Chapter 2 ENERGY BAND THEORY OF SOLIDS

Example 2_1 pgno:49

In [1]:
#exa 2.1
no=1.5*10**10
print "no = ",no,"/cmˆ3" # initializing value of electrons and hole per cmˆ3.
n=1*10**18
print "n = ",n,"/cmˆ3" # initializing value of number of electrons per cmˆ3.
p=(no**2/n)
print "number of holes ,p=(noˆ2/n))= ",p," /cmˆ3" # calculation
#this is solved problem 2.1 of chapter 2

no =  15000000000.0 /cmˆ3
n =  1000000000000000000 /cmˆ3
number of holes ,p=(noˆ2/n))=  225.0  /cmˆ3

Example 2_2 pgno:49

In [2]:
#exa 2.2
from math import sqrt
n=1*10**5
print"n = ",n," /cmˆ3" # initializing value of electrons and hole per cmˆ3.
p=1*10**19
print"p = ",p," /cmˆ3" # initializing value of number of hole per cmˆ3
no=sqrt(n*p)
print"Value of intrinsic concentration ,no=sqrt(n∗p))= ",no," /cmˆ3"# calculation
#this is solved problem 2.2 of chapter 2

n =  100000  /cmˆ3
p =  10000000000000000000  /cmˆ3
Value of intrinsic concentration ,no=sqrt(n∗p))=  1e+12  /cmˆ3

Example 2_3 pgno:49

In [3]:
#exa 2.3
e=1.6*10**-19
print"e = ",e,"columb" # initializing the value of electronic charge .
Ef_Efi =0.309
print"Ef−Efi = ",Ef_Efi," eV" # initializing the value of difference in the energy levels .
no=2.5*10**13
print"no = ",no," /cmˆ3" # initializing value of number of electrons per cmˆ3
T=300
print"T = ",T," K" # initializing value of temperature .
ex=2.718
print"exp = ",ex # initializing the value of exponential .
k=1.38*10**-23
print"k = ”,k,” J/K" # initializing value of boltzmann constant .
n=no*(ex**((Ef_Efi*e)/(k*T)))
print"number of electrons per cmˆ3, n=no∗(exˆ((Ef−Efi)/kT)))= ",n," /cmˆ3" #calculation
#This is solved problem 2.3 of chapter 2.
#The value used for ”Ef−Efi” in the solution is different than provided in the question .
#I have used the value provided in the solution ( i .e Ef Efi =0.309)

e =  1.6e-19 columb
Ef−Efi =  0.309  eV
no =  2.5e+13  /cmˆ3
T =  300  K
exp =  2.718
k = ”,k,” J/K
number of electrons per cmˆ3, n=no∗(exˆ((Ef−Efi)/kT)))=  3.83494867662e+18  /cmˆ3

Example 2_4 pgno:50

In [4]:
#exa 2.4
e=1.6*10**-19
print"e = ",e," columb" # initializing the value of electronic charge .
Ef =0.4065
print "Ef = ",Ef," eV" # initializing the value of fermi level .
n=10**17
print"n = ",n," /cmˆ3" # initializing value of number of electrons per cmˆ3.
T=300
print"T = ",T," K" # initializing value of temperature .
ex=2.718
print"exp = ",ex # initializing the value of exponential .
k=1.38*10**-23
print"k = ",k," J/K" # initializing value of boltzmann constant .
no=n/(ex**((Ef*e)/(k*T)))
print"Number of electrons per cmˆ3, no=n/(exˆ((Ef)/kT) ) )= ",no," electrons /cmˆ3" # calculation
#this is solved problem 2.4 of chapter 2.
#the value used for "n" in the solution is different than provided in the question .
#I have used the value provided in the solution ( i .e n=10ˆ17)

e =  1.6e-19  columb
Ef =  0.4065  eV
n =  100000000000000000  /cmˆ3
T =  300  K
exp =  2.718
k =  1.38e-23  J/K
Number of electrons per cmˆ3, no=n/(exˆ((Ef)/kT) ) )=  15061844796.9  electrons /cmˆ3

