Chapter 6 ELECTRICAL BREAKDOWN IN PN JUNCTIONS

Example 6_2 pgno: 183

In [1]:
#exa 6.2
from math import log
X1 =4.13
print"X1 = ",X1," eV" # initializing value of eldelta Ectron effinity of germanium.
X2 =4.07
print"X2 = ",X2," eV" # initializing value of electron effinity of gallium arsenide .
Eg1 =0.7
print"Eg1 = ",Eg1," eV" # initializing value of energy gap of germanium .
Eg2 =1.43
print"Eg2 = ",Eg2," F/cm" # initializing value of energy gap of gallium arsenide..
Nv1 =6e18
print"Nv1 = ",Nv1," cmˆ−3" # initializing value of density of states in Valence band,Nv for germanium .
Nv2 =7e18
print"Nv2 = ",Nv2," cmˆ−3" # initializing value of density of states in Valence band,Nv for galliminum arsenide .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing valueof thermal voltage . . . Vt = K∗T/e
e=1.6e-19
print"e = ",e," columbs" # initializing value of electronic charge .
no=2.5e13
print"no = ",no," cmˆ−3" # initializingvalue of intrinsic carrier concentration .
Pp=1e17
print"Pp = ",Pp," cmˆ−3" # initializing value of hole concentration on the depletion edge of the N region .
Nd=1e17
print"Nd = ",Nd," cmˆ−3" # initializing value of number of donor ions (which is equal to hole concentration on the depletion edge of the N region).
np=(no**2)/Pp
print"np=",np," cmˆ−3"# calculation
delta_Eg=(Eg2-Eg1)
print"delta Eg=(Eg2−Eg1)=",delta_Eg," eV"#calculation
delta_Ec=(X1-X2)
print"delta Ec=(X1−X2)=",delta_Ec," eV"#calculation
delta_Ev=(delta_Eg-delta_Ec)
print"delta  Ev=(delta  Eg−delta  Ec )=",delta_Ev," eV"# calculation
Vbi=((delta_Ev*1.6*10**-19)/(e))+((Vt*log((Nv1*Nd)/(np*Nv2))))
print"Vbi=((delta Ev∗1.6∗10ˆ−19)/(e))+((Vt∗log((Nv1∗Nd) /(np∗Nv2) ) ) )=",Vbi," V"# calculation

X1 =  4.13  eV
X2 =  4.07  eV
Eg1 =  0.7  eV
Eg2 =  1.43  F/cm
Nv1 =  6e+18  cmˆ−3
Nv2 =  7e+18  cmˆ−3
Vt =  0.0259  eV
e =  1.6e-19  columbs
no =  2.5e+13  cmˆ−3
Pp =  1e+17  cmˆ−3
Nd =  1e+17  cmˆ−3
np= 6250000000.0  cmˆ−3
delta Eg=(Eg2−Eg1)= 0.73  eV
delta Ec=(X1−X2)= 0.06  eV
delta  Ev=(delta  Eg−delta  Ec )= 0.67  eV
Vbi=((delta Ev∗1.6∗10ˆ−19)/(e))+((Vt∗log((Nv1∗Nd) /(np∗Nv2) ) ) )= 1.09563926875  V

Example 6_4 pgno: 184

In [2]:
#exa 6.4
from math import log
Nc=2.8e19
print"Nc = ",Nc," cmˆ−3" # initializing value of effective density of state in the conduction band .
k=-4e15
print"k = ",k," cmˆ4Fˆ−2Vˆ−1" # initializing value of slope of the (1/Cˆ2) versus V curve.
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 dielectric constant of free space.
e=1.6e-19
print"e = ",e," columns" # initializing value of charge of electrons .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
VBI=0.3
print"VBI = ",VBI," V" # initializing value of built in voltage .
E=Eo*Er
print" total permittivity ,E=Eo∗Er =",E," F/cm "# calculation
Nd=((-2)/(e*E)*(1/k))
print"Nd=((−2)/(e∗E)∗(1/k)))=",Nd," cmˆ−3" # c a l c u l a t i o n
Vn=(Vt*(log(Nc/Nd)))
print"Vn=(Vt∗( log (Nc/Nd) ) )=",Vn," V"#calculation
VBn=(VBI+Vn)
print"VBn=(VBI+Vn)=",VBn," V"# calculation
# taking ,... d(1/Cˆ2)/dV as k,... for simlification,

