Chapter - 4 : Bipolar Junction Transistors

Example 4.1 : Page No 223

In [1]:
from __future__ import division
# Given data
I_C= 0.9 # in mA
I_E=1 # in mA
alpha = I_C/I_E 
print "Current gain = %0.1f" %alpha
# Formula I_E= I_B+I_C
I_B= I_E-I_C # in mA
print "The base current = %0.1f mA" %I_B

Current gain = 0.9
The base current = 0.1 mA

Example 4.2 : Page No 224

In [2]:
# Given data
alpha= 0.97 
I_E=1 # in mA
# Formula alpha = I_C/I_E 
I_C= alpha*I_E # in mA
# Formula I_E= I_B+I_C
I_B= I_E-I_C # in mA
print "The base current = %0.2f mA" %I_B

The base current = 0.03 mA

Example 4.3 : Page No 226

In [3]:
# Given data
# Part (i)
a= 0.90 
B=a/(1-a) 
print "At alpha= 0.90, the value of Bita = %0.f" %B
# Part (ii)
a= 0.99 
B=a/(1-a) 
print "At alpha= 0.99, the value of Bita = %0.f" %B

At alpha= 0.90, the value of Bita = 9
At alpha= 0.99, the value of Bita = 99

Example 4.4 : Page No 226

In [4]:
# Given data
bita= 50 
I_E= 10 # in mA
I_B= 200*10**-3 # in mA
alfa= bita/(1+bita)
print "The value of alfa = %0.2f" %alfa
I_C= alfa*I_E # in mA
print "The value of I_C = %0.1f mA using the value of alpha" %I_C
I_C= bita*I_B # in mA
print "The value of I_C = %0.f mA using the value of bita" %I_C

The value of alfa = 0.98
The value of I_C = 9.8 mA using the value of alpha
The value of I_C = 10 mA using the value of bita

Example 4.5 : Page No 233

In [6]:
# Given data
V_BB= 10 # in V
V_CC= 10 # in V
V_BE= 0.7 # in V
R_B= 1 # in MΩ
R_B= R_B*10**6 # in Ω
R_C= 2 # in kΩ
R_C= R_C*10**3 # in Ω
bita= 300 
I_B= (V_BB-V_BE)/R_B # in A
I_C= bita*I_B # in A
V_CE= V_CC-I_C*R_C # in V
P_D= V_CE*I_C # in W
print "The value of I_B = %0.1f µA" %(I_B*10**6)
print "The value of I_C = %0.2f mA" %(I_C*10**3)
print "The value of V_CE = %0.2f volts" %V_CE
print "The value of P_D = %0.1f mW" %(P_D*10**3)

The value of I_B = 9.3 µA
The value of I_C = 2.79 mA
The value of V_CE = 4.42 volts
The value of P_D = 12.3 mW

Example 4.6 : Page No 241

In [8]:
# Given data
bita= 100 
V_BE= 0 # in V
V_BB= 15 # in V
R_B= 470 # in kΩ
R_B= R_B*10**3 # in Ω
V_CC= 15 # in V
R_C= 3.6 # in kΩ
R_C= R_C*10**3 # in Ω
I_B= (V_BB-V_BE)/R_B # in A
I_C= bita*I_B # in A
V_CE= V_CC-I_C*R_C # in V
I_E= I_C+I_B # in A
print "The base current = %0.1f µA" %(I_B*10**6)
print "The collector current = %0.2f mA" %(I_C*10**3)
print "The value of V_CE = %0.2f volts" %V_CE
print "The emitter current = %0.2f mA" %(I_E*10**3)

The base current = 31.9 µA
The collector current = 3.19 mA
The value of V_CE = 3.51 volts
The emitter current = 3.22 mA

Example 4.7 : Page No 242

In [10]:
# Given data
bita= 100 
V_BE= 0.7 # in V
V_BB= 15 # in V
R_B= 470 # in kΩ
R_B= R_B*10**3 # in Ω
V_CC= 15 # in V
R_C= 3.6 # in kΩ
R_C= R_C*10**3 # in Ω
I_B= (V_BB-V_BE)/R_B # in A
I_C= bita*I_B # in A
V_CE= V_CC-I_C*R_C # in V
print "The base current = %0.1f µA" %(I_B*10**6)
print "The collector current = %0.2f mA" %(I_C*10**3)
print "The value of V_CE = %0.2f volts" %V_CE

