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

Marks: 10M

Year: May 2015

For DC analysis all the connected capacitor acts as open circuit as shown in below circuit diagram,

Figure 1: Equivalent circuit diagram when capacitor acts as open circuit.

Vth and Rth can be found using thevenin's theorem at base of BJT.

$Vth = \frac{Vcc \times R_2}{R_1+ R_2} = \frac{12 \times 16K \Omega}{68K \Omega + 16K \Omega}$

$Vth = 2.28V$

$Rth = R_1 || R_2 = 68K \Omega || 16K \Omega$

$Rth = 12.95K \Omega$

Figure 2: Thevenin's equivalent at base terminal

To find $r_e$ apply KVL from Vth to Ground through base emitter.

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

We know that, $I_E = I_C + I_B = (1+\beta)I_B$

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

$I_B = \frac{Vth - Vbe}{Rth + (1+\beta)R_E} = \frac{2.28 - 0.7}{12.95K \Omega + (1+150) \times 1K \Omega}$

$I_B = 9.64 \mu A$

$I_C = \beta I_B = 150 \times 9.64 \mu A$

$I_C = 1.45 mA$ ....(1)

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

Substitute value obtained from equation (1)

$r_e = \frac{25mV}{1.45mA}$

$r_e = 17.29\Omega$

$\beta r_e = 150 \times 17.29 = 2.59K \Omega$

$Z_i, Z_O$ and $A_V$ can be obtained using AC equivalent circuit as shown below:

Figure 3: AC equivalent circuit diagram

Input resistance ($Z_i$):

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

$Z_i = 68K \Omega || 16K \Omega || 2.59K \Omega$

$Z_i = 2.15K \Omega$

Output resistance ($Z_O$):

$Z_O = R_C = 3.3K \Omega$

Voltage Gain($A_V$):

$A_V = - \frac{R_C}{r_e}$

$A_V = - \frac{3.3K \Omega}{17.29}$

$A_V = - 190.86$

Answer:

• $A_V = -190.86$
• $Z_i = 2.15K \Omega$
• $Z_O = 3.3K \Omega$