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+// Exa 2.6
+
+clc;
+clear;
+
+// Given data
+
+// Referring circuit shown in Fig. 2.11(a)
+B=200; // Current gain
+Icq = 100*10^-6; // Amperes
+ADM = 500; // Voltage gain for differential mode signal
+CMRR_db = 80; // in dB(Common mode rejection ratio)
+
+// Solution
+
+// Since gm = Icq/Vt therefore,
+gm = Icq/(25*10^-3); // for Vt = 25 mV
+
+printf('Using Eq. 2.50, we have ADM = -gm*Rc so from this we get Rc as ');
+Rc =abs(- ADM/gm);
+printf(' %d kΩ. \n ',Rc/1000);
+printf('Since CMRR = 80 dB converting it into non dB value so CMRR = ');
+CMRR =  10^(CMRR_db/20);
+printf(' %d. \n ',CMRR);
+printf('Using Eq. 2.55, we get value of Re as ');
+// CMRR = 1+ 2*gm*Re; therefore
+Re = (CMRR-1)/(2*gm);
+printf(' %.2f MΩ. \n ',Re/10^6);