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+//Chapter 3_Semoconductor Devices Fundamentals
+//Caption: Hole Concentration
+//Example3.4: A semiconductor is doped with a concentration of 10^17 atoms/cm^3 of rsenic . What is the equilibrium hole concentration p at 300K. Given ni=1.5*10^10 cm^-3
+// Solution:
+clear;
+clc;
+function p=holeconcentration(ni,Nd)//ni=intrinsiic concentration=1.5*10^10 cm^-3, Nd: donar concentration; since, Nd>>ni, so Nd=n=10^17 atoms/cm^3.
+    p=ni^2/Nd
+    disp('hole concentrartion at 300K is:')
+    disp('per cubic cm',p)
+endfunction
+//holeconcentration(1.5*10^10,10^17);
+\ No newline at end of file