Chapter 1 - Semiconductor Material & Junction Diode

Exa 1.1 Page No 51

In [1]:
# Given data
miu = 0.2# m**2/V-s
V = 100# mV
V = V * 10**-3# V
d = 0.5# mm
d = d * 10**-3# m
# mobility, miu = Vd/E and
E = V/d
# Drift velocity,
Vd = miu*E# m/s
print "The electron drift velocity = %.2f m/s"%Vd
# Time required,
T = d/Vd# sec
T=T*10**6# µs
print "The time required for an electron to move across the thickness = %.2f micro seconds"%T

The electron drift velocity = 40.00 m/s
The time required for an electron to move across the thickness = 12.50 micro seconds

Exa 1.2 Page No 52

In [2]:
# Given data
q = 1.6*10**-19# C
n_i = 2.5*10**19# /m**3
miu_n = 0.38# m**2/V-s
miu_p = 0.18# m**2/V-s
# The intrinsic conductivity for germanium,
sigma_i = q*n_i*(miu_n+miu_p)# (ohm-m)**-1
print "The intrinsic conductivity = %.2f (ohm-m)**-1"%sigma_i

The intrinsic conductivity = 2.24 (ohm-m)**-1

Exa 1.3 Page No 52

In [3]:
# Given data
rho = 0.50# ohm-m
q = 1.6*10**-19# C
miu_n = 0.39# m**2/V-s
miu_p = 0.19# m**2/V-s
sigma = 1/rho# (ohm-m)**-1
#conductivity of a semiconductor, sigma = q*n_i*(miu_p+miu_n) or
n_i = sigma/(q*(miu_n+miu_p))# /m**3
print "The intrinsic carrier concentration = %.2e per m**3"%n_i

The intrinsic carrier concentration = 2.16e+19 per m**3

Exa 1.4 Page No 52

In [4]:
# Given data
N_D = 10**21# /m**3
N_A = 5*10**20# /m**3
NdasD = N_D-N_A# /m**3
n = NdasD# /m**3
miu_n = 0.18# m**2/V-s
q = 1.6*10**-19# C
# The conductivity of silicon,
sigma = q*n*miu_n# (ohm-m)**-1
print "The conductivity of Si sample = %.2f (ohm-m)**-1"%sigma

The conductivity of Si sample = 14.40 (ohm-m)**-1

Exa 1.5 Page No 53

In [5]:
# Given data
At = 63.54## atomic weight of copper
d = 8.9## density = %.2f gm/cm**3
n = 6.023*10**23/At*d# electron/cm**3
q = 1.63*10**-19# C
miu = 34.8# m**2/V-s
# The conductivity of copper,
sigma = n*q*miu# mho/cm
print "The conductivity of copper = %.2e mho/cm"%sigma

The conductivity of copper = 4.79e+05 mho/cm

Exa 1.6 Page No 53

In [6]:
# Given data
sigma = 100# (ohm-m)**-1
n_i = 2.5*10**13# /cm**3
miu_n = 3800# cm**2/V-s
miu_p = 1800# cm**2/V-s
q = 1.6*10**-19# C
# Conductivity of a p-type germanium, sigma = q*p*miu_p or
p = sigma/(q*miu_p)# /cm**3
print "Concentration of holes in a p-type Ge = %.2e /cm**3"%p
# The concentration of electrons = %.2f a p-type Ge
n = (n_i**2)/p# /cm**3
print "The concentration of electrons in a p-type Ge = %.2e /cm**3"%n
#Given for Si
sigma= 0.1# (ohm m)**-1
miu_n= 1300# cm**2/V-sec
n_i= 1.5*10**10# /cm**3
#sigma = q*n*miu_n
n = sigma/(q*miu_n)# /cm**3
print "The concentration of electrons in n-type Si = %.2e /cm**3"%n
# The concentration of holes = %.2f n-type Si
p = (n_i**2)/n# /cm**3
print "The concentration of holes in n-type Si = %.2e /cm**3"%p

Concentration of holes in a p-type Ge = 3.47e+17 /cm**3
The concentration of electrons in a p-type Ge = 1.80e+09 /cm**3
The concentration of electrons in n-type Si = 4.81e+14 /cm**3
The concentration of holes in n-type Si = 4.68e+05 /cm**3

