Chapter 5 Bipolar Transistors

Example 5.1 Page No 125

In [1]:
# given data
V_BB= 10##in V
V_BE= 0.7##in V
V_CC= 20##  V
R_B= 1.5##  MΩ
R_B= R_B*10**6##in Ω
R_C= 5*10**3##in Ω
bita= 125## unit less
I_B= (V_BB-V_BE)/R_B##in A
I_C= bita*I_B##in A
# The dc voltage between the collector and emitter 
V_CE= V_CC-I_C*R_C##in V
print "The dc voltage between the collector and emitter = %.2f volts"%V_CE

The dc voltage between the collector and emitter = 16.12 volts

Example 5.2 Page No 125

In [3]:
from numpy import arange
%matplotlib inline
from matplotlib.pyplot import plot,xlabel,ylabel,title,show
# given data
V_CC= 30##  V
R_C= 1.5##in kΩ
Ver_intercept= V_CC/R_C##in mA
Hor_intercept= V_CC##  V
V_CE= arange(0,0.1+Hor_intercept,0.1) #V
I_C= [(V_CC-VCE)/R_C for VCE in V_CE] ##  mA
# DC load line
plot(V_CE,I_C)
xlabel("V_CE in volts")
ylabel("I_C in mA")
title("DC load line")
show()

Example 5.3 Page No 126

In [4]:
# given data
V_BE= 0.7##in V
V_CC= 30##  V
R_B= 390##  kΩ
R_B= R_B*10**3##in Ω
R_C= 1.5*10**3##in Ω
bita= 80## unit less
I_B= (V_CC-V_BE)/R_B##in A
# The collector current,
I_C= bita*I_B##in A
# The value of V_CE
V_CE= V_CC-I_C*R_C##in V
I_C= I_C*10**3##  mA
print "The value of I_C = %.2f mA"%I_C
print "The value of V_CE = %.2f volts"%V_CE

The value of I_C = 6.01 mA
The value of V_CE = 20.98 volts

Example 5.4 Page No 126

In [6]:
# given data
V_BE= 0.7##  V
V_LED= 2.0##in V
V_CC= 20##  V
R_B= 47##  kΩ
R_B= R_B*10**3##in Ω
R_C= 1*10**3##in Ω
bita= 150## unit less
# The LED current
I_LED= (V_CC-V_LED)/R_C##  A
I_Csat= I_LED##  A
I_Bsat= I_Csat/bita##  A
# The input voltage,
V_IN= I_Bsat*R_B+V_BE##in V
I_LED= I_LED*10**3##  mA
print "The LED current = %.2f mA"%I_LED
print "The value of Vin = %.2f volts"%V_IN

The LED current = 18.00 mA
The value of Vin = 6.34 volts

Example 5.5 Page No 129

In [8]:
# given data
Vz= 10##  V
V_BE= 0.7##  V
V_CC= 30.0##  V
R_E= 5##  kΩ
R_E= R_E*10**3##in Ω
R_C= 4##  kΩ
R_C= R_C*10**3##in Ω
V_E= Vz-V_BE##  V
I_E= V_E/R_E##  A
I_C= I_E##  A
# The collector voltage
V_C= V_CC-I_C*R_C##  V
print "The collector voltage = %.2f volts"%V_C

The collector voltage = 22.56 volts

Example 5.6 Page No 130

In [10]:
# given data
V_BE= 0.7##  V
R2= 1*10**3##in Ω
R1= 3.9*10**3##in Ω
R_E= 100##  Ω
R_C= 150##  kΩ
V_CC= 25.0##  V
Vz= R2*V_CC/(R1+R2)##  V
V_E= Vz-V_BE##  V
I_E= V_E/R_E##  A
I_C= I_E##  A
# The collector voltage 
V_C= V_CC-I_C*R_C##  V
print "The collector voltage = %.2f volts"%V_C

The collector voltage = 18.40 volts

Example 5.7 Page No 130

In [11]:
# given data
R_E= 2*10**3##  Ω
R_C= 1*10**3##  kΩ
V_E= 4.3##in V
V_CC= 15.0##  V
I_E= V_E/R_E##  A
I_C= I_E##in A
# = %.2f the first stage the collector voltage 
V_C= V_CC-I_C*R_C##  A
print "In the first stage the collector voltage = %.2f volts"%V_C
# Second stage
V_E= 2.3##  V
R_E= 220##  Ω
R_C= 470##  Ω
I_E= V_E/R_E##  A
I_C= I_E##in A
# = %.2f the second stage the collector voltage 
V_C= V_CC-I_C*R_C##  A
print "In the second stage the collector voltage = %.2f volts"%V_C

