chapter 5 PN JUNCTION DIODE

Example 5_5 pgno: 142

In [1]:
#exa 5.5
from math import log
Na =10**17
print "Na",Na,"/cmˆ3" # initializing value of medium p doping concentration .
Nd =10**15
print "Nd= ",Nd,"/cmˆ3" # initializing value of light n doping .
no=1.5*10**10
print "no = ",no,"cmˆ−3" # initializing value of intrinsic carrier concentration .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
K=1.38*10**-23
print "K = ",K,"J/k"   #initializing value of boltzmann constant .
T=300
print "T = ",T,"K" # initializing value of temperature .
Efi_Efp=((K*T/e)*log(Na/no))
print "(a)fermi level in the P region , Efi Efp=((KT/e) ∗ log (Na/no ) ) )= ",Efi_Efp,"eV" # calculation .
Efn_Efi=((K*T/e)*log(Nd/no))
print "fermi level in the n region,Efn Efi=((KT/e)∗log (Nd/no) ) )=",Efn_Efi," eV" # calculation
Efn_Efp=(Efi_Efp)+(Efn_Efi)
print"(b) junction potential at room temperature ,Efn Efp=(Efi Efp)+(Efn Efi))=",Efn_Efp," eV" #calculation .

Na 100000000000000000 /cmˆ3
Nd=  1000000000000000 /cmˆ3
no =  15000000000.0 cmˆ−3
e =  1.6e-19 columns
K =  1.38e-23 J/k
T =  300 K
(a)fermi level in the P region , Efi Efp=((KT/e) ∗ log (Na/no ) ) )=  0.406564315296 eV
fermi level in the n region,Efn Efi=((KT/e)∗log (Nd/no) ) )= 0.287405536734  eV
(b) junction potential at room temperature ,Efn Efp=(Efi Efp)+(Efn Efi))= 0.69396985203  eV

Example 5_7 pgno: 143

In [2]:
#exa 5.7
from math import sqrt
from math import exp
Pp =10**18
print "Pp= ",Pp,"/cmˆ3" # initializing value of doping concentration in p region.
Nn =10**15
print "Nn= ",Nn,"/cmˆ3" # initializing value of doping concentration in n region.
tp =7*10** -6
print "tp = ",tp,"s" # initializing value of hole lifetime .
tn =0.2*10** -6
print "tn = ",tn,"s" # initializing value of electron lifetime .
up=800
print "up= ",up,"cm2/Vs" # initializing value of P side mobility .
un=300
print "un= ",un,"cm2/Vs" # initializing value of n side mobility .
no=1.5*(10**10)
print "no = ",no,"cmˆ−3" # initializing value of intrinsic concentration .
Vf=0.6
print "Vf = ",Vf,"V" # initializing value of forward bias voltage .
A=100*(10**-6)
print "A = ",A,"mˆ2"# initializing value of diode cross−sectional area .
e=1.6*(10**-19)
print "e = ",e,"columns" # initializing value of charge of electrons .
K=1.38*(10**-23)
print "K = ",K,"J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T,"K"  # initializing value of temperature .
Vt=(K*T/e)
print "Vt=(K∗T/e))=",Vt,"eV" #calculation .
Dp=Vt*un
print "Dp=Vt∗un=",Dp,"cmˆ−3" # calculation .
Dn=Vt*up
print "Dn=Vt∗up=",Dn,"cmˆ−3" # calculation .
Lp=sqrt(Dp*tp)
print "Lp=(sqrt(Dp∗tp))=",Lp,"cm" # calculation .
Ln=(sqrt(Dn*tn))
print "Ln=(sqrt(Dn∗tn))=",Ln,"cm" # calculation .
npo=(no**2/Pp)
print "npo=(no^2/Pp)=",npo,"cmˆ−3" # calculation .
Ppo=(no**2/Nn)
print "Ppo=(noˆ2/Nn)=",Ppo,"cmˆ−3" #calculation .
Io=(((Dp*Ppo)/(Lp))+((Dn*npo)/(Ln)))*e*A
print "Reverse saturation current ,Io=(((Dp∗Ppo)/(Lp)) + ( ( D n ∗ n p o ) / ( L n ) ) ) ∗ e ∗ A= ",Io," A " #calculation .
If=Io*((exp(Vf/Vt))-1)
print "Diode forward current , If=Io∗((exp(Vf/Vt))−1)=",If,"A" # calculation .
#//the value of Io(reverse saturation current ),after calculation is provided wrong in the book.Due to which If (diode forward current )also differ.

Pp=  1000000000000000000 /cmˆ3
Nn=  1000000000000000 /cmˆ3
tp =  7e-06 s
tn =  2e-07 s
up=  800 cm2/Vs
un=  300 cm2/Vs
no =  15000000000.0 cmˆ−3
Vf =  0.6 V
A =  0.0001 mˆ2
e =  1.6e-19 columns
K =  1.38e-23 J/k
T =  300 K
Vt=(K∗T/e))= 0.025875 eV
Dp=Vt∗un= 7.7625 cmˆ−3
Dn=Vt∗up= 20.7 cmˆ−3
Lp=(sqrt(Dp∗tp))= 0.00737139742518 cm
Ln=(sqrt(Dn∗tn))= 0.00203469899494 cm
npo=(no^2/Pp)= 225.0 cmˆ−3
Ppo=(noˆ2/Nn)= 225000.0 cmˆ−3
Reverse saturation current ,Io=(((Dp∗Ppo)/(Lp)) + ( ( D n ∗ n p o ) / ( L n ) ) ) ∗ e ∗ A=  3.827628972e-15  A 
Diode forward current , If=Io∗((exp(Vf/Vt))−1)= 4.5032547414e-05 A