Exa 1.7 Page No 54

In [7]:
# Given data
miu_n = 3800## cm**2/V-s
miu_p = 1800##  cm**2/V-s
n_i = 2.5*10**13# /cm**3
Nge = 4.41*10**22# /cm**3
q = 1.602*10**-19# C
impurity = 10**8
# The number of donor atoms,
N_D = Nge/impurity##in /cm**3
# The number of holes
p = (n_i**2)/N_D# /cm**3
# Conductivity of an N-type Ge,
sigma = q*N_D*miu_n# (ohm-cm)**-1
# The resistivity of the Ge
rho = 1/sigma# ohm-cm
print "The resistivity of a dopped Ge = %.2f ohm-cm"% rho

The resistivity of a dopped Ge = 3.72 ohm-cm

Exa 1.8 Page No 54

In [8]:
# Given data
Nsi = 4.96*10**22# /cm**3
n_i = 1.52*10**10# /cm**2
q = 1.6*10**-19# C
miu_n = 0.135#  m**2/V-s
miu_n = miu_n * 10**4#  cm**2/V-s
miu_p = 0.048#  m**2/V-s
miu_p = miu_p * 10**4#  cm**2/V-s
# The conductivity of an intrinsic silicon,
sigma = q*n_i*(miu_n+miu_p)# (ohm-cm)**-1
# The resistivity of intrinsic silicon 
rho = 1/sigma# ohm-cm
print "The resistivity of intrinsic silicon = %.2e ohm-cm"%rho

impurity = 50*10**6
# The number of donor atoms,
N_D = Nsi/impurity# /cm**3
# Total free electrons,
n = N_D# /cm**3
# Total holes = %.2f a doped Si,
p = (n_i**2)/n# /cm**3
# Conductivity of a doped Si,
sigma = q*n*miu_n# (ohm-m)**-1
# The resistivity of doped silicon
rho = 1/sigma# ohm-cm
print "The resistivity of doped silicon = %.2f ohm-cm"%rho

The resistivity of intrinsic silicon = 2.25e+05 ohm-cm
The resistivity of doped silicon = 4.67 ohm-cm

Exa 1.9 Page No 55

In [9]:
# Given data
N_D= 5.0*10**28/(2.0*10**8)
# The Fermi level, E_F= E_C if,
N_C= N_D
# Formula N_C= 4.82*10**21*T**(3/2)
T= (N_C/(4.82*10**21.0))**(2.0/3)# K
print "The value of temperature = %.2f K"%T

The value of temperature = 0.14 K

Exa 1.10 Page No 55

In [10]:
import math
# Given data
n_i = 1.5*10**16##m**3
impurity = 10**20
minority = (n_i**2)/impurity# atoms/m**3
q = 1.6*10**-19# C
rho = 2*10**3# ohm-m
# The miniority carrier concentration 
miu_n = 1/(q*rho*n_i*2)##in m**2/V-s
print "The miniority carrier concentration = %.2f m**2/V-s"%miu_n
n = impurity
# The conductivity,
sigma = q*impurity*miu_n# (ohm-m)**-1
# The resistivity 
rho = 1/sigma#  ohm-m
print "The resistivity = %.2f ohm-m"%rho
kT = 0.026# eV
n_o = n
# The position of Fermi level 
E_FdividedEi = kT*math.log(n_o/n_i)# eV
print "The position of Fermi level = %.2f eV"%E_FdividedEi
# Minority carrier concentration 
M = ((n_i*2)**2)/n_o# atoms/cm**3
print "Minority carrier concentration = %.2e atoms/cm**3"%M

The miniority carrier concentration = 0.10 m**2/V-s
The resistivity = 0.60 ohm-m
The position of Fermi level = 0.23 eV
Minority carrier concentration = 9.00e+12 atoms/cm**3

Exa 1.11 Page No 56

In [41]:
# Given data
d = 5.0*10**22# atoms/cm**3
impurity = 10**8# atoms
N_D = d/impurity
n_i = 1.45*10**10
n = N_D
#Low of mass action,  n*p = (n_i**2)
p = (n_i**2)/n# /cm**3
q = 1.6*10**-19# C
miu_n = 1300# cm/V-s
n_i = n
#The Conductivity
sigma = q*miu_n*n_i# (ohm-cm)**-1
# The resistivity
rho = 1/sigma# ohm-cm
print "The resistivity = %.2f ohm-cm"%rho

