# Chapter5:JUNCTIONS IN SEMICONDUCTORS:P-N DIODES¶

## Ex5.1:pg-169¶

In [15]:
Nd = 1.0*10**16 #initialising value of donor atoms in centimeter
Na= 1.0*10**18 #initialising value of accepter atoms in centimeter
Nc= 2.8*10**19 #initialising value of conduction band effective density
Nv= 1.0*10**19 #initialising value of valence band effective density
kbT = 0.026 #initializing value of kbT at 300K
Eg = 1.1 #initializing value of forbidden energy gap
#NOTE: nn=Nd and pp=Na
eVbi = Eg+(kbT*log(Na/Nv))+((kbT*log(Nd/Nc)))
print"built in voltage is ,eVbi =  ","{:.2e}".format(eVbi)," eV"
apsilen = 11.9*8.85*10**-12 #initializing value of relative permitivity
e = 1.6*10**-19 #initializing value of charge of electron
Vbi=eVbi/e
Nd = 1.0*10**22 #initialising value of donor atoms in metrers
Na= 1.0*10**24 #initialising value of accepter atomsin meters
Wp_Vbi = sqrt(((2*apsilen*eVbi)/(e))*(Nd/(Na*(Na+Nd))))
print"depletion width at p side is ,Wp_Vbi =","{:.2e}".format(Wp_Vbi)," m"
Wn_Vo = 100*sqrt(((2*apsilen*eVbi)/(e))*(Nd/(Na*(Na+Nd))))
print"depletion width at n side is ,Wn_Vo = ","{:.2e}".format(Wn_Vo)," m"

built in voltage is ,eVbi =   8.34e-01  eV
depletion width at p side is ,Wp_Vbi = 3.30e-09  m
depletion width at n side is ,Wn_Vo =  3.30e-07  m


## Ex5.2:pg-172¶

In [19]:
Nd = 1.0*10**16 #initialising value of donor atoms in centimeter
Na= 1.0*10**18 #initialising value of accepter atoms in centimeter
ni = 1.5*10**10 #initializing value of intrinsic carrier concentration
#NOTE: nn=Nd and pp=Na
R= 10*10**-6 #initializing value of radius of pn diode
A = math.pi*(R**2)
pn = ni**2/Nd
print"concentration of electron in p type is ,pn = ","{:.2e}".format(pn)," cm**-3"
kbT = 0.026  #initializing value of kbT at 300K
eVbi = (kbT*log(Na/pn))
print"built in voltage is ,eVbi =","{:.2e}".format(eVbi)," V"
apsilen = 11.9*8.84*10**-12 #initializing value of relative permitivity
e = 1.6*10**-19 #initializing value of charge of electron
#NOTE: for reverse bias Vr = 0 V,
Wp_4 = sqrt(((2.0*apsilen*eVbi)/(e))*((Nd*1.0*10**6)/((Na*1.0*10**6)*((Nd*1.0*10**6)+(Na*1.0*10**6)))))
print"depletion width at p side is ,Wp_4 =","{:.2e}".format(Wp_4),"m"
Wn_4 = Wp_4*100
print"depletion width at n side is ,Wn_4 = ","{:.2e}".format(Wn_4),"m"

#for calculation purpose and for differentiating part (I), equating
Vbi_4 = eVbi
Vbi_4=round(Vbi_4,3)
print("")# for spacing
Vbi_2 = Vbi_4 + 2
Vbi_2=round(Vbi_2,3)
#NOTE: for reverse bias Vr = 2 V,
Wp_2 = Wp_4*sqrt(Vbi_2/Vbi_4)
print"depletion width at p side is ,Wp_2 = ","{:.2e}".format(Wp_2),"m"
Wn_2 = Wp_2*100
print"depletion width at n side is ,Wn_2  = ","{:.2e}".format(Wn_2),"m"

print("")# for spacing
Vbi_3 = Vbi_4 + 5
Vbi_3=round(Vbi_3,2)
#NOTE: for reverse bias Vr = 5 V,
Wp_3 = sqrt(((2.0*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_3/Vbi_4)
print"depletion width at p side is ,Wp_3 =  ""{:.2e}".format(Wp_3),"m"
Wn_3 = Wp_3*100
print"depletion width at n side is ,Wn_3 = ","{:.2e}".format(Wn_3),"m"
print("")# for spacing
Vbi_4 = Vbi_4 + 10
Vbi_4=round(Vbi_4,2)
#NOTE: for reverse bias Vr = 10 V,

Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_4/Vbi_4)
print"depletion width at p side is ,Wp_4 =  ","{:.2e}".format(Wp_4),"m"
Wn_4 = Wp_4*100
print"depletion width at n side is ,Wn_4 = ","{:.2e}".format(Wn_4),"m"

print("")# for spacing
Vbi_5 = Vbi_4 - 0.5
Vbi_5=round(Vbi_5,2)
#NOTE: for forward bias Vf = 0.5 V,

Wp_5 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_5/Vbi_4)
print"depletion width at p side is ,Wp_5 =","{:.2e}".format(Wp_5),"m"
Wn_5 = Wp_5*100
print"depletion width at n side is ,Wn_5 = ","{:.2e}".format(Wn_5),"m"

#CALCULATION   FOR    PEAK FIELD :
F =  - e*Nd*(10**6)*Wn_4/apsilen
print"F=","{:.2e}".format(F),"v/m"
F =  - e*Nd*(10**6)*Wn_2/apsilen
print"F=","{:.2e}".format(F),"v/m"
F =  - e*Nd*(10**6)*Wn_3/apsilen
print"F=","{:.2e}".format(F),"v/m"
F =  - e*Nd*(10**6)*Wn_4/apsilen
print"F=","{:.2e}".format(F),"v/m"
F =  - e*Nd*(10**6)*Wn_5/apsilen
print"F=","{:.2e}".format(F),"v/m"
#calculation for
Q = e*(Nd*10**6)*Wn_4*A#charge in depletion region for Vr = 0V
print"Q = ","{:.2e}".format(Q),"C"
#due to approximation taken by author in the textbook .... the values of Vbi_2, Vbi_3, Vbi_4 and
#the values of depletion width(Wp_4, Wp_2,Wp_3, Wp_4, Wn_4, Wn_2, Wn_3, Wn_4) differ from the above solution
# The answer of the textbook can be different due to different precision

concentration of electron in p type is ,pn =  2.25e+04  cm**-3
built in voltage is ,eVbi = 8.17e-01  V
depletion width at p side is ,Wp_4 = 3.26e-09 m
depletion width at n side is ,Wn_4 =  3.26e-07 m

depletion width at p side is ,Wp_2 =  6.06e-09 m
depletion width at n side is ,Wn_2  =  6.06e-07 m

depletion width at p side is ,Wp_3 =  8.71e-09 m
depletion width at n side is ,Wn_3 =  8.71e-07 m

depletion width at p side is ,Wp_4 =   3.26e-09 m
depletion width at n side is ,Wn_4 =  3.26e-07 m

depletion width at p side is ,Wp_5 = 3.19e-09 m
depletion width at n side is ,Wn_5 =  3.19e-07 m
F= -4.96e+06 v/m
F= -9.21e+06 v/m
F= -1.32e+07 v/m
F= -4.96e+06 v/m
F= -4.84e+06 v/m
Q =  1.64e-13 C


