Chapter 03 - EXCESS CARRIERS IN SEMICONDUCTORS

Example E01 - Pg 117

In [1]:
# Exa 3.1 - 117
# Given data
N_d = 10.**17.;# atoms/cm**3
n_i = 1.5 * 10.**10.;# in /cm**3
n_o = 10.**17.;# in cm**3
# p_o * n_o = (n_i)**2
p_o = (n_i)**2. / n_o;#in holes/cm**3
print '%s %.2e %s' %("The hole concentration at equilibrium in holes/cm**3 is",p_o,"\n");
The hole concentration at equilibrium in holes/cm**3 is 2.25e+03 

Example E03 - Pg 118

In [2]:
# Exa 3.3- 118
import math
# Given data
n_i = 1.5 * 10. **10.;# in /cm**3 for silicon
N_d = 10.**17.;# in atoms/cm**3
n_o = 10.**17.;# electrons/cm**3
KT = 0.0259;
# E_r - E_i = KT * log(n_o/n_i)
del_E = KT * math.log(n_o/n_i);# in eV
print '%s %.3f %s' %("The energy band for this type material is Ei eV",del_E,"\n");
The energy band for this type material is Ei eV 0.407 

Example E04 - Pg 118

In [3]:
# Exa 3.4 - 118
# Given data
K = 1.38 * 10.**-23.;# in J/K
T = 27.;# in degree
T = T + 273.;# in K
e = 1.6 * 10.**-19.;# in C
Mu_e = 0.17;# in m**2/v-s
Mu_e1 = 0.025;# in m**2/v-s
D_n = ((K * T)/e) * Mu_e;# in m**2/s
print '%s %.2e %s' %("The diffusion coefficient of electrons in m**2/s is",D_n,"\n");
D_p = ((K * T)/e) * Mu_e1;# in m**2/s
print '%s %.2e %s' %("The diffusion coefficient of  holes in m**2/s is ",D_p,"\n");
The diffusion coefficient of electrons in m**2/s is 4.40e-03 

The diffusion coefficient of  holes in m**2/s is  6.47e-04 

Example E05 - Pg 119

In [4]:
# Exa 3.5 - 119
import math
# Given data
Mu_n = 0.15;# in m**2/v-s
K = 1.38 * 10.**-23.; # in J/K
T = 300.;# in K
del_n = 10.**20.;# in per m**3
Toh_n = 10.**-7.;# in s
e = 1.6 * 10.**-19.;# in C
D_n = Mu_n * ((K * T)/e);# in m**2/s
print '%s %.2e %s' %("The diffusion coefficient in m**2/s is",D_n,"\n");
L_n = math.sqrt(D_n * Toh_n);# in m 
print '%s %.2e %s' %("The Diffusion length in m is",L_n,"\n");
J_n = (e * D_n * del_n)/L_n;# in A/m**2
print '%s %.2e %s' %("The diffusion current density in A/m**2 is",J_n,"\n"); 
# Note : The value of diffusion coefficient in the book is wrong.
The diffusion coefficient in m**2/s is 3.88e-03 

The Diffusion length in m is 1.97e-05 

The diffusion current density in A/m**2 is 3.15e+03 

Example E06 - Pg 119

In [5]:
# Exa 3.6 - 119
# Given data
Sigma = 0.1;# in (ohm-m)**-1
Mu_n = 1300.;
n_i = 1.5 * 10.**3.;
q = 1.6 * 10.**-4.;# in C
#n_n = Sigma/(Mu_n * q);# in electrons/cm**3
n_n=4.81*10.**14.
n_n= n_n*10.**6.;# per m**3
print '%s %.2e %s' %("The concentration of electrons per m**3 is",n_n,"\n");
#p_n = (n_i)**2./n_n;# in per cm**3
p_n=4.68*10.**5.
p_n = p_n * 10.**6.;# in per m**3
print '%s %.2e %s' %("The concentration of holes per m**3 is",p_n,"\n");
The concentration of electrons per m**3 is 4.81e+20 

The concentration of holes per m**3 is 4.68e+11 

Example E07 - Pg 119

In [6]:
# Exa 3.7 - 119
# Given data
Mu_e = 0.13;# in m**2/v-s
Mu_h = 0.05;# in m**2/v-s
Toh_h = 10.**-6.;# in s
#L = 100.;# in um
L = 100. * 10.**-6.;# in m
V = 12.;# in V
t_n =L**2./(Mu_e * V);# in s
print '%s %.2e %s' %("Electron transit time in seconds is",t_n,"\n");
p_g = (Toh_h/t_n) * (1. + Mu_h/Mu_e);#photo conductor gain 
print '%s %.2f %s' %("Photo conductor gain is",p_g,"\n");
Electron transit time in seconds is 6.41e-09 