Nc =  2.8e+19  cmˆ−3
k =  -4e+15  cmˆ4Fˆ−2Vˆ−1
Er =  11.9
Eo =  8.854e-14  F/cm
e =  1.6e-19  columns
Vt =  0.0259  eV
VBI =  0.3  V
 total permittivity ,E=Eo∗Er = 1.053626e-12  F/cm 
Nd=((−2)/(e∗E)∗(1/k)))= 2.96594806886e+15  cmˆ−3
Vn=(Vt∗( log (Nc/Nd) ) )= 0.237056563109  V
VBn=(VBI+Vn)= 0.537056563109  V

Example 6_5 pgno: 184

In [3]:
#exa 6.5
from math import log
Nd =2e17
print"Nd = ",Nd,"/cmˆ−3" # initializing value of donor concentration .
Nc=2.8e19
print"Nc = ",Nc,"/cmˆ−3" # initializing value of effective density of state in the conduction band .
Js =40e-6
print"Js = ",Js,"A/cmˆ2" # initializing value of saturation current density .
T=300
print"T = ",T,"K" # initializing value of absolute temperature .
R=110
print"R = ",R," A/(K−cmˆ2)" #initializing value of richardson ’ s constant .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
VBn=(Vt*(log(R*T**2/Js)))
print"VBn = ",VBn," V" # calculation .
Vn=(Vt*(log(Nc/Nd)))
print"Vn = ",Vn," V" # calculation .
VBI=(VBn-Vn)
print"VBI=(VBn−Vn))=",VBI," V"#calculation
#The value of Vn (after calculation ) is provided wrong in the book,due to which VBI also differ.

Nd =  2e+17 /cmˆ−3
Nc =  2.8e+19 /cmˆ−3
Js =  4e-05 A/cmˆ2
T =  300 K
R =  110  A/(K−cmˆ2)
Vt =  0.0259  eV
VBn =  0.679478119251  V
Vn =  0.127988538746  V
VBI=(VBn−Vn))= 0.551489580505  V

Example 6_6 pgno: 186

In [4]:
#exa 6.6
from math import sqrt
from math import log
Nd =2*10**17
print"Nd = ",Nd," /cmˆ−3" # initializing value of donor concentration .
Dp=30
print"Dp = ",Dp," cmˆ2/s" # initializing value of diffusion cofficient .
Nc=2.8*10**19
print"Nc = ",Nc," /cmˆ−3" # initializing value of effective density of state in the conduction band .
Js =40*10**-6
print"Js = ",Js, "A/cmˆ2" # initializing value of saturation current density .
no=1.5*10**10
print"no = ",no," cmˆ−3" # initializing value of intrinsic concentration of electrons .
tp=10**-6
print"tp = ",tp," s" # initializing value of hole life−time.
T=300
print"T = ",T," K" # initializing value of absolute temperature .
R=110
print"R = ",R," A/(K−cmˆ2)" #initializing value of richardson ’ s constant .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
e=1.6*10**-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=8.854*10**-14
print"Eo = ",Eo," F/cm" # initializing value of dielectric constant of free space.
E=Eo*Er
print" total permittivity ,E=Eo∗Er)=",E," F/cm"# calculation
VBn=(Vt*(log(R*T**2/Js)))
print"VBn = ",VBn," V" # calculation .
Vn=(Vt*(log(Nc/Nd)))
print"Vn = ",Vn," V" # calculation .
VBI=(VBn-Vn)
print"VBI=(VBn−Vn))=",VBI," V"#calculation
W=(sqrt((E*VBI)/(e*Nd)))
print"current density in a metal semiconductor junction ,W = ",W," A" # calculation .
Lp=(sqrt(Dp*tp))
print"Diffusion length ,Lp=(sqrt(Dp∗tp)) = ", Lp," cm" # calculation .
Jpo=(e*Dp*no**2)/(Lp*Nd)
print" saturation hole current density , Jpo=(e∗Dp∗noˆ2) /(Lp∗Nd) ) = ",Jpo," A/cmˆ2" # calculation .
#The value of Vn (after calculation ) is provided wrong in the book,due to which VBI differ and due to VBI ,current density in a metal semiconductor junction (W) gets changed .
#The value of Jpo ( saturation hole current density ),after calculation is also provided wrong in the book .,

