// Exa 2.7 clc; clear; // Given data // Fig. 2.11(a) shows the basic differential amplifier Rc = 2*10^3; // Ω Re = 4.3*10^3; // Ω Vcc = 5 ; // Vcc = |VEE| Bo = 200; Vbe = 0.7; // Volts Vt=25*10^-3; // Volts // Solution printf(' For V1 = V2 = 0, applying KVL for the base emitter loop, we may write,'); printf('\n Vbe+2*(1+Bo)*Ibq*Re-Vee = 0.\n From this we get Ibq as '); Ibq = (Vcc-Vbe)/(2*(1+Bo)*Re); printf(' %.2f μA. \n ',Ibq*10^6); Icq = Bo*Ibq; printf(' The value of Icq = %.3f mA. \n ',Icq*10^3); Vo1 = Vcc - Rc*Icq; printf(' The value of Vo1 = Vo2(due to symmetry) = %.3f V. \n ',Vo1); Vceq = Vo1-(-Vbe); printf(' The value of Vceq = %.3f V. \n ',Vceq); gm = Icq/Vt; r_pi = Bo/gm; // using wq. 2.50 ADM = -gm*Rc; ADM = -gm*Rc; // using equation 2.53(a) Acm can be given as ACM = (-Bo*Rc)/(r_pi+2*(1+Bo)*Re); CMRR = ADM/ACM; CMRR_db = 10*log(CMRR); printf(' The remaining values are as follows: \n ADM = %.2f. \n ACM = %.2f. \n CMRR = %.1f = %.1f dB.\n',ADM,ACM,CMRR,CMRR_db);