# Ch-6 : Bipolar Junction Transistor¶

## Page No. 151 Example 6.1.¶

In [1]:
IE=10
IC=9.8
print "The emitter current is,"
print "IE = IB + IC"
print "10 = IB + 9.8"
IB=IE-IC
print "Therefore,    IB(mA) =%0.2f"%IB

The emitter current is,
IE = IB + IC
10 = IB + 9.8
Therefore,    IB(mA) =0.20


## Page No. 152 Example 6.2.¶

In [3]:
IE=6.28
IC=6.20
print "The common-base d.c. current gain,"
alpha=IC/IE
print "alpha = IC/IE = %0.4f"%alpha

The common-base d.c. current gain,
alpha = IC/IE = 0.9873


## Page No. 155 Example 6.3.¶

In [5]:
alpha=0.967
IE=10
print "The common-base d.c. current gain (alpha) is,"
print "alpha = 0.967 = IC/IE = IC/10"
IC=alpha*IE
print "\nTherefore,    IC = %0.2f mA"%IC
print "The emitter current,    IE = IB + IC"
IB=IE-IC
print "Therefore,    IB =%0.2f mA"%IB

The common-base d.c. current gain (alpha) is,
alpha = 0.967 = IC/IE = IC/10

Therefore,    IC = 9.67 mA
The emitter current,    IE = IB + IC
Therefore,    IB =0.33 mA


## Page No. 156 Example 6.4.¶

In [6]:
IE=10
alpha=0.98
print "The common-base d.c. current gain, alpha = IC/IE"
IC=alpha*IE
print "Therefore,  IC =%0.2f mA"%IC
print "The emitter current, IE = IB + IC"
IB=IE-IC
print "Therefore,  IB =%0.2f mA"%IB

The common-base d.c. current gain, alpha = IC/IE
Therefore,  IC =9.80 mA
The emitter current, IE = IB + IC
Therefore,  IB =0.20 mA


## Page No. 158 Example 6.5.¶

In [8]:
alpha=0.97
print "If alpha=0.97, beta = alpha/(1 - alpha)"
beta=alpha/(1-alpha)
print "beta = %0.4f"%beta
beta1=200
print "If beta=200, alpha = beta/(beta + 1)"
alpha1 =beta1/(beta1+1)
print "alpha =%0.4f"%alpha1

If alpha=0.97, beta = alpha/(1 - alpha)
beta = 32.3333
If beta=200, alpha = beta/(beta + 1)
alpha =0.0000


## Page No. 158 Example 6.6.¶

In [10]:
beta=100.0
IC=40
print "beta = 100 = IC / IB"
IB=IC/beta
print "Therefore,    IB =%0.2f mA"%IB
print "IE = IB + IC"
IE=IB+IC
print "IE =%0.2f mA"%IE

beta = 100 = IC / IB
Therefore,    IB =0.40 mA
IE = IB + IC
IE =40.40 mA


## Page No. 159 Example 6.7.¶

In [12]:
beta=150.
IE=10
alpha=beta/(beta+1)
print "The common-base current gain, alpha = beta / (beta + 1) =%0.4f"%alpha
print "Also,    alpha = IC / IE"
IC=alpha*IE
print "Therefore,    IC = %0.2f mA"%IC
print "the emitter current,    IE = IB + IC"
IB=IE-IC
print "Therefore,    IB =%0.2f mA"%IB

The common-base current gain, alpha = beta / (beta + 1) =0.9934
Also,    alpha = IC / IE
Therefore,    IC = 9.93 mA
the emitter current,    IE = IB + IC
Therefore,    IB =0.07 mA


## Page No. 160 Example 6.8.¶

In [14]:
beta=170.
IC=80
print "We know that (beta),  beta = 170 = IC / IB"
IB=IC/beta
print "Therefore,    IB =%0.2f mA"%IB
IE=IB+IC
print "and    IE = IB + IC =%0.2f mA"%IE

We know that (beta),  beta = 170 = IC / IB
Therefore,    IB =0.47 mA
and    IE = IB + IC =80.47 mA


## Page No. 161 Example 6.9.¶

In [15]:
IB=0.125
beta=200
print "beta = 200 = IC / IB"
IC=beta*IB
print "Therefore,    IC = %0.2f mA"%IC
IE=IB+IC
print "and    IE = IB + IC =%0.2f mA"%IE

beta = 200 = IC / IB
Therefore,    IC = 25.00 mA
and    IE = IB + IC =25.12 mA


## Page No. 162 Example 6.10¶

In [16]:
IE=12.
beta=100
IB=IE/(1+beta)
print "We know that base current,  IB = IE / (1 + beta) =%0.2f mA"%IB
IC=IE-IB
print "and collector current,  IC = IE - IB =%0.2f mA"%IC