Nd =  200000000000000000  /cmˆ−3
Dp =  30  cmˆ2/s
Nc =  2.8e+19  /cmˆ−3
Js =  4e-05 A/cmˆ2
no =  15000000000.0  cmˆ−3
tp =  1e-06  s
T =  300  K
R =  110  A/(K−cmˆ2)
Vt =  0.0259  eV
e =  1.6e-19  columbs
Er =  11.9
Eo =  8.854e-14  F/cm
 total permittivity ,E=Eo∗Er)= 1.053626e-12  F/cm
VBn =  0.679478119251  V
Vn =  0.127988538746  V
VBI=(VBn−Vn))= 0.551489580505  V
current density in a metal semiconductor junction ,W =  4.26124893939e-06  A
Diffusion length ,Lp=(sqrt(Dp∗tp)) =  0.00547722557505  cm
 saturation hole current density , Jpo=(e∗Dp∗noˆ2) /(Lp∗Nd) ) =  9.85900603509e-13  A/cmˆ2

Example 6_8 pgno:186

In [5]:
#exa 6.8
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 .
VBD =20
print"VBD = ",VBD," V" #initializing value of break down voltage .
e=1.6*10**-19
print"e = ",e," columns" # initializing value of charge of electrons .
E=Eo*Er
print" total permittivity ,E=Eo∗Er)=",E," F/cm"# calculation
Emax =5*10**5
print"Emax = ",Emax," V/cm" # initializing value of maximum critical electric field .
ND=(Eo*Er*(Emax**2))/(2*e*VBD)
print"ND=(Eo∗Er∗(Emaxˆ2))/(2∗e∗VBD)=",ND,"cmˆ−3"# calculation
#the formula given in the solution for VBD is somewhat written wrong.The correct formula is ( VBD=(E∗Emaxˆ2/2∗e∗ND)) .

Er =  11.9
Eo =  8.854e-14  F/cm
VBD =  20  V
e =  1.6e-19  columns
 total permittivity ,E=Eo∗Er)= 1.053626e-12  F/cm
Emax =  500000  V/cm
ND=(Eo∗Er∗(Emaxˆ2))/(2∗e∗VBD)= 4.1157265625e+16 cmˆ−3

Example 6_9 pgno: 187

In [6]:
#exa 6.9
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=1.6e-19
print"e = ",e," columns" # initializing value of charge of electrons .
no=1.5e10
print"no = ",no,"cmˆ−3" # initializing value of intrinsic concentration of electrons .
Nd =1e16
print"Nd=",Nd," cmˆ−3"#initializing the value of donor concentration .
Emax =2e5
print"Emax = ",Emax," V/cm" # initializing value of maximum critical electric field .
Na =1e16
print"Na=",Na," cmˆ−3"# initializing the value of acceptor concentration .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
E=Eo*Er
print" total permittivity ,E=Eo∗Er)=",E," F/cm"# calculation
VBI=(Vt*(log(Na*Nd/no**2)))
print"VBI=(Vt∗(log(Na∗Nd/noˆ2))) = ",VBI," V" # calculation .
V=(E*Emax**2)/(e*Nd)
print"breakdown voltage for symetrical abrupt junction ,VBD+VBI=(E∗Emaxˆ2) /( e∗Nd) )=",V,"V" # calculation 
VBD=V-VBI
print"VBD=V−VBI =",VBD," V"# calculation