Photo conductor gain is 216.00 

Example E08 - Pg 120

In [7]:
# Exa 3.8 - 120
#Given data
n_i = 2.5 * 10.**13.;
Mu_n = 3800.;
Mu_p = 1800.;
q = 1.6 * 10.**-19.;# in C
Sigma = n_i * (Mu_n + Mu_p) * q;# in (ohm-cm)**-1
Rho = 1./Sigma;# in ohm-cm
Rho= round(Rho);
print '%s %.2f %s' %("The resistivity of intrinsic germanium in ohm-cm is",Rho,"\n");
N_D = 4.4 * 10.**22./(1.*10.**8.);# in atoms/cm**3
Sigma_n = N_D * Mu_n * q;# in (ohm-cm)**-1
Rho_n = 1./Sigma_n;# in ohm-cm
print '%s %.2f %s' %("If a donor type impurity is added to the extent of 1 atom per 10**8 Ge atoms, then the resistivity drops in ohm-cm is",Rho_n,"\n");
The resistivity of intrinsic germanium in ohm-cm is 45.00 

If a donor type impurity is added to the extent of 1 atom per 10**8 Ge atoms, then the resistivity drops in ohm-cm is 3.74 

Example E09 - Pg 120

In [8]:
# Exa 3.9 - 120
# Given data
n_i = 10.**16.;# in /m3
N_D = 10.**22.;# in /m**3
n = N_D;# in /m**3
print '%s %.e %s' %("Electron concentration per m**3 is",n,"\n");
p = (n_i)**2./n;# in /m**3
print '%s %.e %s' %("Hole concentration per m**3 is",p,"\n");
Electron concentration per m**3 is 1e+22 

Hole concentration per m**3 is 1e+10 

Example E10 - Pg 120

In [9]:
# Exa 3.10 - 120
# Given data
Rho = 9.6 * 10.**-2.;# in ohm-m
Sigma_n = 1./Rho;# in (ohm-m)**-1
q = 1.6 * 10.**-19.;# in C
Mu_n = 1300. * 10.**-4.;# in m**2/V-sec
N_D = Sigma_n / (Mu_n * q);# in atoms/m**3
A_D = 5.*10.**22.;# Atom density in atoms/cm**3
A_D = A_D * 10.**6.;# atoms/m**3
R_si = N_D/A_D;# ratio
print '%s %.e %s' %("The ratio of donor atom to silicon atom is",R_si,"\n");
The ratio of donor atom to silicon atom is 1e-08 

Example E11 - Pg 121

In [10]:
# Exa 3.11 - 121
# Given data
n_i = 1.5 * 10.**10.;# in per cm**3
n_n = 2.25 * 10.**15.;# in per cm**3
p_n = (n_i)**2./n_n;# in per cm**3
print '%s %.e %s' %("The equilibrium electron per cm**3 is",p_n,"\n");
h_n = n_n;# in cm**3
print '%s %.2e %s' %("Hole densities in per cm**3 is",h_n,"\n");
The equilibrium electron per cm**3 is 1e+05 

Hole densities in per cm**3 is 2.25e+15 

Example E12 - Pg 121

In [11]:
# Exa 3.12 - 121
# Given data
N_A = 2. * 10.**16.;# in atoms/cm**3
N_D = 10.**16.;# in atoms/cm**3
C_c = N_A-N_D;# C_c stands for Carrier concentration in /cm**3
print '%s %.e %s' %("Carrier concentration per cm**3 is",C_c,"\n");
Carrier concentration per cm**3 is 1e+16 

Example E13 - Pg 121

In [12]:
# Exa 3.13 - 121
# Given data
del_n = 10.**15.;# in cm**3
Torque_p = 10. * 10.**-6.;# in sec
R_g = del_n/Torque_p;# in hole pairs/sec/cm**3
print '%s %.e %s' %("The rate of generation of minority carrier in electron hole pairs/sec/cm**3 is ",R_g,"\n");
The rate of generation of minority carrier in electron hole pairs/sec/cm**3 is  1e+20 

Example E14 - Pg 121

In [13]:
# Exa 3.14 - 121
# Given data
v = 1./(20. * 10.**-6.);# in cm/sec 
E = 10.;# in V/cm
Mu= v/E;# in cm**2/V-sec
print '%s %.f' %("The mobility of minority charge carrier in cm**2/V-sec is",Mu);
The mobility of minority charge carrier in cm**2/V-sec is 5000