We know that base current,  IB = IE / (1 + beta) =0.12 mA
and collector current,  IC = IE - IB =11.88 mA


## Page No. 162 Example 6.11¶

In [17]:
IB=100*10**-6
IC=2*10**-3
beta=IC/IB
print "(a) To find beta of the transistor "
print "beta = IC / IB = %0.4f"%beta
alpha=beta/(beta+1)
print "(b) To find alpha of the transistor"
print "alpha = beta / (1+beta) = %0.4f"%alpha
IE=IB+IC
IE1=IE*10**3
print "(c) To find emitter current, IE"
print "IE = IB + IC = %0.2f mA"%IE1
# answer in the textbook is wrong
print "(d) To find the new value of beta when delta_IB = 25uA and delta_IC = 0.6mA"
delta_IB=25*10**-6
delta_IC=0.6*10**-3
IB1=IB+delta_IB
IB11=IB1*10**6
IC1=IC+delta_IC
IC11=IC1*10**3
print "Therefore,  IB = %0.2f uA"%IB11
print "            IC = %0.2f mA"%IC11
beta1=IC1/IB1
print "New value of beta of the transistor,"
print "beta = IC / IB =%0.4f"%beta1

(a) To find beta of the transistor
beta = IC / IB = 20.0000
(b) To find alpha of the transistor
alpha = beta / (1+beta) = 0.9524
(c) To find emitter current, IE
IE = IB + IC = 2.10 mA
(d) To find the new value of beta when delta_IB = 25uA and delta_IC = 0.6mA
Therefore,  IB = 125.00 uA
IC = 2.60 mA
New value of beta of the transistor,
beta = IC / IB =20.8000


## Page No. 163 Example 6.12.¶

In [19]:
alpha=0.98
ICO=5*10**-6
ICBO=ICO
IB=100*10**-6
IC=((alpha*IB)/(1-alpha))+(ICO/(1-alpha))
IC1=IC*10**3
print "The collector current is, IC = ((alpha*IB)/(1-alpha))+(ICO/(1-alpha)) = %0.2f mA"%IC1
IE=IB+IC
IE1=IE*10**3
print "The emitter current is, IE = IB + IC =%0.2f mA"%IE1

The collector current is, IC = ((alpha*IB)/(1-alpha))+(ICO/(1-alpha)) = 5.15 mA
The emitter current is, IE = IB + IC =5.25 mA


## Page No. 164 Example 6.13.¶

In [20]:
ICBO=10*10**-6
hFE=50
beta=hFE
IB=0.25*10**-3
IC=(beta*IB)+((1+beta)*ICBO)
IC1=IC*10**3
print "(a) To find the value of collector current when IB = 0.25mA"
print "IC = (beta*IB) + ((1+beta)*ICBO) = %0.2f mA"%IC1
T1=27.
T2=50.
I_CBO = ICBO * (2**((T2-T1)/10))
I_CBO1=I_CBO*10**6
print "(b) To find the value of new collector current if temperature rises to 50 C"
print "I''CBO(beta=50) = ICBO*(2**((T2-T1)/10)) =%0.2f uA"%I_CBO1
IC2=(beta*IB)+((1+beta)*I_CBO)
IC3=IC2*10**3
print "Therefore, the collector current at 50 C is"
print "IC = (beta*IB) + ((1+beta)*I''CBO) =%0.2f mA"%IC3

(a) To find the value of collector current when IB = 0.25mA
IC = (beta*IB) + ((1+beta)*ICBO) = 13.01 mA
(b) To find the value of new collector current if temperature rises to 50 C
I''CBO(beta=50) = ICBO*(2**((T2-T1)/10)) =49.25 uA
Therefore, the collector current at 50 C is
IC = (beta*IB) + ((1+beta)*I''CBO) =15.01 mA


## Page No. 165 Example 6.14.¶

In [21]:
delta_IC=0.99*10**-3
delta_IE=1*10**-3
alpha=delta_IC/delta_IE
print "The current gain of the transistor is alpha = delta_IC/delta_IE = %0.4f"%alpha

