Chapter No 2 : Semiconductor Materials and Their Properties

Example 2.1, Page No 65¶

In [27]:
#initialisation of variables
lambda1 = 11000.0  #in Å
lambda1 = lambda1 * 10**-10
h = 6.625*10**-34
c = 3*10**8
q = 1.6*10**-19    #in C

#CALCULATIONS
E_g = h*c/lambda1  #in J
E_g= E_g/q        #in eV

#RESULTS
print('The energy gap in Si  is = %.2f eV' %E_g)

#Note: The answer in the book is not correct

The energy gap in Si  is = 1.13 eV


Example 2.2, Page No 65¶

In [28]:
import math

#initialisation of variables
E_g = 0.75    #in eV
q=1.6*10**-19   #in C
E_g = E_g*q   #in J
h = 6.63*10**-34   #in J
c = 3*10**8    #in m/s

#CALCULATIONS
lembda = (h*c)/E_g    #in m
lembda = lembda * 10**10   #in Å

#RESULTS
print('The wavelength is =%.f  Å ' %lembda)

The wavelength is =16575  Å


Example 2.3 Page No 81¶

In [29]:
import math
from numpy import *

#initialisation of variables
del_E = 0.3        #value of E_C-E_F in eV
T1 = 330.0          #in K
T = 300            #in K

#CALCULATIONS
del_E1 = del_E*(T1/T)   #value of E_C-E_F in eV

#RESULTS
print('The position of fermi level   is =%.2f eV' %del_E1)
print('Hence the Fermi level will be %.2f eV below the conduction band' %del_E1)

The position of fermi level   is =0.33 eV
Hence the Fermi level will be 0.33 eV below the conduction band


Example 2.4, Page No 81¶

In [30]:
import math

#initialisation of variables
K = 8.63*10**-5
T = 300.0     #in K
N_C = 2.8*10**19     #in cm^-3
del_E = 0.25

#CALCULATIONS
f_F = math.exp( (-del_E)/(K*T) )
n_o = N_C*math.exp( (-del_E)/(K*T) )    #in cm^-3

#RESULTS
print('The probability is =%.6f ' %f_F)
print('The thermal equillibrium electron concentration  is =%.2f X 10^15 cm^-3' %(n_o/(10**15)))
print('Approximation  error ')
# The answer is same as in book only thy have rounded off to 1.8 * 10^15 which is same as above

The probability is =0.000064
The thermal equillibrium electron concentration  is =1.79 X 10^15 cm^-3
Approximation  error


Example 2.5 Page No 82¶

In [31]:
import math

#initialisation of variables
N_V = 1.04*10**19    #in cm^-3
T1 = 400.0        #in K
T2 = 300.0        #in K
del_E = 0.27     #value of E_F-E_V in eV
K = 0.0259

#CALCULATIONS
N_V= N_V*(T1/T2)**(3/2)      #in cm^-3
KT = K*(T1/T2)              #in eV
p_o = N_V*math.exp( (-del_E)/(KT) )    #in /cm^3

#RESULTS
print('The hole concentration is =%.2f x 10^15 per cm^3' %(p_o/(10**15)))
print('Approximationa error ')

The hole concentration is =5.58 x 10^15 per cm^3
Approximationa error


Example 2.6 Page No 86¶

In [32]:
import math

#initialisation of variables
N = 6.02*10**23
A = 63.5            #atomic weight
Rho = 1.7*10**-6    #in ohm cm
d = 8.96           # in gm/cc

#CALCULATIONS
n = (N/A)*d     #in /cc
e = 1.6*10**-19    #in C
Miu_e = 1/(Rho*n*e)    #in cm^2/volt-sec

#RESULTS
print('The mobility of electron  is =%.2f  cm^2/volt.sec' %Miu_e)

The mobility of electron  is =43.28  cm^2/volt.sec


Example 2.7, Page No 87¶

In [33]:
import math
from sympy.mpmath import *
import cmath

#initialisation of variables
d = 8.96   #in gm/cc
At = 63.5   #atomic weight
N_A = 6.02*10**23   #in /gm mole
l = 0.1 #in m
e = 1.6*10**-19   #in C
A = 1.7*10**-6   #in m^2
R = 0.1        #in ohm

