initial commit / add all books

38 files changed, 982 insertions, 0 deletions

diff --git a/3655/CH2/EX2.2/Ex2_2.png b/3655/CH2/EX2.2/Ex2_2.png
new file mode 100644
index 000000000..e0b4caf79
--- /dev/null
+++ b/3655/CH2/EX2.2/Ex2_2.png
diff --git a/3655/CH2/EX2.2/Ex2_2.sce b/3655/CH2/EX2.2/Ex2_2.sce
new file mode 100644
index 000000000..53a742bbb
--- /dev/null
+++ b/3655/CH2/EX2.2/Ex2_2.sce
@@ -0,0 +1,35 @@
+// Example 2.2
+//computation of wavelenth of second emitted photon//
+// Page no. 45
+
+clc;
+clear;
+close;
+
+//Given data
+lambdaA=1400;//wavelength of absorbed photon//
+lambdaB=1850;//wavelength of first emitted photon//
+
+//Calculation for total energy of absorbed photon//
+E=(12400/lambdaA);
+
+//Calculation for energy of emitted photon of wavelenth lambdaB//
+E1=(12400/lambdaB);
+
+//Calculation for energy of second emitted  photon//
+E2=E-E1;
+
+//Calculation for wavelenth of second emitted  photon//
+lambda2=(12400/E2);
+
+//Displaying the result in command window
+printf('\n Total energy of absorbed photon in eV = %0.3f eV',E);
+printf('\n \n Energy of emitted photon of wavelength 1850 A = %0.3f eV',E1);
+printf('\n \n Energy of the second emitted photon= %0.3f eV',E2);
+printf('\n \n Wavelenth of second emitted  photon = %0.0f A',lambda2);
+
+//The answers vary due to roundoff error//
+
+
+
+
diff --git a/3655/CH2/EX2.3/Ex2_3.png b/3655/CH2/EX2.3/Ex2_3.png
new file mode 100644
index 000000000..5b9c84032
--- /dev/null
+++ b/3655/CH2/EX2.3/Ex2_3.png
diff --git a/3655/CH2/EX2.3/Ex2_3.sce b/3655/CH2/EX2.3/Ex2_3.sce
new file mode 100644
index 000000000..fe75e7a8a
--- /dev/null
+++ b/3655/CH2/EX2.3/Ex2_3.sce
@@ -0,0 +1,32 @@
+// Example 2.3
+//computation of number of photons per second emitted by lamp//
+// Page no. 46
+
+clc;
+clear;
+close;
+
+//Given data
+lambda=2537;//wavelength of absorbed photon//
+
+
+//Calculation for total energy of absorbed photon//
+E=(12400/lambda);
+
+//Calculation for radiated power//
+P_radiated=(0.05/(1.60*10^-19));
+
+//Calculation for number of photons per second//
+photon=(P_radiated/E);
+
+//Displaying the result in command window
+printf('\n Total energy per photon = %0.2f eV/photon',E);
+printf('\n \n Radiated Power = %0.2f x 10^17 eV/sec',P_radiated*10^-17);
+printf('\n \n The number of photons per second = %0.2f x 10^16 photons/second',photon*10^-16);
+
+//The answers vary due to roundoff error//
+
+
+
+
+
diff --git a/3655/CH2/EX2.4/Ex2_4.png b/3655/CH2/EX2.4/Ex2_4.png
new file mode 100644
index 000000000..6f6436f79
--- /dev/null
+++ b/3655/CH2/EX2.4/Ex2_4.png
diff --git a/3655/CH2/EX2.4/Ex2_4.sce b/3655/CH2/EX2.4/Ex2_4.sce
new file mode 100644
index 000000000..850e1a6e2
--- /dev/null
+++ b/3655/CH2/EX2.4/Ex2_4.sce
@@ -0,0 +1,43 @@
+// Example 2.4
+//computation of (A) minimum speed of electron travelling (B) minimum frequency of photon//
+// Page no. 47
+
+clc;
+clear;
+close;
+
+//Given data
+ip=21.5;//ionization potential of neon
+e=1.602*10^-19;
+m=9.109*10^-31;
+v_freespace=2.998*10^8;//velocity of light in free space
+planck_const=6.63*10^-34;
+
+
+//..................................(A)......................................//
+
+//Calculation for velocity of the electron//
+v=sqrt((2*ip*e)/m);
+
+//..................................(B)......................................//
+
+//Calculation for wavelength of a photon with energy equal to the ionization potential//
+lambda=12400/ip;
+
+//Calculation for frequency of the photon//
+f=v_freespace/(lambda*10^-10);
+
+//Calculation for frequency of the photon using alternate method//
+f1=(ip*e)/planck_const;
+
+
+//Displaying the result in command window
+printf('\n Velocity of the electron = %0.2f x 10^6 m/sec',v*10^-6);
+printf('\n \n Wavelength of a photon with energy equal to the ionization potential = %0.2f A',lambda);
+printf('\n \n Frequency of the photon = %0.1f x 10^15 Hz',f*10^-15);
+
+
+
+
+
+
diff --git a/3655/CH2/EX2.5/Ex2_5.png b/3655/CH2/EX2.5/Ex2_5.png
new file mode 100644
index 000000000..87e7715af
--- /dev/null
+++ b/3655/CH2/EX2.5/Ex2_5.png
diff --git a/3655/CH2/EX2.5/Ex2_5.sce b/3655/CH2/EX2.5/Ex2_5.sce
new file mode 100644
index 000000000..926c0e82c
--- /dev/null
+++ b/3655/CH2/EX2.5/Ex2_5.sce
