# Chapter 02 - ENERGY BANDS AND CHARGE CARRIERS IN SEMICONDUCTORS¶

## Example E1 - Pg 38¶

In [1]:
# Exa 2.1 - 38
# Given data
lembda = 11000.;#
lembda = lembda * 10.**-10.;# in m
h = 6.625*10.**-34.;# Planck constant
c = 3.*10.**8.;#speed of light in m/s
e = 1.6*10.**-19.;#charge of electron in C
# Energy of the incident photon should at least be, h*v= Eg, so
E_g = (h*c)/(lembda*e);# in eV
print '%s %.2f %s' %("The energy gap in eV is",E_g,"\n");
#Answer in textbook is different due to rounding error

The energy gap in eV is 1.13



## Example E2 - Pg 38¶

In [2]:
# Exa 2.2 -38
# Given data
E_g = 0.75;# in eV
e = 1.6*10.**-19.;# in C
h = 6.63*10.**-34.;# in J
c = 3*10**8.;# in m/s
#Formula E_g = (h*c)/(lembda*e);
lembda = (h*c)/(E_g*e);# in m
lembda = lembda * 10.**10.;# in A
print '%s %.f %s' %("The wavelength in A is",lembda,"\n");

The wavelength in A is 16575



## Example E3 - Pg 53¶

In [3]:
#Exa 2.3 - 53
# Given data
del_E = 0.3;# in eV
T1 = 300.;# in K
T2 = 330.;# in K
# del_E = K * T1 * log(N/N_c) where del_E= E_C-E_F
# del_E1 = K * T2 * log(N/N_c) where del_E1= E_C-E_F at T= 330 degree K
del_E1 = del_E*(T2/T1);# in eV
print '%s %.2f %s' %("The Fermi level will be eV below the conduction band",del_E1,"eV\n")

The Fermi level will be eV below the conduction band 0.33 eV



## Example E04 - Pg 54¶

In [4]:
#Exa 2.4 -54
# Given data
import math
N_c = 2.8 * 10.**19.;# in cm**-3
del_E = 0.25;# fermi energy in eV
KT = 0.0259;# where K is Boltzmann constant
f_F = math.exp(-(del_E)/KT);
print '%s %.2e %s' %("The probability in the conduction band is occupied by an electron is ",f_F,"\n");
# Evaluation of electron concentration
n_o = N_c * math.exp(-(del_E)/KT);# in cm**-3
print '%s %.2e %s' %("The thermal equilibrium electron concentration in cm**-3 is",n_o,"\n");

The probability in the conduction band is occupied by an electron is  6.43e-05

The thermal equilibrium electron concentration in cm**-3 is 1.80e+15



## Example E05 - Pg 54¶

In [5]:
#Exa2.5-54
# Given data
import math
T1 = 300.;# in K
T2 = 400.;# in K
del_E = 0.27;# Fermi level in eV
#KT = (0.0259) * (T2/T1);# in eV
KT=0.03453
N_v = 1.60 * 10.**19.;# in cm**-3
#N_v = N_v * (T2/T1)**(3./2.);# in cm**-3
# Hole concentration
p_o = N_v * math.exp(-(del_E)/KT);# in per cm**3
print '%s %.2e %s' %("The thermal equilibrium hole concentration per cm**3 is",p_o,"\n");

The thermal equilibrium hole concentration per cm**3 is 6.43e+15



## Example E06 - Pg 58¶

In [6]:
# Exa 2.6 - 58
# Given data
At = 63.5;# atomic weight
Rho = 1.7*10.**-6.;# in ohm cm
d = 8.96;# in gm/cc
N_A = 6.02*10.**23.;# in /gm.mole
e = 1.6*10.**-19.;# in C
#Number of atoms of copper persent per unit volume
n = (N_A/At)*d;
Miu_e = 1./(Rho*n*e);# in cm**2/volt.sec
print '%s %.3f %s' %("The electron mobility is",Miu_e,"cm**2/volt-sec\n");