Er =  11.9
Eo =  8.854e-14  F/cm
e =  1.6e-19  columns
no =  15000000000.0 cmˆ−3
Nd= 1e+16  cmˆ−3
Emax =  200000.0  V/cm
Na= 1e+16  cmˆ−3
Vt =  0.0259  eV
 total permittivity ,E=Eo∗Er)= 1.053626e-12  F/cm
VBI=(Vt∗(log(Na∗Nd/noˆ2))) =  0.694640354303  V
breakdown voltage for symetrical abrupt junction ,VBD+VBI=(E∗Emaxˆ2) /( e∗Nd) )= 26.34065 V
VBD=V−VBI = 25.6460096457  V

Example 6_10 pgno: 187

In [7]:
#exa 6.10
from math import log
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=1.6*10**-19
print"e = ",e," columns" # initializing value of charge of electrons .
no=1.5*10**10
print"no = ",no,"cmˆ−3" # initializing value of intrinsic concentration of electrons .
Emax=10**6
print"Emax = ",Emax," V/cm" # initializing value of maximum critical electric field ..
Nd =1*10**18
print"Nd=",Nd," cmˆ−3"#initializing the value of donor concentration .
Na =1*10**18
print"Na=",Na," cmˆ−3"# initializing the value of acceptor concentration .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
VBI=(Vt*(log(Na*Nd/no**2)))
print"VBI=(Vt∗(log(Na∗Nd/noˆ2))) = ",VBI," V" # calculation .
E=Eo*Er
print" total permittivity ,E=Eo∗Er)=",E," F/cm 99"# calculation
V=(E*Emax**2)/(e*Nd)
print"breakdown voltage for symetrical abrupt junction ,VBD+VBI=(E∗Emaxˆ2) /( e∗Nd) )=",V,"V"# calculation
VBD=V-VBI
print"VBD=V−VBI)=",VBD," V"# calculation

Er =  11.9
Eo =  8.854e-14  F/cm
e =  1.6e-19  columns
no =  15000000000.0 cmˆ−3
Emax =  1000000  V/cm
Nd= 1000000000000000000  cmˆ−3
Na= 1000000000000000000  cmˆ−3
Vt =  0.0259  eV
VBI=(Vt∗(log(Na∗Nd/noˆ2))) =  0.933188169937  V
 total permittivity ,E=Eo∗Er)= 1.053626e-12  F/cm 99
breakdown voltage for symetrical abrupt junction ,VBD+VBI=(E∗Emaxˆ2) /( e∗Nd) )= 6.5851625 V
VBD=V−VBI)= 5.65197433006  V

Example 6_11 pgno: 188

In [8]:
#exa 6.11
Nd =1e18
print"Nd = ",Nd," cmˆ−3" # initializing value of donor concentration .
Na = -1e18
print"Na = ",Na," cmˆ3" # initializing value of acceptor concentration .
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 dielectric constant of free space.
e=1.6e-19
print"e = ",e," columns" # initializing value of charge of electrons .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
Vbd=15
print"Vbd = ",Vbd," eV" # initializing value of break down voltage .
W=2e-4
print"W = ",W," cm" # initializing value of the distance over which doping profile varies.
E=Eo*Er
print" total permittivity ,E=Eo∗Er=",E," F/cm"# calculation
a=((Nd-Na)/(W))
print"slope of doping profile curve ,a=((Nd−Na)/(W))= ",a," cmˆ−4"# calculation
Emax=(((Vbd)**2)*9*e*a/(32*E))**(1/3)
print"Emax=(((Vbd)ˆ2)∗9∗e∗a/(32∗E))ˆ(1/3)=",Emax," V/cm"# calculation
## calculation was given wrong in the book

