Power amplifire analog electronics
rocks92
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Oct 21, 2015
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About This Presentation
Power Amplifier circuits.
Output stages of types of power amplifier (class A, class B, class AB, class C, class D)
Distortions( Harmonic and Crossover).
Push-pull amplifier with and without transformer.
Complimentary symmetry and Quasi- complimentary symmetry push pull amplifier.
Slide Content
CONTENTS:-
Power Amplifier circuits.
Output stages of types of power amplifier (class A,
class B, class AB, class C, class D)
Distortions( Harmonic and Crossover).
Push-pull amplifier with and without transformer.
Complimentary symmetry and Quasi-
complimentary symmetry push pull amplifier.
Power Amplifier circuits.
Output stages of types of power amplifier (class A,
class B, class AB, class C, class D)
Distortions( Harmonic and Crossover).
Push-pull amplifier with and without transformer.
Complimentary symmetry and Quasi-
complimentary symmetry push pull amplifier.
POWER AMPLIFIERS:-
Power Amplifier is a circuit which draw power from
D.C. power source and convert into A.C. power and
whose action is controlled by input signal. Power
Amplifier needs large current ,So they are known as
large signal amplifier. In other words we can say that
it is a multistage amplifier which consists a number of
stages that amplify a weak signal until sufficient
power available to operate a speaker or other power
handling devices.
Power Amplifier is a circuit which draw power from
D.C. power source and convert into A.C. power and
whose action is controlled by input signal. Power
Amplifier needs large current ,So they are known as
large signal amplifier. In other words we can say that
it is a multistage amplifier which consists a number of
stages that amplify a weak signal until sufficient
power available to operate a speaker or other power
handling devices.
POWER AMPLIFIERS:-
In power amplifier transformer primary winding is in
series with collector. DC power loss in primary
winding is very less as its resistance is small. Power
transfer to the secondary winding is A.C.
The initial stages which are basically voltage
amplifiers , amplify the voltage level of signal. The
last stage is called power stage and uses a power
amplifiers to amplify the power level of signal.
In power amplifier transformer primary winding is in
series with collector. DC power loss in primary
winding is very less as its resistance is small. Power
transfer to the secondary winding is A.C.
The initial stages which are basically voltage
amplifiers , amplify the voltage level of signal. The
last stage is called power stage and uses a power
amplifiers to amplify the power level of signal.
Signal
pick up
transducer
Voltage
amplifier
Voltage
amplifier
power
amplifier
Output
transducer
Non
Electrical
Energy input
Energy
Output
BLOCK DIAGRAM OF POWER
AMPLIFIERS:-
pick up
transducer
Voltage
amplifier
Voltage
amplifier
power
amplifier
Output
transducer
Non
Electrical
Energy input
Energy
Output
BLOCK DIAGRAM OF POWER
AMPLIFIERS:-
POWER TRANSISTOR:-
The transistor which is employed in power amplifier
is called power transistor. It is different from other
transistors in the following manner :
Base terminal is thicker to reduce the value of
current amplification factor.
Emitter and base layer are heavily doped.
Area of collector region is made large in order to
dissipate the heat developed in transistor during
operation.
The transistor which is employed in power amplifier
is called power transistor. It is different from other
transistors in the following manner :
Base terminal is thicker to reduce the value of
current amplification factor.
Emitter and base layer are heavily doped.
Area of collector region is made large in order to
dissipate the heat developed in transistor during
operation.
TERMS USED IN POWER AMPLIFIER:-
COLLECTOR EFFICIENCY(CONVERSION
EFFICIENCY):-
The effectiveness of power amplifier is measured on the
basis of its ability to convert dc power into ac power.
Power amplifier are designed to provide maximum
collector efficiency.
The ratio of AC output power to DC power supplied by
the DC battery of power amplifier is called Collector
Efficiency.
Collector efficiency=A.C power/D.C. power
COLLECTOR EFFICIENCY(CONVERSION
EFFICIENCY):-
The effectiveness of power amplifier is measured on the
basis of its ability to convert dc power into ac power.