Example 2_5 pgno:50

In [5]:
#exa 2.5
e=1.6*10**-19
print"e = ",e,"columb" # initializing the value of electronic charge .
n=1*10**22
print"n = ",n," /mˆ3" # initializing value of number of electrons per cmˆ3
u=1200*10**-4
print"u = ",u," mˆ2/Vs" # initializing the value of mobility .
L=0.1*10**-2
print"L = ",L," m" # initializing the value of length .
A=100*10**-12
print"A = ",A," mˆ2" # initializing the value of area of cross section .
sigma=n*e*u
print" conductivity , sigma=n∗e∗u)= ",sigma,"siemen/m" # calculation .
p=(1/sigma)
print"Resistivity ,p=(1/sigma))= ",p," ohm metre"#calculation .
R=(p*L/A)
print" resistance ,R=(p∗L/A) )= ",R," ohm" #calculation .
#this is solved problem 2.5 of chapter 2.
#the value used for "A" in the solution is different than provided in the question .
#I have used the value provided in the solution ( i .e A=100∗10ˆ−12)

e =  1.6e-19 columb
n =  10000000000000000000000  /mˆ3
u =  0.12  mˆ2/Vs
L =  0.001  m
A =  1e-10  mˆ2
 conductivity , sigma=n∗e∗u)=  192.0 siemen/m
Resistivity ,p=(1/sigma))=  0.00520833333333  ohm metre
 resistance ,R=(p∗L/A) )=  52083.3333333  ohm

Example 2_6 pgno:50

In [6]:
#exa 2.6
R=52.08*10**3
print"R = ",R,"ohm" # initializing value of Resistance .
V=5
print"V = ",V,"volt" # initializing value of voltage .
I=(V/R)
print" Drift current , I=(V/R) )= ",I," amphere" # calculation
#this is solved problem 2.6 of chapter 2.

R =  52080.0 ohm
V =  5 volt
 Drift current , I=(V/R) )=  9.60061443932e-05  amphere

Example 2_7 pgno:50

In [7]:
#exa 2.7
Eg1 =1.43
print" Energy gap of GaAs = ",Eg1,"eV" # initializing the value of energy gap of GaAs.
Eg2 =2.43
print" Energy gap of GaP = ",Eg2,"eV"# initializing the value of energy gap of Gap.
h=6.624*10**-34
print" Plank constant = ",h," joule "# initializing the value of plank constant .
c=3*10**8
print" Light speed = ",c,"m/s" # initializing the value of speed of light.
x=(Eg2-Eg1)
print"Difference between the energy gap of GaAs and GaP ,x=(Eg2−Eg1) )= ",x," eV"# calculation
g=(0.4*x)
print"Excess energy gap added to GaAs to form GaAsP,(0.4∗x))= ",g," eV "#calculation
Eg=(Eg1+g)
print"Band gap energy GaAsP,Eg=(Eg1+g))= ",Eg ," eV "#calculation
lamda=(c*h/(Eg*1.6*10**-19))
print"wavelength of radiation emitted , lamda=(c∗h/Eg))= ",lamda," metre "
# calculation 19 #this is solved problem 2.7 of chapter 2.

 Energy gap of GaAs =  1.43 eV
 Energy gap of GaP =  2.43 eV
 Plank constant =  6.624e-34  joule 
 Light speed =  300000000 m/s
Difference between the energy gap of GaAs and GaP ,x=(Eg2−Eg1) )=  1.0  eV
Excess energy gap added to GaAs to form GaAsP,(0.4∗x))=  0.4  eV 
Band gap energy GaAsP,Eg=(Eg1+g))=  1.83  eV 
wavelength of radiation emitted , lamda=(c∗h/Eg))=  6.7868852459e-07  metre

Example 2_8 pgno:51

In [8]:
#exa 2.8
Eg1 =1.43
print" Energy gap of GaAs = ",Eg1," eV" # initializing the value of energy gap of GaAs.
Eg2 =2.43
print" Energy gap of GaP = ",Eg2," eV"# initializing the value of energy gap of Gap.
h=6.624*10**-34
print" Plank constant = ",h," joule"# initializing the value of plank constant .
c=3*10**8
print" Light speed = ",c," m/s" # initializing the value of speed of light.
lamda =540*10**6
print" lamda = ",lamda," m" # initializing the value of wavelength .
x=(Eg2-Eg1)
print"Difference between the energy gap of GaAs and GaP ,x=(Eg2−Eg1) )= ",x," eV"# calculation
Eg=((c*h/(lamda*(1.6*10**-19))))
print"Band gap energy ,Eg=(c∗h/lamda∗(1.6∗10ˆ−19)))=",Eg," eV"# calculation
X=Eg-(Eg1)
print"X=Eg−(Eg1)= ",X  # calculation 
#this is solved problem 2.8 of chapter 2.
#the value of Eg(band gap energy )is provided wrong in the book after calculation.Due to this value ofX,alsodiffer.