Nd =  1e+18  cmˆ−3
Na =  -1e+18  cmˆ3
Er =  11.9
Eo =  8.854e-14  F/cm
e =  1.6e-19  columns
Vt =  0.0259  eV
Vbd =  15  eV
W =  0.0002  cm
 total permittivity ,E=Eo∗Er= 1.053626e-12  F/cm
slope of doping profile curve ,a=((Nd−Na)/(W))=  1e+22  cmˆ−4
Emax=(((Vbd)ˆ2)∗9∗e∗a/(32∗E))ˆ(1/3)= 1.0  V/cm

Example 6_12 pgno: 188

In [9]:
#exa 6.12
from math import sqrt
from math import log
Ew =4.55
print"Ew = ",Ew," V" # initializing value of work function of tungusten .
X=4.01
print"X = ",X,"V" # initializing value of electron effinity of silicon .
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=1.6*10**-19
print"e = ",e," columns" # initializing value of charge of electrons .
Nc=2.8*10**19
print"Nc = ",Nc,"/cmˆ−3" # initializing value of effective density of state in the conduction band .
Nd=10**17
print"Nd = ",Nd,"/cmˆ−3" # initializing value of donor concentration .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
VB=(Ew-X)
print" Barrier height ,VB=(Ew−X) = ",VB," V" # calculation .
Ec_Ef=(Vt*log(Nc/Nd))
print"Ec Ef=(Vt∗log(Nc/Nd))=",Ec_Ef," V"#calculation
VBI=(VB-(Ec_Ef))
print"VBI=(VB−(Ec  Ef ) )=",VBI," V"# calculation
xn=sqrt(2*Eo*Er*VBI/(e*Nd))
print"Depletion width ,xn=sqrt(2∗Eo∗Er∗VBI/(e∗Nd))=",xn," cm"# calculation
Emax=(e*Nd*xn/(Eo*Er))
print"maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))=",Emax," V/cm"# calculation

Ew =  4.55  V
X =  4.01 V
Er =  11.9
Eo =  8.854e-14  F/cm
e =  1.6e-19  columns
Nc =  2.8e+19 /cmˆ−3
Nd =  100000000000000000 /cmˆ−3
Vt =  0.0259  eV
 Barrier height ,VB=(Ew−X) =  0.54  V
Ec Ef=(Vt∗log(Nc/Nd))= 0.145941050722  V
VBI=(VB−(Ec  Ef ) )= 0.394058949278  V
Depletion width ,xn=sqrt(2∗Eo∗Er∗VBI/(e∗Nd))= 7.20408525154e-06  cm
maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))= 109398.746827  V/cm

Example 6_13 pgno: 189

In [10]:
#exa 6.13
from math import sqrt
from math import log
Ew =4.5
print"Ew = ",Ew," V" # initializing value of work function of tungusten .
X=4.01
print"X = ",X,"V" # initializing value of electron effinity of silicon .
Er=12
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 .
Vr=3
print"Vr = ",Vr," V" # initializing value of reverse voltage .
e=1.6*10**-19
print"e = ",e," columns" # initializing value of charge of electrons .
Nc=2.8*10**19
print"Nc = ",Nc,"/cmˆ−3" # initializing value of effective density of state in the conduction band .
Nd=10**17
print"Nd = ",Nd,"/cmˆ−3" # initializing value of donor concentration .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
VB=(Ew-X)
print" barrier height ,VB=(Ew−X) = ",VB," V"# calculation .
Ec_Ef=(Vt*log(Nc/Nd))
print"Ec Ef=(Vt∗log(Nc/Nd))=",Ec_Ef," V"#calculation
VBI=(VB-(Ec_Ef))
print"VBI=(VB−(Ec  Ef ) )=",VBI," V"#calculation
xn=sqrt((2*Eo*Er*(VBI+Vr))/(e*Nd))
print"Depletion width ,xn=sqrt(2∗Eo∗Er∗(VBI+Vr)/(e∗Nd))=",xn," cm"#calculation
Emax=(e*Nd*xn/(Eo*Er))
print"maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))=",Emax," V/cm"# calculation
C=sqrt((e*Eo*Er*Nd)/(2*(VBI+Vr)))
print"Capitance per unit area ,C=sqrt (( e∗Eo∗Er∗Nd)/(2∗(VBI+Vr) ) )=",C," F/cmˆ2"# calculation
#the Value of reverse voltage(Vr) provided in the question is different than used in the solution . I have used the value provided in the solution ( i . e Vr=3).
#the value of C (Capitance per unit area) after calculation is provided wrong in the book.

