Chapter 7 Common-Collector Approximations

Example 7.1 Page No 183

In [3]:
from numpy import arange
%matplotlib inline
from matplotlib.pyplot import show,title,ylabel,xlabel,plot
# given data
V_CC= 10##  V
R_E= 430##  Ω
V_BE= 0.7##in V
V_B= 5##in V
# The collector saturation current,
I_Csat= V_CC/R_E##  A
# The collector emitter voltage,
V_CEcutoff= V_CC##  V
# The collector current,
I_C= (V_B-V_BE)/R_E##  A
# The collector emitter voltage,
V_CE= V_CC-(V_B-V_BE)##  V
I_C= I_C*10**3##  mA
print "Q-point = (%.2fV,%.2fmA)"%(V_CE,I_C)
print "DC load line shown in figure"
I_C= I_C*10**-3##  A
V_CE= arange(0,0.1+V_CEcutoff,0.1) #  V
I_C= (V_CC-V_CE)/R_E*10**3##  mA
# The plot of DC load line
plot(V_CE,I_C)#
xlabel("V_CE in volts")
ylabel("I_C in mA")
title("DC load line")
show()

Q-point = (5.70V,10.00mA)
DC load line shown in figure

Example 7.2 Page No 184

In [4]:
# given data
Vin= 100##  mV
Vin= Vin*10**-3##  V
R_E= 430##  Ω
R_L= 1*10**3##  Ω
r_e= 2.5##  Ω
# The ac load resistance,
r_L= R_E*R_L/(R_E+R_L)##  Ω
A= r_L/(r_L+r_e)## unit less
# The output voltage 
Vout= A*Vin##  V
Vout= Vout*10**3##  mV
print "The output voltage = %.2f mV"%Vout

The output voltage = 99.17 mV

Example 7.3 Page No 186

In [6]:
# given data
R_E= 430##  Ω
R_L= 100##  Ω
R1= 10*10**3##  Ω
R2= 10*10**3##  Ω
bita= 200## unit less
r_e= 2.5##  Ω
r_L= R_E*R_L/(R_E+R_L)##  Ω
# The voltge gain 
A= r_L/(r_L+r_e)#
print "The voltge gain = %.2f"%A
Zin_base= bita*(r_L+r_e)##  Ω
# The input impedence 
Zin= R1*R2*Zin_base/(R1*R2+R2*Zin_base+Zin_base*R1)##  Ω
Zin= Zin*10**-3##  k ohm
print "The input impedence = %.2f kΩ"%Zin

The voltge gain = 0.97
The input impedence = 3.85 kΩ

Example 7.4 Page No 187

In [7]:
# given data
R_E= 430##  Ω
R_L= 100##  Ω
R1= 10*10**3##  Ω
R2= 10*10**3##  Ω
bita= 200#
r_e= 2.5##  Ω
# The load resistance
r_L= R_E*R_L/(R_E+R_L)##  Ω
# The power gain 
G= bita*r_L/(r_L+r_e)#
print "The power gain = %.2f"%G

The power gain = 194.01

Example 7.5 Page No 191

In [8]:
# given data
R_C= 5*10**3##  Ω
r_e= 25##  Ω
Vin= 1*10**-3##  V
R_L= 1*10**3##  Ω
A= R_C/r_e#
# Thevenin voltage,
V_TH= A*Vin##  V
# The ac output voltage 
Vout= R_L*V_TH/(R_C+R_L)##  V
Vout= Vout*10**3##  mV
print "The ac output voltage = %.2f mV"%Vout

The ac output voltage = 33.33 mV

Example 7.7 Page No 194

In [9]:
# given data
V_B= 1.8##  V
V_E= 1.1##  V
V_TH= 200*10**-3##  V
I_E= 1*10**-3##  A
r_e= 2.5##in Ω
bita=200#
V_CC= 10##  V
R_C= 5*10**3##  Ω
R_E= 430##  Ω
R_L= 1*10**3##in Ω
I_C= I_E##  A
# The collector voltage,
V_C= V_CC-I_C*R_C##  V
V_E= 4.3##  V
# The emitter current,
I_E= V_E/R_E##  A
# The base current,
I_B= I_E/bita##  A
# The load resistance,
r_L= R_E*R_L/(R_E+R_L)##  Ω
Zin= bita*(r_L+r_e)##  Ω
Vin= Zin*V_TH/(R_C+Zin)##  V
# The ac output voltage
Vout= r_L*Vin/(r_L+r_e)##in V
Vout= Vout*10**3##  mV
print "The ac output voltage = %.2f mV"%Vout

The ac output voltage = 183.21 mV

Example 7.9 Page No 199

In [10]:
# given data
R1= 100##in kΩ
R2= 100##in kΩ
R3= 360##in Ω
bita= 100#
V1= 5##  V
v1= 1.4##  V
v2= 25##  mV
# Voltage at first base
V2= R1/R2*V1##  V
# Emitter current = %.2f second transistor
I_E2= (V2-v1)/R3##  A
I_E2= I_E2*10**3##  mA
# Resistance of second emitter diode,
r_desh_e2= v2/I_E2##  Ω
# Base current
I_B2= I_E2/bita##  mA 
# Emitter current,
I_E1= I_B2##  mA
# First emitter diode resistance
r_desh_e1= v2/I_E1##  Ω
print "The value of r''e2 = %.2f Ω"%r_desh_e2
print "The value of r''e1 = %.2f Ω"%r_desh_e1

The value of r''e2 = 2.50 Ω
The value of r''e1 = 250.00 Ω

Example 7.10 Page No 200

In [11]:
# given data
R_E= 360##  Ω
R_L= 1*10**3##  Ω
R1= 100*10**3##in Ω
R2= 100*10**3##in Ω
r_desh_e1= 250##  Ω
r_desh_e2= 2.5##  Ω
h_FE= 100#
h_fe= 100#
# The load resistance,
r_L= R_E*R_L/(R_E+R_L)##  Ω
Zin1= h_FE*h_fe*r_L##  Ω
Zin= R1*R2*Zin1/(R1*R2+R2*Zin1+Zin1*R1)##  Ω
Zin2= h_FE*(r_L+r_desh_e2)##  Ω
Zin1= h_FE*(Zin2+r_desh_e1)##  Ω
# The input impedence 
Zin= R1*R2*Zin1/(R1*R2+R2*Zin1+Zin1*R1)##  Ω
Zin= Zin*10**-3##  k ohm
print "The input impedence = %.2f kΩ"%Zin

The input impedence = 49.09 kΩ

Example 7.11 Page No 203

In [12]:
# given data
Vin= 20##  V
Vz= 10##  V
Rs= 680##  Ω
V_BE= 0.7##  V
R_L= 15##  Ω
bita= 80#
Is= (Vin-Vz)/Rs##  A
Vout= Vz-V_BE##  V
I_E= Vout/R_L##  A
I_L= I_E##  A
I_B= I_E/bita##  A
# The current through the zener diode 
Iz= Is-I_B##  A
V_CE= Vin-Vout##  V
# The transistor power dissipation 
Po= I_L*(Vin-Vout)##  W
Iz= Iz*10**3##  mA
print "The current through the zener diode = %.2f mA"%Iz
print "The transistor power dissipation = %.2f watt"%Po

The current through the zener diode = -7.75 mA
The transistor power dissipation = 6.63 watt