# Ch-4 : Semiconductor Diodes¶

## Page No. 78 Example 4.1.¶

In :
un1=3800 #mobility of free electrons in pure germanium
up1=1800 #mobility of free holes in pure germanium
un2=1300 #mobility of free electrons in pure silicon
up2=500 #mobility of free holes in pure silicon
q=1.6*10**-19
nig=2.5*10**13
nis=1.5*10**10
sigma1=q*nig*(un1+up1)
print "(i) The intrinsic conductivity for germanium,"
print "sigmai(S/cm) = q*ni*(un+up) = %0.2f"%sigma1
sigma2=q*nis*(un2+up2)
print "(ii) The intrinsic conductivity for silicon,"
print "sigmai(S/cm)= q*ni*(un+np) =%0.2e"%sigma2

(i) The intrinsic conductivity for germanium,
sigmai(S/cm) = q*ni*(un+up) = 0.02
(ii) The intrinsic conductivity for silicon,
sigmai(S/cm)= q*ni*(un+np) =4.32e-06


## Page No. 79 Example 4.6.¶

In :
ni=1.5*10**10
un=1300
up=500
q=1.6*10**-19
nos=5*10**22
print "(a) In intrensic condition, n=p=ni"
print "Hence, sigma_i = q*ni*(un+up)"
sigma_i = q*ni*(un+up)
print "sigma_i = %0.2e S/cm"%sigma_i
print "(b) Number of silicon atoms/cm**3 = 5*10**22"
ND=5*10**22/10**8
print "Hence, ND = %0.2e cm**-3"%ND
print "Further, n = ND"
print "Therefore,    p = ni**2/n = ni**2/ND"
p=ni**2/ND
print "p = %0.2e cm**-3"%p  # wrong answer in textbook
print "Thus p << n. Hence p may be neglected while calculating the conductivity."
print "Hence,      sigma = n*q*un = ND*q*un"
sigma=ND*q*un
print "sigma = %0.2f S/cm"%sigma
NA=(5*10**22)/(5*10**7)
print "(c) NA = %0.2e cm**-3"%NA
print "Further, p = NA"
print "Hence,     n = ni**2/p = ni**2/NA"
n=ni**2/NA
print "n = %0.2e cm**-3"%n
print "Thus p >> n. Hence n may be neglected while calculating the conductivity."
print "Hence,     sigma = p*q*up = NA*q*up"
sigma1=NA*q*up
print "sigma = %0.2f S/cm"%sigma1
print "(d) With both types of impurities present simultaneously, the net acceptor impurity density is,"
Na=NA-ND
print "Na = %0.2e cm**-3"%Na
print "Hence,     sigma = Na*q*up"
sigma2=Na*q*up
print "sigma = %0.2f S/cm"%sigma2

(a) In intrensic condition, n=p=ni
Hence, sigma_i = q*ni*(un+up)
sigma_i = 4.32e-06 S/cm
(b) Number of silicon atoms/cm**3 = 5*10**22
Hence, ND = 5.00e+14 cm**-3
Further, n = ND
Therefore,    p = ni**2/n = ni**2/ND
p = 4.50e+05 cm**-3
Thus p << n. Hence p may be neglected while calculating the conductivity.
Hence,      sigma = n*q*un = ND*q*un
sigma = 0.10 S/cm
(c) NA = 1.00e+15 cm**-3
Further, p = NA
Hence,     n = ni**2/p = ni**2/NA
n = 2.25e+05 cm**-3
Thus p >> n. Hence n may be neglected while calculating the conductivity.
Hence,     sigma = p*q*up = NA*q*up
sigma = 0.08 S/cm
(d) With both types of impurities present simultaneously, the net acceptor impurity density is,
Na = 5.00e+14 cm**-3
Hence,     sigma = Na*q*up
sigma = 0.04 S/cm


