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

Marks: 10M

Year: May 2015

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

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

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

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

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

$I_B = \frac{-0.7+12}{500 + 2(1+100)5K}$

$I_B = 11.18\mu A$

$Icq = \beta I_B = 100 \times 11.18\mu A$

$Icq = 1.118mA$

Apply KVL from VCC to - VEE,

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

Since, $I_E = (1+\beta)I_B = (1+100)11.18\mu A = 1.12mA$

$VCEq = 12 - (1.118m \times1.2K) -(2 \times 1.12m \times 5K) +12$

$VCEq = 11.45V$

Hence Q point of differential amplifier [11.45V, 1.118mA]

Ad (Differential Gain):

Figure 1: AC equivalent circuit

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

but, $r_e = \frac{25mV}{Icq} = \frac{25mV}{1.118mA} = 22.36\Omega$

$A_d = -\frac{100 \times 1.2K}{500+ 100 \times 22.36}$

$A_d = -43.85$

Acm (Common Mode Gain):

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

$Acm = -\frac{100 \times 1.2K}{500 + [100 \times (22.36 + (2 \times 5K))]}$

$Acm = - 0.119$

Answer:

1. Icq = 1.118mA
2. Vceq = 11.45V
3. $A_d = -43.85$
4. $Acm = -0.119$