Example 5_8 pgno: 144

In [3]:
#exa 5.8
Na =4*10**16
print "Na = ",Na,"cmˆ−3" # initializing value of acceptor concentration .
Nd =2*10**19
print "Nd = ",Nd,"cmˆ−3" # initializing value of donor concentration .
no=1.5*10**10
print "no = ",no,"cmˆ−3" # initializing value of intrinsic carrier concentration .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
K=1.38*10**-23
print "K = ",K,"J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T,"K" # initializing value of temperature .
Vbi=((K*T/e)*log((Na*Nd)/(no)**2))
print "Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd) /(no) ˆ2) )=",Vbi,"V" # calculation
#The value used for Nd in the book for solution is different than provided in the question .
#I have used the value provided in the solution(i.eNd=2∗10ˆ19)

Na =  40000000000000000 cmˆ−3
Nd =  20000000000000000000 cmˆ−3
no =  15000000000.0 cmˆ−3
e =  1.6e-19 columns
K =  1.38e-23 J/k
T =  300 K
Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd) /(no) ˆ2) )= 0.926513569765 V

Example 5_9 pgno: 144

In [4]:
#exa 5.9
Na =4*10**16
print "Na = ",Na,"cmˆ−3" # initializing value of acceptor concentration .
Nd =2*10**19
print "Nd = ",Nd,"cmˆ−3" # initializing value of donor concentration .
no=1.8*10**6
print "no = ",no,"cmˆ−3" # initializing value of intrinsic carrier concentration .
e=1.6*10**-19
print "e = ",e,"columns"  # initializing value of charge of electrons .
K=1.38*10**-23
print "K = ",K,"J/k"   # initializing value of boltzmann constant .
T=300
print "T = ",T,"K" # initializing value of temperature .
Vbi=((K*T/e)*log((Na*Nd)/(no)**2))
print "Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd) /(no) ˆ2) )= ",Vbi,"V" # calculation .

Na =  40000000000000000 cmˆ−3
Nd =  20000000000000000000 cmˆ−3
no =  1800000.0 cmˆ−3
e =  1.6e-19 columns
K =  1.38e-23 J/k
T =  300 K
Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd) /(no) ˆ2) )=  1.39371354345 V

Example 5_10 pgno: 144

In [5]:
# exa 5.10
from math import sqrt
Na =10e16
print "Na= ",Na,"/cmˆ3" # initializing value of medium p doping concentration .
Nd =10e18
print "Nd= ",Nd,"/cmˆ3" # initializing value of light n doping .
Vbi =0.64
print "Vbi = ",Vbi,"V" # initializing value of built in voltage .
e=1.6*10**-19
print "e = ",e,"columns"  # initializing value of charge of electrons .
Er=11.9
print "Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854*10**-14
print "Eo = ",Eo," F/cm" # initializing value of permittivity of free space .
E=Eo*Er
print " total permittivity ,E=Eo∗Er= ",E," F/cm"  #calculation .
W=sqrt((2*E*Vbi/e)*((Nd+Na)/(Na*Nd)))
print "W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))=",W," cm"  # calculation .
xn=((W*Na)/(Nd+Na))
print "xn=((W∗Na) /(Nd+Na) ) )=",xn,"cm"  # calculation .
xp=((W*Nd)/(Nd+Na))
print "xp=((W∗Nd) /(Nd+Na) ) )=",xp,"cm"  # calculation .
Emax=(-e*Nd*xn)/E
print "Emax=(−e∗Nd∗xn)/E)=",Emax,"V/cm"  #calculation .
# The value and unit of W(depletion width) ,provided after calculation in the book is wrong.Due to this xn,xp ,Emax also differ.

Na=  1e+17 /cmˆ3
Nd=  1e+19 /cmˆ3
Vbi =  0.64 V
e =  1.6e-19 columns
Er =  11.9
Eo =  8.854e-14  F/cm
 total permittivity ,E=Eo∗Er=  1.053626e-12  F/cm
W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))= 9.22675353524e-06  cm
xn=((W∗Na) /(Nd+Na) ) )= 9.13539953984e-08 cm
xp=((W∗Nd) /(Nd+Na) ) )= 9.13539953984e-06 cm
Emax=(−e∗Nd∗xn)/E)= -138727.017592 V/cm

Example 5_12 pgno: 145

In [6]:
#exa 5.12
Na =10**16
print "Na = ",Na,"/cmˆ3" # initializing value of medium p doping concentration .
Nd =10**18
print "Nd = ",Nd,"/cmˆ3" # initializing value of light n doping .
Vbi =0.64
print "Vbi = ",Vbi,"V" # initializing value of built in voltage .
Vr=20
print "Vr = ",Vr,"V" # initializing value of applied reverse voltage .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
Er=11.9
print "Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854*10**-14
print "Eo = ",Eo," F/cm" # initializing value of permittivity of free space .
E=Eo*Er
print " total permittivity ,E=Eo∗Er= ",E," F/cm" #calculation .
Emax=-(sqrt(((2*e*(Vbi+Vr))/(E))*((Nd*Na)/(Na+Nd))))
print "Emax=−(sqrt(((2∗e∗(Vbi+Vr))/(E))∗((Nd∗Na)/(Na+Nd)))))= ",Emax,"V/cm"  #calculation .