The base current = 30.4 µA
The collector current = 3.04 mA
The value of V_CE = 4.05 volts

Example 4.8 : Page No 255

In [1]:
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
# Given data
V_CC= 15 # in V
V_BE= 0.7 # in V
R_C= 1 # in kΩ
R_C= R_C*10**3 # in Ω
R_E= 2 # in kΩ
R_E= R_E*10**3 # in Ω
R1= 10 # in kΩ
R1= R1*10**3 # in Ω
R2= 5 # in kΩ
R2= R2*10**3 # in Ω
V_CE= np.arange(0,V_CC,0.1)
I_C= (V_CC-V_CE)/(R_C+R_E)*10**3 # in mA
plt.plot(V_CE,I_C) 
plt.plot([0,8.55],[2.15,2.15], '--',)
plt.plot([8.55,8.55],[0,2.15], '--')
plt.xlabel('V_CE in volts')
plt.ylabel('I_C in mA')
plt.title('DC load line') 
V_B= V_CC*R2/(R1+R2) # in V
I_E= (V_B-V_BE)/R_E # in A
I_C= I_E # in A
I_CQ= I_C # in A
V_CE= V_CC-I_C*(R_C+R_E) # in V
print "Q-point is : ",round(V_CE,2)," V",round(I_CQ*10**3,2)," mA"
print "DC load line shown in figure"

Q-point is :  8.55  V 2.15  mA
DC load line shown in figure

Example 4.9 : Page No 258

In [14]:
# Given data
V_BB= 1.8 # in V
V_BE= 0.7 # in V
R1= 10 # in kΩ
R2= 2.2 # in kΩ
R_E= 1 # in kΩ
bita= 200 
R= R1*R2/(R1+R2) # in kΩ
R=R*10**3 # in Ω
R_E= R_E*10**3 # in Ω
I_E= (V_BB-V_BE)/(R_E+R/bita) # in mA
print "The emitter current = %0.2f mA" %(I_E*10**3)
print "This is extremely close to 1.1 mA, the value we get with the simplified analysis."

The emitter current = 1.09 mA
This is extremely close to 1.1 mA, the value we get with the simplified analysis.

Example 4.10 : Page No 261

In [17]:
# Given data
V_CC= 10 # in V
V_BE= 0.7 # in V
V_CE= 5 # in V
bita= 100 
I_C= 5 # in mA
# Applying KVL to collector circuit, V_CC-V_CE-I_C*R_C =0
R_C= (V_CC-V_CE)/I_C # in kΩ
print "The value of R_C = %0.f kΩ" %R_C
I_B= I_C/bita # in mA
print "The value of I_B = %0.f µA" %(I_B*10**3)
# Applying KVL to base circuit, V_CC-I_B*R_B-V_BE= 0
R_B= (V_CC-V_BE)/I_B # in kΩ
print "The value of R_B = %0.f kΩ" %R_B

The value of R_C = 1 kΩ
The value of I_B = 50 µA
The value of R_B = 186 kΩ

Example 4.11 : Page No 261

In [2]:
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
# Given data
V_CC= 6 # in V
V_BE= 0.7 # in V
bita= 100 
R_C= 2 # in kΩ
R_C= R_C*10**3 # in Ω
R_B= 530 # in kΩ
R_B= R_B*10**3 # in Ω
R1= 10 # in kΩ
R1= R1*10**3 # in Ω
R2= 5 # in kΩ
R2= R2*10**3 # in Ω
V_CE= np.arange(0,V_CC,0.1) # in V
I_C= (V_CC-V_CE)/(R_C)*10**3 # in mA
plt.plot(V_CE,I_C) 
plt.xlabel('V_CE in volts') 
plt.ylabel('I_C in mA')
plt.plot([0,4],[1,1], '--',)
plt.plot([4,4],[0,1], '--')
plt.title('DC load line') 
I_B= (V_CC-V_BE)/R_B # in A
I_CQ= I_B*bita # in A
V_CE= V_CC-I_CQ*R_C # in V
print "Q-point is : (",round(V_CE,),"V",round(I_CQ*10**3),"mA )"
print "DC load line shown in figure"

