Chapter 14: Solid State Electronics

Example 14.1, Page 718

In [1]:
#Variable declaration
e = 1.6e-019;   # Charge on an electron, C
mu_h = 0.048;  # Mobility of holes, metre square/volt-s
mu_e = 0.135;  # Mobility of electrons, metre square/volt-s 

#Calculations
# For P-type semiconductor
rho_p = 1e-01;  # Resistivity of P type silicon, omh-m
# As rho_p = 1/(e*N_a*mu_h), solving for N_a
N_a = 1/(e*rho_p*mu_h); # Density of acceptor atoms, per metre cube
# For N-type semiconductor
rho_n = 1e-01;  # Resistivity of N type silicon, omh-m
# As rho_n = 1/(e*N_d*mu_h), solving for N_d
N_d = 1/(e*rho_n*mu_e); # Density of donor atoms, per metre cube

#Results
print "Density of acceptor atoms = %4.2e per metre cube"%N_a
print "Density of donor atoms = %4.2e per metre cube"%N_d   #incorrect answer in textbook

Density of acceptor atoms = 1.30e+21 per metre cube
Density of donor atoms = 4.63e+20 per metre cube

Example 14.2, Page 718

In [2]:
#Variable declaration
e = 1.6e-019;   # Charge on an electron, C
mu_e = 0.36;    # Mobility of an electron, metre square/V-s
mu_h = 0.17;    # Mobility of a hole, metre square/V-s
n_i = 2.5e+018; # Intrinsic concentration of Ge sample, per metre cube

#Calculations
sigma = e*n_i*(mu_h+mu_e);  # Electrical conductivity of Ge sample, mho per metre
rho = 1/sigma;  # Electrical resistivity of Ge, ohm-m

#Results
print "The electrical conductivity of intrinsic germanium sample = %5.3f mho/m"%sigma
print "The electrical resistivity of intrinsic germanium sample = %3.1f ohm-m"%rho

The electrical conductivity of intrinsic germanium sample = 0.212 mho/m
The electrical resistivity of intrinsic germanium sample = 4.7 ohm-m

Example 14.3, Page 719

In [3]:
#Variable declaration
e = 1.6e-019;   # Charge on an electron, C
mu_e = 0.13;    # Mobility of an electron, metre square/V-s
mu_h = 0.05;    # Mobility of a hole, metre square/V-s
n_i = 1.5e+016; # Intrinsic concentration of Si, per metre cube

#Calculations
# Pure Si
sigma = e*n_i*(mu_h+mu_e);  # Electrical conductivity of Si, mho per metre
# Pure Si doped with donor impurity
n_e = 5e+028/1e+09;     # Concentration of electrons, per metre cube
sigma_n = e*n_e*mu_e;   # Electrical conductivity of Si doped with donor impurity, mho per metre
# Pure Si doped with acceptor impurity
n_h = 5e+028/1e+09;     # Concentration of holes, per metre cube
sigma_p = e*n_h*mu_h;   # Electrical conductivity of Si doped with acceptor impurity, mho per metre

#Results
print "The electrical conductivity of pure Si = %4.2e mho/m"%sigma
print "The electrical conductivity of Si doped with donor impurity = %4.2f mho/m"%sigma_n
print "The electrical conductivity of Si doped with acceptor impurity= %4.2f mho/m"%sigma_p

The electrical conductivity of pure Si = 4.32e-04 mho/m
The electrical conductivity of Si doped with donor impurity = 1.04 mho/m
The electrical conductivity of Si doped with acceptor impurity= 0.40 mho/m

Example 14.4, Page 720

In [5]:
from math import *

#Variable declaration
Nd = 1;     # For simplicity assume donor concentration to be unity, per metre cube
Nd_prime = 3*Nd;   # Thrice the donor concentration, per metre cube
dE_CF1 = 0.5;  # Energy difference between normal Fermi level and conduction level, eV
k_BT = 0.03;    # Thermal energy at room temperature, eV

#Calculations
# As Nd_prime/Nd = exp((dE_CF1 - dE_CF2))/k_BT, solving for dE_CF2
dE_CF2 = dE_CF1-k_BT*log(Nd_prime/Nd);  # Energy difference between new postion of Fermi level and conduction level, eV