Na =  10000000000000000 /cmˆ3
Nd =  1000000000000000000 /cmˆ3
Vbi =  0.64 V
Vr =  20 V
e =  1.6e-19 columns
Er =  11.9
Eo =  8.854e-14  F/cm
 total permittivity ,E=Eo∗Er=  1.053626e-12  F/cm
Emax=−(sqrt(((2∗e∗(Vbi+Vr))/(E))∗((Nd∗Na)/(Na+Nd)))))=  -249129.931857 V/cm

Example 5_13 pgno: 146

In [7]:
#exa 5.13
Emax =2*10**5
print "Emax = ",Emax,"V/cm" # initializing value of maximum electric field .
Nd=1*10**18
print "Nd= ",Nd,"/cmˆ3" # initializing value of donor concentration .
Vbi=0.54
print "Vbi = ",Vbi,"V" # initializing value of built in voltage .
Vr=20
print "Vr = ",Vr,"V " #initializing value of applied reverse voltage .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
Er=11.9
print "Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854*10**-14
print "Eo = ",Eo," F/cm" # initializing value of permittivity of free space .
E=Eo*Er
print "total permittivity ,E=Eo∗Er=",E," F/cm"  #calculation .
Na=((Emax**2)*E*Nd)/((2*e*(Vbi+Vr)*Nd)-((Emax**2)*E))
print "Na=(Emaxˆ2∗E∗Nd)/((2∗e∗(Vbi+Vr)∗Nd)−(Emaxˆ2∗E)) )= ",Na,"cmˆ−3" # calculation .

Emax =  200000 V/cm
Nd=  1000000000000000000 /cmˆ3
Vbi =  0.54 V
Vr =  20 V 
e =  1.6e-19 columns
Er =  11.9
Eo =  8.854e-14  F/cm
total permittivity ,E=Eo∗Er= 1.053626e-12  F/cm
Na=(Emaxˆ2∗E∗Nd)/((2∗e∗(Vbi+Vr)∗Nd)−(Emaxˆ2∗E)) )=  6.45341703981e+15 cmˆ−3

Example 5_14 pgno: 146

In [8]:
#exa 5.14
Na =10**16
print "Na = ",Na,"/cmˆ3" # initializing value of acceptor concentration .
Nd =10**18
print "Nd = ",Nd,"/cmˆ3" # initializing value of donor concentration .
A=1
print "A = ",A,"cmˆ2" # initializing value of area for finding junction capacitance per unit area.
Vj =0.54
print "Vj =",Vj,"V" # initializing value of built in voltage .
Va=10
print "Va = ",Va,"V" # initializing value of applied reverse voltage .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
Er=11.9
print "Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854*10**-14
print "Eo = ",Eo," F/cm"  # initializing value of permittivity of free space .
E=Eo*Er
print "total permittivity ,E=Eo∗Er= ",E," F/cm" # calculation .
Cj=sqrt((e*E*A**2/(2*(Va+Vj)))*((Na*Nd)/(Na+Nd)))
print "Cj=sqrt((e∗E∗Aˆ2/(2∗(Va+Vj))∗((Na∗Nd)/(Na+Nd))))=",Cj,"f/cmˆ2" # calculation .

Na =  10000000000000000 /cmˆ3
Nd =  1000000000000000000 /cmˆ3
A =  1 cmˆ2
Vj = 0.54 V
Va =  10 V
e =  1.6e-19 columns
Er =  11.9
Eo =  8.854e-14  F/cm
total permittivity ,E=Eo∗Er=  1.053626e-12  F/cm
Cj=sqrt((e∗E∗Aˆ2/(2∗(Va+Vj))∗((Na∗Nd)/(Na+Nd))))= 8.89830403817e-09 f/cmˆ2

Example 5_15 pgno: 146

In [9]:
#exa 5.15
Na =10**15
print "Na= ",Na,"cmˆ−3"  # initializing value of acceptor concentration .
Nd =10**18
print "Nd= ",Nd," cmˆ−3" # initializing value of donor concentration .
no=1.8*10**6
print "no = ",no," cmˆ−3" # initializing value of intrinsic carrier concentration .
e=1.6*10**-19
print "e = ",e,"columbs" # initializing value of charge of electrons .
K=1.38*10**-23
print "K = ",K," J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T," K" # initializing value of temperature .
Vbi=((K*T/e)*log((Na*Nd)/(no)**2))
print "Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd)/(no)ˆ2))= ",Vbi,"V"  # calculation .

Na=  1000000000000000 cmˆ−3
Nd=  1000000000000000000  cmˆ−3
no =  1800000.0  cmˆ−3
e =  1.6e-19 columbs
K =  1.38e-23  J/k
T =  300  K
Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd)/(no)ˆ2))=  1.220749215 V

Example 5_16 pgno: 146

In [10]:
#5.16
Na =10**18
print "Na= ",Na, "cmˆ−3" # initializing value of acceptor concentration .
Nd =10**18
print "Nd= ",Nd," cmˆ−3" # initializing value of donor concentration .
Vbi =1.4
print "Vbi = ",Vbi  # initializing value of built in voltage .
e=1.6*10**-19
print "e = ",e," columbs"  # initializing value of charge of electrons .
K=1.38*10**-23
print "K = ",K,"J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T," K" # initializing value of temperature .
Vt=0.0259
print "Vt = ",Vt," eV" # initializing value of thermal voltage .
no=sqrt((Na*Nd)/(exp(Vbi/Vt)))
print "no=sqrt ((Na∗Nd)/(exp(Vbi/Vt)))=",no,"cmˆ−3"  #calculation .

