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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Exa 1.29
+clc;
+clear;
+close;
+format('v',6)
+// Given data
+Eta = 1;
+V_T = 0.026;
+// I = Io*( (%e^(V/(Eta*V_T))) - 1 ) and I = -Io;
+// I = -0.9*Io;
+// -0.9*Io = Io*( (%e^(V/(Eta*V_T))) - 1 );
+V = Eta*V_T*log(0.1);// in V
+V = V * 10^3;// in mV
+disp(V,"The voltage in mV is");
+V = 0.05;// in V
+// The ratio of diode current with a forward bias to current with a reverse bias
+If_by_Ir= ( (%e^(V/V_T))-1 )/( (%e^(-V/V_T))-1 );
+disp(If_by_Ir,"The ratio of diode current with a forward bias to current with a reverse bias is");
+Io = 10;// in µA
+V = 0.1;// in V
+// The value of I1
+I1 = Io*( (%e^(V/V_T))-1 );// in µA
+disp(I1,"The value of I1 in µA is");
+V = 0.2;// in V
+// The value of I2
+I2 = Io*( (%e^(V/V_T))-1 );// in µA
+I2 = I2 * 10^-3;// in mA
+disp(I2,"The value of I2 in mA is");
+V = 0.3;// in V
+// The value of I3
+I3 = Io*( (%e^(V/V_T))-1 );// in µA
+I3 = I3 * 10^-6;// in A
+disp(I3,"The value of I3 in A is");