## Page No. 83 Example 4.7.¶

In :
ni=2.5*10**13
un=3800
up=1800
nog=4.4*10**22
q=1.6*10**-19
sigma=q*ni*(un+up)
print "(a) n = p = ni = 2.5*10**13 cm**-3"
print "Therefore,  conductivity,  sigma = q*ni*(un+np) =%0.2f S/cm"%sigma
rho=1/sigma
print "Hence,  resistivity   rho = 1 / sigma =%0.2f (ohm-cm)"%rho
ND=(4.4*10**22)/10**7
print "\n(b) ND = %0.2e cm**-3"%ND
p=ni**2/ND
print "Also, n = ND"
print "Therefore,  p = ni**2 / n = ni**2 / ND =%0.2f (holes/cm**3)"%p
print "Here, as n >> p, p can be neglected."
sigma1=ND*q*un
print "Therefore,  conductivity,    sigma = n*q*un = ND*q*un =%0.2f (S/cm)"%sigma1
rho1=1/sigma1
print "Hence,  resistivity,    rho = 1 / sigma =%0.2f (ohm-cm)"%rho1

NA=(4.4*10**22)/10**8
print "\n(c) NA = %0.2e (cm**-3)"%NA
print "Also, p = NA"
n=ni**2/NA
print "Therefore,  n = ni**2 / p = ni**2 / NA =%0.2f (electrons/cm**3)"%n
sigma2=NA*q*up
print "Here, as p >> n, n may be neglected. Then,"
print "Conductivity,    sigma = p*q*up = NA*q*up =%0.2f (S/cm)"%sigma2
rho2=1/sigma2
print "Hence,  resistivity,    rho = 1 / sigma = %0.2f (ohm-cm)"%rho2

print "\n(d) with both p and n type impurities present,"
print "      ND = 4.4*10**15 cm**-3 and NA = 4.4*10**14 cm**-3"
print "Therefore, the net donor density ND'' is"
Nd=ND-NA
print "ND'' = (ND - NA) =%0.2f (cm**-3)"%Nd
print "Therefore, effective n = ND'' = 3.96*10**15 cm**-3"
p1=ni**2/Nd
print "p = ni**2 / N''D =%0.2f (cm**-3)"%p1
print "Here again p(= ni**2 / N''D) is very small compared with N''D and may be neglected in calculating the effective conductivity."
sigma3=Nd*q*un
print "Therefore,  conductivity,    sigma = ND''*q*un =%0.2f (S/cm)"%sigma3
rho3=1/sigma3
print "Hence,  resistivity,    rho = 1 / sigma =%0.2f (ohm-cm)"%rho3

(a) n = p = ni = 2.5*10**13 cm**-3
Therefore,  conductivity,  sigma = q*ni*(un+np) =0.02 S/cm
Hence,  resistivity   rho = 1 / sigma =44.64 (ohm-cm)

(b) ND = 4.40e+15 cm**-3
Also, n = ND
Therefore,  p = ni**2 / n = ni**2 / ND =142045454545.45 (holes/cm**3)
Here, as n >> p, p can be neglected.
Therefore,  conductivity,    sigma = n*q*un = ND*q*un =2.68 (S/cm)
Hence,  resistivity,    rho = 1 / sigma =0.37 (ohm-cm)

(c) NA = 4.40e+14 (cm**-3)
Also, p = NA
Therefore,  n = ni**2 / p = ni**2 / NA =1420454545454.55 (electrons/cm**3)
Here, as p >> n, n may be neglected. Then,
Conductivity,    sigma = p*q*up = NA*q*up =0.13 (S/cm)
Hence,  resistivity,    rho = 1 / sigma = 7.89 (ohm-cm)

(d) with both p and n type impurities present,
ND = 4.4*10**15 cm**-3 and NA = 4.4*10**14 cm**-3
Therefore, the net donor density ND'' is
ND'' = (ND - NA) =3960000000000000.00 (cm**-3)
Therefore, effective n = ND'' = 3.96*10**15 cm**-3
p = ni**2 / N''D =157828282828.28 (cm**-3)
Here again p(= ni**2 / N''D) is very small compared with N''D and may be neglected in calculating the effective conductivity.
Therefore,  conductivity,    sigma = ND''*q*un =2.41 (S/cm)
Hence,  resistivity,    rho = 1 / sigma =0.42 (ohm-cm)