Na=  1000000000000000000 cmˆ−3
Nd=  1000000000000000000  cmˆ−3
Vbi =  1.4
e =  1.6e-19  columbs
K =  1.38e-23 J/k
T =  300  K
Vt =  0.0259  eV
no=sqrt ((Na∗Nd)/(exp(Vbi/Vt)))= 1829411.05814 cmˆ−3

Example 5_18 pgno: 147

In [11]:
#exa 5.18
Na =10**17
print "Na = ",Na," cmˆ−3"  # initializing value of acceptor concentration .
Nd =5*10**16
print "Nd = ",Nd," cmˆ−3" # initializing value of donor concentration .
e=1.6*10**-19
print "e = ",e," columbs" # initializing value of charge of electrons .
no=1.5*10**10
print "no = ",no," cmˆ3" # initializing value of intrinsic carrier concentration .
T=300
print "T = ",T," K" # initializing value of temperature .
Vt=0.0259
print "Vt = ",Vt," eV" # initializing value of thermal voltage .
Vbi=(Vt*(log(Na*Nd/(no**2))))
print "(a)Vbi=(Vt∗(log(Na∗Nd/(noˆ2))))= ",Vbi," V"  #calculation .
Efi_Efp=(Vt*log(Na/(no)))
print "(b)value of fermi level on each side of junction ,Efi Efp=(Vt∗log(Na/(no)))=",Efi_Efp," V" # calculation .
Efn_Efi=(Vt*log(Nd/(no)))
print "Efn Efi=(Vt∗log(Nd/(no)))=",Efn_Efi," V"  #calculation .
print "(c)The energy band digram is similar to Fig P5 .3"
Vbi=((Efi_Efp)+(Efn_Efi))
print "(d)Vbi=((Efi Efp)+(Efn Efi))/(e)=Vj=",Vbi," V"   # calculation .

Na =  100000000000000000  cmˆ−3
Nd =  50000000000000000  cmˆ−3
e =  1.6e-19  columbs
no =  15000000000.0  cmˆ3
T =  300  K
Vt =  0.0259  eV
(a)Vbi=(Vt∗(log(Na∗Nd/(noˆ2))))=  0.795961750143  V
(b)value of fermi level on each side of junction ,Efi Efp=(Vt∗log(Na/(no)))= 0.40695713106  V
Efn Efi=(Vt∗log(Nd/(no)))= 0.389004619083  V
(c)The energy band digram is similar to Fig P5 .3
(d)Vbi=((Efi Efp)+(Efn Efi))/(e)=Vj= 0.795961750143  V

Example 5_19 pgno: 148

In [12]:
#exa 5.19
from math import log
Na =5*10**17
print "Na = ",Na,"/cmˆ3" # initializing value of medium p doping concentration .
Nd =5*10**17
print "Nd = ",Nd,"/cmˆ3"  # initializing value of light n doping .
no=1.5*10**10
print "no = ",no,"cmˆ−3" # initializing value of intrinsic concentration .
e=1.6*10**-19
print "e = ",e,"columbs"  # initializing value of charge of electrons .
K=1.38*10**-23
print "K = ",K,"J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T,"K" # initializing value of temperature .
Vbi=((K*T/e)*log((Na*Nd)/(no)**2))
print "(a)Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))=",Vbi,"eV" #calculation .
Efi_Efp=((K*T/e)*log(Na/no))
print "(b)fermi level in the P region and N region , Efi Efp=((KT/e)∗log(Na/no)))= ",Efi_Efp,"eV" # calculation .
VBI=2*(Efi_Efp)
print "(c)VBI from the fermi level ,VBI=2∗(Efi Efp))=",VBI,"V"  # calculation .

Na =  500000000000000000 /cmˆ3
Nd =  500000000000000000 /cmˆ3
no =  15000000000.0 cmˆ−3
e =  1.6e-19 columbs
K =  1.38e-23 J/k
T =  300 K
(a)Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))= 0.896417042561 eV
(b)fermi level in the P region and N region , Efi Efp=((KT/e)∗log(Na/no)))=  0.44820852128 eV
(c)VBI from the fermi level ,VBI=2∗(Efi Efp))= 0.896417042561 V

Example 5_20 pgno: 148

In [13]:
#exa 5.20
Nc=2.8*10**19
print "Nc = ",Nc," /cmˆ3"  # initializing value of number of electron in the conduction band .
Nv=1.04*10**19
print "Nv = ",Nv," /cmˆ3" # initializing value of number of electron in the valence band..
no=1.5*10**10
print "no = ",no," cmˆ−3" # initializing value of intrinsic carrier concentration .
e=1.6*10**-19
print "e = ",e," columbs"  # initializing value of charge of electrons .
K=8.62*10**-5
print "K = ",K," J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T," K"  # initializing value of temperature .
Vt=0.0259
print "Vt = ",Vt," eV" # initializing value of thermal voltage .
Ec_Ef =0.21
print "Ec Ef = ",Ec_Ef," eV" # initializing value of energy difference between conduction band and fermi level.
Ef_Ev =0.18
print "Ef Ev = ",Ef_Ev," eV" # initializing value of energy difference between fermi level and valence band .
Nd=(Nc/exp((Ec_Ef)/(K*T)))
print "Nd=(Nc/exp((Ec−Ef)/(K∗T))))= ",Nd," cmˆ−3" #calculation .
Na=(Nv/exp((Ef_Ev)/(K*T)))
print "Na=(Nv/exp (( Ef−Ev) /(K∗T) ) ) )=",Na,"cmˆ−3" # calculation .
Vbi=(Vt*(log(Na*Nd/(no**2))))
print "Built in potential  ,Vbi=(Vt∗(log(Na∗Nd/(noˆ2))))= ",Vbi," V" #calculation .