## Ex5.3:pg-173¶

In [24]:
import cmath
Nd = 1.0*10**16
Na= 1.0*10**18
ni = 1.5*10**10
#NOTE: nn=Nd and pp=Na
R= 10*10**-6
A = math.pi*(R**2)
pn = ni**2/Nd
print"concentration of electron in p type is ,pn = ","{:.2e}".format(pn)," cm**-3"
kbT = 0.026
eVbi = (kbT*log(Na/pn))
print"built in voltage is ,eVbi= ","{:.2e}".format(eVbi)," V"
apsilen = 11.9*8.84*10**-12
e = 1.6*10**-19
#NOTE: for reverse bias Vr = 0 V,
Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))
print"depletion width at p side is ,Wp_4 =  ","{:.2e}".format(Wp_4),"m"
Wn_4 = Wp_4*100
print"depletion width at n side is ,Wn_4  = ","{:.2e}".format(Wn_4),"m"
#for calculation purpose and for differentiating part (I), equating
Vbi_4 = eVbi
print("")# for spacing
Vbi_2 = Vbi_4 + 2
#NOTE: for reverse bias Vr = 2 V,
Wp_2 = Wp_4*sqrt(Vbi_2/Vbi_4)
print"depletion width at p side is ,Wp_2 = ","{:.2e}".format(Wp_2),"m"
Wn_2 = Wp_2*100
print"depletion width at n side is ,Wn_2 =  ","{:.2e}".format(Wn_2),"m"
print("")# for spacing
Vbi_3 = Vbi_4 + 5
#NOTE: for reverse bias Vr = 5 V,
Wp_3 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_3/Vbi_4)
print"depletion width at p side is ,Wp_3 = ","{:.2e}".format(Wp_3),"m"
Wn_3 = Wp_3*100
print"depletion width at n side is ,Wn_3 =  ","{:.2e}".format(Wn_3),"m"
print("")# for spacing
Vbi_4 = Vbi_4 + 10
#NOTE: for reverse bias Vr = 10 V,
Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_4/Vbi_4)
print"depletion width at p side is ,Wp_4 =  ","{:.2e}".format(Wp_4),"m"
Wn_4 = Wp_4*100
print"depletion width at n side is ,Wn_4 =","{:.2e}".format(Wn_4),"m"
print("")# for spacing
Vbi_5 = Vbi_4 - 0.5
#NOTE: for forward bias Vf = 0.5 V,
Wp_5 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_5/Vbi_4)
print"depletion width at p side is ,Wp_5 =","{:.2e}".format(Wp_5),"m"
Wn_5 = Wp_5*100
print"depletion width at n side is ,Wn_5 =","{:.2e}".format(Wn_5),"m"
#CALCULATION   FOR    PEAK FIELD :
Fm =  - e*Nd*(10**6)*Wn_4/apsilen
print"peak Field For Vr = 0V, Fm =  ","{:.2e}".format(Fm),"V/m"
Fm =  - e*Nd*(10**6)*Wn_2/apsilen
print"peak Field for Vr = 2V, Fm =","{:.2e}".format(Fm),"V/m"
Fm =  - e*Nd*(10**6)*Wn_3/apsilen
print"peak Field For Vr = 5V, Fm =  ","{:.2e}".format(Fm),"V/m"
Fm =  - e*Nd*(10**6)*Wn_4/apsilen
print"peak Field For Vr = 10V, Fm =   ","{:.2e}".format(Fm),"V/m"
print"By the appendix B given in the book, the velocity of electron: v = 1*10**7 cm/s"

concentration of electron in p type is ,pn =  2.25e+04  cm**-3
built in voltage is ,eVbi=  8.17e-01  V
depletion width at p side is ,Wp_4 =   3.26e-09 m
depletion width at n side is ,Wn_4  =  3.26e-07 m

depletion width at p side is ,Wp_2 =  6.06e-09 m
depletion width at n side is ,Wn_2 =   6.06e-07 m

depletion width at p side is ,Wp_3 =  8.70e-09 m
depletion width at n side is ,Wn_3 =   8.70e-07 m

depletion width at p side is ,Wp_4 =   3.26e-09 m
depletion width at n side is ,Wn_4 = 3.26e-07 m

depletion width at p side is ,Wp_5 = 3.19e-09 m
depletion width at n side is ,Wn_5 = 3.19e-07 m
peak Field For Vr = 0V, Fm =   -4.96e+06 V/m
peak Field for Vr = 2V, Fm = -9.21e+06 V/m
peak Field For Vr = 5V, Fm =   -1.32e+07 V/m
peak Field For Vr = 10V, Fm =    -4.96e+06 V/m
By the appendix B given in the book, the velocity of electron: v = 1*10**7 cm/s


## Ex5.4:pg-178¶

In [5]:
A= 10**-7
Na=10**18*10**6
Nd=10**16*10**6
Dp1 = 7.8*10**-4
Dn2 = 7.3*10**-4
Tn = 10**-6
Tp = 10**-6
e = 1.6*10**-19
kbT = 0.026
ni = 1.5*10**16
#NOTE: nn=Nd and pp=Na
Lp = sqrt(Dp1*Tp)
print"The hole diffusion length is ,Lp =","{:.2e}".format(Lp),"m"
Ln = sqrt(Dn2*Tn)
print"The electron diffusion length is ,Ln =","{:.2e}".format(Ln),"m"
# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)
# assume that the doants are fully ionised
Io = A*e*((((Dn2)/(Ln))*(ni**2/Na))+(((Dp1)/(Lp))*(ni**2/Nd)))
print"The prefactor current is ,Io =","{:.1e}".format(Io),"A"