#CALCULATIONS
n = (N_A/At)*d   #in /cc
n = n * 10**6   #in /m^3
Rho = (R*A)/l   #in ohm.m
Sigma = 1/Rho   #in mho/m
Miu_e = Sigma/(n*e)     #in m^2/V-sec

#RESULTS
print('The electron mobility is =%.3f X 10^-3 m^2/V-sec ' %(Miu_e*(10**3)))

The electron mobility is =0.043 X 10^-3 m^2/V-sec


Example 2.8 Page No 87¶

In [34]:
import math

#initialisation of variables
N_A = 6.025*10**23    #in /gm mole
d = 10.5             #in gm/cc
At = 108.0         #atomic weight of

#CALCULATIONS
n = (N_A/At)*d   #in /cm^3
n = n * 10**6   #in /m^3
r = 10**-3        #in m
A = math.pi * ((r)**2)   #in m^2
q = 1.6*10**-19
I = 2       #in A
V = I/(n*q*A)      #in m/s

#RESULTS
print('The drift velocity of an electron  is =%.f X 10^-4 m/s' %(V*(10**5)))
print('Approximationa error ')
# answer is same as in book only they have rounded of 0.000068 to 7 * 10^-4

The drift velocity of an electron  is =7 X 10^-4 m/s
Approximationa error


Example 2.9 Page No 88¶

In [35]:
import math

#initialisation of variables
e= 1.6*10**-19   #in C
d= 1.03      #in mm
d= d*10**-3   #in m
R= 6.51    #in ohm
l= 300.0  #in m
n= 8.4*10**28   #per m^3

#CALCULATIONS
r= d/2     #in m (radius)
A= math.pi*r**2   #in m^2
rho= R*A/l     #in ohm meter
sigma=  1/rho    #in mho/m
miu_e= sigma/(n*e)    #m^2/V-sec

#RESULTS
print('The coductivity of copper is =%.2f  mho/m' %sigma)
print('The mobility of charge carriers  is : =%.6f  m^2/V-sec' %miu_e)

The coductivity of copper is =55306469.41  mho/m
The mobility of charge carriers  is : =0.004115  m^2/V-sec


Example 2.9.ii Page No 88¶

In [36]:
import math

#initialisation of variables
Miu_e = 1500.0     #in cm^2/volt.sec
Miu_h = 500.0      #in cm^2/volt.sec
n_i = 1.6*10**10  #in /cm^3
e = 1.6*10**-19   #in C

#CALCULATIONS
Sigma_i = n_i*(Miu_e+Miu_h)*e   #in mho/cm
Sigma = Sigma_i                 #in mho/cm

#RESULTS
print('The conductivity of pure silicon  is =%.7f  mho/cm' %Sigma)

The conductivity of pure silicon  is =0.0000051  mho/cm


Example No 2.10 Page No 89¶

In [37]:
#initialisation of variables
Miu_e = 1500.0  #in cm^2/volt.sec
Miu_h = 500.0  #in cm^2/volt.sec
n_i = 1.6*10**10   #in /cm^3
e = 1.6*10**-19    #in C

#CALCULATIONS
Sigma_i = n_i*(Miu_e+Miu_h)*e    #in mho/cm
Sigma = Sigma_i              #in mho/cm

#RESULTS
print('The conductivity of pure silicon  is =%.2f X 10^-6  mho/cm' %(Sigma*(10**6)))

The conductivity of pure silicon  is =5.12 X 10^-6  mho/cm


Example No 2.11 Page No 90¶

In [38]:
#initialisation of variables
Miu_d = 500.0   #in cm^2/V.S
Rho = 10       #in ohm cm

#CALCULATIONS
e = 1.6*10**-19   #in C
n_d = 1/(Rho*e*Miu_d)     #in /cm^3... correction

#RESULTS
print('The number of donor  is =%.2f X 10^15 atom per cm^3' %(n_d/(10**15)))

The number of donor  is =1.25 X 10^15 atom per cm^3


Example No 2.12 Page No 90¶

In [39]:
#initialisation of variables
d = 5.32         #in gm/cc
N_A = 6.02*10**23   #in atoms/gm.mole
At = 72.6       #atomic weight
Miu = 3800    #in cm^2/v.s