 Energy gap of GaAs =  1.43  eV
 Energy gap of GaP =  2.43  eV
 Plank constant =  6.624e-34  joule
 Light speed =  300000000  m/s
 lamda =  540000000  m
Difference between the energy gap of GaAs and GaP ,x=(Eg2−Eg1) )=  1.0  eV
Band gap energy ,Eg=(c∗h/lamda∗(1.6∗10ˆ−19)))= 2.3e-15  eV
X=Eg−(Eg1)=  -1.43

Example 2_9 pgno:51

In [9]:
#exa 2.9
T1 =500
print" Temperature 1 = ",T1,"K" # initializing the value of temperature 1.
Nv =2*10**19
print" Nv = ",round(Nv,3),"cmˆ−3"# initializing the value of effective density of state for valence band .
T2 =300
print" Temperature 2 = ",T2,"K"# initializing the value of temperature 2.
NV=(Nv*((500/300)**(3/2)))
print"NV at 500K=(Nv((500/300) ˆ(3/2) ) ) )= ",round(NV,3)," cmˆ−3 "#calculation
#this is solved problem 2.9 of chapter 2.

 Temperature 1 =  500 K
 Nv =  2e+19 cmˆ−3
 Temperature 2 =  300 K
NV at 500K=(Nv((500/300) ˆ(3/2) ) ) )=  2e+19  cmˆ−3

Example 2_10 pgno:52

In [10]:
#exa 2.10
Nd =1*10**17
print"Nd = ",Nd,"cmˆ−3" # initializing the value of effective energy density of state.
Ec_Ed =0.045
print" Ec Ed = ",Ec_Ed # initializing the value of donor ionisation level .
Vt =0.0259
print"Vt = ",Vt," eV "# initializing the value of thermal voltage .
Nc=2.8*10**19
print" Nc = ",Nc,"cmˆ−3"# initializing the value of effective density of state of conduction band .
e=2.718
print"exp = ",e # initializing the value of exponential .
N=(((Nc/Nd)*e**((-(Ec_Ed))/Vt))+1)**-1
print "Fraction of electron still in the donor state,(nd/(nd+n)=(((Nc/Nd)∗eˆ((−Ec Ed)/Vt),1)ˆ−1)= ",N  # calculation
#this is solved problem 2.10 of chapter 2.

Nd =  100000000000000000 cmˆ−3
 Ec Ed =  0.045
Vt =  0.0259  eV 
 Nc =  2.8e+19 cmˆ−3
exp =  2.718
Fraction of electron still in the donor state,(nd/(nd+n)=(((Nc/Nd)∗eˆ((−Ec Ed)/Vt),1)ˆ−1)=  0.0198886296934

Example 2_11 pgno:52

In [11]:
#exa 2.11
from math import exp
Na =1*10**16
print"Na = ",Na," cmˆ−3"# initializing the value of acceptor concentration 
Ea_Ev =0.045
print"Ea Ev = ",Ea_Ev # initializing the boron acceptor ionization energy .
Nv=(1.04*10**19)
print"Nv = ",Nv," cmˆ−3"# initializing the value of effective density of state for valence band .
Vt=(0.0259)
print"Vt = ",Vt," eV"# initializing the value of thermal voltage .
p=(1+((Nv/(4*Na))*exp(-(Ea_Ev)/Vt)))**(-1)
print"Fraction of holes that are still in the acceptor state ,(pa/(pa+p))=(1+((Nv/4∗Na)∗exp(−(Ea −Ev)/Vt)))ˆ(−1)= ",p #calculation
#this is solved problem 2.11 of chapter 2

