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authorpriyanka2015-06-24 15:03:17 +0530
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+clear;
+clc;
+//Example 10.4
+Iref=1;
+Io=12*10^-3;
+V1=5;
+V2=-5;
+Vt=0.026;
+Vbe=0.7;
+R1=(V1-Vbe-V2)/Iref;
+printf('\nResistance R1 =%.2fKOhm\n',R1)
+Re=(Vt/Io)*log(Iref/Io);
+printf('\nResistance Re =%.2fKOhm\n',Re)
+Vbe=Io*Re;
+printf('\ndifference between two B-E voltages=%.2f V\n',Vbe)