@@ -0,0 +1,31 @@
+// Example 2.5
+//computation of the energy in jules of the impinging electron after the collision//
+// Page no. 49
+
+clc;
+clear;
+close;
+
+//Given data
+e=1.602*10^-19;
+
+
+//Calculation for the Kinetic energy E1 gained by the electron//
+E1=10*e;
+
+//Calculation for the Kinetic energy E2 needed to be transferred fom the bombarded electron to the atom//
+E2=3.07*e;//(7.73eV-4.66eV)=3.07eV
+
+//Calculation for the Kinetic energy E of the electron after the collision//
+E=E1-E2;
+
+
+//Displaying the result in command window
+printf('\n Kinetic energy E1 gained by the electron = %0.2f x 10^-19 J',E1*10^19);
+printf('\n \n Kinetic energy E2 needed to be transferred fom the bombarded electron to the atom = %0.2f x 10^-19 J',E2*10^19);
+printf('\n \n Kinetic energy E of the electron after the collision = %0.2f x 10^-19 J',E*10^19);
+
+
+
+
+
diff --git a/3655/CH3/EX3.1/Ex3_1.png b/3655/CH3/EX3.1/Ex3_1.png
new file mode 100644
index 000000000..53f0a005c
--- /dev/null
+++ b/3655/CH3/EX3.1/Ex3_1.png
diff --git a/3655/CH3/EX3.1/Ex3_1.sce b/3655/CH3/EX3.1/Ex3_1.sce
new file mode 100644
index 000000000..eb032a131
--- /dev/null
+++ b/3655/CH3/EX3.1/Ex3_1.sce
@@ -0,0 +1,33 @@
+// Example 3.1
+// Computation for mobility of the free electrons in aluminium//
+// Page no.61
+
+clc;
+clear;
+close;
+
+//Given data ;
+d=2.70*10^3//2.70*10^3 kg/m3
+v=3;//3 electrons/atom
+A=26.98;
+M=1.660*10^-27;//1.660*10^-27 kg/atom
+e=1.60*10^-19;
+R=3.44*10^-8;//R=resistivity
+
+//...................................(B)....................................//
+
+//Calculation for concentration of the free electrons in aluminium//
+n=(d*v)/(A*M);
+
+//Calculation for mobility of the free electrons in aluminium//
+mu=10^4/(n*e*R);//mu=mobility of the free electrons
+
+
+
+//Displaying the result in command window
+printf('\n Concentration of the free electrons in aluminium = %0.3f x 10^29 electron/m3',n*10^-29);
+printf('\n \n Mobility of the free electrons in aluminium = %0.2f cm2/V sec',mu);
+
+
+
+//Answers are varying due to round off error//
diff --git a/3655/CH3/EX3.3/Ex3_3.png b/3655/CH3/EX3.3/Ex3_3.png
new file mode 100644
index 000000000..699df4b3e
--- /dev/null
+++ b/3655/CH3/EX3.3/Ex3_3.png
diff --git a/3655/CH3/EX3.3/Ex3_3.sce b/3655/CH3/EX3.3/Ex3_3.sce
new file mode 100644
index 000000000..bc17d25fa
--- /dev/null
+++ b/3655/CH3/EX3.3/Ex3_3.sce
@@ -0,0 +1,29 @@
+// Example 3.3
+// Computation for numerical value of n and EF//
+// Page no.71
+
+clc;
+clear;
+close;
+
+//Given data ;
+A_num=6.02*10^23;//Avogadro number
+atom_wt=184;//Atomic weight
+sg=18.8;//specific gravity of tungstan
+n_elect=2;//number of electrons
+n_atom=1;//number of atom per molecule
+
+
+//Calculation for numerical value of n//
+n=(A_num*(1/atom_wt)*sg*n_elect*n_atom)*10^6;
+
+//Calculation for numerical value of EF//
+EF=3.64*10^-19*n^(2/3);
+
+
+//Displaying the result in command window
+printf('\n Numerical value of n = %0.2f x 10^29 electron/m3',n*10^-29);
+printf('\n \n Numerical value of EF = %0.2f eV',EF);
+
+
+//Answers are varying due to round off error//
diff --git a/3655/CH3/EX3.4/Ex3_4.png b/3655/CH3/EX3.4/Ex3_4.png
new file mode 100644
index 000000000..e92bf6ed8
--- /dev/null
+++ b/3655/CH3/EX3.4/Ex3_4.png
diff --git a/3655/CH3/EX3.4/Ex3_4.sce b/3655/CH3/EX3.4/Ex3_4.sce
new file mode 100644
index 000000000..386820fea
--- /dev/null
+++ b/3655/CH3/EX3.4/Ex3_4.sce
@@ -0,0 +1,24 @@
+// Example 3.4
+// Computation for percentage of electrons leaving a tungstan filament which can surmount a barrier of height 1eV//
+// Page no.78
+
+clc;
+clear;
+close;
+
+//Given data ;
+Vr=1;
+VT=2700/11600;//VT=T/11600
+
+
+//Calculation for percentage of electrons leaving a tungstan filament at 2700 degree K//
+P1=(exp(-Vr/VT))*100;//P1=I/Ith
+
+//Hence only about 1.4% of the electrons have surface-directed energies in excess of 1eV//
+
+
+//Displaying the result in command window
+printf('\n Percentage of electrons leaving a tungstan filament which can surmount a barrier of height 1eV = %0.1f percent',P1);
+
+
+//Answers are varying due to round off error//
diff --git a/3655/CH3/EX3.5/Ex3_5.png b/3655/CH3/EX3.5/Ex3_5.png
new file mode 100644