Na =  10000000000000000  cmˆ−3
Ea Ev =  0.045
Nv =  1.04e+19  cmˆ−3
Vt =  0.0259  eV
Fraction of holes that are still in the acceptor state ,(pa/(pa+p))=(1+((Nv/4∗Na)∗exp(−(Ea −Ev)/Vt)))ˆ(−1)=  0.0213895767669

Example 2_12 pgno:52

In [12]:
#exa 2.12
Nd =1*10**17
print"Nd = ",Nd," cmˆ−3" # initializing the value of donor concentration .
Na=0
print"Na = ",Na," cmˆ−3"# initializing the value of acceptor concentration .
no=1.5*10**10
print"ni = ",no," cmˆ−3"# initializing the value of electron hole per cmˆ3.
n=(-(Na-Nd)+sqrt((Na-Nd)**2+4*no))/2
print"Electron concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= ",n," cmˆ−3"#calculation
p=(no**2/n)
print"Hole concentration ,p)= ",p," cmˆ−3" # calculation
#this is solved problem 2.13 of chapter 2.

Nd =  100000000000000000  cmˆ−3
Na =  0  cmˆ−3
ni =  15000000000.0  cmˆ−3
Electron concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)=  1e+17  cmˆ−3
Hole concentration ,p)=  2250.0  cmˆ−3

Example 2_14 pgno:53

In [13]:
#exa 2.14
Nd =6*10**16
print"Nd = ",Nd," cmˆ−3" # initializing the value of donor concentration .
Na =10**17
print"Na = ",Na," cmˆ−3"# initializing the value of acceptor concentration .
no=1.5*10**10
print"no = ",no," cmˆ−3"# initializing the value of electron and hole per cmˆ3.
p=((Na-Nd)+sqrt((Na-Nd)**2+4*no))/2
print"Hole concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= ",p," cmˆ−3"#calculation
n=(no**2/p)
print "Electron concentration ,n=(noˆ2/p))= ",n # calculation
#this is solved problem 2.14 of chapter 2.
#the value of Na,Nd in the solution is different than provided in the question
#I have used the value used in the solution(i.e Na=10ˆ17 ,Nd=6∗10ˆ16)

Nd =  60000000000000000  cmˆ−3
Na =  100000000000000000  cmˆ−3
no =  15000000000.0  cmˆ−3
Hole concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)=  4e+16  cmˆ−3
Electron concentration ,n=(noˆ2/p))=  5625.0

Example 2_15 pgno:53

In [14]:
#exa 2.15
Nd =6*10**16
print"Nd = ",Nd," cmˆ−3" # initializing the value of donor concentration .
Na =10**17
print"Na = ",Na," cmˆ−3"# initializing the value of acceptor concentration .
no=1.5*10**10
print"no = ",no," cmˆ−3"# initializing the value of electron and hole per cmˆ3.
p=((Na-Nd)+sqrt((Na-Nd)**2+4*no))/2
print"Hole concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)= ",p,"cmˆ−3"#calculation
n=(no**2/p)
print"Electron concentration ,n=(noˆ2/p))= ",n # calculation
#this is solved problem 2.15 of chapter 2.
#the value of Na,Nd in the solution is different than provided in the question
#I have used the value used in the solution(i.e Na=10ˆ17 ,Nd=6∗10ˆ16)

Nd =  60000000000000000  cmˆ−3
Na =  100000000000000000  cmˆ−3
no =  15000000000.0  cmˆ−3
Hole concentration ,n=(−(Na−Nd)+sqrt ((Na−Nd)ˆ2+4∗no))/2)=  4e+16 cmˆ−3
Electron concentration ,n=(noˆ2/p))=  5625.0

