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

Marks: 10M

Year: Dec 2015

Given circuit is Dual Input Balanced Output (DIBO), $r_e$ can be found using Icq so applied input is zero hence given circuit becomes,

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

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

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

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

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

$I_B = 27.96\mu A$

$Icq = \beta I_B = 100 \times 27.96\mu A$

$Icq = 2.796mA$

Ad (Differential Gain):

Figure 1: AC equivalent circuit

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

but, $r_e = \frac{25mV}{Icq} = \frac{25mV}{2.796mA} = 8.94\Omega$

$A_d = -\frac{100 \times 3K}{100+ 100 \times 8.94}$

$A_d = -301.81$

Acm (Common Mode Gain):

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

$Acm = -\frac{100 \times 3K}{100 + [100 \times (8.94 + (2 \times 2K))]}$

$Acm = - 0.747$

CMRR (Common Mode Rejection Ratio):

$CMRR = \frac{A_d}{Acm}$

$CMRR = \frac{-301.81}{-0.747}$

$CMRR = 403.41$

$CMRR$ in db = $20 log (\frac{A_d}{Acm}) = 20 log (403.41)$

$CMRR$ in db = $52.11db$

Answer:

1. CMRR = 403.41
2. CMRR in db = 52.11db
3. $A_d = -301.81$
4. Acm = - 0.747