#CALCULATIONS
n_d = (N_A/At) * d          #in /cm^3
n_d = n_d * 10**-8    #in /cc
e = 1.6*10**-19    #in C
Sigma = n_d * Miu * e           #in mho/cm

#RESULTS
print('The conductivity of specimen is =%.2f  mho/cm' %Sigma)

The conductivity of specimen is =0.27  mho/cm


Example No 2.13 Page No 90¶

In [40]:
#initialisation of variables
Rho = 0.3623*10**-3    #in ohm m
d = 4.42*10**28      #Ge density in atoms/m^3

#CALCULATIONS
Sigma = 1/Rho       #in mho/m
n_d = d*10**-6    #in atoms/m^3
e = 1.6*10**-19    #in C
Miu = Sigma/(n_d*e)    #in m^2/V.sec

#RESULTS
print('The electron mobility is =%.2f  m^2/V-sec' %Miu)

The electron mobility is =0.39  m^2/V-sec


Example No 2.14 Page No 91¶

In [41]:
#initialisation of variables
N_A = 6.025*10**26    #in /kg.Mole
At = 72.59          #atomic weight
d = 5.36*10**3    #in kg/m^3
R = 0.42            #in ohm m
B_i = 10**-6    #rate of boron impurity in %
e = 1.6*10**-19     #in C

#CALCULATIONS
n = (N_A/At)*d    #number of Ge atoms
h = n/10**8      #holes per unit volume
Miu_h = 1/(R*h*e)       #in m^2/V.sec

#RESULTS
print('The Mobility of holes is =%.2f m^2/V-sec' %Miu_h)

The Mobility of holes is =0.03 m^2/V-sec


Example No 2.15 Page No 90¶

In [42]:
#initialisation of variables
n_i = 2*10**19     #in /m^3
Miu_e = 0.36    #in m^2/v.s
Miu_h = 0.17   #in m^2/v.s

#CALCULATIONS
A = 1*10**-4     #in m^2
V = 2            #in Volts
l = 0.3         #in mm
l = l * 10**-3   #in m
e = 1.6*10**-19   #in C
Sigma_i = n_i * e * (Miu_e+Miu_h)   #in mho/m
I = (Sigma_i * V*A)/l        #in amp

#RESULTS
print('The current in amp is =%.2f ' %I)

The current in amp is =1.13


Example No 2.16 Page No 92¶

In [43]:
#initialisation of variables
d = 4.2*10**28   #in atoms/m^3
n_d = d/10**6   #in atoms/m^3
e = 1.6*10**-19   #in C
Miu_e = 0.36    #in m^2/V-sec

#CALCULATIONS
Sigma_n = n_d *e *Miu_e    #in mho/m
Rho_n = 1/Sigma_n          #ohm m

#RESULTS
print('The resistivity in Ωm is =%.2f X 10^-3' %(Rho_n*(10**3)))

The resistivity in Ωm is =0.41 X 10^-3


Example No 2.17 Page No 92¶

In [44]:
#initialisation of variables
n_i = 1*10**19        #in /m^3
Miu_e = 0.36     #in m^2/volt.sec
Miu_h = 0.17      #in m^2/volt.sec
A = 2                #in cm^2
A = A *  10**-4    #in m^2
t = 0.1         #in mm
t = t*10**-3   #in m
V = 4        #in volts
e = 1.6*10**-19   #in C

#CALCULATIONS
Sigma_i = n_i * e * (Miu_e + Miu_h)   #mho/m
J = Sigma_i * (V/t)             #in Amp/m^2
I = J*A                      #in Amp

#RESULTS
print('The current in Amp is =%.2f ' %I)

The current in Amp is =6.78


Example No 2.18 Page No 92¶

In [45]:
#initialisation of variables
Miu_h = 500.0                  #in cm^2/V.s
Miu_e = 1500.0                 #in cm^2/V.s
n_i = 1.6*10**10              #in /cm^3
e = 1.6*10**-19               #in C

#CALCULATIONS
Sigma_i = n_i * e * (Miu_e+Miu_h)       #in mho/cm

#RESULTS
print('The conductivity of pure silicon in mho/cm is =%.2f X 10^-6 ' %(Sigma_i*(10**6)))