The hole diffusion length is ,Lp = 2.79e-05 m
The electron diffusion length is ,Ln = 2.70e-05 m
The prefactor current is ,Io = 1.0e-14 A


## Ex5.5:pg-181¶

In [4]:
Na=10**17
Nd=10**17
Dp1 = 12.5
Dn1 = 35
Dp2 = 10
Dn2 = 220
Tn = 10**-8
Tp = 10**-8
e = 1.6*10**-19
kbT = 0.026
pn1 = 2.25*10**3
np1 = 2.25*10**3
pn2 = 3.38*10**-5
np2 = 3.38*10**-5
#Note : since value of holes and electrons in n- type and p type are not given for silicon and germanium thus we have assume it as above
#NOTE: nn=Nd and pp=Na
Lp1 = sqrt(Dp1*Tp)
Ln1 = sqrt(Dn1*Tn)
Lp2 = sqrt(Dp2*Tp)
Ln2 = sqrt(Dn2*Tn)
# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)
# assume that the doants are fully ionised
Jo1 = e*((((Dn1)/(Ln1))*np1)+(((Dp1)/(Lp1))*pn1))
print"The prefactor current density for silicon is ,Jo1 =","{:.2e}".format(Jo1),"A/cm**2"
Jo2 = e*((((Dn2)/(Ln2))*np2)+(((Dp2)/(Lp2))*pn2))
print"The prefactor current density for GaAs is ,Jo2 =","{:.2e}".format(Jo2),"A/cm**2"

The prefactor current density for silicon is ,Jo1 = 3.40e-11 A/cm**2
The prefactor current density for GaAs is ,Jo2 = 9.73e-19 A/cm**2


## Ex5.6:pg-182¶

In [8]:
Na=5*10**16
Nd=5*10**17
Dp = 15
Dn = 30
Tn = 10**-8
Tp = 10**-7
e = 1.6*10**-19
kbT = 0.026
ni = 1.84*10**6
np=ni**2/Na
print"The electron conc in p type is ,np=","{:.2e}".format(np),"cm**-3"
pn=ni**2/Nd
print"The holes conc in n type is ,pn=","{:.2e}".format(pn),"cm**-3"
Lp = sqrt(Dp*Tp)
print"The hole diffusion length is ,Lp =","{:.2e}".format(Lp),"cm"
Ln = sqrt(Dn*Tn)
print"The electron diffusion length is ,Ln =","{:.2e}".format(Ln),"cm"
Gamma_inj = ((e*Dn*np)/(Ln))/(((e*Dn*np)/(Ln))+((e*Dp*pn)/(Lp)))
print"The efficiency of diode is ,Gamma_inj =",round(Gamma_inj,2)

The electron conc in p type is ,np= 6.77e-05 cm**-3
The holes conc in n type is ,pn= 6.77e-06 cm**-3
The hole diffusion length is ,Lp = 1.22e-03 cm
The electron diffusion length is ,Ln = 5.48e-04 cm
The efficiency of diode is ,Gamma_inj = 0.98


## Ex5.7:pg-182¶

In [11]:
A= 0.1*10**-2
Vf= 1.0
E= 1.43
Na=5.0*10**16
Nd=5.0*10**17
Dp = 15.0
Dn = 30.0
Tn =1*10**-8
Tp = 1.0*10**-7
e = 1.6*10**-19
kbT = 0.026
kni = 1.84*10**6
np=ni**2/Na
pn=ni**2/Nd
Ln = sqrt(Dn*Tn)
In = ((e*A*Dn*np)/Ln)*(exp(Vf/kbT)-1)
print"The electron current is ,In =","{:.1e}".format(In),"A"
In_by_e = In/e
print"The electron generation rate is ,In_by_e = ","{:.1e}".format(In_by_e),"s**-1"
power = In*E
print"The optical power of photon is ,power =","{:.1e}".format(power),"W"

The electron current is ,In = 3.0e-05 A
The electron generation rate is ,In_by_e =  1.9e+14 s**-1
The optical power of photon is ,power = 4.3e-05 W