index 000000000..3d96dccc5
--- /dev/null
+++ b/3655/CH3/EX3.5/Ex3_5.png
diff --git a/3655/CH3/EX3.5/Ex3_5.sce b/3655/CH3/EX3.5/Ex3_5.sce
new file mode 100644
index 000000000..d8ef2e1e9
--- /dev/null
+++ b/3655/CH3/EX3.5/Ex3_5.sce
@@ -0,0 +1,34 @@
+// Example 3.5
+// Computation for percentage increase in zero-external-field thermoinic-emission current because of the schottky effect//
+// Page no.78
+
+clc;
+clear;
+close;
+
+//Given data ;
+V=500;//plate voltage
+ra=1;//anode radius
+rk=0.01;//cathode radius
+r=10^-4;
+T=2500;
+
+
+
+//Calculation for electric field intensity at the surface of the cathode//
+E=(V/(log(ra/rk)))*(1/r);
+
+//Calculation for percentage increase in zero-external-field thermoinic-emission current because of the schottky effect//
+P1=log(%e^((0.434*0.44*E^(1/2))/T));//P1=log(I/Ith)
+P2=10^0.0796;//P2=I/Ith
+
+//I/Ith=1.20,shows that the schottky theory predicts a 20 percent increase over the zero-field emission current//
+
+
+//Displaying the result in command window
+printf('\n Electric field intensity at the surface of the cathode = %0.3f x 10^6 V/m',E*10^-6);
+printf('\n \n log(I/Ith) = %0.4f ',P1);
+printf('\n \n I/Ith = %0.2f ',P2);
+printf("\n \n \n \n I/Ith=1.20,shows that the Schottky theory predicts a 20 percent increase over the zero-field emission current.");
+
+//Answers are varying due to round off error//
diff --git a/3655/CH4/EX4.1/Ex4_1.png b/3655/CH4/EX4.1/Ex4_1.png
new file mode 100644
index 000000000..19eeb31d3
--- /dev/null
+++ b/3655/CH4/EX4.1/Ex4_1.png
diff --git a/3655/CH4/EX4.1/Ex4_1.sce b/3655/CH4/EX4.1/Ex4_1.sce
new file mode 100644
index 000000000..32ad8acdb
--- /dev/null
+++ b/3655/CH4/EX4.1/Ex4_1.sce
@@ -0,0 +1,28 @@
+// Example 4.1
+// Computation for percentage increase in conductivity per degree rise in temperature//
+// Page no.87
+
+clc;
+clear;
+close;
+
+//Given data ;
+T=300;
+EGO=0.785;//forbidden or bandgap energy of germanium at 0 degree K
+k=8.62*10^-5;
+
+
+//Calculation for increase in conductivity per degree rise in temperature//
+conduct_per_temp=(3/(2*T))+(EGO/(2*k*T^2));
+
+//Calculation for percentage increase in conductivity per degree rise in temperature//
+P1=conduct_per_temp*100;
+
+//Thus the conductivity of the intrinsic germanium increases by 5.56 percent per degree kelvin rise in temperature//
+
+//Displaying the result in command window
+printf('\n Increase in conductivity per degree rise in temperature = %0.4f/degree K',conduct_per_temp);
+printf('\n \n percentage increase in conductivity per degree rise in temperature = %0.2f percent',P1);
+
+
+//Answers are varying due to round off error//
diff --git a/3655/CH4/EX4.3/Ex4_3.png b/3655/CH4/EX4.3/Ex4_3.png
new file mode 100644
index 000000000..6477a3841
--- /dev/null
+++ b/3655/CH4/EX4.3/Ex4_3.png
diff --git a/3655/CH4/EX4.3/Ex4_3.sce b/3655/CH4/EX4.3/Ex4_3.sce
new file mode 100644
index 000000000..91749c382
--- /dev/null
+++ b/3655/CH4/EX4.3/Ex4_3.sce
@@ -0,0 +1,44 @@
+// Example 4.3
+// Computation for effective density of states in the conduction band and concentration of electrons in the semiconductor for case(A) and case(B)//
+// Page no.90
+
+clc;
+clear;
+close;
+
+//Given data ;
+mn=1.4*9.1*10^-31;
+k=8.62*10^-5;
+T=300;
+h=6.626*10^-34;
+P=0.25;//p=EC-EF=0.25eV
+T1=400;
+kT=25.9*10^-3;
+
+//.....................................(A)...................................//
+
+//Calculation for effective density of states in the conduction band at T=300K//
+NC=2*(((2*%pi*mn*k*T*(1.60*10^-19))/(h^2))^(3/2));
+
+//Calculation for concentration of electrons in the semiconductor//
+n=NC*exp(-(P)/(kT));
+
+//.....................................(B)...................................//
+
+//Calculation for effective density of states in the conduction band at T=400K//
+NC1=2*(((2*%pi*mn*k*T1*(1.60*10^-19))/(h^2))^(3/2));
+
+//Calculation for concentration of electrons in the semiconductor//
+n1=NC1*exp(-(P)/(kT)); 
+
+//the concentration of electrons and effective density of states in a semiconductor are increased with the increase in the temperature//
+
+//Displaying the result in command window
+printf('\n Effective density of states in the conduction band at T=300K = %0.2f x 10^25 m^-3',NC*10^-25);