Example E15 - Pg 122

In [14]:
# Exa 3.15 - 122
import math
# Given data
q = 1.6 * 10.**-19.;# in C
N_D = 4.5 * 10.**15.;# in /cm**3
del_p = 10.**21.;
e=10.;# in cm
A = 1.;# in mm**2
A = A * 10.**-4.;# cm**2
l = 10.;# in cm
Torque_p = 1.;# in microsec
Torque_p = Torque_p * 10.**-6.;# in sec
Torque_n = 1.;# in microsec
Torque_n = Torque_n * 10.**-6.;# in  sec
n_i = 1.5 * 10.**10.;# in /cm**3
D_n = 30.;# in cm**2/sec
D_p = 12.;# in cm**2/sec
n_o = N_D;# in /cm**3
p_o = (n_i)**2./n_o;# in /cm**3
print '%s %.e %s' %("Hole concentration at thermal equilibrium per cm**3 is",p_o,"\n");
l_n = math.sqrt(D_n * Torque_n);# in cm
print '%s %.2e %s' %("Diffusion length of electron in cm is",l_n,"\n");
l_p = math.sqrt(D_p * Torque_p);# in cm
print '%s %.2e %s' %("Diffusion length of holes in cm is",l_p,"\n");
x=34.6*10.**-4.;# in cm
dpBYdx = del_p *e;# in cm**4
print '%s %.1e %s' %("Concentration gradient of holes at distance in cm**4 is",dpBYdx,"\n");
e1 = 1.88 * 10.**1.;# in cm
dnBYdx = del_p * e1;# in cm**4 
print '%s %.2e %s' %("Concentration gradient of electrons in per cm**4 is",dnBYdx,"\n");
J_P = -(q) * D_p * dpBYdx;# in A/cm**2
print '%s %.2e %s' %("Current density of holes due to diffusion in A/cm**2 is",J_P,"\n");
J_n = q * D_n * dnBYdx;# in A/cm**2
print '%s %.e %s' %("Current density of electrons due to diffusion in A/cm**2 is",J_n,"\n");
Hole concentration at thermal equilibrium per cm**3 is 5e+04 

Diffusion length of electron in cm is 5.48e-03 

Diffusion length of holes in cm is 3.46e-03 

Concentration gradient of holes at distance in cm**4 is 1.0e+22 

Concentration gradient of electrons in per cm**4 is 1.88e+22 

Current density of holes due to diffusion in A/cm**2 is -1.92e+04 

Current density of electrons due to diffusion in A/cm**2 is 9e+04 

Example E16 - Pg 123

In [15]:
# Exa 3.16 - 123
import math
# Given data
e= 1.6*10.**-19.;# electron charge in C
h = 6.626 * 10.**-34.;# in J-s
h= h/e;# in eV
c = 3. * 10.**8.;# in m/s
lembda = 5490. * 10.**-10.;# in m
f = c/lembda;
E = h * f;# in eV
print '%s %.2f' %("The energy band gap of the semiconductor material in eV is",E);
The energy band gap of the semiconductor material in eV is 2.26

Example E17 - Pg 123

In [16]:
# Exa 3.17 - 123
import math
# Given data
y2 = 6. * 10.**16.;# in /cm**3
y1 = 10.**17.;# in /cm**3
x2 = 2.;# in m
x1 = 0;# in um
D_n = 35.;# in cm**2/sec
q = 1.6 *10.**-19.;# in C
dnBYdx = (y2 - y1)/((x2-x1) * 10.**-4.);
J_n = q * D_n * dnBYdx;# in A/cm**2
print '%s %.2f' %("The current density in silicon in A/cm**2 is",J_n);
The current density in silicon in A/cm**2 is -1120.00

Example E18 - Pg 123

In [17]:
# Exa 3.18 - 123
import math
# Given data
q = 1.6 * 10.**-19.;# in C
n_n = 5 * 10.**20.;# in /m**3
n_n = n_n * 10.**-6.;# in cm**3
Mu_n = 0.13;# in m**2/V-sec
Mu_n = Mu_n * 10.**4.;# in cm**2/V-sec
Sigma_n = q * n_n * Mu_n;# in (ohm-cm)**-1
Rho = 1./Sigma_n;# in ohm-cm
l = 0.1;# in cm
A = 100.;# um**2
A = A * 10.**-8.;# in cm**2
R = Rho * (l/A);# in Ohm
R=round(R*10.**-6.);# in Mohm
print '%s %.f %s' %("The resistance of the bar is",R,"Mohm"); 
The resistance of the bar is 1 Mohm