## Ex5.8:pg-183¶

In [13]:
apsilen = 11.9*8.85*10**-14
GL= 1.0*10**22
A= 1.0*10**-4
Vr= 15.0
Na=2.0*10**16
Nd=1.0*10**16
Dp = 12.0
Dn = 20.0
Tn = 1.0*10**-8
Tp = 1.0*10**-8
e = 1.6*10**-19
kbT = 0.026
ni = 1.5*10**10
Ln = sqrt(Dn*Tn)
print"The electron diffusion length is ,Ln =","{:.1e}".format(Ln),"cm"
Lp = sqrt(Dp*Tp)
print"The hole diffusion length is ,Lp =  ","{:.2e}".format(Lp),"cm"
Vbi = kbT*log((Na*Nd)/ni**2)
print"The built in voltage is ,Vbi =","{:.2e}".format(Vbi),"V"
W = sqrt(((2*apsilen)/e)*((Na+Nd)/(Na*Nd))*(Vbi+Vr))
print"The depletion width is ,W =","{:.2e}".format(W),"cm"
IL = e*A*GL*(W+Ln+Lp)
print"The photo current is IL =","{:.2e}".format(IL),"A"

The electron diffusion length is ,Ln = 4.5e-04 cm
The hole diffusion length is ,Lp =   3.46e-04 cm
The built in voltage is ,Vbi = 7.15e-01 V
The depletion width is ,W = 1.76e-04 cm
The photo current is IL = 1.55e-04 A


## Ex5.9:pg-185¶

In [19]:
e = 1.6*10**-19
A=1.0*10**-7
Na=10**18*10**6
Nd=10**16*10**6
Dp = 7.8*10**-4
Dn = 7.3*10**-4
ni = 1.5*10**16
Wln = 5.0*10**-6
Wlp = Wln
#NOTE: nn=Nd and pp=Na
pn = ni**2/Nd
np = ni**2/Na
# from example 5.4 and 5.2 we get the value of diffusion length and zero bias depletion widths
Lp = 27.9*10**-6
print"The electron diffusion length is ,Lp= ","{:.2e}".format(Lp),"m"
Ln = 27*10**-6
print"The electron diffusion length is ,Ln= ","{:.2e}".format(Ln),"m"
Wp = 3.2*10**-10
print"The zero bias depletion widths is ,Wp= ","{:.2e}".format(Wp),"m"
Wn = 0.0000003
print"The zero bias depletion widths is ,Wn ","{:.2e}".format(Wn),"m"
# for short diode the prefactor current is given as
Io = e*A*(((Dp*pn)/(Wln-Wn))+((Dn*np)/abs(Wlp-Wp)))
print"The prefactor current is ,Io = ","{:.2e}".format(Io),"A"
# The prefactor current of short diode is approximately increase by a factor of 5.6 from that of long diode
# Note : due to different precisions taken by me and the author ... my answer differ

The electron diffusion length is ,Lp=  2.79e-05 m
The electron diffusion length is ,Ln=  2.70e-05 m
The zero bias depletion widths is ,Wp=  3.20e-10 m
The zero bias depletion widths is ,Wn  3.00e-07 m
The prefactor current is ,Io =  6.03e-14 A


## Ex5.10:pg-188¶

In [20]:
e = 1.6*10**-19
A= 10**-7
ni = 1.5*10**16
T = 1.0*10**-6
# from example 5.2 we get the value of zero bias depletion widths
W = 0.32*10**-6
Io_GR = (e*A*W*ni)/(2*T)
print"The prefactor of the  is ,generation recombination currentIo_GR = ","{:.2e}".format(Io_GR),"A"
#let V = .2 V
V = 0.2
I_GR = Io_GR*(exp(V/(2*kbT))-1)
print"The diode current is ,I_GR =","{:.2e}".format(I_GR),"A"
#let V = 0.6 V
V = 0.6
I_GR = Io_GR*(exp(V/(2*kbT))-1)
print"The diode current is ,I_GR = ","{:.2e}".format(I_GR),"A"
# The generation-recombination prefactor is much larger than prefactor due to diffusion term
#In forward bias the diffusion current is initially much smaller than the generation recombination term but at high forward bias diffusion current will start to dominate