+printf('\n \n Concentration of electrons in the semiconductor at T=300K = %0.2f x 10^21 m^-3',n*10^-21);
+printf('\n \n Effective density of states in the conduction band at T=400K = %0.2f x 10^25 m^-3',NC1*10^-25);
+printf('\n \n Concentration of electrons in the semiconductor at T=400K = %0.1f x 10^21 m^-3',n1*10^-21);
+
+
+
+//Answers are varying due to round off error//
diff --git a/3655/CH4/EX4.4/Ex4_4.png b/3655/CH4/EX4.4/Ex4_4.png
new file mode 100644
index 000000000..c08d74d0e
--- /dev/null
+++ b/3655/CH4/EX4.4/Ex4_4.png
diff --git a/3655/CH4/EX4.4/Ex4_4.sce b/3655/CH4/EX4.4/Ex4_4.sce
new file mode 100644
index 000000000..f00cee1a7
--- /dev/null
+++ b/3655/CH4/EX4.4/Ex4_4.sce
@@ -0,0 +1,25 @@
+// Example 4.4
+// Computation for position of the intrinsic fermi level in germanium at 300K//
+// Page no.93
+
+clc;
+clear;
+close;
+
+//Given data ;
+kT=0.0259;
+mn=0.55;
+mp=0.37;
+
+
+//Calculation for position of the fermi level with respect to the middle of the bandgap//
+P=-((3*kT)/4)*log(mn/mp);//P=(EF-E_midgap
+
+//the fermi level is located at -7.7 x 10^-3 eV below the middle of the forbidden gap of the intrinsic germanium at 300K//
+
+
+//Displaying the result in command window
+printf('\n Position of the fermi level with respect to the middle of the bandgap = %0.1f x 10^-3 eV',P*10^3);
+
+
+
diff --git a/3655/CH4/EX4.5/Ex4_5.png b/3655/CH4/EX4.5/Ex4_5.png
new file mode 100644
index 000000000..6595919c5
--- /dev/null
+++ b/3655/CH4/EX4.5/Ex4_5.png
diff --git a/3655/CH4/EX4.5/Ex4_5.sce b/3655/CH4/EX4.5/Ex4_5.sce
new file mode 100644
index 000000000..7a1efbf2f
--- /dev/null
+++ b/3655/CH4/EX4.5/Ex4_5.sce
@@ -0,0 +1,65 @@
+// Example 4.5
+// Computation for position of the intrinsic fermi level and the position of the fermi level with respect to the top of the valance band for case(A),the intrinsic carrier concentration of the semiconductor for case(B), effective masses mn & mp of electrons and holes respectively for case(C)//
+// Page no.95
+
+clc;
+clear;
+close;
+
+//Given data ;
+kT=0.0259;
+NC=1.5*10^18;//NC in cm^-3
+NC1=1.5*10^24;//NC1 in m^-3
+NV=1.3*10^19;//NV in cm^-3
+NV1=1.3*10^25;//NV1 in m^-3
+EG=1.43;
+T=300;
+
+//.................................(A)......................................//
+
+//Calculation for position of the intrinsic fermi level with respect to the centre of the bandgap//
+P1=-(kT/2)*log(NC/NV);//P1=(EF-E_midgap)
+
+//Thus fermi level is located at 0.028eV abov the center of the bandgap//
+
+//Calculation for position of the fermi level with respect to the top of the valance band Ev//
+P2=(EG/2)-((kT/2)*log(NC/NV));//P2=EF-EV
+
+//Hence the fermi level is located at 0.743eV above the valance band Ev//
+
+//Calculation for position of the fermi level with respect to the conduction band edge EC//
+P3=-(EG/2)-(kT/2)*log(NC/NV);//P3=EF-EC
+
+//Hence the fermi level is located at 0.687eV below EC//
+
+//.................................(B)......................................//
+
+//Calculation for intrinsic carrier concentration of the semiconductor//
+ni=sqrt(NC*NV*exp(-EG/kT));
+
+//.................................(C)......................................//
+
+//Calculation for effective mass of electrons mn//
+mn=((NC1/(4.82*10^21))^(2/3))*(1/T);
+
+//mn in Kg:
+mn1=mn*9.1*10^-31;
+
+//Calculation for effective mass of holes mp//
+mp=((NV1/(4.82*10^21))^(2/3))*(1/T);
+
+//mp in Kg:
+mp1=mp*9.1*10^-31;
+
+//Displaying the result in command window
+printf('\n Position of the Fermi level with respect to the centre of the bandgap E-midgap= %0.3f eV',P1);
+printf('\n \n Position of the Fermi level with respect to the top of the valance band Ev = %0.3f eV',P2);
+printf('\n \n Position of the Fermi level with respect to the conduction band edge EC = %0.3f eV',P3);
+printf('\n \n Intrinsic carrier concentration = %0.2f x 10^6 cm^-3',ni*10^-6);
+printf('\n \n Effective mass of electrons in meters = %0.3f m',mn);
+printf('\n \n Effective mass of electrons in Kg = %0.2f x 10^-31 Kg',mn1*10^31);
+printf('\n \n Effective mass of holes in meters = %0.3f m',mp);
+printf('\n \n Effective mass of holes in Kg = %0.2f x 10^-31 Kg',mp1*10^31);
+
+//Answer given in textbook for mn is wrong//