The current gain of the transistor is alpha = delta_IC/delta_IE = 0.9900


## Page No. 165 Example 6.15¶

In [23]:
beta_dc=100.
alpha_dc=beta_dc/(1+beta_dc)
print "The d.c. current gain of the transistor in CB mode is, alpha_dc = beta_dc/(1+beta_dc) =%0.2f"%alpha_dc

The d.c. current gain of the transistor in CB mode is, alpha_dc = beta_dc/(1+beta_dc) =0.99


## Page No. 165 Example 6.16.¶

In [24]:
delta_IC=0.995*10**-3
delta_IE=1*10**-3
alpha=delta_IC/delta_IE
print "Common base current gain is, alpha = delta_IC/delta_IE =%0.2f"%alpha
beta=alpha/(1-alpha)
print "Common-emitter current gain is beta = alpha / (1-alpha) =%0.2f"%beta

Common base current gain is, alpha = delta_IC/delta_IE =0.99
Common-emitter current gain is beta = alpha / (1-alpha) =199.00


## Page No. 166 Example 6.17.¶

In [26]:
beta=49.
alpha=beta/(1+beta)
print "We know that,  alpha = beta/(1+beta)"
print "Therefore, the common base current gain is, alpha =%0.2f"%alpha
print "We also know that, alpha = IC / IE"
IE=3*10**-3
IC=alpha*IE
IC1=IC*10**3
print "Therefore,  IC = alpha * IE = %0.2f mA"%IC1

We know that,  alpha = beta/(1+beta)
Therefore, the common base current gain is, alpha =0.98
We also know that, alpha = IC / IE
Therefore,  IC = alpha * IE = 2.94 mA


## Page No. 166 Example 6.18.¶

In [32]:
IB=15*10**-6
beta=150.
IC=beta*IB
IC1=IC*10**3
print "The collector current, IC(mA) = beta * IB =%0.2f mA"%IC1
IE=IC+IB
IE1=IE*10**3
print "The emitter current, IE(mA) = IC + IB =%0.2f mA"%IE1
alpha=beta/(1+beta)
print "Common-base current gain, alpha = beta/(1+beta) = %0.2f"%alpha

The collector current, IC(mA) = beta * IB =2.25 mA
The emitter current, IE(mA) = IC + IB =2.26 mA
Common-base current gain, alpha = beta/(1+beta) = 0.99


## Page No. 167 Example 6.19.¶

In [34]:
print "Referring to fig.6.18, the base current is,"
VBB=4
VBE=0.7
RB=200*10**3
IB=(VBB-VBE)/RB
IB1=IB*10**6
print "IB = (VBB - VBE) / RB =%0.2f uA"%IB1
beta=200
IC=beta*IB
IC1=IC*10**3
print "The collector current is, IC = beta*IB =%0.2f mA"%IC1
IE=IC+IB
IE1=IE*10**3
print "The emitter current is, IE = IC + IB =%0.2f mA"%IE1
VCC=10
RC=2*10**3
VCE=VCC-(IC*RC)
print "Therefore, VCE = VCC - IC*RC =%0.2f V"%VCE

Referring to fig.6.18, the base current is,
IB = (VBB - VBE) / RB =16.50 uA
The collector current is, IC = beta*IB =3.30 mA
The emitter current is, IE = IC + IB =3.32 mA
Therefore, VCE = VCC - IC*RC =3.40 V


## Page No. 167 Example 6.20.¶

In [35]:
alpha_dc=0.99
ICBO=5*10**-6
IB=20*10**-6
IC=((alpha_dc*IB)/(1-alpha_dc))+(ICBO/(1-alpha_dc))
IC1=IC*10**3
print IC1,"IC = ((alpha_dc*IB)/(1-alpha_dc)) + (ICBO/(1-alpha_dc)) =%0.2f mA"%IC1
IE=IB+IC
IE1=IE*10**3
print "Therefore,  IE = IB + IC =%0.2f mA"%IE1

2.48 IC = ((alpha_dc*IB)/(1-alpha_dc)) + (ICBO/(1-alpha_dc)) =2.48 mA
Therefore,  IE = IB + IC =2.50 mA


## Page No. 168 Example 6.21.¶

In [37]:
ICBO=0.2*10**-6
ICEO=18*10**-6
IB=30*10**-3
print "The leakage current  ICBO = 0.2 uA"
print "                     ICEO = 18 uA"
print "Assume that          IB = 30 mA"
print "IE = IB + IC"
print "IC = IE - IB = (beta*IB)+((1+beta)*ICBO)"
print "We know that,  ICEO = ICBO/(1-alpha) = (1+beta)*ICBO"
beta=(ICEO/ICBO)-1
print "beta = (ICEO / ICBO)-1 =%0.2f"%beta
IC=(beta*IB)+((1+beta)*ICBO)
print "IC = (beta*IB) + ((1+beta)*ICBO) =%0.2f A"%IC
alpha_dc=1-(ICBO/ICEO)
print "alpha_dc = 1 - (ICBO / ICEO) =%0.2f"%alpha_dc
beta_dc=(IC-ICBO)/(IB-ICEO)
print "beta_dc = (IC-ICBO) / (IB-ICEO) =%0.2f"%beta_dc

