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+//Exa 4.16.3

+clc;

+clear;

+close;

+// Given data

+Eta = 1;

+V_T = 26;// in mV

+V_T= V_T*10^-3;// in V

+// I = I_o * (%e^(V/(Eta*V_T)) - 1) and I = -(0.9) * I_o;

+V= log(1-0.9)*V_T;// in V

+disp(V,"The voltage in volts is : ")

+// Part (ii)

+V1=0.05;// in V

+V2= -0.05;// in V

+ratio= (%e^(V1/(Eta*V_T))-1)/(%e^(V2/(Eta*V_T))-1)

+disp(ratio,"The ratio of the current for a forward bias to reverse bias is : ")

+// Part (iii)

+Io= 10;// in µA

+Io=Io*10^-3;// in mA

+//For 

+V=0.1;// in V

+I = Io * (%e^(V/(Eta*V_T)) - 1);// in mA

+disp(I,"For V=0.1 V , the value of I in mA is : ")

+//For 

+V=0.2;// in V

+I = Io * (%e^(V/(Eta*V_T)) - 1);// in mA

+disp(I,"For V=0.2 V , the value of I in mA is : ")

+//For 

+V=0.3;// in V

+I = Io * (%e^(V/(Eta*V_T)) - 1);// in mA

+disp(I*10^-3,"For V=0.3 V , the value of I in A is : ")

+disp("From three value of I, for small rise in forward voltage, the diode current increase rapidly")