+//Answers are varying due to round off error//
diff --git a/3655/CH4/EX4.6/Ex4_6.png b/3655/CH4/EX4.6/Ex4_6.png
new file mode 100644
index 000000000..fdb07829e
--- /dev/null
+++ b/3655/CH4/EX4.6/Ex4_6.png
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//
diff --git a/3655/CH4/EX4.7/Ex4_7.png b/3655/CH4/EX4.7/Ex4_7.png
new file mode 100644
index 000000000..5d39d1beb
--- /dev/null
+++ b/3655/CH4/EX4.7/Ex4_7.png
diff --git a/3655/CH4/EX4.7/Ex4_7.sce b/3655/CH4/EX4.7/Ex4_7.sce
new file mode 100644
index 000000000..16f778a97
--- /dev/null
+++ b/3655/CH4/EX4.7/Ex4_7.sce
@@ -0,0 +1,48 @@
+// Example 4.7
+// Computation for Calculation for concentration of holes and electrons in a p-type germanium at 300 degreeK for case(A) and position of fermi level with respect to the edge of the conduction band for case(B)//
+// Page no.101
+
+clc;
+clear;
+close;
+
+//Given data ;
+sigma=100;//100 ohmcm^-1
+mu_p=1800;
+e=1.60*10^-19;
+ni=2.5*10^13;
+kT=0.0259;
+NV=6.0*10^19;
+EG=0.72;
+
+
+//.................................(A)......................................//
+
+//Calculation for concentration of holes in a p-type germanium at 300 degreeK//
+Pp=sigma/(mu_p*e);
+
+//Calculation for concentration of electrons in a p-type germanium at 300 degreeK//
+np=ni^2/Pp;
+
+//.................................(B)......................................//
+
+//Calculation for P1//
+P1=kT*log(NV/Pp);//P1=EF-EV
+
+//the fermi level is located at 0.133eV above the edge of the valance band//
+
+//Calculation for position of EF with respect to conduction band//
+P2=P1-EG;
+
+//the fermi level is located at -0.587eV below the edge of the conduction band//
+
+
+//Displaying the result in command window
+printf('\n Concentration of holes of the given germanium = %0.2f x 10^17 cm^-3',Pp*10^-17);
+printf('\n \n Concentration of electrons of the given germanium = %0.2f x 10^9 cm^-3',np*10^-9);
+printf('\n \n EF-EV = %0.3f eV',P1);
+printf("\n (This shows that the fermi level is located at 0.133eV above the edge of the valance band)");
+printf('\n \n Position of EF with respect to conduction band = %0.3f eV',P2);
+printf("\n (This shows that the fermi level is located at -0.587eV below the edge of the conduction band)");
+
+//Answers are varying due to round off error//
diff --git a/3655/CH4/EX4.8/Ex4_8.png b/3655/CH4/EX4.8/Ex4_8.png
new file mode 100644
index 000000000..fdf787a7d
--- /dev/null
+++ b/3655/CH4/EX4.8/Ex4_8.png
diff --git a/3655/CH4/EX4.8/Ex4_8.sce b/3655/CH4/EX4.8/Ex4_8.sce
new file mode 100644
index 000000000..b0597c00b
--- /dev/null
+++ b/3655/CH4/EX4.8/Ex4_8.sce
@@ -0,0 +1,25 @@
+// Example 4.8
+// Computation for temperature at which fermi level conincides with the edge of the conduction band//
+// Page no.101
+
+clc;
+clear;
+close;
+
+//Given data ;
+NC=2.5*10^20;
+
+
+//Calculation for concentration of donor atoms per cm3//
+ND=(1/(2*10^8))*(5*10^22);
+
+//Calculation for desired temperature//
+T=(NC/(4.82*10^21))^(2/3);
+
+
+//Displaying the result in command window
+printf('\n Concentration of donor atoms per cm3 = %0.1f x 10^14 atoms/cm3',ND*10^-14);
+printf('\n \n Desired temperature = %0.2f degree K',T);
+
+
+//Answers are varying due to round off error//
diff --git a/3655/CH4/EX4.9/Ex4_9.png b/3655/CH4/EX4.9/Ex4_9.png
new file mode 100644
index 000000000..52d7036df
--- /dev/null
+++ b/3655/CH4/EX4.9/Ex4_9.png
diff --git a/3655/CH4/EX4.9/Ex4_9.sce b/3655/CH4/EX4.9/Ex4_9.sce
new file mode 100644
index 000000000..c2515340a
--- /dev/null
+++ b/3655/CH4/EX4.9/Ex4_9.sce
@@ -0,0 +1,79 @@
+// Example 4.9
+// Computation for the magnitude and polarity of the Hall voltage between the terminal 1 & 2,the value of Hall coefficient for case(A) and case(B),Hall coefficient and mobility of the majority carrier of the semiconductor for case(D)//
+// Page no.108
+
+clc;
+clear;
+close;
+
+//Given data ;
+e=1.60*10^-19;
+nn=2.5*10^15;
+B=6*10^-5;
+I=10*10^-3;
+w=5*10^-1
+p1=4.0*10^-4;
+Pp=2.5*10^15;
+VH3=6*10^-3;
+w3=0.5;
+d=0.4;
+Vd=5;
+l=1.2;
+
+
+//...................................(A)....................................//
+
+//Calculation for charge density p//
+p=-e*nn;
+
+//negative sign denotes the tyoe of majority carrier in the conduction of current in the semiconductor is eletron//