Q-point is : ( 4.0 V 1.0 mA )
DC load line shown in figure

Example 4.12 : Page No 265

In [23]:
# Given data
V_CC= 12 # in V
V_BE= 0.7 # in V
bita= 100 
R_C= 10 # in kΩ
R_C=R_C*10**3 # in Ω
R_B= 100 # in Ω
R_B=R_B*10**3 # in Ω
I_BQ= (V_CC-V_BE)/((1+bita)*R_C+R_B) # in A
I_CQ= bita*I_BQ # in A
V_CEQ= V_CC-(I_CQ+I_BQ)*R_C # in volts
print "Q-Point value for the circuit =",round(V_CEQ,3),"V and",round(I_CQ*10**3,3),"mA"
# For dc load line when 
I_C=0 
V_CE= V_CC-(I_C+I_BQ)*R_C # in V
print "At I_C=0, the value of V_CE = %0.2f volts" %V_CE
# When
V_CE= 0 
I_C= (V_CC-I_BQ*R_C)/R_C # in A
print "At V_CE=0, the value of I_C = %0.1f mA" %(I_C*10**3)

Q-Point value for the circuit = 1.718 V and 1.018 mA
At I_C=0, the value of V_CE = 11.90 volts
At V_CE=0, the value of I_C = 1.2 mA

Example 4.13 : Page No 266

In [25]:
# Given data
V_BE= 0.7 # in V
V_CC= 15 # in V
V_CE= 5 # in V
I_C= 5 # in mA
I_C=I_C*10**-3 # in A
bita= 100 
I_B= I_C/bita # in A
# Applying KVL to collector circuit, V_CC= (I_C+I_B)*R_C+V_CE
R_C= (V_CC-V_CE)/(I_C+I_B) # in Ω
# Applying KVL to base circuit, V_CC= (I_C+I_B)*R_C+I_B*R_B+V_BE
R_B= (V_CC-V_BE-R_C*(I_C+I_B))/I_B # in Ω
print "The value of R_C = %0.2f kΩ" %(R_C*10**-3)
print "The value of R_B = %0.f kΩ" %(R_B*10**-3)

The value of R_C = 1.98 kΩ
The value of R_B = 86 kΩ

Example 4.14 : Page No 267

In [27]:
# Given data
I_B= 20*10**-6 # in A
V_CE= 7.3 # in V
V_BE= 0.6 # in V
V_E= 2.1 # in V
R_E= 0.68*10**3 # in Ω
R_C= 2.7*10**3 # in Ω
I_E= V_E/R_E # in A
I_C= I_E # in A (approx)
bita= round(I_C/I_B) 
V_CC= V_CE+I_C*R_C+I_E*R_E # in V
# From V_CC= I_B*R_B+V_BE+V_E
R_B= (V_CC-(V_BE+V_E))/I_B # in Ω
print "The value of bita = %0.f" %bita
print "The value of V_CC = %0.1f volts" %V_CC
print "The value of R_B = %0.f kΩ" %(R_B*10**-3)

# Note: In  the book, there is an error to calculate the value of R_B, hence the value of R_B in the book is wrong.

The value of bita = 154
The value of V_CC = 17.7 volts
The value of R_B = 752 kΩ

Example 4.15 : Page No 268

In [29]:
# Given data
V_CC = 18 # in V
bita = 90 
R_C = 2.2 * 10**3 # in ohm
R_E = 1.8*10**3 # in ohm
R_B = 510*10**3 # in ohm
I_B = V_CC/( (bita*(R_C+R_E))+R_B ) # in A
I_C = bita*I_B # in A
print "The value of I_C = %0.1f mA" %(I_C*10**3)
V_CE = I_B*R_B # in V
print "The value of V_CE = %0.1f V" %V_CE

The value of I_C = 1.9 mA
The value of V_CE = 10.6 V

Example 4.16 : Page No 269

In [31]:
# Given data
bita = 50 
V_CC = 12 # in V
V_BE = 0.7 # in V
R_B = 240 # in kohm
R_B = R_B*10**3 # in ohm
I_C = 2.35 * 10**-3 # in A
R_C = 2.2 # in kohm
R_C = R_C * 10**3 # in ohm
I_BQ = (V_CC - V_BE)/R_B # in A
print "The value of I_BQ = %0.2f µA" %(I_BQ*10**6)
I_CQ = bita*I_BQ # in A
print "The value of I_CQ = %0.2f mA" %(I_CQ*10**3)
V_CEQ = V_CC - (I_C*R_C) # in V
print "The value of V_CEQ = %0.2f V" %V_CEQ
V_B = V_BE # in V
print "The value of V_B = %0.1f V" %V_B
V_BC = V_B -V_CEQ # in V
print "The voltage = %0.2f V" %V_BC

