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

Marks: 10M

Year: Dec 2016

Given circuit is Dual Input Unbalanced Output (DIUBO), to find Q-point applied input is zero hence given circuit becomes,

$- I_B R_S - VBE - 2I_ER_E - (-VEE) = 0$

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

$-I_BR_S - VBE - 2 (1+\beta)I_BR_E + VEE = 0$

$I_B = \frac{-VBE + VEE}{R_S + 2(1+\beta)R_E}$

$I_B = \frac{-0.7+12}{150 + 2(1+100)8.2K}$

$I_B = 6.82\mu A$

$Icq = \beta I_B = 100 \times 6.82\mu A$

$Icq = 0.682mA$

Apply KVL from VCC to - VEE,

$VCEq = VCC - Icq R_C - 2 I_ER_E - (-VEE)$

Since, $I_E = (1+\beta)I_B = (1+100)6.82\mu A = 0.688mA$

$VCEq = 12 - (0.682m \times 3.3K) -(2 \times 0.688m \times 8.2K) +12$

$VCEq = 10.44V$

Hence Q point of differential amplifier [10.44V, 0.682mA]

$A_d $(Differential Gain):

Figure 1: AC equivalent circuit

$A_d = -\frac{\beta R_C}{2(R_S + \beta r_e)}$

but, $r_e = \frac{25mV}{Icq} = \frac{25mV}{0.682mA} = 36.65\Omega$

$A_d = -\frac{100 \times 3.3K}{2(150+ 100 \times 36.65)}$

$A_d = -43.25$

Answers:

1. Icq = 0.682mA
2. Vceq = 10.44V
3. $A_d = -43.25$