diff options
Diffstat (limited to '3655/CH4/EX4.6/Ex4_6.sce')
| -rw-r--r-- | 3655/CH4/EX4.6/Ex4_6.sce | 116 |
1 files changed, 116 insertions, 0 deletions
diff --git a/3655/CH4/EX4.6/Ex4_6.sce b/3655/CH4/EX4.6/Ex4_6.sce new file mode 100644 index 000000000..f61254ee5 --- /dev/null +++ b/3655/CH4/EX4.6/Ex4_6.sce @@ -0,0 +1,116 @@ +// 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// |