Nc =  2.8e+19  /cmˆ3
Nv =  1.04e+19  /cmˆ3
no =  15000000000.0  cmˆ−3
e =  1.6e-19  columbs
K =  8.62e-05  J/k
T =  300  K
Vt =  0.0259  eV
Ec Ef =  0.21  eV
Ef Ev =  0.18  eV
Nd=(Nc/exp((Ec−Ef)/(K∗T))))=  8.32539212771e+15  cmˆ−3
Na=(Nv/exp (( Ef−Ev) /(K∗T) ) ) )= 9.86510951303e+15 cmˆ−3
Built in potential  ,Vbi=(Vt∗(log(Na∗Nd/(noˆ2))))=  0.68954178887  V

Example 5_21 pgno: 149

In [14]:
#exa 5.21
from math import sqrt
from math import exp
Vbi =1.2
print "Vbi = ",Vbi,"/cmˆ3" # initializing value of built in voltage .
no=1.8*10**6
print "no = ",no,"cmˆ−3”" # initializing value of intrinsic concentration .
Vt =0.0259
print "Vt = ",Vt,"eV" # initializing value of thermal voltage .
Er =13.1
print "Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854*10**-14
print "Eo = ",Eo," F/cm" # initializing value of permittivity of free space .
e =1.6*(10**-19)
print "e = ",e," columbs" # initializing value of charge of electrons .
E=Eo*Er
print " total permittivity ,E=Eo∗Er=",E," F/cm"  # calculation .
NaNd=((no**2)*(exp(Vbi/Vt)))
print "(a)NaNd=((noˆ2)∗(exp(Vbi/Vt)))= ",NaNd," /cmˆ6"  # calculation .
Na=(sqrt(NaNd/(4)))
print "Na=(sqrt(NaNd/(4)))=",Na," /cmˆ3"  # calculation .
Nd=4*Na
print "(b)Nd=4∗Na= ",Nd," /cmˆ3"   # calculation.
W=sqrt((2*E*Vbi/e)*((Nd+Na)/(Na*Nd)))
print "(c)W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))= ",W," cm"  # calculation .
xn=0.2*W
print "(d)xn=0.2∗W= ",xn," cm"  # calculation .
xp=0.8*W
print "xp=0.8∗W= ",xp," cm"  # calculation .
Emax=(-e*Nd*xn)/E
print "(e)Emax=(−e∗Nd∗xn)/E)= ",Emax,"V/cm"  # calculation .
#The value of Na after calculation is provided wrong in the book.Due to which value of W,xn,xp and Emax differ ,than the answer provided in the book .

Vbi =  1.2 /cmˆ3
no =  1800000.0 cmˆ−3”
Vt =  0.0259 eV
Er =  13.1
Eo =  8.854e-14  F/cm
e =  1.6e-19  columbs
 total permittivity ,E=Eo∗Er= 1.159874e-12  F/cm
(a)NaNd=((noˆ2)∗(exp(Vbi/Vt)))=  4.28841757806e+32  /cmˆ6
Na=(sqrt(NaNd/(4)))= 1.03542474112e+16  /cmˆ3
(b)Nd=4∗Na=  4.14169896446e+16  /cmˆ3
(c)W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))=  4.58296745959e-05  cm
(d)xn=0.2∗W=  9.16593491919e-06  cm
xp=0.8∗W=  3.66637396767e-05  cm
(e)Emax=(−e∗Nd∗xn)/E)=  -52367.8167292 V/cm

Example 5_22 pgno: 150

In [15]:
#exa 5.22
from math import log
Na =10**16
print "Na = ",Na,"cmˆ−3" # initializing value of acceptor concentration .
Nd =5*10**15
print "Nd = ",Nd,"cmˆ−3" # initializing value of donor concentration .
no=1.5*10**10
print "no = ",no,"cmˆ−3"  # initializing value of intrinsic concentration .
Vbi =0.676
print "Vbi = ",Vbi,"V" # initializing value of built in voltage .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
K=1.38*10**-23
print "K = ",K,"J/k" # initializing value of boltzmann constant .
T=(Vbi*(e/K)*(1/(log((Na*Nd)/(no**2)))))
print "T=(Vbi∗(e/K)∗(1/(log((Na∗Nd)/(noˆ2))))))= ",T,"K"  # calculation .

Na =  10000000000000000 cmˆ−3
Nd =  5000000000000000 cmˆ−3
no =  15000000000.0 cmˆ−3
Vbi =  0.676 V
e =  1.6e-19 columns
K =  1.38e-23 J/k
T=(Vbi∗(e/K)∗(1/(log((Na∗Nd)/(noˆ2))))))=  299.984615257 K

Example 5_23 pgno: 150

In [16]:
#exa 5.23
Na =5*10**17
print "Na = ",Na,"cmˆ−3" # initializing value of acceptor concentration .
Nd =10**17
print "Nd = ",Nd,"cmˆ−3" # initializing value of donor concentration .
no=1.5*10**10
print "no = ",no,"cmˆ−3" # initializing value of intrinsic concentration .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
K=1.38*10**-23
print "K = ",K,"J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T,"K"  # initializing value of temperature .
VBI=0.847
print "VBI = ",VBI,"V" # initializing value of VBI when VBI is reduced by 1%.
Vbi=((K*T/e)*log((Na*Nd)/(no)**2))
print "(a)Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))= ",Vbi,"V" # calculation .
T=(e*VBI/K)*((log(Na*Nd/(no**2)))**-1)
print "(b)T=(VBI∗(e/K)∗(1/(log((Na∗Nd)/(noˆ2))))))= ",T,"K"  # calculation .
#the answer for part (b) is not provided in the book .