# Note: In the book, there is a calculation error to evaluating the value of V_CEQ. So the answer in the book is wrong

The value of I_BQ = 47.08 µA
The value of I_CQ = 2.35 mA
The value of V_CEQ = 6.83 V
The value of V_B = 0.7 V
The voltage = -6.13 V

Example 4.17 : Page No 269

In [33]:
# Given data
V_CC = 18 # in V
V_BE = 0.7 # in V
R_C = 3.3 # in kohm
R_C = R_C * 10**3 # in ohm
R_B = 210 # in kohm
R_B  = R_B * 10**3 # in ohm
bita = 75 
R_C = 3.3 # in kohm
R_C = R_C * 10**3 # in ohm
R_E = 510 # in ohm
I_B = (V_CC-V_BE)/( R_C+R_B+bita*(R_C+R_E) ) # A
print "The value of I_B = %0.f µA" %round(I_B*10**6)
I_C = bita*I_B # in A
print "The value of I_C = %0.1f mA" %(I_C*10**3)
V_C = V_CC - (I_C*R_C) # in V
print "The voltage = %0.2f V" %V_C

The value of I_B = 35 µA
The value of I_C = 2.6 mA
The voltage = 9.42 V

Example 4.18 : Page No 271

In [35]:
# Given data
V_BE = 0.7 # in V
I_B = 40 * 10**-6 # in A
V_CC = 20 # in V (From the load line)
print "The voltage = %0.f V" %V_CC
I_C = 8 # in mA
R_C = V_CC/I_C # in kohm
print "The resistance = %0.1f kohm" %R_C
R_B = (V_CC - V_BE)/I_B # in ohm
print "The resistance = %0.1f kohm" %(R_B*10**-3)

The voltage = 20 V
The resistance = 2.5 kohm
The resistance = 482.5 kohm

Example 4.19 : Page No 271

In [37]:
# Given data
R1 = 47 # in kohm
R1=  R1*10**3 # in ohm
R2 = 10 # in kohm
R2= R2*10**3 # in ohm
R_E = 1.1 # in kohm
R_E = R_E * 10**3 # in ohm
R_C = 2.4 # in kohm
R_C = R_C * 10**3 # in ohm
V_CC = -18 # in V
V_B = (R2*V_CC)/(R1+R2) # in V
V_BE = -0.7 # in V
V_E = V_B - V_BE # in V
I_E = abs(V_E)/R_E # in A
V_CE = V_CC + (I_E)*(R_C+R_E) # in V
print "The value of V_B = %0.2f volts" %V_B
print "The value of I_E = %0.2f mA" %(I_E*10**3)
print "The value of V_CE = %0.2f V" %V_CE

The value of V_B = -3.16 volts
The value of I_E = 2.23 mA
The value of V_CE = -10.18 V

Example 4.20 : Page No 273

In [40]:
# Given data
V_BE = 0.8 # in V
V_CE = 0.2 # in V
V1 = 5 # in V
R_B = 50 # in kohm
R_B= R_B*10**3 # in ohm
R_C = 3 # in K ohm
R_C = R_C * 10**3 # in ohm
bita = 100 
R_E = 2 # in kohm
R_E= R_E*10**3 # in ohm
I_B = (V1-V_BE)/(R_B+(1+bita)*R_E) # in A
print "The value of I_B = %0.2f µA" %(I_B*10**6)
V_CC = 10 # in V
I_Csat = (V_CC - V_CE - (I_B*R_E))/(R_C+R_E) #in A
print "The value of I_C(sat) = %0.3f mA" %(I_Csat*10**3)
I_Bmin = I_Csat /bita # in A
print "The minimum value of I_B = %0.3f µA" %(I_Bmin*10**6)

# Note: There is calculation error to evaluate the value of I_Csat in the book, so the answer in the book is wrong

The value of I_B = 16.67 µA
The value of I_C(sat) = 1.953 mA
The minimum value of I_B = 19.533 µA

