Chapter 4 Bipolar Junction Transistor

Exa 4.1 page No 211

In [2]:
from __future__ import division
# Given data
Alpha = 0.98
Vo = 4.9# V
R_L = 5# k ohm
R_L = R_L * 10**3# ohm
I_C = Vo/R_L# A
I_C = I_C * 10**3# mA
# Alpha = I_C/I_E
I_E = I_C/Alpha# mA
# The base current 
I_B = I_E-I_C# mA
print "The base current = %.2f mA"%I_B
The base current = 0.02 mA

Exa 4.2 page No 211

In [3]:
from __future__ import division
# Given data
Alpha = 0.95
I_CBO = 5* 10**-3# mA
I_E = 3# mA
# The collector current 
I_C = (Alpha*I_E)+I_CBO# mA
print "The collector current = %.2f mA"%I_C
# The base current 
I_B = I_E-I_C# mA
print "The base current = %.2f mA"%I_B
The collector current = 2.85 mA
The base current = 0.15 mA

Exa 4.3 page No 211

In [5]:
from __future__ import division
# Given data
Alpha = 0.9
# For alpha = 0.9, the value of Beta 
Beta = Alpha/(1-Alpha)
print "For alpha = 0.9, the value of Beta = %.2f"%Beta
Alpha = 0.99
# For alpha = 0.99, the value of Beta 
Beta = Alpha/(1-Alpha)
print "For alpha = 0.99, the value of Beta = %.2f"%Beta
Alpha = 0.98
# For alpha = 0.98, the value of Beta 
Beta = Alpha/(1-Alpha)
print "For alpha = 0.98, the value of Beta = %.2f"%Beta
For alpha = 0.9, the value of Beta = 9.00
For alpha = 0.99, the value of Beta = 99.00
For alpha = 0.98, the value of Beta = 49.00

Exa 4.4 page No 212

In [6]:
from __future__ import division
# Given data
R_C = 2# k ohm
R_C = R_C * 10**3# ohm
V = 2# V
# The collector current
I_C = V/R_C# A
Beta = 50
# The base current 
I_B = I_C/Beta# A
I_B= I_B*10**6# µA
print "The base current = %.2f µA"%I_B
The base current = 20.00 µA

Exa 4.5 page No 212

In [9]:
from __future__ import division
# Given data
Beta = 49
I_E = 12# mA
I_B = 240# µA
I_B = I_B * 10**-3# mA
Alpha = Beta/(1+Beta)
#Using alpha rating, the value of collector current,
I_C = Alpha*I_E# mA
print "The value of alpha is : ",Alpha
print "Using alpha rating, the value of I_C = %.2f mA"%I_C
#Using beta rating, the value of collector current,
I_C = Beta*I_B# mA
print "Using bita rating, the value of I_C = %.2f mA"%I_C
The value of alpha is :  0.98
Using alpha rating, the value of I_C = 11.76 mA
Using bita rating, the value of I_C = 11.76 mA

Exa 4.6 page No 212

In [12]:
from __future__ import division
# Given data
Alpha = 0.975
Beta = Alpha/(1-Alpha)
# The value of beta,
print "The value of Beta : ",Beta
Beta = 200
# The value of alpha,
Alpha = Beta/(1+Beta)
print "The value of Alpha : %.4f"%Alpha
The value of Beta :  39.0
The value of Alpha : 0.9950

Exa 4.7 page No 213

In [13]:
from __future__ import division
# Given data
Alpha = 0.98
I_CO = 5# µA
I_CO = I_CO * 10**-3# mA
I_B = 100# µA
I_B = I_B * 10**-3# mA
Beta = Alpha/(1-Alpha)
# The collector current 
I_C = Beta*I_B + (1+Beta)*I_CO# mA
print "The collector current = %.2f mA"%I_C
# The emitter current 
I_E = I_C+I_B# mA
print "The emitter current = %.2f mA"% I_E
The collector current = 5.15 mA
The emitter current = 5.25 mA