The conductivity of pure silicon in mho/cm is =5.12 X 10^-6


Example No 2.19 Page No 96¶

In [46]:
#initialisation of variables
Si_density = 10.5                #in gm/cc
N_A = 6.025*10**23
At = 108.0                  #atomic weight
B = 0.8                   #in Tesla
w = 0.50               #in cm
w = w * 10**-2          #in m
t = 0.10              #in mm
t = t * 10**-3       #in m
A = w*t                 #in m^2
q = 1.6*10**-19          #in C
I = 2                    #in ampere

#CALCULATIONS
n = (N_A/At) * Si_density  #in /cc
n = n * 10**6              #in /m^3
V_H = (B*I*t)/(n*q*A)           #in volts

#RESULTS
print('The hall voltage produced is =%.3f X 10^-7 volts' %(V_H*(10**7)))

The hall voltage produced is =0.341 X 10^-7 volts


Example No 2.20 Page No 96¶

In [47]:
#initialisation of variables
Sigma = 5.8*10**7        #in mho/m
l = 1                     #in m
d = 1                  #in cm
d = d * 10**-2          #in m
W = 1               #in mm
W = W*10**-3       #in m
I = 1             #in Amp
B = 1            #in Tesla
V_H = 0.074*10**-6        #in Volts
A = 10**-2 * 10**-3          #in m^2

#CALCULATIONS
R_H = (V_H*A)/(B*I*d)              #in m^3/c
Miu = Sigma * R_H                    #in m^2/volt.sec

#RESULTS
print('Hall coefficient  is =%.1f X 10^-11  m^3/c' %(R_H*(10**11)))
print('The mobility of electron is = %.2f X 10^-3 m^2/volt.sec ' %(Miu*(10**3)))

Hall coefficient  is =7.4 X 10^-11  m^3/c
The mobility of electron is = 4.29 X 10^-3 m^2/volt.sec


Example No 2.21 Page No 97¶

In [48]:
#initialisation of variables
n_i = 1.4*10**18             #in /m^3
n_D = 1.4*10**24             #in /m^3
n = n_D                      #in /m^3

#CALCULATIONS
p = n_i**2/n               #in /m^3
R_e = n/p              #Ratio of electron

#RESULTS
print('Concentration of is =%.2f X 10^12 holes per m^3 ' %(p/(10**12)))
print('Ratio of electron to hole concentration is =%.f 10^12 ' %(R_e/(10**12)))

Concentration of is =1.40 X 10^12 holes per m^3
Ratio of electron to hole concentration is =1 10^12


Example No 2.22 Page No 97¶

In [49]:
#initialisation of variables
B = 0.48                #in Wb/m^2
R_H = 3.6 * 10**-4           #in m^3/c
R = 9*10**-3           #in ohm-m

#CALCULATIONS
Sigma = 1/R                #in (ohm-m)^-1
Rho = 1/R_H                #in coulomb/m^3
e = 1.6*10**-19            #in C
n = Rho/e                  #in /m^3
Miu = Sigma * R_H          #in m^2/volt-s

#RESULTS
print('The mobility of electron  is =%.2f  m^2/volt-s' %Miu)

The mobility of electron  is =0.04  m^2/volt-s


Example No 2.23 Page No 104¶

In [50]:
#initialisation of variables
e = 1.6*10**-19           #in C
R_H = 0.0145              #in m^3/coulomb
Miu_e = 0.36              #m^2/v-s
E = 100                  #V/m

#CALCULATIONS
n = 1/(e*R_H)             #in /m^3
J= n*e*Miu_e*E            #in A/m^2

#RESULTS
print('The current density  is =%.2f  A/m^2 ' %J)

The current density  is =2482.76  A/m^2


Example No 2.24 Page No 105¶

In [51]:
import math

#initialisation of variables
e = 1.6*10**-19            #in C
Miu_e = 7.04*10**-3           #in m^2/volt-sec
n = 5.8*10**28              #number of electron/m^3
m = 9.1*10**-31
E_F = 5.5                    # in eV

#CALCULATIONS
Torque = (Miu_e/e)*m                # in sec
Rho = 1/(n*e*Miu_e)                 #in ohm cm
V_F = math.sqrt( (2*E_F*e)/m )           #in m/s