## Page No. 87 Example 4.8¶

In :
un=1250
up=475
q=1.6*10**-19
sigma_i=1/(25*10**4)
format(9)
ni=1/((25*10**4)*(1.6*10**-19)*(1250+475))
print "sigma_i = qni(un+up) = 1 / 25*10**4"
print "\nTherefore,  ni = sigma_i / q(un+up) =%0.2f"%ni

ND=(4*10**10)-10**10
print "\nNet donor density,   ND(= n) = %0.2e (cm**-3)"%ND
p=ni**2/ND
print "Hence,  p = ni**2 / ND =%0.2f (cm**-3)"%p
sigma=(1.6*10**-19)*((1250*3*10**10)+(475*0.7*10**10))
print "Hence,  sigma = q*(n*un + p*up) =%0.2f"%sigma

J=6.532*4*10**-6
print "Therefore, total conduction current density, J = sigma*E =%0.2f A/cm**2"%J

sigma_i = qni(un+up) = 1 / 25*10**4

Therefore,  ni = sigma_i / q(un+up) =14492753623.19

Net donor density,   ND(= n) = 3.00e+10 (cm**-3)
Hence,  p = ni**2 / ND =7001330252.75 (cm**-3)
Hence,  sigma = q*(n*un + p*up) =0.00
Therefore, total conduction current density, J = sigma*E =0.00 A/cm**2


## Page No. 92 Example 4.9.¶

In :
ni=1.5*10**10
un=1300
up=500
q=1.6*10**-19
sigma=300
print "(a) Concentration in N-type silicon"
format(10)
n=sigma/(q*un)
print "The conductivity of an N-type Silicon is sigma = q*n*un"
print "Concentratoin of electrons, n = sigma / q*un =%0.2e (cm**-3)"%n
p=ni**2/n
print "Hence,  concentration of holes, p = ni**2 / n =%0.2e (cm**-3)"%p
print "(b) Concentration in P-type silicon"
p=sigma/(q*up)
print "The conductivity of a P-type Silicon is sigma = q*p*up"
print "Hence,  concentratoin of holes, p  = sigma / q*up =%0.2e (cm**-3)"%p
n=ni**2/p
print "and concentration of electrons, n = ni**2 / p = %0.2f (cm**-3)"%n

(a) Concentration in N-type silicon
The conductivity of an N-type Silicon is sigma = q*n*un
Concentratoin of electrons, n = sigma / q*un =1.44e+18 (cm**-3)
Hence,  concentration of holes, p = ni**2 / n =1.56e+02 (cm**-3)
(b) Concentration in P-type silicon
The conductivity of a P-type Silicon is sigma = q*p*up
Hence,  concentratoin of holes, p  = sigma / q*up =3.75e+18 (cm**-3)
and concentration of electrons, n = ni**2 / p = 60.00 (cm**-3)


## Page No. 93 Example 4.10.¶

In :
ND=(4.2*10**28)/10**6
print "Density of added impurity atoms is, ND = %0.2e (atoms/m**3)"%ND
ni=2.5*10**19
p=ni**2/ND
print "Also, n = ND"
print "Therefore,  p = ni**2 / n = ni**2 / ND =%0.2e (m**-3)"%p
print "Here, as p << n, p may be neglected."
q=1.6*10**-19
un=0.38
sigma=q*ND*un
print "Therefore,  sigma = q*ND*un =%0.2f (S/m)"%sigma

rho=1/sigma
print "Therefore,  resistivity, rho = 1 / sigma =%0.2f ohm-m"%rho

L=5*10**-3
A=5*10**-6
R=(rho*L)/A**2
R1=R*10**-3
print "Resistance,  R = rho*L / A =%0.2f kohm"%R1
I=10**-6
V=R*I
V1=V*10**3
print "Voltage drop,  V = RI =%0.2f mV"%V1

Density of added impurity atoms is, ND = 4.20e+22 (atoms/m**3)
Also, n = ND
Therefore,  p = ni**2 / n = ni**2 / ND =1.49e+16 (m**-3)
Here, as p << n, p may be neglected.
Therefore,  sigma = q*ND*un =2553.60 (S/m)
Therefore,  resistivity, rho = 1 / sigma =0.00 ohm-m
Resistance,  R = rho*L / A =78.32 kohm
Voltage drop,  V = RI =78.32 mV


