Chapter 31 : Power Amplifiers

Example No. 31_1 Page No. 1019

In [2]:
%matplotlib inline
from matplotlib.pyplot import plot,show,title,xlabel,ylabel
# Calculate the following dc quantities Icq, Vceq, Pd, Ic(sat) and Vce(off). Also draw the dc load line

# Given Data

R1 = 18.*10**3#       # Resistor 1=18k Ohms
R2 = 2.7*10**3#      # Resistor 2=2.7k Ohms
Vcc = 20.#           # Supply Voltage(Collector)=20 Volts
Vbe = 0.7#          # Voltage Base-Emitter=0.7 Volts
Re = 240.#           # Emitter Resistor=240 Ohms
Rc = 1.*10**3#        # Collector Resistor=1k Ohms

Vb = Vcc*(R2/(R1+R2))#

Ve = Vb-Vbe#

#Ie = Ic#

Icq = Ve/Re#
print 'The value of Icq = %0.2e Amps'%Icq
print 'i.e Approx 7.91 mAmps'

Vceq = Vcc-Icq*(Rc+Re)#
print 'The value of Vceq = %0.2f Volts'%Vceq

Pd = Vceq*Icq#
print 'The Power Dissipation = %0.2e Watts'%Pd
print 'i.e 80.6 mWatts'

Icsat = Vcc/(Rc+Re)#
print 'The value of Ic(sat) = %0.2e Amps'%Icsat
print 'i.e 16.1 mAmps'

Vceoff = Vcc#
print 'The value of Vce(off) = %0.2f Volts'%Vceoff

# For DC load line

Vce1=[Vceoff ,Vceq, 0]
Ic1=[0, Icq, Icsat]

#To plot DC load line

print "Q(%f,%f)\n"%(Vceq,Icq)
plot(Vce1, Ic1)
plot(Vceq,Icq)
plot(0,Icq)
plot(Vceq,0)
plot(0,Icsat)
plot(Vceoff,0)
xlabel("Vce in volt")
ylabel("Ic in Ampere")
title("DC Load-line for Common-Emitter Class A Amplifier Circuit")
show()

The value of Icq = 7.95e-03 Amps
i.e Approx 7.91 mAmps
The value of Vceq = 10.14 Volts
The Power Dissipation = 8.06e-02 Watts
i.e 80.6 mWatts
The value of Ic(sat) = 1.61e-02 Amps
i.e 16.1 mAmps
The value of Vce(off) = 20.00 Volts
Q(10.138406,0.007953)

Example No. 31_2 Page No. 1024

In [5]:
%matplotlib inline
from matplotlib.pyplot import plot,show,title,xlabel,ylabel

# Claculate the following AC quantities Av, Vout, Pl, Pcc and percent efficiency. Also calculate the endpoints of ac loadline

# Given data

Icq = 7.91*10**-3#       # Collector Currect(Q-point)=7.91 mAmps
Rl = 1.5*10**3#          # Load Resistor=1.5 kOhms
Rc = 1.*10**3#            # Collector Resistor=1 kOhms
Vin = 25.*10**-3#         # Input Voltage=25 mVolts(p-p)
R1 = 18.*10**3#           # Resistor 1=18 kOhms
R2 = 2.7*10**3#          # Resistor 2=2.7 kOhms
Vcc = 20.#               # Supply Voltage(Collector)=20 Volts
Vceq = 10.19#           # Voltage Colector-Emitter(Q-point)=10.19 Volts

rc = (25.*10**-3)/Icq#
rl = (Rc*Rl)/(Rc+Rl)

Av = rl/rc#
print 'The Voltage Gain Av =%0.2f'%Av
print 'Approx 190'

Vout = Av*Vin#
print 'The Output Voltage = %0.2f Volts'%Vout

Pl = (Vout*Vout)/(8*Rl)#
print 'The Load Power = %0.2e Watts'%Pl
print 'i.e Approx 1.88 mWatts'

Ivd = Vcc/(R1+R2)#
# Ic = Icq
Icc = Ivd+Icq#

Pcc = Vcc*Icc#
print 'The Dc Input Power = %0.2e Watts'%Pcc
print 'i.e Approx 177.4 mWatts'

efficiency = ((Pl/Pcc)*100)#
print 'The Efficiency in %% =%0.2f'%efficiency
print 'Approx 1%'

# Endpoints of AC load line

icsat = Icq+(Vceq/rl)#
print 'The Y-axis Value of AC Load-line is ic(sat) = %0.2e Amps'%icsat
print 'i.e 24.89 mAmps'

vceoff = Vceq+Icq*rl#
print 'The X-axis value of AC Load-line is vce(off) = %0.2f Volts'%vceoff

# For AC load line

Vce1=[vceoff, Vceq, 0]
Ic1=[0 ,Icq ,icsat]

#To plot AC load line

print "Q(%f,%f)\n"%(Vceq,Icq)
plot(Vce1, Ic1)
plot(Vceq,Icq)
plot(0,Icq)
plot(Vceq,0)
plot(0,icsat)
plot(vceoff,0)
xlabel("Vce in volt")
ylabel("Ic in Ampere")
title("AC Load-line for Common-Emitter Class A Amplifier Circuit")
show()