The prefactor of the  is ,generation recombination currentIo_GR =  3.84e-11 A
The diode current is ,I_GR = 1.76e-09 A
The diode current is ,I_GR =  3.94e-06 A


## Ex5.11:pg-189¶

In [20]:
A= 1.0*10**-8
Na=1.0*10**23
Nd=1.0*10**23
Dp = 10.0*10**-4
Dn = 30.0*10**-4
Tn = 1.0*10**-7
Tp = 1.0*10**-7
tau = 1.0*10**-8
e = 1.6*10**-19
kbT = 0.026
ni = 1.5*10**16
apsilen = 11.9*8.85*10**-12
#NOTE: nn=Nd and pp=Na
Lp = sqrt(Dp*Tp)
print"The hole diffusion length is ,Lp =  ","{:.2e}".format(Lp),"m"
Ln = sqrt(Dn*Tn)
print"The electron diffusion length is ,Ln = ","{:.2e}".format(Ln),"m"
# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)
np=ni**2/Na
print"The electron conc in p type is ,np=","{:.2e}".format(np),"m**-3"
pn=ni**2/Nd
print"The holes conc in n type is ,pn=","{:.2e}".format(pn),"m**-3"
Vbi = kbT*log((Na*Nd)/ni**2)
print"The built in voltage is ,Vbi =","{:.2e}".format(Vbi),"V"
Io = e*((((Dn)*np)/(Ln))+((Dp*pn)/(Lp)))
print"The prefactor in the ideal diode  is ,Io =","{:.2e}".format(Io),"A"
#let Vf = 0.5 V
Vf = 0.5
W = sqrt((2*apsilen/e)*((Na+Nd)/Nd/Na)*(Vbi-Vf))
print"The depletion width is ,W =","{:.2e}".format(W),"m"
Io_GR  = e*A*W*ni/(2*tau)
print"prefactor for recombination generation current, Io_GR = ","{:.2e}".format(Io_GR),"A"
I = (Io*exp(Vf/kbT))+(Io_GR*exp(Vf/(2*kbT)))
I_V1 = I
print"Current, I =","{:.2e}".format(I_V1),"A"
#let V = 0.6 V
Vf = 0.6
W = sqrt((2*apsilen/e)*((Na+Nd)/Nd/Na)*(Vbi-Vf))
print"The depletion width is ,W =","{:.2e}".format(W),"m"
Io_GR  = e*A*W*ni/2/tau
print"prefactor for recombination generation current, Io_GR = ","{:.1e}".format(Io_GR),"A"
I = (Io*exp(Vf/kbT))+(Io_GR*exp(Vf/(2*kbT)))
I_V2 = I
print"Current, I =","{:.2e}".format(I_V2),"A"
V1 = 0.5
V2 = 0.6
n = e*(V2-V1)/kbT/log(I_V2/I_V1)
print"Ideallity factor ,n =","{:.2e}".format(n)
#note: in the text book the value of
#-+"prefactor of ideal diode equation, Io"
#calculated by author is wrong thus it efect the overall calculation of the solution ,so the answer in the are wrong of some of the values

The hole diffusion length is ,Lp =   1.00e-05 m
The electron diffusion length is ,Ln =  1.73e-05 m
The electron conc in p type is ,np= 2.25e+09 m**-3
The holes conc in n type is ,pn= 2.25e+09 m**-3
The built in voltage is ,Vbi = 8.17e-01 V
The prefactor in the ideal diode  is ,Io = 9.84e-08 A
The depletion width is ,W = 9.14e-08 m
prefactor for recombination generation current, Io_GR =  1.10e-10 A
Current, I = 2.21e+01 A
The depletion width is ,W = 7.56e-08 m
prefactor for recombination generation current, Io_GR =  9.1e-11 A
Current, I = 1.04e+03 A
Ideallity factor ,n = 1.60e-19


## Ex5.12:pg-195¶

In [11]:
apsilen = 11.9*8.85*10**-14
Na=1.0*10**19
Nd=1.0*10**16
e = 1.6*10**-19
Fcrit1 = 4.0*10**5
Fcrit2 = 1.0*10**7
VBD_Si = (apsilen*Fcrit1**2)/(2*e*Nd)
print"The breakdown field for silicon is ,VBD_Si =","{:.2e}".format(VBD_Si)," V"
VBD_C = (apsilen*Fcrit2**2)/(2*e*Nd)
print"The breakdown field for diomond is ,VBD_C =","{:.2e}".format(VBD_C)," V"
# Note : In the textbook answer of breakdown voltage of silicon is wrong due to which breakdown voltage of diomand also differ