The leakage current  ICBO = 0.2 uA
ICEO = 18 uA
Assume that          IB = 30 mA
IE = IB + IC
IC = IE - IB = (beta*IB)+((1+beta)*ICBO)
We know that,  ICEO = ICBO/(1-alpha) = (1+beta)*ICBO
beta = (ICEO / ICBO)-1 =89.00
IC = (beta*IB) + ((1+beta)*ICBO) =2.67 A
alpha_dc = 1 - (ICBO / ICEO) =0.99
beta_dc = (IC-ICBO) / (IB-ICEO) =89.05


## Page No. 168 Example 6.22.¶

In [39]:
alpha_dc=0.99
ICBO=50*10**-6
IB=1*10**-3
IC=((alpha_dc*IB)/(1-alpha_dc))+(ICBO/(1-alpha_dc))
IC1=IC*10**3
print "Assume that,  IB = 1 mA"
print "IC = ((alpha_dc*IB) / (1-alpha_dc)) + (ICBO/(1-alpha_dc)) = %0.2f mA"%IC1
IE=IC+IB
IE1=IE*10**3
print "IE = IC + IB = %0.2f mA"%IE1

Assume that,  IB = 1 mA
IC = ((alpha_dc*IB) / (1-alpha_dc)) + (ICBO/(1-alpha_dc)) = 104.00 mA
IE = IC + IB = 105.00 mA


## Page No. 169 Example 6.23.¶

In [43]:
%matplotlib inline
from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend
print "Maximum VCE = VCC = 24V"
IC=24/(8*10**3)  #in Ampere
x1=IC*10**3  #in mA
print "Maximum  IC = VCC / RC = %0.2f mA"%x1
print "(ii) For fixing the optimum operating point Q, mark the middle of the d.c. load line AB and the corresponding VCE and IC values can be found"
VCEQ=24./2
print "Here,    VCEQ(V) = VCC / 2 = %0.2f V"%VCEQ #in volts
print "         ICQ = 1.5 mA"
print ""
Rac=(8*24.)/(8+24) #in k-ohm
print "AC load,   R_a.c. = RC || RL = %0.2f kohm"%Rac
VCE=12+((1.5*10**-3)*(6*10**3)) #in Volts
print "Therefore, maximum VCE(V) = VCEQ + ICQ*R_a.c. = %0.2f "%VCE
print "This locates the point D(OD = 21V) on the VCE axis"
IC=(1.5*10**-3)+(12/(6*10**3))  #in Ampere
x3=IC*10**3  #in mA
print "Maximum IC =  ICQ + VCEQ/R_a.c. = %0.2f mA"%x3
print "This locates the point C(OC = 3.5mA) on the IC axis. By joining points C and D a.c. load line CD is constructed. "
x=[24,0]
y=[0,3]
plot(x,y)
x1=[21,0]
y1=[0,3.5]
plot(x1,y1)
title("Fig.6.22(b)")
xlabel("VCE(V)")
ylabel("IC(mA)")
show()

(i) DC load line:
Maximum VCE = VCC = 24V
Maximum  IC = VCC / RC = 0.00 mA
(ii) For fixing the optimum operating point Q, mark the middle of the d.c. load line AB and the corresponding VCE and IC values can be found
Here,    VCEQ(V) = VCC / 2 = 12.00 V
ICQ = 1.5 mA

AC load,   R_a.c. = RC || RL = 6.00 kohm
Therefore, maximum VCE(V) = VCEQ + ICQ*R_a.c. = 21.00
This locates the point D(OD = 21V) on the VCE axis
Maximum IC =  ICQ + VCEQ/R_a.c. = 1.50 mA
This locates the point C(OC = 3.5mA) on the IC axis. By joining points C and D a.c. load line CD is constructed.