Example 4.21 : Page No 274

In [45]:
# Given data
R1 = 5 # in kohm
R1= R1*10**3 # in ohm
R2 = 5 # in kohm
R2= R2*10**3 # in ohm
R_B = R1*R2/(R1+R2) # in ohm
R_E = 1 # in kohm
R_E = R_E * 10**3 # in ohm
V_EE = 3 # in V
V_Th = (R2*V_EE)/(R1+R2) # in V
V_BE = 0.7 # in V
bita = 44 
I_B = (V_EE - V_BE - V_Th)/( ((1+bita)*R_E)+R_B) # in A
I_BQ = I_B # in A
print "The value of I_BQ = %0.2f µA" %(I_BQ*10**6)
I_C = bita*I_BQ # in A
print "The value of I_C = %0.2f mA" %(I_C*10**3)
I_E = (1+bita)*I_B # in A
print "The value of I_E = %0.3f mA" %(I_E*10**3)
V_EC = (I_E*R_E)-V_EE # in V
print "The value of V_EC = %0.3f V" %V_EC
print "Q-point = (",round(V_EC,3),"V",round(I_C*10**3,2),"mA )"

The value of I_BQ = 16.84 µA
The value of I_C = 0.74 mA
The value of I_E = 0.758 mA
The value of V_EC = -2.242 V
Q-point = ( -2.242 V 0.74 mA )

Example 4.22 : Page No 275

In [49]:
# Given data
V_BE = 0.7 # in V
V_BB = 5 # in V
R_B = 100 # in kohm
R_B = R_B * 10**3 # in ohm
R_E = 2 # in kohm
R_E = R_E * 10**3 # in ohm
bita = 100 
I_B = (V_BB-V_BE)/( R_B+((1+bita)*R_E) ) # in A
print "The value of I_B = %0.3f mA" %(I_B*10**3)
V_B = V_BB-(I_B*10**-3*R_B) # in V
I_C = bita*I_B # in A
print "The value of I_C = %0.1f mA" %(I_C*10**3)
V_CC = 10 # in V
V_C = V_CC-(I_C*R_E) # in V
print "The voltage = %0.1f V" %V_C
print "Transistor is in active region is valid"

The value of I_B = 0.014 mA
The value of I_C = 1.4 mA
The voltage = 7.2 V
Transistor is in active region is valid

Example 4.23 : Page No 276

In [51]:
# Given data
V_CC = 20 # in V
V_BE = 0.7 # in V
R_B = 430 # in kohm
R_B = 430 * 10**3 # in ohm
bita = 50 
R_E = 1 # in kohm
R_E = R_E * 10**3 # in ohm
R_C = 2 # in kohm
R_C = R_C * 10**3 # in ohm
I_B = (V_CC - V_BE)/(R_B +(1+bita)*R_E) # in A
print "The base current = %0.1f µA" %(I_B*10**6)
I_C = bita*I_B # in A
print "The collector current = %0.2f mA" %(I_C*10**3)
V_CE = V_CC - I_C*(R_C+R_E) # in V
print "The value of V_CE = %0.2f V" %V_CE
V_C = V_CC - (I_C*R_C) # in V
print "The value of V_C = %0.2f V" %V_C
V_E = V_C - V_CE # in V
print "The value of V_E = %0.2f V" %V_E
V_B = V_BE+V_E # in V
print "The value of V_B = %0.2f V" %V_B
V_BC = V_B-V_C # in V
print "The value of V_BC = %0.2f V" %V_BC

The base current = 40.1 µA
The collector current = 2.01 mA
The value of V_CE = 13.98 V
The value of V_C = 15.99 V
The value of V_E = 2.01 V
The value of V_B = 2.71 V
The value of V_BC = -13.28 V