#Result
print "The new postion of Fermi level when donor concentration is trebled = %5.3f eV"%dE_CF2

The new postion of Fermi level when donor concentration is trebled = 0.467 eV

Example 14.5, Page 721

In [6]:
from math import *

#Variable declaration
e = 1.6e-019;   # Charge on an electron, C
T = 300;    # Room temperature, K
J0 = 300e-03;   # Saturation current density of the pn junction diode, A/metre square
J = 1e+05;  # Forward current density of pn junction diode, A/metre square
k_B = 1.38e-023;    # Boltzmann constant, J/K
eta = 1;    # Ideality factor for Ge diode

#Calculations
# As J = J0*exp(e*V/(eta*k_B*T)), solving for V
V = eta*k_B*T/e*log(J/J0);   # Voltage required to cause a forward current density in pn junction diode, volt

#Results
print "The voltage required to cause a forward current density in pn junction diode = %5.3f V"%V

The voltage required to cause a forward current density in pn junction diode = 0.329 V

Example 14.6, Page 721

In [7]:
from math import *

#Variable declaration
e = 1.6e-019;   # Charge on an electron, C
T = 300;    # Room temperature, K
J0 = 200e-03;   # Saturation current density of the pn junction diode, A/metre square
J = 5e+04;  # Forward current density of pn junction diode, A/metre square
k_B = 1.38e-023;    # Boltzmann constant, J/K
eta = 1;    # Ideality factor for Ge diode

#Calculations
# As J = J0*exp(e*V/(eta*k_B*T)), solving for V
V = eta*k_B*T/e*log(J/J0);   # Voltage required to cause a forward current density in pn junction diode, volt

#Result
print "The voltage required to cause a forward current density in pn junction diode = %5.3f V"%V

The voltage required to cause a forward current density in pn junction diode = 0.322 V

Example 14.7, Page 722

In [8]:
from math import *

#Variable declaration
e = 1.6e-019;   # Charge on an electron, C
T = 300;    # Room temperature, K
J0 = 300e-03;   # Saturation current density of the pn junction diode, A/metre square
J = 1e+05;  # Forward current density of pn junction diode, A/metre square
k_B = 1.38e-023;    # Boltzmann constant, J/K
eta = 2;    # Ideality factor for Ge diode

#Calculations
# As J = J0*exp(e*V/(eta*k_B*T)), solving for V
V = eta*k_B*T/e*log(J/J0);   # Voltage required to cause a forward current density in pn junction diode, volt

#Result
print "The voltage required to cause a forward current density in Si iode = %5.3f V"%V

The voltage required to cause a forward current density in Si iode = 0.658 V

Example 14.8, Page 723

In [9]:
#Variable declaration
I = 55e-03;     # Forward current through Si diode, A
V = 3;          # Forward bias across Si diode, V
eta = 2;        # Ideality factor for Si diode

#Calculations
R_dc = V/I;     # Static diode resistance, ohm
R_ac = 0.026*eta/I;    # Dynamic diode resistance, ohm

#Results
print "The static diode resistance = %4.1f ohm"%R_dc
print "The dynamic diode resistance = %5.3f ohm"%R_ac

The static diode resistance = 54.5 ohm
The dynamic diode resistance = 0.945 ohm

Example 14.9, Page 723

In [10]:
from math import *

#Variable declaration
R_L = 1000; # Load resistance across HWR, ohm
V_rms = 200;    # Rms value of voltage supply, V

#Calculations
V0 = sqrt(2)*V_rms; # Peak value of voltage, V
I0 = V0/(R_L*1e-03);    # Peak value of current, mA
I_dc = I0/pi;  # Average value of current, mA
I_rms = I0/2;   # Rms value of current, mA
V_dc = I_dc*R_L/1e+03;    # Dc output voltage, V
PIV = V0;   # Peak inverse voltage, V

#Results
print "The average value of current = %2d mA"%I_dc
print "The rms value of current = %5.1f mA"%I_rms
print "The dc output voltage = %2d V"%(V_dc/1)
print "PIV = %5.1f V"%PIV