Exa 4.8 page No 213

In [14]:
from __future__ import division
# Given data
I_CBO = 10# µA
I_CBO = I_CBO * 10**-6# A
Beta = 50
h_FE = Beta
I_B = 0.25# mA
I_B = I_B * 10**-3# A
# The collector current 
I_C = (Beta*I_B) + ((1+Beta)*I_CBO)##in A
I_C = I_C * 10**3# mA
print "The collector current = %.2f mA"%I_C
T2 = 50# degree C
T1 = 25# degree C
I_CBOat25 = 10# µA
I_CBOat50 = I_CBOat25 * (2**((T2-T1)/10))# µA
I_CBOat50 = I_CBOat50 * 10**-6# A
#The new collector current 
I_C = (Beta*I_B) + ((1+Beta)*I_CBOat50)# A
I_C = I_C * 10**3# mA
print "The new collector current = %.2f mA"%I_C
The collector current = 13.01 mA
The new collector current = 15.38 mA

Exa 4.9 page No 215

In [15]:
from __future__ import division
# Given data
V_BE = 0.6# V
V_CC = 10# V
Beta = 60
R_B = 20#  k ohm
R_B = R_B * 10**3# ohm
R_C = 300# ohm
# V_CC - (I_B*R_B)-V_BE = 0
I_B = (V_CC-V_BE)/R_B# A
# The collector current 
I_C = Beta*I_B# A
I_C= I_C*10**3# mA
print "The collector current = %.2f mA"%I_C
# V_CC - (I_C*R_C) -V_CE = 0
# The collector emitter voltage 
V_CE = V_CC - (I_C*10**-3*R_C)# V
print "The collector emitter voltage = %.2f V"%V_CE
The collector current = 28.20 mA
The collector emitter voltage = 1.54 V

Exa 4.10 page No 216

In [17]:
from __future__ import division
%matplotlib inline 
from matplotlib.pyplot import plot,xlabel,ylabel,show,title
# Given data
V_BE = 0.6# V
V_CE = 0# V
R_C = 2# k ohm
R_C = R_C * 10**3# ohm
V_CC = 9# V
Beta= 2
R_B = 8# k ohm
R_B = R_B * 10**3# ohm
# V_CC - (I_C*R_C) - V_CE = 0
I_C = V_CC/R_C# A
I_C= I_C*10**3# mA
V_CE = V_CC# V
plot([V_CE,0],[0,I_C])
xlabel("V_CE in volts")
ylabel("I_C in mA")
title("DC load line")
show()
I_C= I_C*10**-3# A
I_B = (V_CC-V_BE)/R_B# A
#Collector emitter voltage, V_CE = V_CC - (I_C*R_C) = V_CC - (Beta*10**-3*I_B*R_C)
V_CE = V_CC - (Beta*I_B*R_C)# V
# Collector current,
I_C =Beta*I_B# A
I_C= I_C*10**3# mA
print "The operating point is : (%.2fV,%.2fmA)"%(V_CE,I_C)
The operating point is : (4.80V,2.10mA)

Exa 4.11 page No 216

In [19]:
from __future__ import division
# Given data
V_CC = 12# V
Beta = 100
R = 600# k ohm
R = R * 10**3# ohm
R1 = 600# ohm
#Applying KVL to input side, V_CC -(I_E*R1) - (I_B*R) - V_BE = 0
I_B = V_CC/(R+(Beta*R1))# A
I_C = Beta*I_B# A
I_E = I_C+I_B# A
# Applying KVL to output side
V_CE = V_CC - (I_E*R1)# V
I_E= I_E*10**3# mA
print "The emitter current = %.2f mA"%I_E
print "The collector voltage = %.2f V"%V_CE
The emitter current = 1.84 mA
The collector voltage = 10.90 V