The electron mobility is 43.281 cm**2/volt-sec



## Example E07 - Pg 59¶

In [7]:
# Exa 2.7 -59
# Given data
l = 0.1;# in m
A = 1.7;# in mm**2
A = A * 10.**-6.;# in m**2
R = 0.1;# in ohm
At = 63.5;# atomic weight
N_A = 6.02*10.**23.;
d = 8.96;# in gm/cc
n = (N_A/At)*d;# in /cc
n = n * 10.**6.;# in /m**3
e = 1.6*10.**-19.;#electron charge in C
# Resistivity of copper
#Formula R = Rho*(l/A);
Rho = (R*A)/l;# in ohm m
# Conductivity of copper
Sigma = 1./Rho;# in mho/m
# Formula Sigma = n*e*Miu_e
Miu_e = Sigma/(n*e);# in m**2/V.sec
print '%s %.6f %s' %("The mobility in m**2/V-sec is",Miu_e,"\n");

The mobility in m**2/V-sec is 0.000043



## Example E08 - Pg 59¶

In [8]:
# Exa 2.8- 59
# Given data
import math
d = 10.5;# in gm/cc
At = 108.;# atomic weight
N_A = 6.025*10.**23.;# in /gm mole
r = 10**-3.;# in m
q = 1.6*10.**-19.;# in C
# The number of electrons per unit volume
n = (N_A/At)*d;# in /cm**3
n = n * 10.**6.;# in /m**3
A = math.pi*((r)**2.);# in m**2
I = 2.;# in A
# Evaluation of drivt velocity with the help of current
# I = q*n*A*V;
V = I/(n*q*A);# in m/s
print '%s %.e %s' %("The drift velocity in m/s is",V,"\n");

# Note: Calculation in the book is wrong, so the answer in the book is wrong.

The drift velocity in m/s is 7e-05



## Example E09 - Pg 59¶

In [9]:
# Exa 2.9-59
import math
# Given data
d = 1.03;# in mm
d = d *10.**-3.;# in m
r = d/2.;# in m
R = 6.51;# in ohm
l = 300.;# in mm
e = 1.6*10.**-19.;# electron charge in C
n = 8.4*10.**28.;# in /m**3
A = math.pi*r**2.;# cross section area
#Formula  R = Rho*(l/A);
Rho = (R* A)/l;#in ohm m
Sigma = 1./Rho;# in mho/m
print '%s %.2e %s' %("The conductivity of copper in mho/m is",Sigma,"\n");
# Evaluation of mobility
#Formula sigma = n*e*Miu_e
Miu_e = Sigma/(n*e);# in m**2/V.sec
print '%s %.2e %s' %("The mobility in m**2/V-sec is",Miu_e,"\n");

The conductivity of copper in mho/m is 5.53e+07

The mobility in m**2/V-sec is 4.12e-03



## Example E10 - Pg 61¶

In [10]:
# Exa 2.10-61
# Given data
Mu_e = 1500.;# in cm**2/volt sec
Mu_h = 500.;# in cm**2/volt sec
n_i = 1.6 * 10.**10.;# in per cm**3
e = 1.6 * 10.**-19.;# in C
# The conductivity of pure semiconductor
Sigma = n_i * (Mu_e + Mu_h) * e;# in mho/cm
print '%s %.2e %s' %("The conductivity of pure semiconductor in mho/cm is",Sigma,"\n");

The conductivity of pure semiconductor in mho/cm is 5.12e-06



## Example E11 - Pg 61¶

In [11]:
# Exa 2.11-61
# Given data
Rho = 10.;# in ohm-cm
Mu_d = 500.;# in cm**2/v.s.
e = 1.6*10.**-19.;# electron charge in C
# The number of donor atom
n_d = 1./(Rho * e * Mu_d);# in per cm**3
print '%s %.2e %s' %("The number of donor atom per cm**3 is ",n_d,"\n");

The number of donor atom per cm**3 is  1.25e+15



## Example E12 - Pg 62¶

In [12]:
#Exa 2.12-62
#Given data
AvagadroNumber = 6.02 * 10.**23.;# in atoms/gm.mole
at_Ge = 72.6;# atom weight of Ge
e = 1.6 * 10.**-19.;# in C
D_Ge = 5.32;# density of Ge in gm/c.c
Mu = 3800.;# in cm**2/v.s.
C_Ge = (AvagadroNumber/at_Ge) * D_Ge;# concentration of Ge atoms in per cm**3
n_d = C_Ge/10.**8.;# in per cc
Sigma = n_d * Mu * e;# in mho/cm
print '%s %.3f %s' %("The conductivity in mho/cm is",Sigma,"\n");