+
+//Calculation for magnitude of the Hall voltage//
+VH=(B*I)/(p1*w);
+
+//the polarity of Hall voltage at terminal1 is negative with respect to the terminal2//
+
+//Calculation for Hall coefficient//
+RH=1/p;
+
+//...................................(B)....................................//
+
+//Calculation for the charge density//
+p2=e*Pp;
+
+//Calculation for magnitude of the Hall voltage//
+VH2=(B*I)/(p2*w);
+
+//Calculation for Hall coefficient//
+RH2=1/p2;
+
+//The holes are deflected in the -Y direction as electron and hence the terminal1 will be positive with respect to the termninal2//
+
+//...................................(D)....................................//
+
+//Calculation for Hall coefficient//
+RH3=(VH3*w3)/(I*B);
+
+//Calculation for concentration of holes in semiconductor//
+Pp3=1/(e*RH3);
+
+//Calculation for mobility of the holes//
+mu_p=(l/d)*(VH3/(B*Vd));
+
+
+//Displaying the result in command window
+printf("........................Part (A)..........................");
+printf('\n \n Charge density = %0.1f x 10^-4 C/cm3',p*10^4);
+printf('\n \n Magnitude of the Hall voltage = %0.1f x 10^-3 V',VH*10^3);
+printf('\n \n Hall coefficient = %0.1f x 10^3 cm3/C',RH*10^-3);
+printf("\n \n ........................Part (B)..........................");
+printf('\n \n Charge density = %0.1f x 10^-4 C/cm3',p2*10^4);
+printf('\n \n Magnitude of the Hall voltage = %0.0f mV',VH2*10^3);
+printf('\n \n Hall coefficient = %0.1f x 10^3 cm3/C',RH2*10^-3);
+printf("\n \n ........................Part (D)..........................");
+printf('\n \n Hall coefficient = %0.0f x 10^3 cm3/C',RH3*10^-3);
+printf('\n \n Concentration of holes in semiconductor = %0.2f x 10^15 cm^-3',Pp3*10^-15);
+printf('\n \n Mobility of the holes = %0.0f cm2/V sec',mu_p);
+
+
+//Answers are varying due to round off error//
diff --git a/3655/CH9/EX9.1/Ex9_1.png b/3655/CH9/EX9.1/Ex9_1.png
new file mode 100644
index 000000000..0c9825851
--- /dev/null
+++ b/3655/CH9/EX9.1/Ex9_1.png
diff --git a/3655/CH9/EX9.1/Ex9_1.sce b/3655/CH9/EX9.1/Ex9_1.sce
new file mode 100644
index 000000000..1f794c6e1
--- /dev/null
+++ b/3655/CH9/EX9.1/Ex9_1.sce
@@ -0,0 +1,154 @@
+// Example 9.1
+// Computation for ICQ and VCEQ for case(a),o/p voltage Vo, overall gain of amplifier, gain by the transistor, i/p resistance at base, total i/p resistance, overall current gain & o/p resistance for case(b),minimum value of Re2 ofr case(c) and minimum values of capacitors Cb, Cc and Ce for case(d)//
+// Page no.366
+
+clc;
+clear;
+close;
+
+//Given data
+R1_A=33*10^3;
+R2_A=10*10^3;
+VCC=12;
+VBEQ=0.7;
+hfe_A=100
+Re1_A=1*10^3;
+Re2_A=470;;
+Rc=2.2*10^3;
+R1_B=33*10^3;
+R2_B=10*10^3;
+Rs_B=600;
+Rc_B=2.2*10^3;
+Re1_B=1*10^3;
+Re2_B=0;
+RL1_B=47*10^3;
+Vs=10*10^-3;
+hfe_B=150;
+fs_min=2*10^3;
+
+//...................................(A).....................................//
+
+//Calculation for VBB//
+VBB=(R2_A*VCC)/(R1_A+R2_A);
+
+//Calculation for Rb//
+Rb=(R1_A*R2_A)/(R1_A+R2_A);
+
+//Calculation for collector current//
+//for Re2=0:
+ICQ1=hfe_A*((VBB-VBEQ)/(Rb+(hfe_A*(Re1_A))));//IBEQ=(VBB-VBEQ)/(Rb+(hfe*(Re1+Re2)))
+
+//for Re2=470 ohm:
+ICQ2=hfe_A*((VBB-VBEQ)/(Rb+(hfe_A*(Re1_A+Re2_A))));
+
+//Calculation for quiescent collector to emitter voltage//
+//for Re2=0:
+VCEQ1=VCC-ICQ1*(Rc+Re1_A);
+
+//for Re2=470 ohm:
+VCEQ2=VCC-ICQ2*(Rc+Re1_A+Re2_A);
+
+//Calculation for re_dash//
+//when ICQ=1.94mA:
+re_dash1=(26*10^-3)/ICQ1;//IEQ=ICQ
+//when ICQ=1.35mA:
+re_dash2=(26*10^-3)/ICQ2;//for Re2=470 ohm
+
+//...................................(B)....................................//
+
+//Calculation for resistance of the parallel combination of Rc and Rl1//
+Rl1c=(Rc_B*RL1_B)/(Rc_B+RL1_B);
+Rl2e=0;
+
+//Calculation for Vth//
+Vth=((R1_B*R2_B)/(R1_B+R2_B))/(Rs_B+(R1_B*R2_B)/(R1_B+R2_B))*Vs;
+
+//Calculation for Rth//
+Rth=1/((1/Rs_B)+(1/R1_B)+(1/R2_B));
+
+//Calculation for Ib//
+Ib=Vth/(Rth+(hfe_B*(re_dash1+Rl2e)));
+
+//Calculation for the output voltage of circuit//
+Vo=-Rl1c*hfe_B*Ib;
+
+//Calculation for overall voltage gain of the circuit//
+Avs=abs(Vo/Vs);
+
+//Calculation for gain of the transistor//
+Av=abs(Rl1c/(re_dash1+Rl2e));
