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+// Determine concentration of electrons and holes
+// Basic Electronics
+// By Debashis De
+// First Edition, 2010
+// Dorling Kindersley Pvt. Ltd. India
+// Example 1-26 in page 52
+
+clear; clc; close;
+
+// Data given
+N_D=2*10^14; // Number of donor atoms in atoms/cm^2
+N_A=3*10^14; // Number of acceptor atoms in atoms/cm^2
+ni=2.5*10^19; // number of intrinsic atoms in atoms/cm^2
+
+// Calculation
+p_0=(0.5*10^14)+sqrt(0.25*10^28 + 6.25*10^26);
+n_0=-(0.5*10^14)+sqrt(0.25*10^28 + 6.25*10^26);
+printf("(a)Concentration of free electrons is %0.3e cm^-3\n",n_0);
+printf("(b)Concentration of holes is %0.3e cm^-3\n",p_0);
+printf("since p_0>n_0 the sample is p-type\n");
+printf("When N_A=N_D=10^15,\n n_0=p_0 from the neutrality equation\n");
+printf("Thus the germanium sample in this question is intrinsic by compensation");
+printf("When N_D=10^16,\n");
+p_0=(6.25*10^26)/10^16;
+printf("(c)p_0=%0.2e cm^-3\n",p_0);
+printf("Since n_0>p_0,germanium sample in this case is n-type");
+
+// Result
+// (a) Number of free electrons are 0.058*10^14 cm^-3 
+// (b) Number of holes are 1.058*10^14 cm^-3
+// Semiconductor can be made intrinsic without doping or by equal doping
+\ No newline at end of file