The resistivity = 9.62 ohm-cm

Exa 1.12 Page No 57

In [11]:
# Given data
d = 5.0*10**22# atoms/cm**3
impurity = 10**8# atoms
N_D = d/impurity
n_i = 1.45*10**10
n = N_D
#Low of mass action,  n*p = (n_i**2)
p = (n_i**2)/n# /cm**3
q = 1.6*10**-19# C
miu_n = 1300# cm/V-s
n_i = n
#The Conductivity
sigma = q*miu_n*n_i# (ohm-cm)**-1
# The resistivity
rho = 1/sigma# ohm-cm
print "The resistivity = %.2f ohm-cm"%rho

The resistivity = 9.62 ohm-cm

Exa 1.14 Page No 58

In [40]:
import math
# Given data
n_i = 1.5*10**10# electrons/cm**3
N_D = 10**17# electrons/cm**3
n = N_D# electrons/cm**3
# The minority carrier concentration
p = (n_i**2)/n# holes/cm**3
print "The minority carrier concentration = %.2e holes/cm**3"%p
kT = 0.026
# The location of Fermi level 
E_FminusEi = kT*math.log(N_D/n_i)# eV
print "The location of Fermi level = %.3f eV"%E_FminusEi

The minority carrier concentration = 2.25e+03 holes/cm**3
The location of Fermi level = 0.409 eV

Exa 1.15 Page No 59

In [13]:
# Given data
V = 1# V
I = 8# mA
I = I * 10**-3# A
R = V/I# ohm
l = 2# mm
l = l * 10**-1# cm
b = 2# mm
b = b * 10**-1# cm
A = l*b# cm**2
L = 2# cm
# R = (rho*L)/A
sigma = L/(R*A)# (ohm-cm)**-1
# n = N_D
miu_n = 1300# cm**2/V-s
q = 1.6 * 10**-19# C
# sigma = n*q*miu_n
N_D = sigma/( miu_n*q )# /cm**3
print "The doping level = %.2e /cm**3"%N_D
d = 2.0
E = V/d
# The drift velocity 
Vd = miu_n * E# cm/s
print "The drift velocity = %.2f cm/sec"%Vd

The doping level = 1.92e+15 /cm**3
The drift velocity = 650.00 cm/sec

Exa 1.17 Page No 60

In [14]:
import math
# Given data
l = 1000# ft
l = l * 12*2.54# cm
R = 6.51# ohm
rho = R/l# ohm/cm
# The conductivity 
sigma = 1/rho# mho/cm
sigma = sigma * 10**2# mho/m
D= 1.03*10**-3# m
A= math.pi*D**2/4# m**2
print "The conductivity = %.2e mho/m"%sigma
q = 1.6*10**-19# C
n = 8.4*10**28# electrons/m**3
# sigma = n*q*miu
miu = sigma/(n*q)# m**2/V-s
print "The mobility = %.2e m**2/V-s"%miu
T = 2
# The drift velocity 
V = T/(n*q*A)# m/s
print "The drift velocity = %.2e m/s"%V

The conductivity = 4.68e+05 mho/m
The mobility = 3.48e-05 m**2/V-s
The drift velocity = 1.79e-04 m/s

Exa 1.18 Page No 61

In [15]:
# Given data
N_D = 2*10**16# /cm**3
N_A = 5*10**15# /cm**3
# The concentration of holes 
Pp = N_D-N_A# /cm**3
print "The concentration of holes = %.2e /cm**3"%Pp
n_i = 10**12
# The concentartion of electrons 
n_p = (n_i**2)/Pp# /cm**3
print "The concentartion of electrons = %.2e /cm**3"%n_p

The concentration of holes = 1.50e+16 /cm**3
The concentartion of electrons = 6.67e+07 /cm**3

Exa 1.19 Page No 62

In [16]:
import math
# Given data
rho = 0.005# ohm-m
Bz = 0.48# Wb/m**2
R_H = 3.55*10**-4# m**3/C
ExByJx= rho
# R_H = Ey/(Bz*Jx)
EyByJx= R_H*Bz
# The hall angle 
theta_H = math.degrees(math.atan(EyByJx/ExByJx))# °
print "The hall angle = %.2f degree"%theta_H

The hall angle = 1.95 degree

Exa 1.20 Page No 63

In [17]:
# Given data
R_H = 3.55 * 10**-4# m**3/C
Ix = 15# mA
Ix = Ix * 10**-3# A
A = 15*1# mm
A = A * 10**-6# m**2
Bz = 0.48# Wb/m**2
Jx = Ix/A# A/m**2
# R_H = Ey/(Bz*Jx)
Ey = R_H*Bz*Jx# V/m
# voltage between contacts 
Voltage = Ey*Ix# V
print "The voltage between contacts = %.4f V"%Voltage