Example 2_16 pgno:53

In [15]:
#exa 2.16
from math import exp
Nv1 =1.04*10**19
print"Nv = ",Nv1," cmˆ−3"# initializing the value of valence band concentration at 300K.
Ef_Ev =0.3
print"Ef Ev = ",Ef_Ev," eV"# initializing the value of boron acceptor ionization energy.
T1 =300
print"T = ",T1,"K"# initializing the value of temperature 1.
T2 =500
print"T = ",T2,"K"# initializing the value of temperature 2.
Vt1 =0.0259
print"Vt1 = ",Vt1,"eV"# initializing the value of thermal voltage at 300K.
k=1.38*10**-23
print"k = ",k,"J/K" # initializing value of boltzmann constant .
e=1.6*10**-19
print"e = ",e,"columb" # initializing the value of electronic charge .
K1=(Nv1/((T1)**(3/2)))
print"Value of constant ,K1=(Nv/((T) ˆ(3/2) ) )= ",K1," cmˆ−3 K(−2/3)"# calculation
Nv2=K1*T2**(3/2)
print"Value of valence band concentration at 500K,Nv =K1∗T(3/2)= ",Nv2," cmˆ−3"# calculation
VT=(k*T2/e)
print"Value of parameter VT at 500K,VT=(K∗T/e)= ",VT," cmˆ−3"# calculation
p=(Nv2*(exp(-(Ef_Ev)/(VT))))
print"Hole concentration ,p=(Nv∗(exp(Ef Ev)/(VT)))= ",p," cmˆ−3"# calculation
#this is solved problem 2.16 of chapter 2.

Nv =  1.04e+19  cmˆ−3
Ef Ev =  0.3  eV
T =  300 K
T =  500 K
Vt1 =  0.0259 eV
k =  1.38e-23 J/K
e =  1.6e-19 columb
Value of constant ,K1=(Nv/((T) ˆ(3/2) ) )=  3.46666666667e+16  cmˆ−3 K(−2/3)
Value of valence band concentration at 500K,Nv =K1∗T(3/2)=  1.73333333333e+19  cmˆ−3
Value of parameter VT at 500K,VT=(K∗T/e)=  0.043125  cmˆ−3
Hole concentration ,p=(Nv∗(exp(Ef Ev)/(VT)))=  1.65083278171e+16  cmˆ−3

Example 2_17 pgno:54

In [16]:
#exa 2.17
from math import sqrt
Nv =7*10**18
print"Nv = ",Nv,"cmˆ−3"# initializing the value of valence band concentration at 300K.
Nc=4.7*10**17
print"Nc = ",Nc,"cmˆ−3"# initializing the value of conduction band concentration at 300K.
T1 =300
print"T = ",T1,"K"# initializing the value of temperature 1.
T2 =450
print"T = ",T2,"K"# initializing the value of temperature 2.
Vt1 =0.0259
print"Vt1 = ",Vt1,"eV"# initializing the value of thermal voltage at 300K.
Vt2 =0.03881
print"Vt2 = ",Vt2,"eV"# initializing the value of thermal voltage at 450K.
Eg=1.42
print"Eg = ",Eg,"eV"# initializing the value of thermal voltage .
no=(sqrt(Nc*Nv*(exp(-Eg/Vt1))))
print"intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg/Vt1))))= ",no #calculation
K1=(Nc/((T1)**(3/2)))
print"Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= ",K1 # calculation
k1=(K1*T2**(3/2))
print"Value of constant k1 at 450K ,k1=(K1∗T2ˆ(3/2))= ",k1# calculation
K2=(Nv/((T1)**(3/2)))
print"Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= ",K2# calculation
k2=(K2*T2**(3/2))
print"Value of constant k2 at 450K ,k2=(K2∗T2ˆ(3/2))= ",k2 # calculation
K=k1*k2
print"Value of constant K,= ",K # calculation
no1=(sqrt(K*(exp(-Eg/Vt2))))
print"intrinsic concentration at 450K,no=(sqrt(K∗(exp(−Eg/Vt2) ) ) )= ",no1," cmˆ3"# calculation
#this is solved problem 2.17 of chapter 2.

Nv =  7000000000000000000 cmˆ−3
Nc =  4.7e+17 cmˆ−3
T =  300 K
T =  450 K
Vt1 =  0.0259 eV
Vt2 =  0.03881 eV
Eg =  1.42 eV
intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg/Vt1))))=  2255422.87974
Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )=  1.56666666667e+15
Value of constant k1 at 450K ,k1=(K1∗T2ˆ(3/2))=  7.05e+17
Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )=  23333333333333333
Value of constant k2 at 450K ,k2=(K2∗T2ˆ(3/2))=  10499999999999999850
Value of constant K,=  7.4025e+36
intrinsic concentration at 450K,no=(sqrt(K∗(exp(−Eg/Vt2) ) ) )=  30874193378.4  cmˆ3