The conductivity in mho/cm is 0.268



## Example E13 - Pg 62¶

In [13]:
# Exa2.13-62
# Given data
Rho = 0.3623 * 10.**-3.;# in Ohm m
Sigma = 1/Rho;#in mho/m
D = 4.42 * 10.**28.;# Ge density in atom/m**3
n_d = D / 10.**6.;# in atom/m**3
e = 1.6 * 10.**-19.;# in C
# The mobility of electron in germanium
Mu = Sigma/(n_d * e);# in m**2/V.sec
print '%s %.2f %s' %("The mobility of electron in germanium in m**2/V.sec is",Mu,"\n");

The mobility of electron in germanium in m**2/V.sec is 0.39



## Example E14 - Pg 62¶

In [14]:
#Exa 2.14-62
# Given data
AvagadroNumber = 6.025 * 10.**26.;# in kg.Mole
W = 72.59;# atomic weight of Ge
D = 5.36 * 10.**3.;#density of Ge in kg/m**3
Rho = 0.42;# resistivity in Ohm m
e = 1.6 * 10.**-19.;# in C
Sigma = 1./Rho;# in mho/m
n = (AvagadroNumber/W) * D;# number of Ge atoms present per unit volume
# Holes per unit volume, H = n*10**-6%
H= n*10.**-8.;
a=H;
# Formula sigma= a*e*Mu_h
Mu_h = Sigma/(a * e);# in m**2/V.sec
print '%s %.4f %s' %("Mobility of holes in m**2/V.sec is",Mu_h,"\n");

Mobility of holes in m**2/V.sec is 0.0334



## Example E15 - Pg 63¶

In [16]:
#Exa 2.15-63
# Given data
e = 1.6 * 10.**-19.;# in C
n_i = 2 * 10.**19.;# in /m**3
Mu_e = 0.36;# in m**2/v.s
Mu_h = 0.17;# in m**2/v.s
A = 1. * 10.**-4.;# in  m**2
V = 2.;#in volts
l = 0.3;# in mm
l = l * 10.**-3.;# in m
E=V/l;# in volt/m
Sigma = n_i * e * (Mu_e + Mu_h);# in mho/m
# J = I/A = Sigma * E
I= Sigma*E*A;
print '%s %.2f %s' %("The current produced in a small germanium plate in amp is",I,"\n");

The current produced in a small germanium plate in amp is 1.13



## Example E16 - Pg 63¶

In [17]:
# Exa 2.16-63
# Given data
D = 4.2 * 10.**28.;#density of Ge atoms per m**3
N_d = D / 10.**6.;# per m**3
e = 1.6 * 10.**-19.;# in C
Mu_e = 0.36;# in m**2/V-sec
# Donor concentration is very large as compared to intrinsic carrier concentration
Sigma_n = N_d *  e * Mu_e;# in mho/m (intrinsic concentration can be neglected)
Rho_n = 1./Sigma_n;# in ohm m
print '%s %.3e %s' %("The resistivity of drop Ge in ohm m is ",Rho_n,"\n");

The resistivity of drop Ge in ohm m is  4.134e-04



## Example E17 - Pg 64¶

In [18]:
# Exa 2.17-64
# given data
e = 1.6 * 10.**-19.;# in C
n_i = 1 * 10.**19.;# in per m**3
Mu_e = 0.36;# in m**2/volt.sec
Mu_h = 0.17;# in m**2/volt.sec
A = 2.;# in cm**2
A = A * 10.**-4.;# im m**2
t = 0.1;# in mm
t = t * 10.**-3.;# in m
V = 4.;# in volts
Sigma_i = n_i * e * (Mu_e + Mu_h);# in mho/m
J = Sigma_i * (V/t);# in Amp/m**2
# Current produced, I= J*A
I = J * A;# in Amp
print '%s %.3f %s' %("The current produced in a Ge sample in Amp is",I,"\n");

The current produced in a Ge sample in Amp is 6.784



## Example E18 - Pg 64¶

In [19]:
#Exa 2.18-64
# Given data
e = 1.6 * 10.**-19.;# in C
Mu_h = 500.;# in cm**2/V.s.
Mu_e = 1500.;# in cm**2/V.s.
n_i = 1.6 * 10.**10.;# in per cm**3
# Conductivity of pure silicon at room temperature
Sigma_i = n_i * e * ( Mu_h + Mu_e);# in mho/cm
print '%s %.2e %s' %("Conductivity of pure silicon at room temperature in mho/cm is",Sigma_i,"\n");