The voltage between contacts = 0.0026 V

Exa 1.21 Page No 63

In [18]:
# Given data
A = 0.001# cm**2
l = 20# µm
l = l * 10**-4# cm
V = 20# V
I = 100# mA
I = I * 10**-3# A
R = V/I# ohm
# R = l/(sigma*A)
sigma = l/(R*A)# (ohm-cm)**-1
miu_n = 1350# cm**2/V-s
q = 1.6*10**-19# C
# sigma = n*q*miu_n or
# The concentration of donor atoms 
n = sigma/(q*miu_n)# cm**-3
print "The concentration of donor atoms = %.3e cm**-3"%n

The concentration of donor atoms = 4.630e+13 cm**-3

Exa 1.22 Page No 64

In [19]:
# Given data
R = 2# k ohm
R = R * 10**3# ohm
L = 200# µm
L = L * 10**-4# cm
A = 10**-6# cm**2
miu_n = 8000# cm**2/V-s
q = 1.6*10**-19# C
n = '0.9*N_D'
# R = (rho*l)/A= (1/(n*q*miu_n))*(l/A)
# rho = L/(R*q*miu_n*A)
n = L/(R*q*miu_n*A)# /cm**-3
# The doping needed 
Nd= n/0.9
print "The doping needed = %.3e cm**-3"%Nd

The doping needed = 8.681e+15 cm**-3

Exa 1.23 Page No 65

In [20]:
import math
# Given data
KT = 26*10**-3
Nd = 10**15
n_i = 1.5*10**10
# The position of the Fermi level 
E_FminusE_Fi = KT*math.log(abs( Nd/n_i ))# eV
print "The position of the Fermi level = %.2f eV"%E_FminusE_Fi

The position of the Fermi level = 0.29 eV

Exa 1.24 Page No 65

In [21]:
import math
# Given data
Na = 5 * 10**15# cm**-3
Nc = 2.8 * 10**19# cm**-3
E_CminusE_F = 0.215# eV
KT = 26* 10**-3# eV
# The concentration of donors atoms 
Nd = Na + Nc * (math.exp( -E_CminusE_F/KT ))# cm**-3
print "The concentration of donors atoms = %.4e cm**-3"%Nd

The concentration of donors atoms = 1.2176e+16 cm**-3

Exa 1.25 Page No 65

In [22]:
# Given data
Nd = 10**18
R = 10# ohm
A =10**-6# cm**2
L = 10# mm
L = L * 10**-4# cm
miu_n = 800# cm**2/V-s
q = 1.6*10**-19# C
#Formula used, n = L/(q*miu_n*A*R)
n = L/(q*miu_n*A*R)# cm**-3
# The percentage doping efficiency 
doping = (n/Nd)*100## % doping efficiency in %
print "The percentage doping efficiency = %.2f %%"%doping

The percentage doping efficiency = 78.12 %

Exa 1.26 Page No 66

In [23]:
import math
# Given data
Io = 2*10**-7# A
V = 0.1# V
# Current through the diode under forward bias,
I = Io*( (math.exp(40*V))-1 )# A
I = I * 10**6# µA
print "The current through the diode under forward bias = %.2f µA"%I

# Note: Calculated value of I in the book is wrong.

The current through the diode under forward bias = 10.72 µA

Exa 1.28 Page No 67

In [24]:
import math
# Given data
T = 125.0# degree C
T = T + 273.0# K
V_T = T/11600.0
Io = 30# µA
Io = Io * 10**-6# A
V = 0.2# V
# The dynamic resistance = %.2f forward direction,
r_f = V_T/( Io * (math.exp(V/V_T)) )# ohm
print "The dynamic resistance in forward direction = %.2f ohm"%r_f
r_f = V_T/( Io * (math.exp(-V/V_T)) )# ohm
# The dynamic resistance = %.2f reverse direction 
r_f = r_f * 10**-6# Mohm
print "The dynamic resistance in reverse direction = %.2f Mohm"%r_f

The dynamic resistance in forward direction = 3.36 ohm
The dynamic resistance in reverse direction = 0.39 Mohm