Ew =  4.5  V
X =  4.01 V
Er =  12
Eo =  8.854e-14  F/cm
Vr =  3  V
e =  1.6e-19  columns
Nc =  2.8e+19 /cmˆ−3
Nd =  100000000000000000 /cmˆ−3
Vt =  0.0259  eV
 barrier height ,VB=(Ew−X) =  0.49  V
Ec Ef=(Vt∗log(Nc/Nd))= 0.145941050722  V
VBI=(VB−(Ec  Ef ) )= 0.344058949278  V
Depletion width ,xn=sqrt(2∗Eo∗Er∗(VBI+Vr)/(e∗Nd))= 2.10742608187e-05  cm
maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))= 317359.548508  V/cm
Capitance per unit area ,C=sqrt (( e∗Eo∗Er∗Nd)/(2∗(VBI+Vr) ) )= 5.04160031586e-08  F/cmˆ2

Example 6_13 pgno: 190

In [11]:
#exa 6.14
from math import log
Ew =4.28
print"Ew = ",Ew," V" # initializing value of work function of tungusten .
X=4.01
print"X = ",X,"V" # initializing value of electron effinity of silicon .
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=1.6*10**-19
print"e = ",e," columns" # initializing value of charge of electrons .
Nc=2.8*10**19
print"Nc = ",Nc,"/cmˆ−3" # initializing value of effective density of state in the conduction band .
Nd=10**15
print"Nd = ",Nd,"/cmˆ−3" # initializing value of donor concentration .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
VB=(Ew-X)
print" barrier height ,VB=(Ew−X) = ",VB," V"# calculation .
Ec_Ef=(Vt*log(Nc/Nd))
print"Ec Ef=(Vt∗log(Nc/Nd))=",Ec_Ef," V"#calculation
VBI=(VB-(Ec_Ef))
print"VBI=(VB−(Ec  Ef ) )=",VBI," V"#calculation
xn=sqrt(2*Eo*Er*VBI/(e*Nd))
print"Depletion width ,xn=sqrt(2∗Eo∗Er∗VBI/(e∗Nd))=",xn," cm"# calculation
Emax=(e*Nd*xn/(Eo*Er))
print"maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))=",Emax," V/cm"# calculation
#the Value of donor concentration (Nd) provided in the question is different than used in the solution . I have used the value provided in the question(i.e Nd=10ˆ15). ,i.e answer differs than provided in the book .

Ew =  4.28  V
X =  4.01 V
Er =  11.9
Eo =  8.854e-14  F/cm
e =  1.6e-19  columns
Nc =  2.8e+19 /cmˆ−3
Nd =  1000000000000000 /cmˆ−3
Vt =  0.0259  eV
 barrier height ,VB=(Ew−X) =  0.27  V
Ec Ef=(Vt∗log(Nc/Nd))= 0.265214958539  V
VBI=(VB−(Ec  Ef ) )= 0.00478504146083  V
Depletion width ,xn=sqrt(2∗Eo∗Er∗VBI/(e∗Nd))= 7.93854843013e-06  cm
maximum electric field ,Emax=(e∗Nd∗xn/(Eo∗Er))= 1205.52050616  V/cm