The average value of current = 90 mA
The rms value of current = 141.4 mA
The dc output voltage = 90 V
PIV = 282.8 V

Example 14.10, Page 724

In [17]:
from math import *

#Variable declaration
R_L = 980; # Load resistance across FWR, ohm
R_F = 20.;   # Internal resistance of two crystal diodes in FWR, ohm
V_rms = 50;    # Rms value of voltage supply, V

#Calculations
V0 = sqrt(2)*V_rms; # Peak value of voltage, V
I0 = V0/((R_L+R_F)*1e-03);    # Peak value of current, mA
I_dc = 2*I0/pi;  # Average value of current, mA
I_rms = I0/sqrt(2);   # Rms value of current, mA
V_dc = I_dc*R_L/1e+03;    # Dc output voltage, V
eta = 81.2/(1+R_F/R_L);     # Rectification efficiency
PIV = 2*V0;   # Peak inverse voltage, V

#Results
print "The average value of current = %2d mA"%I_dc
print "The rms value of current = %2d mA"%I_rms
print "The dc output voltage = %4.1f V"%(V_dc/1)
print "The rectification efficiency = %4.1f percent"%eta
print "PIV = %5.1f V"%PIV

The average value of current = 45 mA
The rms value of current = 50 mA
The dc output voltage = 44.1 V
The rectification efficiency = 79.6 percent
PIV = 141.4 V

Example 14.11, Page 725

In [12]:
#Variable declaration
delta_IC = 1e-03;   # Change in collector current, A
delta_IB = 50e-06;   # Change in base current, A

#Calculations
bta = delta_IC/delta_IB;  # Base current amplification factor
alpha = bta/(1+bta);    # Emitter current amplification factor

#Results
print "Alpha of BJT = %4.2f"%alpha
print "Beta of BJT = %2d"%bta

Alpha of BJT = 0.95
Beta of BJT = 20

Example 14.12, Page 725

In [13]:
#Variable declaration
I_E = 2;    # Emitter current, mA
alpha = 0.88;   # Emitter current amplification factor

#Calculations
I_C = alpha*I_E;    # Collector current, mA
I_B = I_E - I_C;        # Base current of BJT in CB mode, mA

#Result
print "The base current of BJT in CB mode = %4.2f mA"%I_B

The base current of BJT in CB mode = 0.24 mA

Example 14.13, Page 725

In [14]:
#Variable declaration
I_CBO = 12.5e-03;   # Reverse saturation current, mA
I_E = 2;    # Emitter current, mA
I_C = 1.97; # Collector current, mA

#Calculations
# As I_C = alpha*I_E+I_CBO, solving for alpha
alpha = (I_C - I_CBO)/I_E;  # Emitter current gain
I_B = I_E - I_C;    # Base current, mA

#Results
print "The emitter current gain = %5.3f"%alpha
print "The base current = %4.2f mA"%I_B

The emitter current gain = 0.979
The base current = 0.03 mA

Example 14.14, Page 726

In [15]:
#Variable declaration
alpha = 0.98;   # Emitter current amplification factor
I_CBO = 5e-06;  # Reverse saturation current, A

#Calculations
bta = alpha/(1-alpha);  # Emitter current amplification factor
I_CEO = 1/(1-alpha)*I_CBO;  # Leakage current of BJT in CE mode, mA

#Results
print "The base current gain = %2g"%bta
print "The leakage current of BJT in CE mode = %4.2f mA"%(I_CEO/1e-03)

The base current gain = 49
The leakage current of BJT in CE mode = 0.25 mA

Example 14.15, Page 726

In [16]:
#Variable declaration
R_i = 50;       # Dynamic input resistance of PNP transistor, ohm
R_L = 5e+03;    # Load resistance in collector circuit, ohm
alpha = 0.96;   # Emitter current amplification factor

#Calculations
A_v = alpha*R_L/R_i;    # Voltage gain
A_p = alpha*A_v;    # Power gain

#Results
print "The voltage gain = %2g"%A_v
print "The power gain = %2d"%A_p

The voltage gain = 96
The power gain = 92