Exa 4.12 page No 217

In [20]:
from __future__ import division
# Given data
I_C = 6.4# mA
I_C = I_C * 10**-3# A
V_CE = 8.4# V
Beta = 80
V_CC= 10# V
R = 250# ohm
R_E = 500# ohm
I_B = I_C/Beta# A
#Applying KVL to the input side, V_CC - (I_B*R_B) - V_BE - (I_E*R1) = 0 or
R_B = (V_CC-(Beta+1)*R_E*I_B)/I_B
R_B = R_B * 10**-3# k ohm
print "The value of R_B = %.2f k ohm"%R_B
The value of R_B = 84.50 k ohm

Exa 4.13 page No 218

In [21]:
from __future__ import division
# Given data
Beta = 44
R_L = 1.51 * 10**3# ohm
R_E = 270# ohm
V_CC = 4.5# V
R2 = 2.7 * 10**3# ohm
R1 = 27 * 10**3# ohm
R_C = 1.5 * 10**3# ohm
Vth = V_CC * (R2/(R1+R2))# V
Rth = (R1*R2)/(R1+R2)# ohm
# Applying KVL to input circuit, Vth - (I_B*Rth) - V_BE - (I_E*R_E) = 0 or
I_C = (Vth*Beta)/(Rth + (Beta*R_E))# A     (On putting I_C= Beta*I_B and V_BE=0)
#Applying KVL to output side, V_CC - (I_C*R_C) - V_CE - (I_E*R_E) = 0
V_CE = V_CC - (I_C*(R_C+R_E))# V
S = (Beta+1)*( (1+(Rth/R_E))/(Beta+1+(Rth/R_E)) )
I_C= I_C*10**3# mA
print "The stability factor : %.2f"%S
print "The quiescent points : (%.2fmA,%.2fV)"%(I_C,V_CE)

# Note: In the book the calculated value of V_CE is correct as well as coding output, but at last they print wrong value (4.01 V)
The stability factor : 8.39
The quiescent points : (1.26mA,2.28V)

Exa 4.15 page No 220

In [22]:
from __future__ import division
# Given data
S = 12
V_CC = 16# V
R_C = 1.5 * 10**3# ohm
V_CE = 8# V
V_BE= 0.2# V
I_C = 4# mA
I_C = I_C * 10**-3# A
Beta = 50
# Vth = V_CC*(R2/(R1+R2)) and Rth = (R1*R2)/(R1+R2)
#Applying KVL to input side, Vth - I_B*Rth - V_BE - I_E*R_E = 0            (i)
I_B = I_C/Beta# A
I_E = I_B*(1+Beta)# A
# Applying KVL to output section,
R_E = (V_CC - (I_C*R_C) - V_CE)/I_E# ohm
#S = ((Rth+R3)*(1+Beta))/(Rth + ((1+Beta)*R_E))                (ii)
Rth= R_E*(1+Beta-S-Beta*S)/(S-1-Beta)# ohm
V_BN= V_BE+I_E*R_E##in V
Vth= V_BN+I_B*Rth# V
# Vth= V_CC*R2/(R1+R2) and Rth= R2*R1/(R1+R2), So Vth= V_CC*Rth/R1
R1= V_CC*Rth/Vth# ohm
R2= R1*Rth/(R1-Rth)# ohm
R1= round(R1*10**-3)# k ohm
R2= R2*10**-3# k ohm
R_E= R_E*10**-3# k ohm
print "The value of R1 = %.2f kΩ"%R1
print "The value of R2 = %.2f kΩ"%R2
print "The value of R_E = %.2f kΩ"%R_E
The value of R1 = 41.00 kΩ
The value of R2 = 8.52 kΩ
The value of R_E = 0.49 kΩ