Example 2_18 pgno:55

In [17]:
#exa 2.18
from math import sqrt
from math import exp
Nv=1.04*10**19
print"Nv = ",Nv,"cmˆ−3"# initializing the value of valence band concentration at 300K.
Nc=2.8*10**19
print"Nc = ",Nc,"cmˆ−3"# initializing the value of conduction band concentration at 300K.
T1 =300
print"T = ",T1,"K"# initializing the value of temperature 1.
T2 =550
print"T = ",T2,"K"# initializing the value of temperature 2.
Vt1 =0.0259
print"Vt1 = ",Vt1,"eV"# initializing the value of thermal voltage at 300K.
Vt2 =0.0474
print"Vt2 = ",Vt2,"eV"# initializing the value of thermal voltage at 550K.
Eg1=1.12
print"Eg1 = ",Eg1,"eV"# initializing the value of thermal voltage .
no=(sqrt(Nc*Nv*(exp(-Eg1/Vt1))))
print"intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg1/Vt1))))= ",no #calculation
K1=(Nc/((T1)^(3/2)))
print"Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= ",K1  # calculation
k1=(K1*T2**(3/2))
print"Value of constant k1 at 550K ,k1=(K1∗T2ˆ(3/2))= ",k1 # calculation 
K2=(Nv/((T1)**(3/2)))
print"Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= ",K2 # calculation
k2=(K2*T2**(3/2))
print"Value of constant k2 at 550K ,k2=(K2∗T2ˆ(3/2))= ",k2  # calculation
K=k1*k2
print"Value of constant K,= ",K # calculation
no1=(sqrt(K*(exp(-Eg1/Vt2))))
print"Intrinsic concentration at 550K,no=(sqrt(K∗(exp(−Eg1/Vt2))))= ",no1," cmˆ3"# calculation
Nd=(4*(no1**2)/(1.2))
print"Donor concentration at which intrinsic concentration is 10% of the total electron concentration ,Nd=(4∗(no1ˆ2) /(1.2) )= ",Nd," cmˆ3"# calculation
#this is solved problem 2.18 of chapter 2.
#the value of temperature and % of the intrinsic carrier concentration given in the question is different than used in the solution .
#I have used the value provided in the solution (i.e T2=550 and % of the intrinsic carrier concentration =10%)
#the value of Donor concentration at which intrinsic concentration is 10% of the total electron concentration (Nd) , is provided wrong in the book after calculation .

Nv =  1.04e+19 cmˆ−3
Nc =  2.8e+19 cmˆ−3
T =  300 K
T =  550 K
Vt1 =  0.0259 eV
Vt2 =  0.0474 eV
Eg1 =  1.12 eV
intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg1/Vt1))))=  6949358641.26
Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )=  9.3023255814e+16
Value of constant k1 at 550K ,k1=(K1∗T2ˆ(3/2))=  5.11627906977e+19
Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )=  3.46666666667e+16
Value of constant k2 at 550K ,k2=(K2∗T2ˆ(3/2))=  1.90666666667e+19
Value of constant K,=  9.75503875969e+38
Intrinsic concentration at 550K,no=(sqrt(K∗(exp(−Eg1/Vt2))))=  2.31051731905e+14  cmˆ3
Donor concentration at which intrinsic concentration is 10% of the total electron concentration ,Nd=(4∗(no1ˆ2) /(1.2) )=  1.77949676054e+29  cmˆ3

Example 2_19 pgno:55

In [18]:
#exa 2.19
Ec_Ef =0.2
print"Ec Ef = ",Ec_Ef," eV" # initializing the value of difference in the energy levels.
Nc=2.8*10**19
print"Nc =",Nc," cmˆ−3"# initializing the conduction band concentration .
Na =3*10**16
print"Na =",Na," cmˆ−3"# initializing the acceptor concentration .
Vt =0.0259
print"Vt =",Vt," eV"# initializing the thermal voltage at 300K.
Nd=(Nc*(exp(-(Ec_Ef)/(Vt))))+(Na)
print"Donor concentration ,Nd=(Nc∗(exp(−(Ec Ef)/(Vt))) )+(Na)= ",Nd," cmˆ−3"# calculation
#this is solved problem 2.19 of chapter 2.