+
+//Calculation for input resistance at the base//
+Rin_base=hfe_B*(re_dash1+Rl2e);
+
+//Calculation for total input resistance//
+Rin_total=((1/R1_B)+(1/R2_B)+(1/Rin_base))^-1;
+
+//Calculation for overall current gain//
+AIs=((hfe_B*(Rs_B+Rin_total))/(Rth+(hfe_B*(re_dash1+Rl2e))))*(((R1_B*R2_B)/(R1_B+R2_B))/(Rs_B+(R1_B*R2_B)/(R1_B+R2_B)));
+
+//Calculation for output resistance //
+Rout=2.2*10^3;//Rout=Rout1=Rc
+
+//...................................(C)....................................//
+
+//Calculation for a//
+a=260*10^-3/(hfe_A*(VBB-VBEQ));
+
+//Calculation for minimum value of Re2//
+Re2_min=(a/(1-(a*hfe_A)))*(Rb+(hfe_A*Re1_A));
+
+//Calculation for re_dash3//
+re_dash3=Re2_min/10;
+
+//Calculation for overall voltage gain of the amplifier//
+Avs_dash1=abs((hfe_B/(Rth+hfe_B*(Re2_min+re_dash3)))*4.27*10^3*((R1_B*R2_B)/(R1_B+R2_B))/(Rs_B+(R1_B*R2_B)/(R1_B+R2_B)));//Rl2e=Re2=Re2_min
+
+//Calculation for overall voltage gain of the amplifier by neglecting re_dash//
+Avs_dash2=abs((hfe_B/(Rth+hfe_B*(Re2_min)))*4.27*10^3*((R1_B*R2_B)/(R1_B+R2_B))/(Rs_B+(R1_B*R2_B)/(R1_B+R2_B)));//Rl2e=Re2=Re2_min
+
+//...................................(D)....................................//
+
+//Calculation for the minimum value Cb_min of the blocking capacitor Cb//
+Cb_min=1.59/(fs_min*(Rs_B+Rin_total));
+
+//Calculation for the minimum value Cc_min of the coupling capacitor Cc//
+Cc_min=1.59/(fs_min*(Rc+RL1_B));
+
+//Calculation for the minimum value Ce1_min of the bypass capacitor Ce//
+Ce1_min=1.59/(fs_min*Re1_A);
+
+//Displaying the result in command window
+printf("\n ......................(A)............................");
+printf('\n VBB = %0.2f V',VBB);
+printf('\n Rb = %0.2f K',Rb*10^-3);
+printf('\n For Re2=0, Collector current = %0.2f mA',ICQ1*10^3);
+printf('\n For Re2=470 ohm, Collector current = %0.2f mA',ICQ2*10^3);
+printf('\n For ICQ=1.94 mA, Quiescent collector to emitter voltage = %0.2f V',VCEQ1);
+printf('\n For ICQ=1.35 mA, Quiescent collector to emitter voltage = %0.2f V',VCEQ2);
+printf('\n For ICQ=1.94 mA and Re2=0,the value of re_dash= %0.2f K',re_dash1);
+printf('\n For ICQ=1.35 mA and Re2=470 ohm,the value of re_dash = %0.2f ohm',re_dash2);
+printf("\n \n ......................(B)............................");
+printf('\n Resistance of parallel combination of Rc and RL1 = Rl1c = %0.1f K',Rl1c*10^-3);
+printf('\n Resistance of parallel combination of Re2 and RL2 = Rl2e = %0.0f ',Rl2e);
+printf('\n Vth = %0.2f mV',Vth*10^3);
+printf('\n Rth = %0.0f ohm',Rth);
+printf('\n ac base current = Ib = %0.2f microA',Ib*10^6);
+printf('\n Output voltage of circuit = Vo = %0.2f V',Vo);
+printf('\n Overall voltage gain of the circuit = %0.0f ',Avs);
+printf('\n Gain of the transistor = %0.0f ',Av);
+printf('\n Input resistance at the base = %0.0f K',Rin_base*10^-3);
+printf('\n Total input resistance = %0.2f K',Rin_total*10^-3);
+printf('\n Overall current gain = %0.0f ',AIs);
+printf("\n \n ......................(C)............................");
+printf('\n a = %0.4f ',a);
+printf('\n Minimum value of Re2 = %0.0f ohm',Re2_min);
+printf('\n re_dash = %0.1f ohm',re_dash3);
+printf('\n Overall voltage gain of the amplifier = %0.0f ',Avs_dash1);
+printf('\n Overall voltage gain of the amplifier by neglecting re_dash = %0.0f ',Avs_dash2);
+printf("\n \n ......................(D)............................");
+printf('\n The minimum value Cb_min of the blocking capacitor Cb = %0.2f microF',Cb_min*10^6);
+printf('\n The minimum value Cc_min of the coupling capacitor Cc = %0.2f nF',Cc_min*10^9);
+printf('\n The minimum value Ce1_min of the bypass capacitor Ce = %0.1f microF',Ce1_min*10^6);
+
+//Answers are varying due to round off error//
diff --git a/3655/CH9/EX9.2/Ex9_2.png b/3655/CH9/EX9.2/Ex9_2.png
new file mode 100644
index 000000000..66d606068
--- /dev/null
+++ b/3655/CH9/EX9.2/Ex9_2.png
diff --git a/3655/CH9/EX9.2/Ex9_2.sce b/3655/CH9/EX9.2/Ex9_2.sce
new file mode 100644
index 000000000..8b7acb700
--- /dev/null
+++ b/3655/CH9/EX9.2/Ex9_2.sce
@@ -0,0 +1,82 @@
+// Example 9.2
+// Computation for VCEQ and re' for case(a),o/p overall gain, gain,i/p resistance, overall current gain & o/p resistance for case(b),minimum value of Ce for case(c)//