Exa 4.19 page No 224

In [25]:
from __future__ import division
# Given data
h_FE = 100
R = 50# k ohm
R = R * 10**3# ohm
R1 = 3# k ohm
R1 = R1 * 10**3# ohm
V1 = 10# V
V2 = 5# V
V_BE = 0.8# V
V_CE = 0.2# V
#Applying KVL = %.2f input side, V2 - (R*I_B)-V_BE = 0
I_B = (V2-V_BE)/R# A
I_B = I_B * 10**3# mA
#Applying KVL to output side, V1 - (R1*I_C) - V_CE = 0
I_C = (V1-V_CE)/R1# A
I_C = I_C * 10**3# mA
I_Bmin = I_C/h_FE# mA
if I_B>I_Bmin:
    print "As the value of I_B (%.2f mA)"%I_B,"is greater than the value of I_Bmin (%.2f mA)"%(I_Bmin)
    print "So the transistor will be in saturation region."
As the value of I_B (0.08 mA) is greater than the value of I_Bmin (0.03 mA)
So the transistor will be in saturation region.

Exa 4.20 page No 225

In [34]:
from __future__ import division
from numpy import mat
from numpy.linalg import solve
# Given data
V_BE= 0.8# V
V_CE= 0.2# V
Beta = 100
h_FE = Beta
V1 = 3# V
V2 = -10# V
R_B = 7*10**3# ohm
R_E = 3*10**3# ohm
R_C = 500# ohm
#Applying KVL to input side, V1 - (I_B+I_C)*Rc - V_BEsat - (R1*I_B) = 0 or I_B*(R_B+R_C)+I_C*R_C= V1-V_BE              (i)
# Applying KVL to output side, V2+I_C*R_E+V_CE+R_C*(I_B+I_C) = 0 or   I_B*R_C+I_C*(R_C+R_E)= -V2-V_CE              (ii)
A= mat([[(R_B+R_C), R_C],[R_C, (R_C+R_E)]])
B= mat([[V1-V_BE], [-V2-V_CE]])
C=solve(A,B) # Solving eq(i) and (ii) by matrix method

I_B= C[0,0]# A
I_C= C[1,0]# A
I_B=I_B*10**3# mA
I_C=I_C*10**3# mA
I_Bmin= I_C/h_FE##in mA
if I_B>I_Bmin:
    print "Part (a) :"
    print "As the value of I_B (%.2f"%(I_B),"mA) is greater than the value of I_Bmin (%.2f"%(I_Bmin),"mA),"
    print "So the transistor will be in saturation region."

Vo= -V_CE-(I_B+I_C)*10**-3*R_C# V
print "Part (b) : The value of Vo = %.2f volts"%Vo
V_BEactive= 0.7# V
V_BC= -0.5
# V_BN= (1+Beta)*(V1-V_BEactive)*R_C/(R_B+(1+Beta)*R_C)       (iii)
# V_CN= Beta*R_E*(V1-V_BEactive)/(R_B+(1+Beta)*R_C)                (iv)
# and V_BC= V_BN-V_CN, so from eq(iii) and (iv)
R_C= (V_BC*R_B+Beta*R_E*(V1-V_BEactive))/((1+Beta)*(V1-V_BEactive)-V_BC*(1+Beta))# ohm
print "Part (c) :The value of R_C = %.2f ohm"%R_C

# Note : 1  = %.2f the book, the calculated value of I_B i.e 0.14mA is wrong.
# Note: 2   = %.2f the book the calculated value of R_C i.e 819 Ω is wrong. we can easily check that on putting 0.819 kΩ = %.2f the last step.
Part (a) :
As the value of I_B (0.11 mA) is greater than the value of I_Bmin (0.03 mA),
So the transistor will be in saturation region.
Part (b) : The value of Vo = -1.65 volts
Part (c) :The value of R_C = 2427.51 ohm

Exa 4.21 page No 226

In [36]:
from __future__ import division
%matplotlib inline 
from matplotlib.pyplot import plot,xlabel,ylabel,show,title