## Page No. 170 Example 6.24.¶

In [45]:
%matplotlib inline
from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend
print "Maximum VCE = VCC = 12V, which locates the point B(OB = 12V) of the d.c. load line"
IC=12/(2*10**3)  #in Ampere
x1=IC*10**3  #in mA
print "Maximum  IC = VCC / (RC+RE) = %0.2f mA"%x1

print "(ii) Operating point Q"
V2=((4*10**3)/(12*10**3))*12  #in V
print "Therefore,    V2 = %0.2f V"%V2
print "       V2 = VBE + IE*RE"
IE=(4-0.7)/(1*10**3)  #in Ampere
x2=IE*10**3  #in mA
print "Therefore,  IE = V2-VBE / RE = %0.2f mA"%x2
IC=x2 #in mA
print "        IC  = IE =%0.2f mA"%IC
VCE=12-((3.3*10**-3)*(2*10**3)) #in volts
print "VCE = VCC - IC(RC+RE) = %0.2f V"%VCE
Rac=1.5/2.5 #in k-ohm
print "AC load,   Ra.c.(k-ohm) = RC || RL = %0.2f kohm"%Rac
VCE=5.4+((3.3*10**-3)*(0.6*10**3)) #in Volts
print "Therefore, maximum VCE = VCEQ + ICQ*Ra.c. = %0.2f V"%VCE
IC=(3.3*10**-3)+(5.4/(0.6*10**3))  #in Ampere
x3=IC*10**3  #in mA
print "Maximum  IC(mA) = ICQ + VCEQ/Ra.c. = %0.2f mA"%x3
x=[7.38,0]
y=[0,12.3]
plot(x,y)
x1=[12,0]
y1=[0,6]
plot(x1,y1)
title("Fig.6.23(b)")
xlabel("VCE(V) -->")
ylabel("IC(mA) -->")
show()

(i) DC load line:
Maximum VCE = VCC = 12V, which locates the point B(OB = 12V) of the d.c. load line
Maximum  IC = VCC / (RC+RE) = 0.00 mA
(ii) Operating point Q
Therefore,    V2 = 0.00 V
V2 = VBE + IE*RE
Therefore,  IE = V2-VBE / RE = 3.30 mA
IC  = IE =3.30 mA
VCE = VCC - IC(RC+RE) = 5.40 V
AC load,   Ra.c.(k-ohm) = RC || RL = 0.60 kohm
Therefore, maximum VCE = VCEQ + ICQ*Ra.c. = 7.38 V
Maximum  IC(mA) = ICQ + VCEQ/Ra.c. = 12.30 mA


## Page No. 171 Example 6.25.¶

In [49]:
ICQ=1*10**-3
VCEQ=6.
VCC=10.
beta=100.
VBE=0.7
RC=(VCC-VCEQ)/ICQ
RC1=RC*10**-3
RC2=round(RC1)
print "The collector resistance is, RC = (VCC - VCEQ) / ICQ =%0.2f kohm"%RC2
IBQ=ICQ/beta
IBQ1=IBQ*10**6
print "The base current is, IBQ = ICQ / beta =%0.2f uA"%IBQ1
RB=(VCC-VBE)/IBQ
RB1=RB*10**-6
print "The base resistance is, RB = (VCC - VBE(on)) / IBQ =%0.2f Mohm"%RB1

The collector resistance is, RC = (VCC - VCEQ) / ICQ =4.00 kohm
The base current is, IBQ = ICQ / beta =10.00 uA
The base resistance is, RB = (VCC - VBE(on)) / IBQ =0.93 Mohm


## Page No. 171 Example 6.26.¶

In [50]:
beta=100
VBE=0.7
VCC=10
RB=20*10**3
RC=0.4*10**3
RE=0.6*10**3
VBB=5
print "VBB = IB*RB + VBE(on) + IE*RE"
print "Also,  IE = IB + IC = IB + beta*IB = (1 + beta)*IB"
IB=(VBB-VBE)/(RB+((1+beta)*RE))
IB1=IB*10**6
print "The base current,  IB = (VBB - VBE(on)) / (RB + ((1+beta)*RE)) =%0.2f uA"%IB1
IC=beta*IB
IC1=IC*10**3
print "Therefore,  IC = beta*IB =%0.2f mA"%IC1
IE=IC+IB
IE1=IE*10**3
print "IE = IC + IB = %0.2f mA"%IE1
VCE=VCC-(IC*RC)-(IE*RE)
print "VCE = VCC - (IC*RC) - (IE*RE) =%0.2f V"%VCE
print "The Q point is at"
print "VCEQ = %0.2f V"%VCE
print "and ICQ(mA) = %0.2f mA"%IC1