## Page No. 94 Example 4.11.¶

In :
from math import log
q=1.6*10**-19
ni=2.5*10**13
up=1800
un=3800
VT=0.026
rho=6
format(9)
NA=1/(6*q*up)
print "(a) Resistivity,  rho = 1 / sigma = 1 / NA*q*up = 6 ohm-cm"
print "Therefore,  NA = 1 / 6*q*up =%0.2e (1/cm**3)"%NA
ND=1/(4*q*un)
print "Similarly,  ND(1/cm**3) = 1 / 4*q*un =%0.2e (1/cm**3)"%ND
Va=VT*log((ND*NA)/ni**2)
print "Therefore, Va = VT*ln(ND*NA / ni**2) = %0.2f V"%Va
print "Hence,  Eo = %0.2f eV"%Va
Va1=0.026*log((2*ND*2*NA)/ni**2)
print "\n(b) Vo = 0.026*ln(2*ND*2*NA / ni**2) =%0.2f V"%Va1
print "Therefore,  Eo(eV) = %0.2f eV"%Va1

(a) Resistivity,  rho = 1 / sigma = 1 / NA*q*up = 6 ohm-cm
Therefore,  NA = 1 / 6*q*up =5.79e+14 (1/cm**3)
Similarly,  ND(1/cm**3) = 1 / 4*q*un =4.11e+14 (1/cm**3)
Therefore, Va = VT*ln(ND*NA / ni**2) = 0.15 V
Hence,  Eo = 0.15 eV

(b) Vo = 0.026*ln(2*ND*2*NA / ni**2) =0.19 V
Therefore,  Eo(eV) = 0.19 eV


## Page No. 97 Example 4.12.¶

In :
from math import exp
Ia=0.3*10**-6
VF=0.15
I=Ia*((exp(40*VF))-1)
I1=I*10**6
print "The current flowing through the PN diode under forward bias is,"
print "I = Io*(e**40*VF - 1) =%0.2f uA"%I1

The current flowing through the PN diode under forward bias is,
I = Io*(e**40*VF - 1) =120.73 uA


## Page No. 97 Example 4.13.¶

In :
from math import exp
VF=0.6
T=298
Io=10**-5
eta=2
VT=T/11600.0
print "The volt-equivalent of the temperature(T) is,"
print "VT(V) = T / 11600 = %0.2f"%VT
I=Io*((exp((VF/(eta*VT))))-1)
print "Therefore, the diode current, I = Io*e**((VF/eta*VT)-1) =%0.2f A"%I

The volt-equivalent of the temperature(T) is,
VT(V) = T / 11600 = 0.03
Therefore, the diode current, I = Io*e**((VF/eta*VT)-1) =1.18 A


## Page No. 98 Example 4.16.¶

In :
I=5*10**-3
T=300
print "Forward resistance of a PN junction diode, rf = (eta*VT)/I  where VT = T/11600 and eta = 2 for silicon"
print "Therefore,  rf = 2*(T/11600) / 5*10**-3"
eta=2 #for silicon
rf=600/(11600*5*10**-3)
print "rf = %0.2f ohm"%rf

Forward resistance of a PN junction diode, rf = (eta*VT)/I  where VT = T/11600 and eta = 2 for silicon
Therefore,  rf = 2*(T/11600) / 5*10**-3
rf = 10.34 ohm


## Page No. 100 Example 4.17.¶

In :
Io1=7.5*10**-6
T1=27
T2=127
print "The saturation current at 400 K is,"
print "Io2 = Io1 * 2**((T2-T1)/10)"
print "    = 7.5*10**-6 * 2**(127-27/10)"
Io2=Io1*(2**((T2-T1)/10))
I=Io2*10**3
print "Io2 = %0.2f mA"%I

The saturation current at 400 K is,
Io2 = Io1 * 2**((T2-T1)/10)
= 7.5*10**-6 * 2**(127-27/10)
Io2 = 7.68 mA