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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Exa 5.14.9
+clc;
+clear;
+close;
+//Given data
+bita = 100;
+V_CE = 0.2;//in V
+V_BE = 0.8;// in V
+R_C= 500;// in Ω
+R_B= 44*10^3;// in Ω
+R_E= 1*10^3;// in Ω
+V_CC= 15;// in V
+V_GE= -15;// in V
+// Applying KVL to collector circuit
+// V_CC-V_GE - I_Csat*R_C-V_CE-I_E*R_E=0, but I_Csat= bita*I_Bmin and I_E= 1+bita
+I_Bmin= (V_CC-V_GE-V_CE)/(R_C*bita+(1+bita)*R_E);// in A
+// Applying KVL to the base emitter circuit
+// V_BB-I_Bmin*R_B-V_BE-I_E*R_E + V_CC=0
+V_BB= I_Bmin*R_B + V_BE + (1+bita)*I_Bmin*R_E-V_CC;// in V
+disp(I_Bmin*10^3,"The value of I_B(min) in mA is : ")
+disp(V_BB,"The value of V_BB in volts is : ")