+// Page no.370
+
+clc;
+clear;
+close;
+
+//Given data
+hFE=100;
+hfe=150;
+Rc_A=2.2*10^3;
+VBB=2.79;
+VBEQ=0.7;
+Rb=7.67*10^3;
+Re1=0;
+Re2=2.2*10^3;
+VCC=12;
+Rc_B=0;
+RL=47*10^3;
+R1=33*10^3;
+R2=10*10^3;
+Rs=600;
+fs_min=1000;
+
+//..................................(A).....................................//
+
+//Calculation for collector current//
+//for Re2=0:
+ICQ=hFE*((VBB-VBEQ)/(Rb+(hFE*(Re1+Re2))));
+
+//Calculation for quiescent collector to emitter voltage//
+VCEQ=VCC-ICQ*Re2;
+
+//Calculation for re_dash//
+re_dash=(26*10^-3)/ICQ;//IEQ=ICQ
+
+//...................................(B)....................................//
+
+//Calculation for resistance of the parallel combination of Rc and Rl1//
+Rl2e=(Re2*RL)/(Re2+RL);
+
+//Calculation for Rth//
+Rth=1/((1/Rs)+(1/R1)+(1/R2));
+
+//Calculation for overall voltage gain of the amplifier//
+Av2s=((hfe*Rl2e)/(Rth+hfe*(re_dash+Rl2e)))*(((R1*R2)/(R1+R2))/(Rs+(R1*R2)/(R1+R2)));
+
+//Calculation for input resistance at the base//
+Rin_base=hfe*re_dash;
+
+//Calculation for total input resistance//
+Rin_total=((1/R1)+(1/R2)+(1/Rin_base))^-1;
+
+//Calculation for overall current gain//
+AIs=((hfe*(Rs+Rin_total))/(Rth+(hfe*(re_dash+Rl2e))))*(((R1*R2)/(R1+R2))/(Rs+(R1*R2)/(R1+R2)));
+
+//Calculation for output resistance//
+Rout=Re2*(Rth+hfe*re_dash)/(Rth+hfe*(re_dash+Re2));
+
+//...................................(C)....................................//
+
+//Calculation for the minimum value Ce1_min of the coupling capacitor Ce//
+Ce_min=1.59/(fs_min*(Rout+RL));
+
+//Displaying the result in command window
+printf("\n ......................(A)............................");
+printf('\n Collector current = ICQ = %0.2f mA',ICQ*10^3);
+printf('\n Quiescent collector to emitter voltage = VCEQ = %0.2f V',VCEQ);
+printf('\n Value of re_dash = %0.2f K',re_dash);
+printf("\n \n ......................(B)............................");
+printf('\n Overall voltage gain of the amplifier = %0.2f ',Av2s);
+printf('\n Rl2e = %0.1f K',Rl2e*10^-3);
+printf('\n Rth = %0.0f ohm',Rth);
+printf('\n Input resistance at the base = %0.2f K',Rin_base*10^-3);
+printf('\n Total input resistance = %0.2f K',Rin_total*10^-3);
+printf('\n Overall current gain = AIs = %0.2f ',AIs);
+printf('\n Output resistance = Rout = %0.1f ohm',Rout);
+printf("\n \n ......................(C)............................");
+printf('\n The minimum value Ce1_min of the coupling capacitor Ce = %0.1f nF',Ce_min*10^9);
+
+//Answers are varying due to round off error//
diff --git a/3655/CH9/EX9.4/Ex9_4.png b/3655/CH9/EX9.4/Ex9_4.png
new file mode 100644
index 000000000..7d186df8e
--- /dev/null
+++ b/3655/CH9/EX9.4/Ex9_4.png
diff --git a/3655/CH9/EX9.4/Ex9_4.sce b/3655/CH9/EX9.4/Ex9_4.sce
new file mode 100644
index 000000000..754af667c
--- /dev/null
+++ b/3655/CH9/EX9.4/Ex9_4.sce
@@ -0,0 +1,55 @@
+// Example 9.4
+// Computation for various gains & i/p &o/p impedances//
+// Page no.387
+
+clc;
+clear;
+close;
+
+//Given data
+hfe=50;
+hoe=25*10^-6;
+RL=1*10^3;
+hie=1100;
+hre=2.5*10^-4;
+Rs=1000;
+
+
+//..................................(A).....................................//
+
+//Calculation for current gain//
+AI=-hfe/(1+hoe*RL);
+
+//Calculation for input resistance//
+Ri=hie+(hre*AI*RL);
+
+//Calculation for voltage gain//
+Av=(AI*RL)/Ri;
+
+//Calculation for overall voltage gain//
+Avs=(Av*Ri)/(Ri+Rs);
+
+//Calculation for overall current gain//
+AIs=(AI*Rs)/(Ri+Rs);
+
+//Calculation for output admittance//
+Yo=hoe-((hfe*hre)/(hie+Rs));
+
+//Calculation for output impedance//
+Zo=1/Yo;
+
+//Calculation for the power gain//
+Ap=Av*AI;
+
+//Displaying the result in command window
+
+printf('\n Current gain = AI = %0.1f ',AI);
+printf('\n \n Input resistance = Ri = %0.0f ohm',Ri);
+printf('\n \n Voltage gain = Av = %0.1f ',Av);
+printf('\n \n Overall voltage gain = Avs = %0.1f ',Avs);
+printf('\n \n Overall current gain = AIs = %0.1f ',AIs);
+printf('\n \n Output admittance = Yo = %0.1f x 10^-6 mho',Yo*10^6);
+printf('\n \n Output impedance = Zo = %0.1f K',Zo*10^-3);
+printf('\n \n The power gain = Ap =Av*AI = %0.0f ',Ap);
+
+//Answers are varying due to round off error//