# Given data
V_CC = 10# V
R_C = 10# k ohm
R_C = R_C * 10**3# ohm
R_B = 100*10**3# ohm
Beta_dc = 100
V_BE = 0.7# V
# Applying KVL to input section, V_CC = (I_E*R_E) + (I_B*R_B) + V_BE or
I_C = (V_CC-V_BE)/( R_C + (R_B/Beta_dc) )# A
V_CE =V_CC - (I_C*R_C)# V
I_C=I_C*10**3# mA
print "DC load line shown in figure."
print "Q-points : %.2f"%(V_CE),"V, %.2f"%(I_C)+"mA"
I_Csat =V_CC/R_C*10**3# mA
V_CEcutoff = V_CC# V
plot([V_CEcutoff ,0],[0,I_Csat])
xlabel("V_CE in volts")
ylabel("I_C in mA")
title("DC load line")
show()
DC load line shown in figure.
Q-points : 1.55 V, 0.85mA

Exa 4.23 page No 228

In [37]:
from __future__ import division
# Given data
V_CC = 24# V
V_CE= 12# V
Rc = 5# k ohm
Rc = Rc * 10**3# ohm
Beta = 100
I_C = 2# mA
I_C = I_C * 10**-3# A
#Applying KVL to the output section, V_CC = (I_C*R_C) + V_CE + (I_E*R_E) or
# (V_CC-V_CE)/I_C =Rc + R_E*(1+(1/Beta))   (on putting I_E= I_C+I_B and I_B = I_C/Beta)
R_E = (((V_CC-V_CE)/I_C)-Rc)*(1/(1+(1/Beta)))# ohm
print "The value of R_E = %.2f ohm"%R_E
The value of R_E = 990.10 ohm

Exa 4.24 page No 229

In [38]:
from __future__ import division
# Given data
S = 3
Beta = 100
I_C = 2# mA
I_C = I_C * 10**-3# A
R_E = 990# ohm
V_CC = 24# V
V_BE= 0.65# V
I_B = I_C/Beta# A
I_E= I_B+I_C# A
# S = ((Beta+1)*(Rth+R_E))/(Rth+(R_E*(1+Beta))), where Rth= R1*R2/(R1+R2)
Rth = ((R_E*Beta) - (S*R_E) - (S*R_E*Beta) + R_E)/(S-Beta-1)# ohm
Vth= I_B*Rth+V_BE+I_E*R_E# V
# Vth= V_CC*R2/(R1+R2) or 
R1= V_CC*Rth/Vth# ohm
R1= R1*10**-3# k ohm
R2= Vth*R1/(V_CC-Vth)# k ohm
print "The value of R1 = %.2f k ohm"%R1
print "The value of R2 = %.2f k ohm"%R2
The value of R1 = 18.20 k ohm
The value of R2 = 2.30 k ohm

Exa 4.25 page No 230

In [39]:
from __future__ import division
# Given data
I_C =5 * 10**-3# A
V_CE = 8# V
V_E = 6# V
S = 10
h_fc = 200
Beta = h_fc
V_CC = 20# V
V_BE = 0.6# V
I_B =I_C/Beta# A
I_E = I_C+I_B# A
# I_C*R_C = V_CC - V_CE - V_E
R_C = (V_CC - V_CE - V_E)/I_C# ohm
R_C = R_C * 10**-3# k ohm
print "The value of R_C = %.2f k ohm"%R_C
R_C = R_C * 10**3# ohm
#Voltage at point E, V_E =I_E*R_E
R_E = V_E/I_E# ohm
R_E = R_E * 10**-3# k ohm
print "The value of R_E = %.2f k ohm"%R_E
R_E = R_E * 10**3# ohm
# S = ((Beta+1)*(R_B+R_E))/( R_B+(R_E*(1+Beta)) ), where R_B= R1*R2/(R1+R2)
R_B = ((R_E*(1+Beta))-(S*R_E*(1+Beta)))/( S-(1+Beta) )# ohm
# Vth = V_CC*(R2/(R1+R2)) = V_CC*(R_B/R1)
# Applying KVL we get, Vth= I_B*R_B+V_BE+V_E or
Vth = (I_B*R_B) + V_BE + V_E# V
R1 =(V_CC/Vth)*R_B# ohm
R1= R1*10**-3# k ohm
print "The value of R1 = %.2f k ohm"%R1
R2 = (R1*Vth)/(V_CC-Vth)# k ohm
print "The value of R2 = %.2f k ohm"%R2
The value of R_C = 1.20 k ohm
The value of R_E = 1.19 k ohm
The value of R1 = 32.86 k ohm
The value of R2 = 17.24 k ohm

