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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+// Exa 2.35
+clc;
+clear;
+close;
+// Given data
+N_A= 4.4*10^22/10^8;// in /m^3
+N_D= 10^3*N_A;// in /m^3
+ni= 2.5*10^13;// /cm^3
+Vt= 26;// in mV
+Vt= Vt*10^-3;// in V
+Vj= Vt*log(N_A*N_D/ni^2);// in V
+disp(Vj,"The junction potential in volts is : ")