The Voltage Gain Av =189.84
Approx 190
The Output Voltage = 4.75 Volts
The Load Power = 1.88e-03 Watts
i.e Approx 1.88 mWatts
The Dc Input Power = 1.78e-01 Watts
i.e Approx 177.4 mWatts
The Efficiency in % =1.06
Approx 1%
The Y-axis Value of AC Load-line is ic(sat) = 2.49e-02 Amps
i.e 24.89 mAmps
The X-axis value of AC Load-line is vce(off) = 14.94 Volts
Q(10.190000,0.007910)

Example No. 31_3 Page No. 1025

In [6]:
#Calculate the following quantities: Pl, Pcc, Pdmax & percent efficiency

# Given data

Vin = 20.#       # Input Voltage=20 Volts(p-p)
Vopp = 20.#      # Output Voltage(p-p)=20 Volts(p-p)
Vcc = 24.#       # Supply Voltage(Collector)=24 Volts
Vop = 10.#       # Output Voltage(peak)=10 Volts
Rl = 8.#         # Load Resistor=8 Ohms

Vopp1 = Vopp*Vopp#
Pl = (Vopp1/(8*Rl))#
print 'The Load Power = %0.2f Watts'%Pl

Icc = ((Vop/Rl)*0.318)#

Pcc = Vcc*Icc
print 'The DC Input Power = %0.2f Watts'%Pcc

eff = ((Pl/Pcc)*100)#
print 'The Efficiency in %% =%0.2f'%eff

Pd = (Vcc*Vcc)/(40*Rl)#
print 'The Maximum Power Dissipation = %0.2f Watts'%Pd

The Load Power = 6.25 Watts
The DC Input Power = 9.54 Watts
The Efficiency in % =65.51
The Maximum Power Dissipation = 1.80 Watts

Example No. 31_4 Page No. 1037

In [14]:
# Calculate the following quantities Pl, Pcc & percent efficiency

# Given data

Rl = 8#             # Load Resistor=8 Ohms
Vopp = 50#          # Output Voltage(p-p)=50 Volts(p-p)
Vcc = 30#           # Supply Voltage(Collector)=30 Volts
Vopk = Vopp/2#      # Output Voltage(peak)

Pl = (Vopp*Vopp)/(8*Rl)#
print 'The Load Power = %0.2f Watts'%Pl

Pcc = Vcc*0.636*(Vopk/Rl)#
print 'The DC Input Power = %0.2f Watts'%Pcc

efficiency = ((Pl/Pcc)*100)#
print 'The Efficiency in %% = %0.2f'%efficiency

The Load Power = 39.00 Watts
The DC Input Power = 57.24 Watts
The Efficiency in % = 68.13

Example No. 31_5 Page No. 1038

In [13]:
from math import sqrt
# Calculate the fr of LC tank circuit and dc bias voltage at base

# Given data

L = 100*10**-6#      # Inductor=100 uHenry
C = 63.325*10**-12#  # Capacitor=63.325 pFarad
Vin = 1.5#          # Input Voltage(peak)=1.5 Volts
Vbe = 0.7#          # Voltage Base-Emitter=0.7 Volts

A = sqrt(L*C)#
fr = 1./(2*3.14*A)#
print 'The Resonant Frequency = %0.2e Hertz'%fr
print 'i.e 2 MHz'

Vdc = (Vin-Vbe)#
print 'The DC Bias Voltage at Base = %0.2f Volts'%Vdc

The Resonant Frequency = 2.00e+06 Hertz
i.e 2 MHz
The DC Bias Voltage at Base = 0.80 Volts

Example No. 31_6 Page No. 1039

In [11]:
# Calculate the minimum base reisitance Rb, necessary to provide clamping action

# Given data

C = 0.01*10**-6#     # Capacitor=0.01 uFarad
fr = 2.*10**6#        # Resonant Frequency=2 MHertz

fin = fr
T = 1/fin

Rb = 10*T/C
print 'The Minimum Base Reisitance Rb to Provide Clamping Action = %0.f Ohms'%Rb

The Minimum Base Reisitance Rb to Provide Clamping Action = 500 Ohms

Example No. 31_7 Page No. 1040

In [8]:
# Calculate the Bandwidth

# Given data

L = 100*10**-6#      # Inductor=100 uHenry
fr = 2*10**6#        # Resonant Frequency=2 MHertz
ri = 12.56#         # Resistance of Coil=12.56 Ohms
Rp = 100*10**3#      # Rp=100 kOhms

Xl = 2*3.14*fr*L#
Qcoil = Xl/ri#
Ztank = Qcoil*Xl#

A = Ztank#
B = Rp#
C = (A*B)/(A+B)#
Qckt = C/Xl#

BW = fr/Qckt#
print 'The Bandwidth = %0.f Hertz'%BW
print 'i.e Approx 45 kHz'

The Bandwidth = 45120 Hertz
i.e Approx 45 kHz