# Note : In the book, the calculated value of collector voltage in first stage is not accurate.

In the first stage the collector voltage = 12.85 volts
In the second stage the collector voltage = 10.09 volts

Example 5.8 Page No 133

In [12]:
# given data
V_BE= 0.7##in V
V_CC= 30##  V
R_B= 3*10**6##  Ω
bitamin= 100## unit less
bitamax= 300## unit less
I_B= (V_CC-V_BE)/R_B##  A
# The minimum value of collector current 
I_Cmin= bitamin*I_B##  A
# The maximum value of collector current 
I_Cmax= bitamax*I_B##  A
I_Cmin= I_Cmin*10**3##  mA
I_Cmax= I_Cmax*10**3##  mA
print "The minimum value of collector current = %.2f mA"%I_Cmin
print "The maximum value of collector current = %.2f mA"%I_Cmax

The minimum value of collector current = 0.98 mA
The maximum value of collector current = 2.93 mA

Example 5.9 Page No 139

In [13]:
# given data
V_BE= 0.7##in V
V_CC= 15##  V
R_E= 100##  Ω
R_C= 910##  Ω
R_B= 430*10**3##  Ω
bita= 300## unit less
# The collector current,
I_C= (V_CC-V_BE)/(R_E+R_B/bita)##  A
I_C= I_C*10**3##  mA
print "The value of I_C = %.2f mA"%I_C
I_C= I_C*10**-3##  A
# The collector to emitter voltage,
V_CE= V_CC-I_C*(R_C+R_E)##  V
print "The value of V_CE = %.2f volts"%V_CE

The value of I_C = 9.33 mA
The value of V_CE = 5.58 volts

Example 5.10 Page No 140

In [14]:
# given data
V_BE= 0.7##in V
V_CC= 15##  V
R_C= 1*10**3##  Ω
R_B= 200*10**3##  Ω
bita= 300## unit less
# The collector current,
I_C= (V_CC-V_BE)/(R_C+R_B/bita)##  A
I_C=I_C*10**3##  mA
print "The value of I_C = %.2f mA"%I_C
I_C=I_C*10**-3##  A
# The collector to emitter voltage,
V_CE= V_CC-I_C*R_C##  V
print "The value of V_CE = %.2f volts"%V_CE

The value of I_C = 8.58 mA
The value of V_CE = 6.42 volts

Example 5.11 Page No 140

In [15]:
# given data
V_BE= 0.7##in V
V_CC= 15##  V
V_EE= 15##  V
R_E= 10*10**3##  Ω
R_C= 5.1*10**3##  Ω
I_E= (V_EE-V_BE)/R_E##  A
# The collector current,
I_C= I_E##  A
V_C= V_CC-I_C*R_C##  A
V_E= -V_BE##  V
V_CE= V_C-V_E##  V
# The collector to emitter voltage,
V_CE= V_CC+V_EE-I_C*(R_C+R_E)
I_C= I_C*10**3##  mA
print "The value of I_C = %.2f mA"%I_C
print "The value of V_CE = %.2f volts"%V_CE

The value of I_C = 1.43 mA
The value of V_CE = 8.41 volts

Example 5.12 Page No 142

In [16]:
# given data
V_BE= 0.7##in V
V_CC= 30##  V
Vz= 6##  V
R_E= 3*10**3##  Ω
R_C= 4*10**3##  Ω
I_E= (Vz-V_BE)/R_E##  A
I_C= I_E##  A
# For first stage the collector voltage to ground 
V_C= V_CC-I_C*R_C##  v
print "For first stage the collector voltage to ground = %.2f volts"%V_C
Vz= 10##  V
R_E= 2*10**3##in Ω
R_C= 3*10**3##  Ω
I_E= (Vz-V_BE)/R_E##  A
I_C= I_E##  A
# For second stage the collector voltage to ground 
V_C= I_C*R_C##  v
print "For second stage the collector voltage to ground = %.2f volts"%V_C

For first stage the collector voltage to ground = 22.93 volts
For second stage the collector voltage to ground = 13.95 volts