Na =  500000000000000000 cmˆ−3
Nd =  100000000000000000 cmˆ−3
no =  15000000000.0 cmˆ−3
e =  1.6e-19 columns
K =  1.38e-23 J/k
T =  300 K
VBI =  0.847 V
(a)Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))=  0.854772836577 V
(b)T=(VBI∗(e/K)∗(1/(log((Na∗Nd)/(noˆ2))))))=  297.271964113 K

Example 5_24 pgno: 150

In [17]:
#exa 5.24
from math import log
from math import sqrt
Na =4e12
print "Na = ",Na," /cmˆ3" # initializing value of medium p doping concentration .
Nd =4e16
print "Nd = ",Nd," /cmˆ3" # initializing value of light n doping.
no=1.5*10e10
print "no = ",no," /cmˆ3" # initializing value of intrinsic carrier concentration .
K=1.38e-23
print "K = ",K," J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T," K" #  initializing value of temperature .
e=1.6e-19
print "e = ",e," columbs" # initializing value of charge of electrons .
Er=11.9
print "Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854e-14
print "Eo = ",Eo," F/cm" # initializing value of permittivity of free space .
E=Eo*Er
print "total permittivity ,E=Eo∗Er=",E," F/cm" # calculation .
Vbi=((K*T/e)*log((Na*Nd)/(no)**2))
print "Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd)/(no)ˆ2))=",Vbi," eV" #calculation .
W=sqrt((2.*E*Vbi/e)*((Nd+Na)/(Na*Nd)))
print"W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))=",W," cm"  # calculation .
xn=((W*Na)/(Nd+Na))
print "xn=((W∗Na) /(Nd+Na) ) )=",xn,"cm" # calculation .
xp=((W*Nd)/(Nd+Na))
print "xp=((W∗Nd) /(Nd+Na) ) )=",xp,"cm" #calculation .
Emax=(e*Nd*xn)/E
print "Emax=(e∗Nd∗xn)/E)=",Emax," V/cm" #calculation .
#the value of W( depletion width) , after calculation is provided wrong in the book,due to this xn,xp ,Emax also differ.(also,the value of Nd+Na substitute in the formula for for xn,xp is wrong )

Na =  4e+12  /cmˆ3
Nd =  4e+16  /cmˆ3
no =  1.5e+11  /cmˆ3
K =  1.38e-23  J/k
T =  300  K
e =  1.6e-19  columbs
Er =  11.9
Eo =  8.854e-14  F/cm
total permittivity ,E=Eo∗Er= 1.053626e-12  F/cm
Built in potential potential ,Vbi=((K∗T/e)∗log((Na∗Nd)/(no)ˆ2))= 0.408234249531  eV
W=sqrt((2∗E∗Vbi/e)∗((Nd+Na)/(Na∗Nd))))= 0.00115943039897  cm
xn=((W∗Na) /(Nd+Na) ) )= 1.15931446752e-07 cm
xp=((W∗Nd) /(Nd+Na) ) )= 0.00115931446752 cm
Emax=(e∗Nd∗xn)/E)= 704.19794046  V/cm

Example 5_25 pgno: 151

In [18]:
#exa 5.25
Na =4*10**17
print "Na = ",Na,"/cmˆ3" # initializing value of donor concentration .
Nd =4*10**15
print "Nd = ",Nd,"/cmˆ3" # initializing value of light n doping.
no=1.5*10**10
print "no = ",no,"cmˆ−3" # initializing value of intrinsic concentration .
Emax =300*10**3
print "Emax = ",Emax,"/cmˆ3" # initializing value of maximum electric field .
K=1.38*10**-23
print "K = ",K,"J/k" # initializing value of boltzmann constant .
T=300
print "T = ",T,"K" # initializing value of temperature .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
Er=11.9
print " Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854*10**-14
print "Eo = ",Eo," F/cm" # initializing value of permittivity of free space .
E=Eo*Er
print " total permittivity ,E=Eo∗Er=",E," F/cm" # calculation .
Vbi=((K*T/e)*log((Na*Nd)/(no)**2))
print "Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))=",Vbi," V" # calculation .
xn=(E*Emax/(Nd*e))
print "xn=(E∗Emax) /( e∗Nd) )=",xn," cm" #calculation .
W=(xn*(Nd+Na)/Na)
print "W=(xn(Nd+Na)/Na))=",W," cm" #calculation .
Vr=((W**2*e/(2*E))*((Na*Nd)/(Na+Nd)))-(Vbi)
print "Vr=(Wˆ2∗e/(2∗E))∗((Na∗Nd)/(Na+Nd))−(Vbi))=",Vr," V" # calculation .

