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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Exa 5.11.4
+clc;
+clear;
+close;
+// Given data
+V_CC = 10;// in V
+R_C = 3;// in k Ω
+R_C= R_C*10^3;// in Ω
+bita = 100;
+I_CO = 20;// in nA
+I_CO = I_CO * 10^-9;// in A
+V_BB = 5;// in V
+R_B = 200;// in kΩ
+R_B= R_B*10^3;// in Ω
+V_BE = 0.7;// in V
+// Applying KVL to the base circuit, V_BB= I_B*R_B+V_BE
+I_B = (V_BB - V_BE)/R_B;// in A
+disp(I_B*10^6,"The base current in µA is");
+I_C = (bita * I_B) + I_CO;// in A
+disp(I_C*10^3,"The collector current in mA is");
+I_E = I_C + I_B;// in A
+disp(I_E*10^3,"Emitter current in mA is");
+V_CE = V_CC - (I_C * R_C);// in V
+disp(V_CE,"Collector emitter voltage in V is");
+