VBB = IB*RB + VBE(on) + IE*RE
Also,  IE = IB + IC = IB + beta*IB = (1 + beta)*IB
The base current,  IB = (VBB - VBE(on)) / (RB + ((1+beta)*RE)) =53.35 uA
Therefore,  IC = beta*IB =5.33 mA
IE = IC + IB = 5.39 mA
VCE = VCC - (IC*RC) - (IE*RE) =4.63 V
The Q point is at
VCEQ = 4.63 V
and ICQ(mA) = 5.33 mA


## Page No. 172 Example 6.27.¶

In [54]:
%matplotlib inline
from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend
IC=6/(3*10**3) #in Ampere
x1=IC*10**3  #in mA
print "When  VCE = 0, IC = VCC/RC = %0.2f mA"%x1
print ""
print "(ii) Operating point Q:"
IB=(6-0.7)/(530*10**3)
x2=IB*10**6
print "Therefore,              IB = VCC-VBE / RB = %0.2f uA"%x2
IC=100*10*10**-6 # in Ampere
x3=IC*10**3 # in mA
print "Therefore,              IC(mA) = beta*IB = %0.2f mA"%x3
VCE=6-((1*10**-3)*(3*10**3)) # in volts
print "                           VCE = VCC - IC*RC = %02.f V"%VCE
print "Therefore operating point is VCEQ = 3 V and ICQ = 1 mA"
print ""
print "(iii) Stability factor:    S = 1 + beta = 1 + 100 = 101"
x=[6,0]
y=[0,2]
plot(x,y)
xlabel("VCE (V) --->")
ylabel("IC (mA) --->")
show()

(i) DC load line:
When  VCE = 0, IC = VCC/RC = 0.00 mA

(ii) Operating point Q:
Therefore,              IB = VCC-VBE / RB = 10.00 uA
Therefore,              IC(mA) = beta*IB = 1.00 mA
VCE = VCC - IC*RC = 03 V
Therefore operating point is VCEQ = 3 V and ICQ = 1 mA

(iii) Stability factor:    S = 1 + beta = 1 + 100 = 101


## Page No. 172 Example 6.28.¶

In [55]:
VCC=12.
beta=100.
VBE=0.7
print "(a) To determine RB :"
VCE=7
IC=1*10**-3
RC=(VCC-VCE)/IC
RC1=RC*10**-3
print "RC = (VCC - VCE) / IC =%0.2f kohm"%RC1
IB=IC/beta
IB1=IB*10**6
print "IB = IC / beta = %0.2f uA"%IB1
RB=(VCC-VBE-(IC*RC))/IB
RB1=RB*10**-3
print "RB = (VCC - VBE - (IC*RC)) / IB = %0.2f kohm"%RB1
S=(1+beta)/(1+(beta*(RC/(RC+RB))))
print "(b) Stability factor, S =(1 + beta) / (1 + (beta*(RC / (RC+RB)))) =%0.2f"%S
beta1=50
print "(c) VCC = (beta*IB*RC) + (IB*RB) + VBE"
print "        = IB * ((beta*RC) + RB) + VBE"
IB=(VCC-VBE)/((beta1*RC)+RB)
IB1=IB*10**6
print "IB = (VCC-VBE) / ((beta*RC)+RB) =%0.2f uA"%IB1
IC=beta1*IB
IC1=IC*10**3
print "IC = beta*IB =%0.2f mA"%IC1
VCE=VCC-(IC*RC)
print "VCE = VCC - (IC*RC) =%0.2f V"%VCE
print "Therefore the coordinates of new operating point are :"
print "VCEQ(V) =%0.2f V"%VCE
print "ICQ =%0.2f mA"%IC1

(a) To determine RB :
RC = (VCC - VCE) / IC =5.00 kohm
IB = IC / beta = 10.00 uA
RB = (VCC - VBE - (IC*RC)) / IB = 630.00 kohm
(b) Stability factor, S =(1 + beta) / (1 + (beta*(RC / (RC+RB)))) =56.51
(c) VCC = (beta*IB*RC) + (IB*RB) + VBE
= IB * ((beta*RC) + RB) + VBE
IB = (VCC-VBE) / ((beta*RC)+RB) =12.84 uA
IC = beta*IB =0.64 mA
VCE = VCC - (IC*RC) =8.79 V
Therefore the coordinates of new operating point are :
VCEQ(V) =8.79 V
ICQ =0.64 mA


