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+// Example 4.6
+// Computation for concentration of free electrons and holes for case(A),case(B),case(C) and case(D)//
+// Page no.98
+
+clc;
+clear;
+close;
+
+//Given data ;
+NA=3*10^14;
+ND=2*10^14;
+ni=2.5*10^13;
+mu_n=3800;//3800cm2/V sec
+e=1.60*10^-19;
+mu_p=1800;//1800cm2/V sec
+NA1=10^15;
+ND1=10^15;
+T1=400;
+T2=300;
+EGO=0.785;
+kT2=0.0259;
+NA2=0;
+ND2=10^15;
+
+//.................................(A)......................................//
+
+//Calculation for concentration of electrons//
+n=(1/2)*((ND-NA)+(sqrt((NA-ND)^2+(4*ni^2))));
+
+//Calculation for concentration of holes//
+p=(ni^2)/n;
+
+//Calculation for conductivity of electrons//
+sigma_n=n*mu_n*e;
+
+//Calculation for conductivity of holes//
+sigma_p=p*mu_p*e;
+
+//Calculation for P1//
+P1=sigma_p/sigma_n;
+
+//Calculation for P2//
+P2=mu_p/mu_n;
+
+//Thus P1 is greater than P2 which implies that the conductivity is primarily due to holes in the given sample. Hence the sample is a p-type germanium//
+
+//.................................(B)......................................//
+
+//Calculation for concentration of electrons//
+n1=(1/2)*((ND1-NA1)+(sqrt((NA1-ND1)^2+(4*ni^2))));
+
+//Calculation for concentration of holes//
+p1=(ni^2)/n1;
+
+//Calculation for conductivity of electrons//
+sigma_n1=n1*mu_n*e;
+
+//Calculation for conductivity of holes//
+sigma_p1=p1*mu_p*e;
+
+//Calculation for P3//
+P3=sigma_p1/sigma_n1;
+
+//Hence the sample is an intrinsic germanium//
+
+//.................................(C)......................................//
+
+//Calculation for ni1 at T=400 degreeK//
+ni_400=(((T1/T2)^3)*(exp((EGO/kT2)*(1-(T2/T1)))*(2.5*10^13)^2))^(1/2);
+
+//Calculation for concentration of electrons//
+n2=(1/2)*((ND-NA)+(sqrt((NA-ND)^2+(4*ni_400^2))));
+
+//Calculation for concentration of holes//
+p2=(ni_400^2)/n2;
+
+//ratio of conductivities due to holes and electrons is given by sigma_P/sigma_n=sigma_pi/sigma_ni, hence the germanium sample under consideration is essentially intrinsic//
+
+//.................................(D)......................................//
+
+//Calculation for concentration of electrons//
+n3=(1/2)*((ND2-NA2)+(sqrt((NA2-ND2)^2+(4*ni^2))));
+
+//Calculation for concentration of holes//
+p3=(ni^2)/n3;
+
+//ratio of conductivities due to holes and electrons is given by sigma_P/sigma_n=sigma_pi/sigma_ni, hence the coductivity due to holes is almost negligible as compared to that of electrons in the sample, the sample is essentialy n-type//
+
+
+
+//Displaying the result in command window
+printf(".........................Part (A)..................................");
+printf('\n \n Concentration of electrons = %0.2f x 10^12 cm^-3',n*10^-12);
+printf('\n \n Concentration of holes = %0.2f x 10^14 cm^-3',p*10^-14);
+printf('\n \n Conductivity of electrons = %0.4f (ohm cm)^-1',sigma_n);
+printf('\n \n Conductivity of holes = %0.4f (ohm cm)^-1',sigma_p);
+printf('\n \n Ratio of Conductivity of holes to the Conductivity of electrons = %0.2f ',P1);
+printf('\n \n Ratio of Conductivity of holes to the Conductivity of electrons in an intrinsic germanium = %0.2f ',P2);
+printf("\n \n .........................Part (B)..................................");
+printf('\n \n Concentration of electrons = %0.1f x 10^13 cm^-3',n1*10^-13);
+printf('\n \n Concentration of holes = %0.1f x 10^13 cm^-3',p1*10^-13);
+printf('\n \n Conductivity of electrons = %0.4f (ohm cm)^-1',sigma_n1);
+printf('\n \n Conductivity of holes = %0.4f (ohm cm)^-1',sigma_p1);
+printf('\n \n Ratio of Conductivity of holes to the Conductivity of electrons = Ratio of Conductivity of holes to the Conductivity \n of electrons in an intrinsic germanium = %0.2f ',P3);
+printf("\n (Hence the sample is an intrinsic germanium)");
+printf("\n \n .........................Part (C)..................................");
+printf('\n \n Intrinsic Concentration at T=400 degreeK = %0.1f x 10^15 cm^-3',ni_400*10^-15);
+printf('\n \n Concentration of electrons = %0.2f x 10^15 cm^-3',n2*10^-15);
+printf('\n \n Concentration of holes = %0.2f x 10^15 cm^-3',p2*10^-15);
+printf("\n \n .........................Part (D)..................................");
+printf('\n \n Concentration of electrons = %0.4f x 10^15 cm^-3',n3*10^-15);
+printf('\n \n Concentration of holes = %0.4f x 10^11 cm^-3',p3*10^-11);
+
+
+
+//Answers are varying due to round off error//