Na =  400000000000000000 /cmˆ3
Nd =  4000000000000000 /cmˆ3
no =  15000000000.0 cmˆ−3
Emax =  300000 /cmˆ3
K =  1.38e-23 J/k
T =  300 K
e =  1.6e-19 columns
 Er =  11.9
Eo =  8.854e-14  F/cm
 total permittivity ,E=Eo∗Er= 1.053626e-12  F/cm
Built in potential potential ,Vbi=((K∗T/e)∗log ((Na∗Nd)/(no)ˆ2))= 0.765710585218  V
xn=(E∗Emax) /( e∗Nd) )= 0.0004938871875  cm
W=(xn(Nd+Na)/Na))= 0.000498826059375  cm
Vr=(Wˆ2∗e/(2∗E))∗((Na∗Nd)/(Na+Nd))−(Vbi))= 74.058198321  V

Example 5_26 pgno: 152

In [19]:
#exa 5.26
Na =5*10**15
print "Na = ",Na,"cmˆ−3" # initializing value of acceptor concentration .
Nd =10**18
print "Nd = ",Nd,"cmˆ−3" # initializing value of donor concentration .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
Vr=10
print "Vr = ",Vr,"V" # initializing value reverse voltage .
Er=11.9
print "Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854*10**-14
print "Eo = ",Eo,"F/cm" # initializing value of permittivity of free space .
E=Eo*Er
print " total permittivity ,E=Eo∗Er=",E," F/cm"  # calculation .
Emax=10**6
print "Emax = ",Emax,"V/cm" # initializing value of maximum electric field .
W=(2.*Vr/(Emax))
print "W = ",W,"cm" # calculation .
Nd=(Emax*E)/(W*e)
print "Nd=(Emax∗e)/(W∗e))=",Nd,"cmˆ−3" # calculation 

Na =  5000000000000000 cmˆ−3
Nd =  1000000000000000000 cmˆ−3
e =  1.6e-19 columns
Vr =  10 V
Er =  11.9
Eo =  8.854e-14 F/cm
 total permittivity ,E=Eo∗Er= 1.053626e-12  F/cm
Emax =  1000000 V/cm
W =  2e-05 cm
Nd=(Emax∗e)/(W∗e))= 3.29258125e+17 cmˆ−3

Example 5_27 pgno: 152

In [20]:
#exa 5.27
from math import sqrt
from math import log
Na =5*10**15
print "Na = ",Na,"cmˆ3" # initializing value of acceptor concentration .
Nd =10**18
print "Nd = ",Nd,"cmˆ3"  # initializing value of donor concentration .
no=1.5*10**10
print "no = ",no,"cmˆ−3" # initializing value of intrinsic carrier concentration .
Vr1=0
print "Vr1 = ",Vr1,"V" # initializing value of built in voltage .
Vr2=5
print "Vr2 = ",Vr2,"V" # initializing value of applied reverse voltage .
A=3*10**-5
print "A = ",A,"cmˆ2" # initializing value of cross sectional area .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
Er=11.9
print "Er = ",Er # initializing value of relative dielectric permittivity constant .
Eo=8.854*10**-14
print "Eo = ",Eo," F/cm" # initializing value of permittivity of free space .
E=Eo*Er
print "total permittivity ,E=Eo∗Er=",E," F/cm"   # calculation .
Vt=0.0259
print "Vt=",Vt," V" # initializing the value of thermal voltage .
Vbi=((Vt)*log((Na*Nd)/(no)**2))
print "Built in potential ,Vbi=(Vt∗log ((Na∗Nd)/(no)ˆ2) )= ",Vbi," V"  # calculation .
Cj1=sqrt((e*E*(A**2)/(2*(Vr1+Vbi)))*((Na*Nd)/(Na+Nd)))
print "Cj1=sqrt((e∗E∗(Aˆ2)/(2∗(Vr1+Vbi))∗((Na∗Nd)/(Na+Nd))))=",Cj1," F" #calculation .
Cj2=sqrt((e*E*(A**2)/(2*(Vr2+Vbi)))*((Na*Nd)/(Na+Nd)))
print "Cj2=sqrt((e∗E∗(Aˆ2)/(2∗(Vr2+Vbi))∗((Na∗Nd)/(Na+Nd))))=",Cj2," F"  #calculation .
# the value of Vr2 use for calculating answer of Cj2 is different than provided in question .
# I have used the value provided in the solution ( i .e. Vr2=5)

Na =  5000000000000000 cmˆ3
Nd =  1000000000000000000 cmˆ3
no =  15000000000.0 cmˆ−3
Vr1 =  0 V
Vr2 =  5 V
A =  3e-05 cmˆ2
e =  1.6e-19 columns
Er =  11.9
Eo =  8.854e-14  F/cm
total permittivity ,E=Eo∗Er= 1.053626e-12  F/cm
Vt= 0.0259  V
Built in potential ,Vbi=(Vt∗log ((Na∗Nd)/(no)ˆ2) )=  0.795961750143  V
Cj1=sqrt((e∗E∗(Aˆ2)/(2∗(Vr1+Vbi))∗((Na∗Nd)/(Na+Nd))))= 6.88597370389e-13  F
Cj2=sqrt((e∗E∗(Aˆ2)/(2∗(Vr2+Vbi))∗((Na∗Nd)/(Na+Nd))))= 2.55181216611e-13  F