## Page No. 173 Example 6.29.¶

In [56]:
VCC=12.
RC=250.
IB=0.25*10**-3
beta=100.
VCEQ=8.
RB=VCEQ/IB
RB1=RB*10**-3
print "RB = VCEQ / IB =%0.2f kohm"%RB1
S=(1+beta)/(1+(beta*(RC/(RC+RB))))
print "Stability factor, S = (1+beta) / 1 + (beta*(RC/RC+RB)) =%0.2f"%S

RB = VCEQ / IB =32.00 kohm
Stability factor, S = (1+beta) / 1 + (beta*(RC/RC+RB)) =56.90


## Page No. 173 Example 6.30.¶

In [59]:
VCC=16
RC=3*10**3
RE=2*10**3
R1=56*10**3
R2=20*10**3
alpha=0.985
VBE=0.3
print "For a germanium transistor, VBE=0.3V. As alpha=0.985"
beta=alpha/(1-alpha)
beta1=round(beta)
print "beta = alpha / ( 1 - alpha) =%0.2f"%beta1
print "(a) To find the coordinates of the operating point"
print "Referring to fig. 6.29,"
VT=(R2/(R1+R2))*VCC
print "Thevenin voltage,     VT = (R2 / (R1+R2)) * VCC =%0.2f V"%VT
RB=(R1*R2)/(R1+R2)
RB1=RB*10**-3
print "Thevenin resistance, RB = (R1 * R2) / (R1 + R2) = %0.2f kohm"%RB1
IC=(VT-VBE)/((RB/beta)+(RE/beta)+RE)
IC1=IC*10**3
print "Therefore,   IC =%0.2f mA"%IC1
print "Since IB is very small, IC ~ IE = 1.73 mA"
IE=IC
VCE=VCC-(IC*RC)-(IE*RE)
print "Therefore,    VCE = VCC - (IC*RC) - (IE*RE) =%0.2f V"%VCE
print "Therefore, the coordinates of the operating point are :"
print "IC = %0.2f mA"%IC1
print "VCE =%0.2f V"%VCE
print "(b) To find the stability factor S"
S = (1+beta)*((1+(RB/RE))/(1+beta+(RB/RE)))
print "S =%0.2f"%S

For a germanium transistor, VBE=0.3V. As alpha=0.985
beta = alpha / ( 1 - alpha) =66.00
(a) To find the coordinates of the operating point
Referring to fig. 6.29,
Thevenin voltage,     VT = (R2 / (R1+R2)) * VCC =0.00 V
Thevenin resistance, RB = (R1 * R2) / (R1 + R2) = 14.74 kohm
Therefore,   IC =-0.13 mA
Since IB is very small, IC ~ IE = 1.73 mA
Therefore,    VCE = VCC - (IC*RC) - (IE*RE) =16.67 V
Therefore, the coordinates of the operating point are :
IC = -0.13 mA
VCE =16.67 V
(b) To find the stability factor S
S =7.24


## Page No. 174 Example 6.31.¶

In [60]:
VCE=12
IC=2*10**-3
VCC=24
VBE=0.7
beta=50
RC=4.7*10**3
S=5.1
print "(a) To determine RE,"
print "VCE = VCC - (IC*RC) - (IE*RE)"
RE = (VCC - (IC*RC) - VCE)/IC  #IC=IE
RE1=RE*10**-3
print "Therefore,  RE = %0.2f kohm"%RE1

print "(b) To determine R1 and R2,"
RB=((RE*beta)/(((1+beta)/S)-1))-RE
RB1=(RB*10**-3)
print "Therefore,  RB(k-ohm) = ((RE*beta) / (((1+beta)/S)-1)) - RE = %0.2f kohm"%RB1
R2=0.1*beta*RE
R2_1=R2*10**-3
print "Therefore,    R2 = %0.2f kohm"%R2_1
R1=(5.9*10**3*R2)/(R2-(5.9*10**3)) #RB=5.9
R1_1=round(R1*10**-3)
print "Therefore, R1 = R2 / ((R2/RB)-1)= %0.2f kohm"%R1_1

(a) To determine RE,
VCE = VCC - (IC*RC) - (IE*RE)
Therefore,  RE = 1.30 kohm
(b) To determine R1 and R2,
Therefore,  RB(k-ohm) = ((RE*beta) / (((1+beta)/S)-1)) - RE = 5.92 kohm
Therefore,    R2 = 6.50 kohm
Therefore, R1 = R2 / ((R2/RB)-1)= 64.00 kohm