Example 4.24 : Page No 277

In [52]:
# Given data
V_CC = 20 # in V
V_BE = 0.7 # in V
R_B = 680 # in kohm
R_B = R_B * 10**3 # in ohm
R_C = 4.7 # in kohm
R_C = R_C * 10**3 # in ohm
bita = 120 
I_B = (V_CC - V_BE)/(R_B+bita*R_C) # in A
I_CQ = bita*I_B # in A
print "The value of I_CQ = %0.2f mA" %(I_CQ*10**3)
V_CEQ = V_CC - (I_CQ*R_C) # in V
print "The value of V_CEQ = %0.2f V" %V_CEQ
V_B = V_BE # in V
V_C = 11.26 # in V
V_E = 0 # in V
print "The value of V_E = %0.f V" %V_E
V_BC = V_B - V_C # in V
print "The value of V_BC = %0.2f V" %V_BC

The value of I_CQ = 1.86 mA
The value of V_CEQ = 11.25 V
The value of V_E = 0 V
The value of V_BC = -10.56 V

Example 4.25 : Page No 278

In [53]:
# Given data
V_CC = 16 # in V
V_BE = 0.7 # in V
R_B = 470 # in kohm
R_B= R_B*10**3 # in ohm
bita = 120 
R_C = 3.6 # in kohm
R_C= R_C*10**3 # in ohm
R_E = 0.51 # in kohm
R_E= R_E*10**3 # in ohm
I_B = (V_CC - V_BE)/(R_B+bita*(R_C+R_E)) # in A
print "The base current = %0.2f µA" %(I_B*10**6)
I_C  = bita*I_B # in A
print "The collector current = %0.2f mA" %(I_C*10**3)
V_C = V_CC - I_C*R_C # in V
print "The collector voltage = %0.2f V" %V_C

The base current = 15.88 µA
The collector current = 1.91 mA
The collector voltage = 9.14 V

Example 4.26 : Page No 279

In [55]:
# Given data
V_CC = 10 # in V
V_BE = 0.7 # in V
R_B = 250 # in kohm
R_B= R_B*10**3 # in ohm
bita = 90 
R_C = 4.7 # in kohm
R_C= R_C*10**3 # in ohm
R_E = 1.2 # in kohm
R_E= R_E*10**3 # in ohm
I_BQ = (V_CC - V_BE)/(R_B + bita*(R_C+R_E)) # in A
print "The base current at Q-point = %0.2f µA" %(I_BQ*10**6)
I_CQ = bita*I_BQ # in A
print "The collector current at Q-point = %0.2f mA" %(I_CQ*10**3)
V_CEQ = V_CC - (I_CQ*(R_C+R_E)) # in V
print "Collector emitter voltage at Q point = %0.3f V" %V_CEQ

The base current at Q-point = 11.91 µA
The collector current at Q-point = 1.07 mA
Collector emitter voltage at Q point = 3.677 V

Example 4.27 : Page No 281

In [58]:
# Given data
V_CC = 12 # in V
V_BE = 0.7 # in V
R_B = 150 # in kohm
R_B= R_B*10**3 # in ohm
bita = 180 
R_C = 4.7 # in kohm
R_C= R_C*10**3 # in ohm
R_E = 3.3 # in kohm
R_E= R_E*10**3 # in ohm
I_B = (V_CC-V_BE)/(R_B + bita*(R_C+R_E)) # in A
print "The base current = %0.2f µA" %(I_B*10**6)

The base current = 7.11 µA

Example 4.28 : Page No 282

In [62]:
# Given data
V_B = 4 # in V
V_BE = 0.7 # in V
R_E = 1.2 # in kohm
R_E= R_E*10**3 # in ohm
V_E = V_B-V_BE # in V
R_C = 2.2 # in kohm
R_C= R_C*10**3 # in ohm
R_B= 330 # in kohm
R_B= R_B*10**3 # in ohm
bita = 180 
I_B = 7.11 * 10**-6 # in A
V_CC = 18 # in V
print "Part (a)"
print "The value of V_E = %0.1f V" %V_E
I_C = V_E/R_E # in A
print "Part (b)"
print "The value of I_C = %0.2f mA" %(I_C*10**3)
V_C =V_CC - (I_C*R_C) # in V
print "Part (c)"
print "The value of V_C = %0.2f V" %V_C
V_CE = V_C-V_E # in V
print "Part (d)"
print "The value of V_CE = %0.2f V" %V_CE
I_B = (V_CC - (I_C*R_C) - V_BE - V_E)/R_B # in A
print "Part (e)"
print "Base current = %0.2f µA" %(I_B*10**6)
bita = I_C/I_B 
print "Part (f)"
print "Current gain = %0.f" %bita