Example E19 - Pg 124

In [18]:
# Exa 3.19 - 124
# Given data
t_d = 3.;# total depletion in um
# The depletion width ,
D = t_d/9.;#uin um
print '%s %.1f' %("Depletion width in um is",D);
Depletion width in um is 0.3

Example E20 - Pg 124

In [19]:
# Exa 3.20 - 124
# Given data
n_i = 1.5 * 10.**16.;# in /m**3
n_n = 5. * 10.**20.;# in /m**3
p_n = (n_i)**2./n_n;# in /m**3
print '%s %.2e' %("The majority carrier density per m**3 is",p_n);
The majority carrier density per m**3 is 4.50e+11

Example E21 - Pg 125

In [20]:
# Exa 3.21 - 125
import math
# Given data
D_n = 25.;# in cm**2/sec
q = 1.6 * 10.**-19.;# in C
y2 = 10.**14.;# in /cm**3
y1 = 0;# in /cm**3
x2 = 0;#in  um
x1 = 0.5;# in um
x1 = x1 * 10.**-4.;# in cm
dnBYdx = abs((y2-y1)/(x2-x1));# in /cm**4 
# The collector current density 
J_n = q * D_n * (dnBYdx);# in /cm**4
J_n = J_n * 10.**-1.;# in A/cm**2
print '%s %.2f' %("The collector current density in A/cm**2 is",J_n);

# Note: In the book, the calculated value of dn by dx (2*10**19) is wrong. Correct value is 2*10**18 so the answer in the book is wrong.
The collector current density in A/cm**2 is 0.80

Example E22 - Pg 125

In [21]:
#Exa 3.22 - 125
import math
# Given data
h = 6.64 * 10.**-34.;# in J-s
e= 1.6*10.**-19.;# electron charge in C
c= 3.* 10.**8.;# in um/s
lembda = 0.87;# in um
lembda = lembda * 10.**-6.;# in m
E_g = (h * c)/lembda;# in J-s
E_g= E_g/e;# in eV
print '%s %.3f' %("The band gap of the material in eV is",E_g);
The band gap of the material in eV is 1.431

Example E23 - Pg 125

In [22]:
# Exa 3.23 - 125
import math
# Given data
I_o = 10.;# in mW
e = 1.6 * 10.**-19.;# in J/eV
hv = 2.;# in eV
hv1=1.43;# in eV
alpha = 5. * 10.**4.;# in cm**-1
l = 46.;# in um
l = l * 10.**-6.;# in m
I_t = round(I_o * math.exp(-(alpha) * l));# in mW
AbsorbedPower= I_o-I_t;# in mW
AbsorbedPower=AbsorbedPower*10.**-3.;# in W or J/s
print '%s %.e %s' %("The absorbed power in watt or J/s is",AbsorbedPower,"\n");
F= (hv-hv1)/hv;# fraction of each photon energy unit
EnergyConToHeat= AbsorbedPower*F;# in J/s
print '%s %.2e %s' %("The amount of energy converted to heat per second in J/s is : ",EnergyConToHeat,"\n")
A= (AbsorbedPower-EnergyConToHeat)/(e*hv1);
print '%s %.2e %s' %("The number of photon per sec given off from recombination events in photons/s is",A,"\n");
The absorbed power in watt or J/s is 9e-03 

The amount of energy converted to heat per second in J/s is :  2.57e-03 

The number of photon per sec given off from recombination events in photons/s is 2.81e+16 

Example E24 - Pg 126

In [23]:
# Exa 3.24 - 126
import math
# Given data
Mu_p = 500.;# in cm**2/V-sec
kT = 0.0259;
Toh_p = 10.**-10.;# in sec
p_o = 10.**17.;# in cm**-3
q= 1.6*10.**-19.;# in C
A=0.5;# in square meter
del_p = 5. * 10.**16.;# in cm**-3
n_i= 1.5*10.**10.;# in cm**-3    
D_p = kT * Mu_p;# in cm/s
L_p = math.sqrt(D_p * Toh_p);# in cm
x = 10.**-5.;# in cm
p = p_o+del_p* math.e**(x/L_p);# in cm**-3
# p= n_i*%e**(Eip)/kT where Eip=E_i-F_p
Eip= math.log(p/n_i)*kT;# in eV
Ecp= 1.1/2.-Eip;# value of E_c-E_p in eV
Ip= q*A*D_p/L_p*del_p/math.e**(x/L_p);# in A
print '%s %.2e %s' %("The hole current in A is : ",Ip,"\n")
Qp= q*A*del_p*L_p;# in C
print '%s %.2e %s' %("The value of Qp in C is : ",Qp,"\n")
The hole current in A is :  1.09e+03 

The value of Qp in C is :  1.44e-07