Exa 4.26 page No 231

In [40]:
from __future__ import division
# Given data
V_CEsat = 0.1# V
V_BEsat = 0.6# V
h_fc = 50
Beta = h_fc
V_CC = 12# V
R_C = 1# k ohm
R_C = R_C * 10**3# ohm
R_B = 10# k ohm
R_B = R_B * 10**3# ohm
# The collector current,
I_C = (V_CC-V_CEsat)/R_C# A
I_B = I_C/Beta# mA
# The value of V_BB to saturate the transistor 
V_BB = (I_B*R_B) + V_BEsat# V
print "The value of V_BB to saturate the transistor = %.2f V"%V_BB
The value of V_BB to saturate the transistor = 2.98 V

Exa 4.27 page No 233

In [41]:
from __future__ import division
# Given data
V_CC = 30# V
R1 = 90# k ohm
R1 = R1 * 10**3# ohm
R2 = 45# k ohm
R2 = R2 * 10**3#  ohm
R_C = 5#  k ohm
R_C = R_C * 10**3# ohm
R_E = 5# k ohm
R_E =R_E * 10**3# ohm
V_BE = 0.6# V
h_FE = 100
Beta = h_FE
I_CBO = 10# µA
I_CBO = I_CBO * 10**-6# A
V_BB = V_CC*(R2/(R1+R2))# V
Vth = V_BB# V
R_BB = (R1*R2)/(R1+R2)# ohm
Rth= R_BB# ohm
I_C = (V_CC-V_BE)/( (Rth/Beta)+R_E )# A
I_CEO = (Beta+1)*I_CBO# A
# The value of I_C 
I_C = I_C-I_CEO# A
I_C= I_C*10**3# mA
print "The value of I_C = %.2f mA"%I_C
# The value of Vo
Vo =V_CC-(I_C*10**-3*R_C)# V
print "The value of Vo = %.2f V"%Vo

# Note: In the book, the putted value of V_CC (i.e 20 V ) to evaluate the value of I_C is wrong, so the value of I_C = %.2f the book is wrong and due to this the value of Vo is also wrong.
The value of I_C = 4.54 mA
The value of Vo = 7.31 V

Exa 4.28 page No 233

In [42]:
from __future__ import division
# Given data
Beta = 130
V_BE = 0.7# V
V_CC = 18# V
R1 = 510# k ohm
R1 = R1 * 10**3# ohm
R2 = 510# k ohm
R2 = R2 * 10**3# ohm
R_C = 9.1# k ohm
R_C = R_C * 10**3# ohm
R_E = 7.5*10**3# ohm
Vth = V_CC*(R2/(R1+R2))# V
Rth = (R1*R2)/(R1+R2)# ohm
#Applying KVL = %.2f the input loop, Vth = (I_B*Rth)+V_BE+(I_E*R_E) or
I_CQ = (Vth-V_BE)/( (Rth/Beta)+R_E+(R_E/Beta) )# A         (On putting I_E = I_C+I_B and I_C = Beta*I_B)
I_CQ= I_CQ*10**3# mA
print "The value of I_CQ = %.2f mA"%I_CQ
# Applying KVL = %.2f the output loop, V_CC = (I_C*R_C) + V_CEQ + (I_E*R_E)
V_CEQ = V_CC - (I_CQ*10**-3*( R_C+R_E+(R_E/Beta) ))# V
print "The value of V_CEQ = %.2f V"%V_CEQ
The value of I_CQ = 0.87 mA
The value of V_CEQ = 3.48 V