Part (a)
The value of V_E = 3.3 V
Part (b)
The value of I_C = 2.75 mA
Part (c)
The value of V_C = 11.95 V
Part (d)
The value of V_CE = 8.65 V
Part (e)
Base current = 24.09 µA
Part (f)
Current gain = 114

Example 4.29 : Page No 284

In [63]:
# Given data
I_E = 10 # in mA
I_C = 9.95 # in mA
I_B = I_E-I_C # in mA
print "The base current = %0.2f mA" %I_B

The base current = 0.05 mA

Example 4.30 : Page No 284

In [64]:
# Given data
I_C = 10 # in mA
I_B = 0.1 # in mA
bita = I_C/I_B 
print "The current gain = %0.f" %bita

The current gain = 100

Example 4.31 : Page No 284

In [66]:
# Given data
V_BE = 0.7 # in V
V_BB = 10 # in V
R_B = 470 # in kohm
R_B = R_B * 10**3 # in ohm
I_B = (V_BB-V_BE)/R_B # in A
print "The base current = %0.2f µA" %(I_B*10**6)

The base current = 19.79 µA

Example 4.32 : Page No 285

In [69]:
# Given data
V_BB = 10 # in V
V_BE = 0 # in V
R_B = 470 # in kohm
R_B = R_B * 10**3 # in ohm 
I_B = (V_BB - V_BE)/R_B # in A
bita = 200 
I_C = bita*I_B # in A
V_CC = 10 # in V
R_C = 820 # in ohm
V_CE = V_CC  - (I_C*R_C) # in V
print "Part (a) : For ideal approximation"
print "The collector emitter voltage = %0.2f V" %V_CE
P_D = V_CE * I_C # in W
print "Power dissipation = %0.2f mW" %(P_D*10**3)
print "Part (b) : For second approximation"
V_BE = 0.7 # in V
I_B = (V_BB-V_BE)/R_B # in A
I_C = bita*I_B # in A
V_CE = V_CC - (I_C*R_C) # in V
print "The collector emitter voltage = %0.2f V" %V_CE
P_D = V_CE * I_C # in W
print "Power dissipation = %0.2f mW" %(P_D*10**3)

Part (a) : For ideal approximation
The collector emitter voltage = 6.51 V
Power dissipation = 27.70 mW
Part (b) : For second approximation
The collector emitter voltage = 6.75 V
Power dissipation = 26.73 mW

Example 4.33 : Page No 286

In [71]:
# Given data
V_BE = 0 # in V
V_BB = 12 # in V
R_B = 680 # in kohm
R_B = R_B * 10**3 # in ohm
I_B = (V_BB-V_BE)/R_B # in A
beta_dc = 175 
I_C = beta_dc*I_B # in A
V_CC = 12 # in V
R_C = 1.5 # in kohm
R_C = R_C * 10**3 # in ohm
V_CE = V_CC - (I_C*R_C) # in V
print "Part (a) For ideal approximation"
print "The collector emitter voltage = %0.2f V" %V_CE
P_D = V_CE * I_C # in mW
print "Transistor power = %0.2f mW" %(P_D*10**3)
print "Part (b) For second approximation"
V_BE1 = 0.7 # in V
I_B = (V_BB-V_BE1)/R_B # in A
I_C = beta_dc * I_B # in A
V_CE = V_CC - (I_C*R_C) # in V
print "Collector emitter voltage = %0.2f V" %V_CE
P_D = V_CE * I_C # in W
print "Power dissipation = %0.2f mW" %(P_D*10**3)

Part (a) For ideal approximation
The collector emitter voltage = 7.37 V
Transistor power = 22.75 mW
Part (b) For second approximation
Collector emitter voltage = 7.64 V
Power dissipation = 22.21 mW

Example 4.34 : Page No 288

In [3]:
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
# Given data
V_CC = 20 # in V
R_C = 3.3 # in kohm
R_C = R_C * 10**3 # in ohm
I_C = V_CC/R_C # in A
print "Collector current = %0.2f mA" %(I_C*10**3)
V_CE = V_CC # in V
print "Collector emitter voltage = %0.f V" %V_CE
V_CE=np.arange(0,20,0.1)  # in V
I_C= (V_CC-V_CE)/(R_C*10**-3) # in mA
plt.plot(V_CE,I_C) 
plt.xlabel('V_CE in volts')
plt.ylabel('I_C in mA')
plt.title('DC load line')
print "DC load line shown in figure"