Example 5_28 pgno: 153

In [4]:
#exa 5.28
from math import sqrt
Na =1*10**16
print "Na = ",Na,"cmˆ3" # initializing value of acceptor concentration. 
Nd = 5*10**16
print "Nd = ",Nd,"cmˆ3" # initializing value of donor concentration .
no=1.5*10**10
print "no = ",no,"cmˆ−3" # initializing value of intrinsic concentration .
Dn=25
print "Dn = ",Dn,"cmˆ2/sec" # initializing value of diffusion cofficient on the P side.
Dp=10
print "Dp = ",Dp,"cmˆ2/sec" # initializing value of diffusion cofficient on the N side .
tp =5*10** -7
print "tn = ",tp,"s" # initializing value of hole lifetime .
tn =5*10** -7
print "tp = ",tn,"s" # initializing value of electron lifetime .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
Pno=(no**2/Nd)
print "Pno=(noˆ2/Nd))= ",Pno,"cmˆ−3" # calculation .
Npo=(no**2/Na)
print "Npo=(noˆ2/Na))= ",Npo,"cmˆ−3" # calculation .
Lp=(sqrt(Dp*tp))
print "Lp=(sqrt(Dp∗tp)))= ",Lp,"cm" # calculation .
Ln=(sqrt(Dn*tn))
print "Ln=(sqrt(Dn∗tn)))= ",Ln,"cm"  # calculation .
Jo=((e*((Dp*Pno/(Lp))+(Dn*Npo)/(Ln))))
print "Jo=((e ∗((Dp∗Pno/(Lp) )+(Dn∗Npo) /(Ln) ) ) )= ",Jo," A/cmˆ2" # calculation .

Na =  10000000000000000 cmˆ3
Nd =  50000000000000000 cmˆ3
no =  15000000000.0 cmˆ−3
Dn =  25 cmˆ2/sec
Dp =  10 cmˆ2/sec
tn =  5e-07 s
tp =  5e-07 s
e =  1.6e-19 columns
Pno=(noˆ2/Nd))=  4500.0 cmˆ−3
Npo=(noˆ2/Na))=  22500.0 cmˆ−3
Lp=(sqrt(Dp∗tp)))=  0.0022360679775 cm
Ln=(sqrt(Dn∗tn)))=  0.00353553390593 cm
Jo=((e ∗((Dp∗Pno/(Lp) )+(Dn∗Npo) /(Ln) ) ) )=  2.86757820103e-11  A/cmˆ2

Example 5_29 pgno: 154

In [2]:
#exa 5.29
from math import sqrt
Na =10**15
print "Na = ",Na,"cmˆ3" # initializing value of acceptor concentration .
Nd =10**15
print "Nd = ",Nd,"cmˆ3" # initializing value of donor concentration .
no=1.5*10**10
print "no = ",no,"cmˆ−3" # initializing value of intrinsic carrier concentration .
Dn=50
print "Dn = ",Dn,"cmˆ2/sec" # initializing value of built in voltage .
Dp=20
print "Dp = ",Dp,"cmˆ2/sec" # initializing value of applied reverse voltage .
tp =5*10** -7
print "tn = ",tp,"s" # initializing value of hole lifetime .
tn =5*10** -7
print "tp = ",tn,"s" # initializing value of electrons lifetime .
e=1.6*10**-19
print "e = ",e,"columns" # initializing value of charge of electrons .
Pno=(no**2/Nd)
print "Pno=(noˆ2/Nd))= ",Pno,"cmˆ−3" # calculation .
Npo=(no**2/Na)
print "Npo=(noˆ2/Na))= ",Npo,"cmˆ−3" # calculation .
Lp=(sqrt(Dp*tp))
print "Lp=(sqrt(Dp∗tp)))= ",Lp,"cm"  # calculation .
Ln=(sqrt(Dn*tn))
print "Ln=(sqrt(Dn∗tn)))= ",Ln,"cm" # calculation .
Jo=((e*((Dp*Pno/(Lp))+(Dn*Npo)/(Ln))))
print "Jo=((e ∗((Dp∗Pno/(Lp) )+(Dn∗Npo) /(Ln))))=",Jo,"A/cmˆ2" # calculation .
# the value of tp , tn provided in the question , is different than that provided in the solution.
# I have used the value ,provided in the solution(i. e . tp=tn =5∗10ˆ7)

Na =  1000000000000000 cmˆ3
Nd =  1000000000000000 cmˆ3
no =  15000000000.0 cmˆ−3
Dn =  50 cmˆ2/sec
Dp =  20 cmˆ2/sec
tn =  5e-07 s
tp =  5e-07 s
e =  1.6e-19 columns
Pno=(noˆ2/Nd))=  225000.0 cmˆ−3
Npo=(noˆ2/Na))=  225000.0 cmˆ−3
Lp=(sqrt(Dp∗tp)))=  0.00316227766017 cm
Ln=(sqrt(Dn∗tn)))=  0.005 cm
Jo=((e ∗((Dp∗Pno/(Lp) )+(Dn∗Npo) /(Ln))))= 5.87683991532e-10 A/cmˆ2

Example 5_30 pgno: 154

In [23]:
#exa 5.30
Eg = -1.1
print "Eg = ",Eg,"V" # initializing value of energy gap .
Vf1 =0.6
print "Vf1 = ",Vf1,"V" # initializing value of forward voltage for case 1.
T1 =300
print "T1 = ",T1,"K" # initializing value of temperature for case 1.
T2 =310
print "T2 = ",T2,"K" # initializing value of temperature for case 2 .
Vf2=(((Eg+Vf1)*T2)/(T1))-Eg
print "Forward voltage for case 2,Vf2=((Eg+Vf1)∗T2)/( T1)+Eg)= ",Vf2," V"  # calculation .

Eg =  -1.1 V
Vf1 =  0.6 V
T1 =  300 K
T2 =  310 K
Forward voltage for case 2,Vf2=((Eg+Vf1)∗T2)/( T1)+Eg)=  0.583333333333  V