Conductivity of pure silicon at room temperature in mho/cm is 5.12e-06



## Example E19 - Pg 67¶

In [20]:
#Exa 2.19-67
#Given data
l= 0.50*10.**-2.;# width of ribbon in m
d= 0.10*10.**-3.;# thickness of ribbon in m
A= l*d;# area of ribbon in m**2
B = 0.8;# in Tesla
D = 10.5;#density in gm/cc
I = 2.;# in amp
q = 1.6 * 10.**-19.;# in C
n=6.*10.**28.;# number of elec. per m**3
V_H = ( I * B * d)/(n * q * A);# in volts
print '%s %.2e %s' %("The hall Voltage produced in volts is",V_H,"\n");

The hall Voltage produced in volts is 3.33e-08



## Example E20 - Pg 68¶

In [21]:
#Exa 2.20-68
# Given data
l = 1.;# in m
d = 1.;# in cm
d = d * 10.**-2.;# in m
W = 1.;# in mm
W = W * 10.**-3.;# in m
A = d * W;# in m**2
I= 1.;# in A
B = 1.;# Tesla
V_H = 0.074 * 10.**-6.;# in volts
Sigma = 5.8 * 10.**7.;# in mho/m
# The hall coefficient
R_H = (V_H * A)/(B*I*d);# in m**3/c
print '%s %.1e %s' %("The hall coefficient in m**3/c is",R_H,"\n");
# Mobility  of electrons in copper
Mu = Sigma * R_H;# in m**2/volt-sec
print '%s %.1e %s' %("The mobility  of electrons in copper in m**2/volt-sec is ",Mu,"\n");

The hall coefficient in m**3/c is 7.4e-11

The mobility  of electrons in copper in m**2/volt-sec is  4.3e-03



## Example E21 - Pg 69¶

In [22]:
#Exa2.2169
# Given data
n_i = 1.4 * 10.**18.;# in /m**3
n_D = 1.4 * 10.**24.;# in /m**3
# Concentration of electrons
n=n_D;# in /m**3
p = n_i**2./n;# in /m**3
# The ratio of electrons to hole concentration
R = n/p;
print '%s %.e %s' %("The ratio of electrons to hole concentration is",R,"\n");

The ratio of electrons to hole concentration is 1e+12



## Example E22 - Pg 69¶

In [23]:
#Exa 2.22- 69
#Given data
R = 9. * 10.**-3.;# in ohm-m
R_H = 3.6 * 10.**-4.;# in m**3
e = 1.6 * 10.**-19.;# in C
Sigma = 1./R;# in (ohm-m)**-1
#Rho = 1./R_H;# in coulomb/m**3
Rho=2778.
# Density of charge carriers
n = Rho/e;# in /m**3
print '%s %.5e %s' %("Density of charge carriers per m**3 is",n,"\n");
# Mobility of charge carriers
Mu = Sigma * R_H;# in m**2/v-s
print '%s %.2f %s' %("Mobility of charge carriers in m**2/V-s is",Mu,"\n");

Density of charge carriers per m**3 is 1.73625e+22

Mobility of charge carriers in m**2/V-s is 0.04



## Example E23 - Pg 77¶

In [24]:
#Exa 2.23-77
# Given data
e = 1.6 * 10.**-19.;# in C
R_H = 0.0145;# in m**3/coulomb
Mu_e = 0.36;# in m**2/v-s
E = 100.;# in V/m
n = 1./(e * R_H);# in /m**3
# The current density of specimen
J = n * e * Mu_e * E;# in A/m**2
print '%s %.2e %s' %("The current density of specimen in A/m**2 is",J,"\n");

The current density of specimen in A/m**2 is 2.48e+03



## Example E24 - Pg 77¶

In [25]:
#Exa 2.24-77
import math
#Given data
Mu_e = 7.04 * 10.**-3.;# in m**2/v-s
m = 9.1 * 10.**-31.;
E_F = 5.5;# in eV
n = 5.8 * 10.**28.;
e = 1.6 * 10.**-19.;# in C
# Relaxation Time
Torque = (Mu_e/e) * m;# in sec
print '%s %.1e %s' %("Relaxation Time in sec is ",Torque,"\n");
# Resistivity of conductor
Rho = 1. /(n * e * Mu_e);# in ohm-m
print '%s %.3e %s' %("Resistivity of conductor in ohm-m is ",Rho,"\n");
# Velocity of electrons with fermi-energy
V_F = math.sqrt((2 * E_F * e)/m);# in m/s
print '%s %.4e %s' %("Velocity of electrons with fermi-energy in m/s is",V_F,"\n");

#Note: The calculated value of Resistivity of conductor is wrong.