Exa 4.29 page No 235

In [43]:
from __future__ import division
# Given data
V_CC = 18# V
V_BE = 0.7# V
R1 = 510# k ohm
R1 = R1 * 10**3# ohm
Beta = 130
R_E = 7.5# k ohm
R_E = R_E * 10**3# ohm
R_C = 9.1# k ohm
R_C = R_C * 10**3# ohm
# The value of I_CQ 
I_CQ = (V_CC-V_BE)/( (R1/Beta)+R_E+(R_E/Beta) )# A
I_CQ= I_CQ*10**3# mA
print "The value of I_CQ = %.2f mA"%I_CQ
# The value of V_CEQ 
V_CEQ = V_CC - I_CQ*10**-3*(R_C+R_E+(R_E/Beta))# V
print "The value of V_CEQ = %.2f V"%V_CEQ
The value of I_CQ = 1.51 mA
The value of V_CEQ = -7.10 V

Exa 4.30 page No 236

In [45]:
from __future__ import division
# Given data
V_CC = 10# V
Beta = 90
V_BE = 0.7# V
R_E = 4.7# k ohm
R_E = R_E * 10**3# ohm
R_B = 250# k ohm
R_B = R_B * 10**3# ohm
R = 1.2# k ohm
R = R * 10**3# ohm
#Applying KVL for input loop, V_CC = R_E*I_E + R_B*I_B + V_BE  +I_E*R or
I_CQ = (Beta*(V_CC-V_BE))/( ((1+Beta)*(R_E+R))+R_B )# A     (On putting I_E = I_C+I_B and I_B = I_C/Beta)
I_CQ=I_CQ*10**3# mA
print "The value of I_CQ = %.2f mA"%I_CQ
I_CQ=I_CQ*10**-3# A
#Applying KVL for output loop, V_CC = ((I_CQ+(I_CQ/Beta))*R_E)+V_CEQ + ((I_CQ+(I_CQ/Beta))*R)
V_CEQ = V_CC - ( (I_CQ+(I_CQ/Beta)) * (R_E+R) )# V
print "The value of V_CEQ = %.2f V"%V_CEQ
# If beta is increased by 50% i.e
bita = Beta+Beta/2
I_deshCQ = (bita*(V_CC-V_BE))/( ((1+bita)*(R_E+R))+R_B )# A
I_deshCQ=I_deshCQ*10**3# mA
print "The value of I''CQ = %.2f mA"%I_deshCQ
V_deshCEQ = V_CC - ( (I_CQ+(I_CQ/bita)) * (R_E+R) )# V
print "The value of V''CEQ = %.2f V"%V_deshCEQ

# Note: In the book, there is calculation error to evaluate the value of V'CEQ, So the answer = %.2f the book is wrong.
The value of I_CQ = 1.06 mA
The value of V_CEQ = 3.65 V
The value of I''CQ = 1.19 mA
The value of V''CEQ = 3.68 V

Exa 4.34 page No 238

In [47]:
from __future__ import division
# Given data
I_B = 20# µA
I_B = I_B * 10**-6# A
R_C = 2.7# k ohm
R_C = R_C * 10**3# ohm
V_CE = 7.3# V
V_BE = 0.7# V
R_E = 0.68# k ohm
R_E = R_E * 10**3# ohm
V_E = 2.1# V
I_E = V_E/R_E# A
I_C =I_E-I_B# A  (as I_E = I_C+I_B)
Beta = I_C/I_B
print "The value of beta = %.2f"%Beta
# Applying KVL to the output loop, 
V_CC = (I_C*R_C) +V_CE + (I_E*R_E)# V
print "The value of V_CC = %.2f V"%V_CC
# Applying KVL to the output loop, V_CC = (I_B*R_B)+V_BE+(I_E*R_E) or
R_B = (V_CC-V_BE-(I_E*R_E))/I_B# ohm
R_B= R_B*10**-3## k ohm
print "The value of R_B = %.2f k ohm"%R_B
The value of beta = 153.41
The value of V_CC = 17.68 V
The value of R_B = 744.21 k ohm