Example 6_15 pgno: 191

In [12]:
#exa 6.15
from math import sqrt
from math import log
Ew =5.1
print"Ew = ",Ew," V" # initializing value of work function of tungusten .
X=4.01
print"X = ",X,"V" # initializing value of electron effinity of silicon .
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=1.6*10**-19
print"e = ",e," columns" # initializing value of charge of electrons .
Nc=2.8*10**19
print"Nc = ",Nc,"/cmˆ−3" # initializing value of effective density of state in the conduction band .
Nd =5*10**15
print"Nd = ",Nd,"/cmˆ−3" # initializing value of donor concentration .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
Vr=5
print"Vr = ",Vr," V" # initializing value of reverse voltage .
A=1*10**-4
print"A = ",A," cmˆ2" # initializing valueof area of the gold silicon junction diode..
VB=(Ew-X)
print" barrier height ,VB=(Ew−X) = ",VB," V"# calculation .
Ec_Ef=(Vt*log(Nc/Nd))
print"Ec Ef=(Vt∗log(Nc/Nd))=",Ec_Ef," V"#calculation
VBI=(VB-(Ec_Ef))
print"VBI=(VB−(Ec  Ef ) )=",VBI," V"#calculation
C1=sqrt((e*Eo*Er*Nd)/(2*(VBI+Vr)))
print"Capitance per unit area ,C1=sqrt (( e∗Eo∗Er∗Nd)/(2∗(VBI+Vr) ) )=",C1," F/cmˆ2"#calculation
C=C1*A
print"total junction capatiance ,C=C1∗A=",C,"F"# calculation

Ew =  5.1  V
X =  4.01 V
Er =  11.9
Eo =  8.854e-14  F/cm
e =  1.6e-19  columns
Nc =  2.8e+19 /cmˆ−3
Nd =  5000000000000000 /cmˆ−3
Vt =  0.0259  eV
Vr =  5  V
A =  0.0001  cmˆ2
 barrier height ,VB=(Ew−X) =  1.09  V
Ec Ef=(Vt∗log(Nc/Nd))= 0.223530516607  V
VBI=(VB−(Ec  Ef ) )= 0.866469483393  V
Capitance per unit area ,C1=sqrt (( e∗Eo∗Er∗Nd)/(2∗(VBI+Vr) ) )= 8.4758805431e-09  F/cmˆ2
total junction capatiance ,C=C1∗A= 8.4758805431e-13 F

Example 6_17 pgno: 191

In [13]:
#exa 6.17
from math import pi
from math import sqrt
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," columns" # initializing value of charge of electrons .
Emax =30*10**3
print"Emax = ",Emax," V/cm" # initializing value of maximum critical electric field ..
E=Eo*Er
print" total permittivity ,E=Eo∗Er)=",E," F/cm"# calculation
V=sqrt(e*Emax/(4*pi*E))
print"lowering of the barrier height ,V=sqrt(e∗Emax/(4∗pi∗E) )=",V," V"# calculation
Xmax=sqrt(e/(16*pi*E*Emax))
print" position of the maximum barrier height ,Xmax=sqrt(e/(16∗%pi∗E∗Emax))=",Xmax," cm"# calculation

Er =  13.1
Eo =  8.854e-14  F/cm
e =  1.6e-19  columns
Emax =  30000  V/cm
 total permittivity ,E=Eo∗Er)= 1.159874e-12  F/cm
lowering of the barrier height ,V=sqrt(e∗Emax/(4∗pi∗E) )= 0.0181472273453  V
 position of the maximum barrier height ,Xmax=sqrt(e/(16∗%pi∗E∗Emax))= 3.02453789089e-07  cm

