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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Exa 1.10
+clc;
+clear;
+close;
+// Given data
+format('v',7)
+V_CC= 9;// in volt
+V_EE= 9;// in volt
+V_BE= 0.7;// in volt (Assuming value)
+R_C= 47;// in k ohm
+R_C= R_C*10^3;// in ohm
+R_E= 43;// in k ohm
+R_E= R_E*10^3;// in ohm
+Ri_1= 20;// in ohm
+Ri_2= Ri_1;// in ohm
+v_in1= 2.5;// in mv
+v_in1=v_in1*10^-3;// in volt
+Bita_1= 75;
+Bita_2= Bita_1;
+I_CQ = (V_EE-V_BE)/(2*R_E+Ri_1/Bita_1);// in amp
+I_E= I_CQ;// in amp
+V_CEQ= V_CC + V_BE - I_CQ*R_C;// in volt
+re_desh= (26*10^-3)/I_E;// in ohm
+// However, voltage gain of single-input, unbalanced-output differential amplifier is given by so
+A_d = R_C/(2*re_desh);
+v_out= A_d*v_in1;// in volt
+disp(v_out,"Output voltage in volt")
+