Exa 1.29 Page No 68

In [25]:
import math
# Given data
Eta = 1
V_T = 0.026
# I = Io*( (exp(V/(Eta*V_T))) - 1 ) and  I = -Io
# I = -0.9*Io
# -0.9*Io = Io*( (exp(V/(Eta*V_T))) - 1 )
V = Eta*V_T*math.log(0.1)# V
V = V * 10**3# mV
print "The voltage = %.2f mV"%V
V = 0.05# V
# The ratio of diode current with a forward bias to current with a reverse bias 
If_by_Ir= ( (math.exp(V/V_T))-1 )/( (math.exp(-V/V_T))-1 )
print "The ratio of diode current with a forward bias to current with a reverse bias = %.3f"%If_by_Ir
Io = 10# µA
V = 0.1# V
# The value of I1 
I1 = Io*( (math.exp(V/V_T))-1 )# µA
print "The value of I1 = %.2f µA"%I1
V = 0.2# V
# The value of I2
I2 = Io*( (math.exp(V/V_T))-1 )# µA 
I2 = I2 * 10**-3# mA
print "The value of I2 = %.2f mA"%I2
V = 0.3# V
# The value of I3
I3 = Io*( (math.exp(V/V_T))-1 )# µA
I3 = I3 * 10**-6# A
print "The value of I3 = %.2f A"%I3

The voltage = -59.87 mV
The ratio of diode current with a forward bias to current with a reverse bias = -6.842
The value of I1 = 458.13 µA
The value of I2 = 21.90 mA
The value of I3 = 1.03 A

Exa 1.30 Page No 69

In [26]:
import math
# Given data
# Io150 = Io25 * 2**((150-25)/10)
#Io150 = 5800*Io25
T = 150# degree C
T = T  + 273# K
V_T = 8.62*10**-5 * T# V
V = 0.4# V
Eta = 2
Vt = 0.026# V 
# The factor  by which current will get multiplied 
I150byI25= 5800*math.exp(V/(Eta*V_T))/math.exp(V/(Eta*Vt))
print "The factor  by which current will get multiplied = %.3f"%I150byI25

The factor  by which current will get multiplied = 638.025

Exa 1.31 Page No 69

In [27]:
# Given data
R = 1# ohm
V = 5# V
V1 = 0.5# V
R1 = 1# k ohm
R1 = R1 * 10**3# ohm
# V-(I_D*R1)-(I_D*R) - V1 = 0
I_D = (V-V1)/(R1+R)# A
I_D = I_D * 10**3# mA
V_D = (I_D*10**-3*R) + V1# V
print "The operating point of the diode is : (%.2fV,%.2fmA)"%(V_D,I_D)

The operating point of the diode is : (0.50V,4.50mA)

Exa 1.32 Page No 70

In [28]:
import math
# Given data
Eta = 1
kT = 26# meV
# (%e**((e*V1)/kT)) = 2 or
#The voltage drop across the forward biased diode
V1 = math.log(2)*kT# mV
V1= V1*10**-3# V
print "The voltage drop across the forward biased diode, = %.4f V"%V1

The voltage drop across the forward biased diode, = 0.0180 V

Exa 1.33 Page No 71

In [29]:
import math
# Given data
epsilon_Ge = 16/(36*math.pi*10**11)# F/cm
d = 2*10**-4# cm
A = 1# mm**2
A = A * 10**-2# cm**2
epsilon_o = epsilon_Ge# F/cm
# The space charge capacitance 
C_T = (epsilon_o*A)/d# F
C_T = C_T * 10**12# pF
print "The space charge capacitance = %.2f pF"%C_T

The space charge capacitance = 70.74 pF

Exa 1.34 Page No 71

In [30]:
import math 
# Given data
D = 0.102# cm 
A = (math.pi*(D**2))/4# cm**2
sigma_p = 0.286# (ohm-cm)**-1
q = 1.6*10**-19# C
miu_p = 500
# Formula used, sigma_p = q*miu_p*N_A
N_A = sigma_p/(q*miu_p)# atoms/cm**3
V1 = 5# V
V2 = 0.35# V
Vb = V1+V2# V
# The transition capacitance,
C_T = 2.92*10**-4*((N_A/Vb)**(1./2))*A# pF/cm**2
print "The value of C_T = %.2f pf/cm**2"%C_T

The value of C_T = 61.68 pf/cm**2

Exa 1.35 Page No 71

In [31]:
# Given data
C_T1 = 15# pF
Vb1 = 8# V
Vb2 = 12# V
# C_T1/C_T2 = (Vb2/Vb1)**(1/2)
C_T2 = C_T1 * ((Vb1/Vb2)**(1/2))# pF
print "The value of C_T for reverse bias = %.2f pF"%C_T2