Collector current = 6.06 mA
Collector emitter voltage = 20 V
DC load line shown in figure

Example 4.35 : Page No 289

In [73]:
# Given data
V_BB = 10 # in V
V_BE = 0.7 # in V
R_B = 1 # in kohm
R_B = 1 * 10**6 # in ohm
I_B = (V_BB-V_BE)/R_B # in A
print "The base current = %0.1f µA" %(I_B*10**6)
beta_dc = 200 
I_C = beta_dc * I_B # in A
print "The collector current = %0.2f mA" %(I_C*10**3)
V_CC = 20 # in V
R_C = 3.3 # in kohm
R_C = R_C * 10**3 # in ohm
V_CE = V_CC - I_C*R_C # in V
print "The collector voltage = %0.3f V" %V_CE

The base current = 9.3 µA
The collector current = 1.86 mA
The collector voltage = 13.862 V

Example 4.36 : Page No 290

In [75]:
# Given data
V_BB = 5 # in V
V_BE = 0.7 # in V
R_B = 680 # in kohm
R_B = 680*10**3 # in ohm
I_B = (V_BB-V_BE)/R_B # in A
print "The base current = %0.2f µA" %(I_B*10**6)
beta_dc= 150 
I_C = beta_dc * I_B # in A
print "The collector current = %0.2f mA" %(I_C*10**3)
V_CC = 5 # in V
R_C = 470 # in ohm
V_CE = V_CC-(I_C*R_C) # in V
print "Voltage between collector and ground = %0.2f V" %V_CE

The base current = 6.32 µA
The collector current = 0.95 mA
Voltage between collector and ground = 4.55 V

Example 4.37 : Page No 291

In [77]:
# Given data
V_BB = 2.5 # in V
V_BE = 0.7 # in V
V_E = V_BB-V_BE # in V
print "The emitter voltage = %0.1f V" %V_E
R_E = 1.8 # in kohm
R_E = R_E * 10**3 # in ohm
I_E = V_E/R_E # in A
I_C= I_E # in A
V_CC = 20 # in V
R_C = 10 # in kohm
R_C = R_C * 10**3 # in ohm
V_C = V_CC-(I_C*R_C) # in V
print "The collector voltage = %0.f V" %V_C

The emitter voltage = 1.8 V
The collector voltage = 10 V

Example 4.38 : Page No 292

In [78]:
# Given data
V_CC = 25 # in V
R2 = 2.2 # in kohm
R1 = 10 # in kohm
V_BB = (V_CC * R2)/(R1+R2) # in V
V_BE = 0.7 # in V
V_E = V_BB - V_BE # in V
print "The emitter voltage = %0.1f V" %V_E
R_E = 1 # in kohm
R_E = R_E * 10**3 # in ohm
I_E = V_E/R_E # in A
I_C= I_E # in A
V_CC = 25 # in V
R_C = 3.6 # in kohm
R_C = R_C * 10**3 # in ohm
V_C = V_CC - (I_C*R_C) # in V
print "Collector voltage = %0.2f V" %V_C

The emitter voltage = 3.8 V
Collector voltage = 11.29 V

Example 4.39 : Page No 293

In [80]:
# Given data
V_BB = 4.50 # in V
V_E = 3.8 # in V
V_C = 11.32 # in V
I_C = 3.8 # in mA
I_C=I_C*10**-3 # in A
V_BE = 0.7 # in V
R1 = 10 # in kohm
R2 = 2.2 # in kohm
R_B = (R1*R2)/(R1+R2) # in kohm
R_B = R_B * 10**3 # in ohm
I_B = (V_BB-V_BE)/R_B # in A
print "The base current = %0.2f mA" %(I_B*10**3)
V_CE = V_C-V_E # in V
print "Collector emitter voltage = %0.2f V" %V_CE
print "Thus the Q-point is :",round(V_CE,2),"V",round(I_B*10**3,2),"mA"

# Note: There is calculation error to evaluate the value of I_B. So the answer in the book is wrong.

The base current = 2.11 mA
Collector emitter voltage = 7.52 V
Thus the Q-point is : 7.52 V 2.11 mA