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

Marks: 10M

Year: May 2016

For Q Point all connected capacitor acts as open circuit. Hence given circuit becomes,

Figure below shows the equivalent circuit and thevenin's equivalent circuit diagram.

Figure 1: Equivalent circuit and Thevenin's equivalent circuit

Vth and Rth can be found using thevenin's theorem at base terminal,

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

$Vth = \frac{12 \times 12K \Omega}{47K \Omega + 12K \Omega}$

$Vth = 2.44V$

$Rth = R_1 || R_2 = 47K \Omega || 12K \Omega$

$Rth = 9.55K \Omega$

Apply KVL from Vth to ground through base emitter terminal,

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

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

$I_B = \frac{Vth - Vbe}{Rth + (1+\beta) R_E} = \frac{2.44-0.7}{9.55K \Omega + (1+100) 1K \Omega}$

$I_B = 15.73 \mu A$

$I_Cq = \beta I_B = 100 \times 15.73 \mu A$

$I_Cq = 1.57mA$

Apply KVL from Vcc to ground through collector emitter,

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

But, $I_E \approx I_C$

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

$Vceq = 12 - 1.57mA (2.2K \Omega + 1K \Omega)$

$Vceq = 6.976V$

Hence Q point of BJT (6.976V, 1.57mA).

DC load line:

For DC load line apply KVL from Vcc to ground through collector emitter,

$Vcc - I_C R_C - Vce - I_E R_E = 0$

$Vcc - I_C R_C - Vce - (I_C + I_B) R_E = 0$

But, $I_C \gt\gt I_B, I_C + I_B \approx I_C$

$Vcc - I_C R_C - Vce - I_C R_E = 0$

$I_C = \frac{Vcc - Vce}{R_C + R_E}$

$I_C = \frac{Vcc}{R_C + R_E} - \frac{Vce}{R_C + R_E}$ .......(1)

$y = c + mx$

Put $I_C = 0$ in equation (1)

$Vce = Vcc = 12V$

Put $Vce = 0$ in equation (1)

$I_C = \frac{Vcc}{R_C + R_E}$

$I_C = \frac{12}{2.2K \Omega + 1K \Omega} = 3.75mA$

Figure below shows the DC load line,

Figure 2: DC load line