Ec Ef =  0.2  eV
Nc = 2.8e+19  cmˆ−3
Na = 30000000000000000  cmˆ−3
Vt = 0.0259  eV
Donor concentration ,Nd=(Nc∗(exp(−(Ec Ef)/(Vt))) )+(Na)=  4.24031697774e+16  cmˆ−3

Example 2_20 pgno:56

In [19]:
#exa 2.20
from math import sqrt
from math import exp
Nv =6*10**18
print"Nv = ",Nv,"cmˆ−3"# initializing the value of valence band concentration at 300K.
Nc=1.04*10**19
print"Nc = ",Nc,"cmˆ−3"# initializing the value of conduction band concentration at 300K.
T1 =300
print"T1 = ",T1,"K"# initializing the value of temperature 1.
T2 =200
print"T2 = ",T2,"K"# initializing the value of temperature 2.
Vt1 =0.0259
print"Vt1 = ",Vt1,"eV"# initializing the value of thermal voltage at 300K.
Vt2 =0.0173
print"Vt2 = ",Vt2,"eV"# initializing the value of thermal voltage at 200K.
Eg1=0.60
print"Eg1 = ",Eg1,"eV"# initializing the value of thermal voltage used for 300K .
no=(sqrt(Nc*Nv*(exp(-Eg1/Vt1))))
print"intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg1/Vt1))))= ",no  #calculation
Eg2=0.66
print"Eg2 = ",Eg2,"eV"# initializing the value of thermal voltage used for 200K.
K1=(Nc/((T1)**(3/2)))
print"Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )= ",K1  # calculation
k1=(K1*T2**(3/2))
print"Value of constant k1 at 200K ,k1=(K1∗T2ˆ(3/2))= ",k1 # calculation
K2=(Nv/((T1)**(3/2)))
print"Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )= ",K2 # calculation
k2=(K2*T2**(3/2))
print"Value of constant k2 at 200K ,k2=(K2∗T2ˆ(3/2))= ",k2 # calculation
K=k1*k2
print"Value of constant K,= ",K # calculation
no1=(sqrt(K*(exp(-Eg2/Vt2))))
print"intrinsic concentration at 200K,no=(sqrt(K∗(exp(−Eg2/Vt2))))= ",round(no1,2)," cmˆ3"# calculation
#this is solved problem 2.20 of chapter 2.
#The answer of intrinsic concentration at 300K,(no) is provided wrong in the book.

Nv =  6000000000000000000 cmˆ−3
Nc =  1.04e+19 cmˆ−3
T1 =  300 K
T2 =  200 K
Vt1 =  0.0259 eV
Vt2 =  0.0173 eV
Eg1 =  0.6 eV
intrinsic concentration at 300K,no=(sqrt(Nc∗Nv∗(exp(−Eg1/Vt1))))=  7.36468677124e+13
Eg2 =  0.66 eV
Value of constant ,K1=(Nc/((T) ˆ(3/2) ) )=  3.46666666667e+16
Value of constant k1 at 200K ,k1=(K1∗T2ˆ(3/2))=  6.93333333333e+18
Value of constant ,K2=(Nv/((T1) ˆ(3/2) ) )=  20000000000000000
Value of constant k2 at 200K ,k2=(K2∗T2ˆ(3/2))=  4000000000000000000
Value of constant K,=  2.77333333333e+37
intrinsic concentration at 200K,no=(sqrt(K∗(exp(−Eg2/Vt2))))=  27369762834.6  cmˆ3

Example 2_21 pgno:56

In [20]:
#exa 2.21
Eg1=2
print"Eg1 = ",Eg1," eV" # initializing the value of band energy gap for semiconductor1.
Eg2 =2.2
print"Eg2 = ",Eg2," eV"# initializing the value of band energy gap for semiconductor2.
Vt =0.0259
print"Vt = ",Vt," eV"# initializing the value of thermal voltage at 300K.
No=sqrt(exp((-Eg1/Vt)-(-Eg2/Vt)))
print"Ratio of their intrinsic concentration at 300K,(no1/no2)=sqrt(exp((−Eg1/Vt)−(−Eg2/Vt)))= ",No  # calculation

Eg1 =  2  eV
Eg2 =  2.2  eV
Vt =  0.0259  eV
Ratio of their intrinsic concentration at 300K,(no1/no2)=sqrt(exp((−Eg1/Vt)−(−Eg2/Vt)))=  47.5130239084