The breakdown field for silicon is ,VBD_Si = 5.27e+01  V
The breakdown field for diomond is ,VBD_C = 3.29e+04  V


## Ex5.13:pg-199¶

In [30]:
import numpy as np
print"Let the intercept of the 1/c**2 Vs V plot is represented by Icv, which is the built in voltage"
Icv = 0.68
Vbi = Icv
print"Let the slope of the intercept of the 1/c**2 Vs V plot is represented by dIcv"
dIcv=2.1*10**23
C = 7.0*10**-13
#NOTE: The above mentioned values are taken from the figure given in the question in textbook
apsilen = 11.9*8.85*10**-12
e = 1.6*10**-19
kbT = 0.026
A = 10**-7
ni = sqrt(2.25*10**20)
Neff = 2/(A**2*e*apsilen*dIcv)
print"The thickness of n region is ,Neff =","{:.2e}".format(Neff)," m**-3"
Neff = Neff/10**6
print"The thickness of n region is ,Neff =","{:.2e}".format(Neff)," cm**-3"
NaNd = exp(Vbi/kbT)*ni**2
print"NaNd =","{:.2e}".format(NaNd)," cm**-6"
# solving for Na and Nd by creating a quadratic equation using the equations mentioned in the book
p1 = [Neff*NaNd, -NaNd, Neff]
#Neff*NaNd - NaNd*X + Neff*X**2
p1=p1[0]
#p2=p1[1]
print(p1)
R= roots(p1)
#s=roots(p2)
Na= R
#Nd= s
#format ('e',10)
#print"Na = ""{:.2e}".format(Na),"cm**-3"
#print"Nd = ""{:.2e}".format(Nd),"cm**-3"
W = (apsilen*A)/C
print"The thickness of n region is ,W =","{:.1e}".format(W)," m"

Let the intercept of the 1/c**2 Vs V plot is represented by Icv, which is the built in voltage
Let the slope of the intercept of the 1/c**2 Vs V plot is represented by dIcv
The thickness of n region is ,Neff = 5.65e+19  m**-3
The thickness of n region is ,Neff = 5.65e+25  cm**-3
NaNd = 5.14e+31  cm**-6
The thickness of n region is ,W = 1.5e-05  m


## Ex5.14:pg-201¶

In [7]:
import cmath
e = 1.6*10**-19
I= 1*10**-3
kbT = 0.026
Tp = 10**-6
Gs = (I)/(kbT)
print"The diode conductance is Gs = ","{:.2e}".format(Gs),"A/V"
Cdiff = (I*Tp)/(2*kbT)
print"The diffusion capacitance is Cdiff =  ","{:.2e}".format(Cdiff),"F"
# The diffusion capacitance is much larger than junction capacitance hence neglecting junction capacitance
Y = Gs+(1j*2*math.pi*10**6*Cdiff)
print"The admittance of the diode is Y =",Y,"A/V"
# Note : due to different precisions taken by me and the author ... my answer differ

The diode conductance is Gs =  3.85e-02 A/V
The diffusion capacitance is Cdiff =   1.92e-08 F
The admittance of the diode is Y = (0.0384615384615+0.120830486677j) A/V


## Ex5.15:pg-207¶

In [1]:
Vr= 10.0
R= 10.0*10**3
#The junction capacitance is 20pF at zero bias and 10 pF at full reverse bias so
Cavg= ((20+10)/2)
Tp = 10**-7
Ir = (Vr)/(R)
Tsd = Tp*log(2)
print"The storage delay time is Tsd = ","{:.2e}".format(Tsd)," s"
Tt = 2.3*R*Cavg*10**-12
print"The time Tt =  ","{:.2e}".format(Tt)," s"
T = Tsd+Tt
print"The total diode recovery time is T =  ","{:.2e}".format(T)," s"
# Note : due to different precisions taken by me and the author ... my answer differ

The storage delay time is Tsd =  6.93e-08  s
The time Tt =   3.45e-07  s
The total diode recovery time is T =   4.14e-07  s