#RESULTS
print('Relaxtion time is =%.2f X 10^-15 sec ' %(Torque*(10**15)))
print('Resistivity of conductor  is =%.2f X 10^-18 in Ωm' %(Rho*(10**8)))
print('Velocity of electron with the fermi energy  is =%.2f X 10^-6 in m/s' %(V_F/(10**6)))

# Note: The calculation of Part (ii) is wrong also the unit of resistivity of conductor is wrong

Relaxtion time is =40.04 X 10^-15 sec
Resistivity of conductor  is =1.53 X 10^-18 in Ωm
Velocity of electron with the fermi energy  is =1.39 X 10^-6 in m/s


Example No 2.25 Page No 105¶

In [52]:
import math
#initialisation of variables

E= 5.95           #in eV
EF= 6.25          #in eV
delE= 0.01

#CALCULATIONS
#delE= 1-1/(1+exp((E-EF)/KT))
K=1.38*10**-23                      #Boltzman Constant in J/K
T = ((E-EF)/math.log(1/(1-delE) -1)*1.6*10**-19)/K               #in K

#RESULTS
print('The temperature is =%.2f K' %T)

The temperature is =756.95 K


Example No 2.26 Page No 105¶

In [53]:
import math
#initialisation of variables

T1 = 400.0       #in K
T2 = 300.0       #in K
N_V = 1.04*10**19       #in cm^-3

#CALCULATIONS
N1 = N_V*((T1/T2)**(3/2))          #in cm^-3
KT = 0.0259*(T1/T2)           #in eV
FermiLevel= 0.27              #in eV
P_O = N1*math.exp( (-FermiLevel)/KT )          #in cm^-3

#RESULTS
print('The thermal equillibrium hole concentration  is =%.2f X 10^15 cm^-3' %(P_O/(10**15)))
print('Approximationa error ')
# Answer in the book is wrong

The thermal equillibrium hole concentration  is =5.58 X 10^15 cm^-3
Approximationa error


Example No 2.27 Page No 107¶

In [54]:
import math
#initialisation of variables
T1 = 550.0              #in K
T2 = 300.0           #in K
N1 = 1.04*10**19

#CALCULATIONS
N_V = N1*((T1/T2)**(3))
N_C = 2.8*10**19
E_g = -1.12
KT = 0.0259*(T1/T2)
n_i = math.sqrt(N_C*N_V*math.exp(E_g/KT))   #in cm^-3
Nd= math.sqrt(n_i**2/((1.05-1/2.0)**2-(1/2.0)**2))

#RESULTS
print('The value of n_i  is =%.2f X 10^14 cm^-3' %(n_i/(10**14)))
print('The value of N_d is =%.2f X 10^15 cm^-3' %(Nd/(10**15)))
#Formula n_o= Nd/2+sqrt((Nd/2)^2+n_i^2) and n_o = 1.05*N_d;

The value of n_i  is =3.20 X 10^14 cm^-3
The value of N_d is =1.40 X 10^15 cm^-3


Example No 2.28 Page No 106¶

In [55]:
import math
#initialisation of variables
n_o = 10**15         #in cm^-3
n_i = 10**10         #in cm^-3
p_o = 10**5       #in cm^-3
del_p = 10**13    #in cm^-3

#CALCULATIONS
del_n = del_p         #in cm^-3
KT= 0.0259          #in eV
Fermi_level1= KT*math.log(n_o/n_i)           #in eV
Fermi_level2= KT*math.log((n_o+del_n)/n_i)        #in eV
Fermi_level3= KT*math.log((p_o+del_p)/n_i)        #in eV

#RESULTS
print('Fermi level for thermal equillibrium  is : =%.2f eV ' %Fermi_level1)
print('Quasi-Fermi level for electrons in non equillibrium  is : =%.2f eV' %Fermi_level2)
print('Quasi-Fermi level for holes in non equillibrium  is: =%.2f eV' %Fermi_level3)

Fermi level for thermal equillibrium  is : =0.30 eV
Quasi-Fermi level for electrons in non equillibrium  is : =0.30 eV
Quasi-Fermi level for holes in non equillibrium  is: =0.18 eV