icq (R_C + R_E) = Vceq$

$10 - 2.92m \times (1.2K\Omega + 500\Omega) = Vceq$

$Vceq = 5.036V$

AC Equivalent circuit to find Ri and Ro:

Output Resistance ($R_O$):

$R_O = R_C = 1.2K\Omega$

Input Resistance ($R_i$):

From AC equivalent diagram,

$R_i = R_1 || R_2 || \beta r_e$

$r_e = \frac{25mV}{I_Cq}$

$r_e = \frac{25mV}{2.92mA} = 8.56\Omega$

$\beta r_e = 150 \times 8.56 = 1.28K\Omega$

$R_i = 33K || 10K || 1.28K $

$R_i = 1.097K\Omega$

Answer:

1. $I_Cq = 2.92mA$
2. $Vceq = 5.036V$
3. $R_i = 1.097K\Omega$
4. $R_O = 1.2K\Omega$