## Page No. 174 Example 6.32.¶

In [61]:
R1=56*10**3
R2=12.2*10**3
RC=2*10**3
RE=400
VCC=10
VBE=0.7
beta=150
RTH=(R1*R2)/(R1+R2)
RTH1=round(RTH*10**-3)
print "RTH = R1 || R2 = %0.2f kohm"%RTH1
VTH=(R2/(R1+R2))*VCC
print "VTH  = (R2 / (R1+R2)) * VCC = %0.2f V"%VTH
IBQ=(VTH-VBE)/(RTH+((1+beta)*RE))
IBQ1=IBQ*10**6
print "IBQ = (VTH-VBE(on)) / (RTH + ((1+beta)*RE)) =%0.2f uA"%IBQ1
ICQ=beta*IBQ
ICQ1=ICQ*10**3
print "Therefore,    ICQ = beta * IBQ =%0.2f mA"%ICQ1
IEQ=IBQ+ICQ
IEQ1=IEQ*10**3
print "IEQ = IBQ + ICQ = %0.2f mA"%IEQ1
VCEQ=VCC-(ICQ*RC)-(IEQ*RE)
print "VCEQ = VCC - (ICQ*RC) - (IEQ*RE) = %0.2f V"%VCEQ
print "The Q point is at :"
print "VCEQ = %0.2f V"%VCEQ
print "ICQ = %0.2f mA"%ICQ1

RTH = R1 || R2 = 10.00 kohm
VTH  = (R2 / (R1+R2)) * VCC = 1.79 V
IBQ = (VTH-VBE(on)) / (RTH + ((1+beta)*RE)) =15.46 uA
Therefore,    ICQ = beta * IBQ =2.32 mA
IEQ = IBQ + ICQ = 2.33 mA
VCEQ = VCC - (ICQ*RC) - (IEQ*RE) = 4.43 V
The Q point is at :
VCEQ = 4.43 V
ICQ = 2.32 mA


## Page No. 174 Example 6.33.¶

In [64]:
VCC=22
RC=2*10**3
beta=60
VBE=0.6
R1=100*10**3
R2=5*10**3
RE=100
a=VCC-VBE-((R1*VCC)/(R1+R2))
c=(((R1*R2)/(R1+R2))+((1+beta)*RE))
IB=a/c
IB1=IB*10**6
print "IB = %0.2f uA"%IB1
IC=beta*IB
IC1=IC*10**3
print "IC = %0.2f mA "%IC1
VCE = VCC - (IC*RC) - ((1+beta)*IB*RE)
print "VCE = %0.2f V"%VCE
print "To find stability factor, (S):"
print "Stability factor for voltage divider bias is"
RB=(R1*R2)/(R1+R2)
S=(1+beta)/(1+(beta*(RE/(RE+RB))))
print "S =(1+beta)/(1+(beta*(RE/(RE+RB)))) =        where RB = R1 || R2 = %0.f"%S

IB = 128.90 uA
IC = 7.73 mA
VCE = 5.75 V
To find stability factor, (S):
Stability factor for voltage divider bias is
S =(1+beta)/(1+(beta*(RE/(RE+RB)))) =        where RB = R1 || R2 = 61


## Page No. 175 Example 6.34.¶

In [68]:
VCC=10
RC=2*10**3
beta=50
RB=100*10**3
VBE=0.7  #collector to base resistor
IB=(VCC-VBE)/(RB+(beta*RC))
IB1=IB*10**6
print "IB = %0.2f uA"%IB1
IC=beta*IB
IC1=IC*10**3
print "Hence, IC = beta * IB =%0.2f mA"%IC1
VCE=VCC-(IC*RC)
print "VCE = VCC - IC*RC = %0.2f V"%VCE
print "Therefore,the co-ordinates of the new operating point are:"
print "VCEQ = %0.2f V"%VCE
print "ICQ = %0.2f mA"%IC1
print "To find the stability factor S"
S=(1+beta)/(1+(beta*(RC/(RC+RB))))
print "S = (1+beta) / (1 + (beta*[RC/(RC+RB)])) =%0.2f"%S

IB = 46.50 uA
Hence, IC = beta * IB =2.33 mA
VCE = VCC - IC*RC = 5.35 V
Therefore,the co-ordinates of the new operating point are:
VCEQ = 5.35 V
ICQ = 2.33 mA
To find the stability factor S
S = (1+beta) / (1 + (beta*[RC/(RC+RB)])) =51.00