Relaxation Time in sec is  4.0e-14

Resistivity of conductor in ohm-m is  1.531e-08

Velocity of electrons with fermi-energy in m/s is 1.3907e+06



## Example E25 - Pg 78¶

In [26]:
#Exa 2.25-78
import math
# Given data
E= 5.95;# in eV
EF= 6.25;# in eV
delE= 0.01;
# delE= 1-1/(1+exp((E-EF)/KT))
K=1.38*10.**-23.;# Boltzmann Constant in J/K
# The temperature at which there is a 1 % probability that a state 0.30 eV below the Fermi energy level
T = ((E-EF)/math.log(1./(1.-delE) -1.)*1.6*10.**-19.)/K;# in K
print '%s %.f %s' %("The temperature in K is : ",T,"\n")


The temperature in K is :  757



## Example E26 - Pg 78¶

In [27]:
#Exa 2.26-78
# Given data
import math
N_V = 1.04 * 10.**19.;# in cm**-3
T1 = 300.;# in K
T2 = 400.;# in K
del_E = 0.27;# in eV
# The value of N_V at T=400 K,
N_V = N_V * (T2/T1)**1.5;# in cm**-3
KT = (0.0259) * (T2/T1);# in eV
# The thermal equilibrium hole concentration in silicon
P_o = N_V * math.exp(-(del_E)/KT);# in cm**-3
print '%s %.2e %s' %("The thermal equilibrium hole concentration in silicon in cm**-3 is ",P_o,"\n");

The thermal equilibrium hole concentration in silicon in cm**-3 is  6.44e+15



## Example E27 - Pg 78¶

In [28]:
#Exa 2.27-78
import math
#Given data
N_c = 2.8 * 10.**19.;
N_V = 1.04 *10.**19.;
T1 = 550.;# in K
T2 = 300.;# in K
E_g = 1.12;
KT = (0.0259) ;
n_i = math.sqrt(N_c *N_V *(T1/T2)**3.* math.exp(-(E_g)/KT*T2/T1));# in cm**-3
# n_o = N_d/2 + sqrt((N_d/2)**2 + (n_i)**2)
# 1.05*N_d -N_d/2= sqrt((N_d/2)**2 + (n_i)**2)
# Minimum donor concentration required
N_d=math.sqrt((n_i)**2./((0.55)**2.-1./4.));
print '%s %.2e %s' %("Minimum donor concentration required in cm**-3 is",N_d,"\n");

Minimum donor concentration required in cm**-3 is 1.40e+15



## Example E28 - Pg 79¶

In [29]:
#Exa 2.28-79
import math
#Given data
T = 300.;# in K
n_o = 10.**15.;# in cm**-3
n_i = 10.**10.;# in cm**-3
p_o = 10.**5.;# in cm**-3
del_n = 10.**13.;# in cm**-3
del_p = del_n;# in cm**-3
KT = 0.0259;# in eV
delta_E1= KT*math.log(n_o/n_i);# value of E_F-E_Fi in eV
delta_E2= KT*math.log((n_o+del_n)/n_i);# value of E_Fn-E_Fi in eV
delta_E3= KT*math.log((p_o+del_p)/n_i);# value of E_Fi-E_Fp in eV
print '%s %.4f %s' %("The Fermi level for thermal equillibrium in eV is : ",delta_E1,"\n")
print '%s %.4f %s' %("The quase-Fermi level for electrons in non equillibrium in eV is : ",delta_E2,"\n")
print '%s %.3f %s' %("The quasi-Fermi level for holes in non equillibrium in eV is : ",delta_E3,"\n")
print '%s' %("The quasi-Fermi level for electrons is above E_Fi ")
print '%s' %("While the quasi-Fermi level for holes is below E_Fi")

The Fermi level for thermal equillibrium in eV is :  0.2982

The quase-Fermi level for electrons in non equillibrium in eV is :  0.2984

The quasi-Fermi level for holes in non equillibrium in eV is :  0.179

The quasi-Fermi level for electrons is above E_Fi
While the quasi-Fermi level for holes is below E_Fi