Example 6_18 pgno: 192

In [14]:
#exa 6.18
from math import exp
A=10**-4
print"A = ",A," cmˆ−2" # initializing value of cross sectional area .
VBn =0.55
print"VBn = ",VBn,"V" # initializing value of barrier height .
T=300
print"T = ",T,"K" # initializing value of absolute temperature .
R=110
print"R = ",R," A/(K−cmˆ2)" #initializing value of richardson ’ s constant .
Vt=0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
V=0.25
print"V = ",V," V" # initializing value of forward bias voltage .
Io=A*R*T**2*exp(-VBn/Vt)
print"reverse saturation current , Io=A∗R∗Tˆ2∗exp(−VBn/Vt) = ",Io," A" # calculation .
I=Io*((exp(V/Vt))-1)
print"diode current , I=Io(exp(V/Vt)−1)=",I,"A"# calculation

A =  0.0001  cmˆ−2
VBn =  0.55 V
T =  300 K
R =  110  A/(K−cmˆ2)
Vt =  0.0259  eV
V =  0.25  V
reverse saturation current , Io=A∗R∗Tˆ2∗exp(−VBn/Vt) =  5.93151320618e-07  A
diode current , I=Io(exp(V/Vt)−1)= 0.00922931077027 A

Example 6_19 pgno:192

In [15]:
#exa 6.19
from math import log
Io1 =10**-9
print"Io1 = ",Io1," A" # initializing value of reverse saturation current of silicon SBD.
Io2 =10**-14
print"Io2 = ",Io2,"A" # initializing value of reverse saturation current of a PN junction .
Vt =0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
I=100*10**-6
print"I = ",I," A" # initializing value of required current .
VfSBD=Vt*((log(I/Io1+1)))
print"forward Voltage for silicon SBD,VfSBD=Vt∗(( log(I/Io1+1)))= ",VfSBD," V" # calculation
VfPN=Vt*((log(I/Io2+1)))
print"forward Voltage for silicon SBD,VfPN=Vt∗((log(I/Io2+1)))=",VfPN," V"#calculation

Io1 =  1e-09  A
Io2 =  1e-14 A
Vt =  0.0259  eV
I =  0.0001  A
forward Voltage for silicon SBD,VfSBD=Vt∗(( log(I/Io1+1)))=  0.298185028541  V
forward Voltage for silicon SBD,VfPN=Vt∗((log(I/Io2+1)))= 0.596369539088  V

Example 6_20 pgno: 193

In [16]:
#exa 6.20
from math import log
Io1 =10*10**-7
print"Io1 = ",Io1," A" # initializing value of reverse saturation current of silicon SBD.
Io2 =10*10**-7
print"Io2 = ",Io2,"A" # initializing value of reverse saturation current of a PN junction .
Vt =0.0259
print"Vt = ",Vt," eV" # initializing value of thermal voltage .
I=1*10**-3
print"I = ",I," A" # initializing value of forward current .
V=0.25
print"V = ",V," V" # initializing value of difference in the forward voltage of the two diode .
VfSBD=Vt*((log(I/Io1+1)))
print"forward Voltage for silicon SBD,VfSBD=Vt∗(( log(I/Io1+1)))= ",VfSBD," V" # calculation 109
VfPN=(V+VfSBD)
print"forward volage applied across the PN Diode ,VfPN=(V+VfSBD)=",VfPN," V"#calculation  
Io=(I/((exp(VfPN/Vt))-1))
print"reverse saturation current of the PN junction Diode,Io=(I/((exp(VfPN/Vt))−1))=",Io," A" # calculation

Io1 =  1e-06  A
Io2 =  1e-06 A
Vt =  0.0259  eV
I =  0.001  A
V =  0.25  V
forward Voltage for silicon SBD,VfSBD=Vt∗(( log(I/Io1+1)))=  0.178936748784  V
forward volage applied across the PN Diode ,VfPN=(V+VfSBD)= 0.428936748784  V
reverse saturation current of the PN junction Diode,Io=(I/((exp(VfPN/Vt))−1))= 6.41998882039e-11  A