The value of C_T for reverse bias = 15.00 pF

Exa 1.36 Page No 72

In [32]:
import math
# Given data
V_T = 0.026# V
Eta = 1
I = '-0.9*Io'
# T = Io*((%e**(V/(Eta*V_T)))-1 )
# I = Io*((%e**(V/(Eta*V_T)))-1 )
V = math.log(0.1)*V_T# V 
V = V * 10**3# mV
print "The voltage = %.2f mV"%V

The voltage = -59.87 mV

Exa 1.37 Page No 72

In [33]:
# Given data
Vin = 20# V
Vgamma = 0.7# V
R = 20# k ohm
R = R * 10**3# ohm
# Vin-(I_D*Vin) - Vgamma = 0 or
# The value of I_D,
I_D = (Vin-Vgamma)/R# A
I_D = I_D * 10**3# mA
print "Part (a) : The value of I_D for first circuit = %.2f mA"%I_D

# Part (b)
Vin= 10.# V
Vgamma = 0.7# V
R = 100# k ohm
# Drain current,
I_D= Vin/R# mV
print "Part (b) : The value of I_D for second circuit = %.2f mA"%I_D

Part (a) : The value of I_D for first circuit = 0.97 mA
Part (b) : The value of I_D for second circuit = 0.10 mA

Exa 1.38 Page No 73

In [34]:
# Given data
R1 = 1# k ohm
R1 = R1 * 10**3# ohm
R2 = 2# k ohm
R2 = R2 * 10**3# ohm
V = 10# V
V1 = 0.7# V 
# V * (I_D*R1) - (R2*I_D) - V1 = 0
I_D = (V-V1)/(R1+R2)# A
I_D = I_D * 10**3# mA
print "The value of I_D = %.2f mA"%I_D
# The output voltage,
Vo = (I_D*10**-3 * R2) +V1# V
print "The value of Vo = %.2f V"%Vo

The value of I_D = 3.10 mA
The value of Vo = 6.90 V

Exa 1.39 Page No 73

In [35]:
# Given data
V = 10.# V
R = 10# ohm
# Current through resistance,
I = V/R# A
print "Part (a): The current through resistance = %.2f A"%I
print "Part (b) : Current through 10 ohm resistance will be Zero"
print "Part (c): Current will be zero"
print "Part (d): The diode will be ON and current = %.2f A"%I

Part (a): The current through resistance = 1.00 A
Part (b) : Current through 10 ohm resistance will be Zero
Part (c): Current will be zero
Part (d): The diode will be ON and current = 1.00 A

Exa 1.40 Page No 74

In [36]:
# Given data
Vth= 0.5# V
R_F= 1*10**3# ohm
V= 5# V
# Applying KVL for loop, V-Vd-R_F*Ii= 0    (i)
# When Ii=0
Vd= V# V
# When Vd= 0
Ii= V/R_F# A
# From eq(i)
Ii= (V-Vth)/R_F# A
Vd= V-R_F*Ii# V
print "The operating point is  : (%.2fV,%.2fmA)"%(Vd,Ii*1000)

The operating point is  : (0.50V,4.50mA)

Exa 1.43 Page No 76

In [37]:
# Given data
V_CC = 6# V
Vr = 0.6# V
V1= V_CC##in V
V2 = V1-Vr# V
print "The voltage at V1 = %.2f volts"%V1
print "The voltage at V2 = %.2f volts"%V2

The voltage at V1 = 6.00 volts
The voltage at V2 = 5.40 volts

Exa 1.44 Page No 76

In [38]:
# Given data
V_T = 0.7# V
V = 5# V
R = 2# k ohm
R = R * 10**3# ohm
Vs = 0.7
Vx = Vs+V_T# V
# The value of I1 
I1 = (V-Vx)/R# A
I1 = I1 * 10**3# mA
print "The value of I1 = %.2f mA"%I1
# The value of I2 
I2 = I1# mA
print "The value of I2 = %.2f mA"%I2

The value of I1 = 1.80 mA
The value of I2 = 1.80 mA

Exa 1.45 Page No 77

In [39]:
# Given data
Rf = 300.# ohm
V = 0.5# V
R = 600.# ohm
Vi = 2.# V
# The output voltage 
Vo = (Vi-V)*( R/(R+Rf) )# V
print "The value of Vo = %.2f V"%Vo

The value of Vo = 1.00 V