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

Marks: 10M

Step 1 : Finding ICq and VCEq

For DC analysis consider input of differential amplifier as zero. Consider figure 2 as shown below,

Apply KVL from base of Q1 to -12V through emitter,

-Vbe – 2 Ie Re – (-12) = 0

$-0.7 + 12 = 2\times Ie\times3.3k$

Ie = 1.712mA

But, ICq1=ICq2=Ie/2,

ICq1=ICq2=0.856mA

Apply KVL from Vcc to –Vee through collector emitter from Q1 transistor

$12 – Icq1 \times Rc – VCEq – 2Re \times Ie – (-12) = 0$

$24 + (0.856mA \times 1.2k) – (2 \times 3.3k \times 1.712mA) = VCEq$

VCEq = 11.67V

Step 2 : Finding Ad, ACM and CMRR

$Ad=-\frac{\beta \times Rc}{Rc+\beta re}$

Since, Rs=0

$Ad=-\frac{Rc}{re}$

$re=\frac{26mV}{ICq}$

$re=\frac{26mV}{0.856mA}=30.37\Omega$

$Ad=-\frac{Rc}{re}=-\frac{1.2k}{30.37}=-39.51$

Ad = - 39.51

Finding ACM,

$ACM=-\frac{\beta \times Rc}{Rs+\beta(re+2Re)}$

Since, Rs=0

$ACM=-\frac{Rc}{re+2Re}=-\frac{1.2k}{30.37+(2\times 3.3k)}$

ACM = - 0.1809

Finding CMRR,

$CMRR=\frac{Ad}{Ac}=\frac{(re+2Re)}{re}$

$CMRR=\frac{30.37+(2\times 3.3k)}{30.37}$

CMRR = 218.31