Exa 4.36 page No 239

In [3]:
from __future__ import division
%matplotlib inline 
from matplotlib.pyplot import plot,xlabel,ylabel,show,title

# Given data
V_CC = 20# V
R_C = 3.3# k ohm
R_C = R_C * 10**3# ohm
R_B = 1# Mohm
R_B = R_B * 10**6# ohm
V_CE = V_CC# V
I_C = V_CC/R_C# A
I_C=I_C*10**3# mA
# Plotting of the DC load line,
plot([V_CE,0],[0,I_C])
xlabel("V_CE in volts.")
ylabel("I_C in mA.")
title("DC load line.")
print "At saturation, the value of I_C = %.2f mA"%I_C
print "At cut off, the value of V_CE = %.2f volts"%V_CE
print "DC load line shown in figure."
At saturation, the value of I_C = 6.06 mA
At cut off, the value of V_CE = 20.00 volts
DC load line shown in figure.

Exa 4.37 page No 240

In [50]:
from __future__ import division
# Given data
Beta = 100
Rth = 1.8# k ohm
V_CC = 10# V
R2 = 2.2# k ohm
R2 = R2 * 10**3# ohm
R1 = 10# k ohm
R1 = R1 * 10**3# ohm
R_C=3.6*10**3# ohm
R_E = 1# k ohm
R_E = R_E * 10**3# ohm
V_BE = 0.7# V
Vth = (R2/(R1+R2))*V_CC# V
Rth = (R1*R2)/(R1+R2)# ohm
# Applying KVL for input loop, Vth = (I_B*Rth)+V_BE+(I_E*R_E) or 
I_E = (Vth-V_BE)/( R_E+(Rth/(Beta+1)) )# A   (On putting I_E = (Beta+1)*I_B)
I_E= I_E*10**3# mA
print "The value of I_E = %.2f mA"%I_E
I_B =I_E/(Beta+1)# mA
I_C= I_E-I_B# mA
#Applying KVL for output loop, V_CC = (I_C*R_C) + V_CE + (I_E*R_E)
V_CE = V_CC - (I_C*10**-3*R_C) - (I_E*10**-3*R_E)# V
print "The value of V_CE = %.2f V"%V_CE
The value of I_E = 1.08 mA
The value of V_CE = 5.05 V

Exa 4.38 page No 242

In [52]:
from __future__ import division
# Given data
h_fe = 110
h_ie = 1.6# k ohm
h_ie = h_ie * 10**3# ohm
h_re = 2*10**-4
h_oe = 20*10**-6# A/V
R_B = 470# k ohm
R_B = R_B * 10**3# ohm
R_C = 4.7# k ohm
Ri= 1.51*10**3# ohm
R_C =R_C * 10**3# ohm
Rin = h_ie - ( (h_fe*h_re*R_C)/(1+(h_oe*R_C)) )# ohm
# The value of Zi,
Zi = (R_B*Ri)/(R_B+Ri)# ohm
Zi= Zi*10**-3# k ohm
print "The value of Zi = %.2f k ohm"%Zi
Zi= Zi*10**3# ohm
R_L = R_C# ohm
# The voltag gain,
Av = -h_fe/( Zi*(h_oe+(1/R_L)) )
print "The value of Av = %.2f"%Av
Rs = 0# ohm
# The value of Zo 
Zo = 1/( h_oe-( (h_fe*h_re)/(Zi+Rs) ) )# ohm
Zo= Zo*10**-3## k ohm
print "The value of Zo = %.2f k ohm"%Zo
The value of Zi = 1.51 k ohm
The value of Av = -313.97
The value of Zo = 185.75 k ohm