Power amplifier are designed to provide maximum
collector efficiency.
The ratio of AC output power to DC power supplied by
the DC battery of power amplifier is called Collector
Efficiency.
Collector efficiency=A.C power/D.C. power
The ratio of AC output power to DC power
supplied by the DC battery of power
amplifier is termed as collector efficiency.
%collector efficiency=AC power delivered
to load/DC power supplied to load
=(Pac/Pdc)100
supplied by the DC battery of power
amplifier is termed as collector efficiency.
%collector efficiency=AC power delivered
to load/DC power supplied to load
=(Pac/Pdc)100
DISTORTION:-
Characteristics of power transistor is very non-linear.
Due to this non-linearity the wave shape of output
signal becomes different from waveform of input
signal.
Distortion is defined as the change of output wave
shape from the input wave shape of the amplifier.
Characteristics of power transistor is very non-linear.
Due to this non-linearity the wave shape of output
signal becomes different from waveform of input
signal.
Distortion is defined as the change of output wave
shape from the input wave shape of the amplifier.
POWER DISSIPATION CAPACITY
(COLLECTOR DISSIPATION):-
To keep the temperature within limit the transistor
must dissipate this heat to its surroundings. For this
Heat Sink is attached.
The ability of transistor to dissipate heat developed
in it is called as power dissipation capacity.
(COLLECTOR DISSIPATION):-
To keep the temperature within limit the transistor
must dissipate this heat to its surroundings. For this
Heat Sink is attached.
The ability of transistor to dissipate heat developed
in it is called as power dissipation capacity.
COMPARISION OF VOLTAGE COMPARISION OF VOLTAGE
AMPLIFIER AND POWER AMPLIFIER:-AMPLIFIER AND POWER AMPLIFIER:-
Difference between them is as follow:-
AMPLIFIER AND POWER AMPLIFIER:-AMPLIFIER AND POWER AMPLIFIER:-
Difference between them is as follow:-
S.NOVoltage
amplifier
Power
amplifier
1
Transistor chosen should have high
value of β about 100
Transistor should have small value of
β about 20 to 50.
2
Load resistance Rc has high value
about 10KΩ
Load has small value 10Ω to 20Ω
3
Input voltage is low approx few mVInput voltage is high about few volt
4
It has low power output & high
voltage output.
It has high power output and low
voltage output .
5
Collector current has low value
100mA.
Collector current has high value.
6
Output impedance of voltage
amplifier has high value.
Output impedance has low value.
7
Usually R-C coupling is used. Transformer or tuned circuit is always
used
amplifier
Power
amplifier
1
Transistor chosen should have high
value of β about 100
Transistor should have small value of
β about 20 to 50.
2
Load resistance Rc has high value
about 10KΩ
Load has small value 10Ω to 20Ω
3
Input voltage is low approx few mVInput voltage is high about few volt
4
It has low power output & high
voltage output.
It has high power output and low
voltage output .
5
Collector current has low value
100mA.
Collector current has high value.
6
Output impedance of voltage
amplifier has high value.
Output impedance has low value.
7
Usually R-C coupling is used. Transformer or tuned circuit is always
used
CLASSIFICATION OF POWER
AMPLIFIER:-
Power amplifier
Primary
class
according to
freq.)
According to mode of
operation
Based on driving output
Audio amp.
Radio amp.
Class A
Class B
Class AB
Class C
Class D
Single ended P.A.
Double ended P.A.
Push-Pull P.A.
Complementary &
Symmetry Push
Pull P.A.
Quasi Symmetry
Push Pull PA
AMPLIFIER:-
Power amplifier
Primary
class
according to
freq.)
According to mode of
operation
Based on driving output
Audio amp.
Radio amp.
Class A
Class B
Class AB
Class C
Class D
Single ended P.A.
Double ended P.A.
Push-Pull P.A.
Complementary &
Symmetry Push
Pull P.A.
Quasi Symmetry
Push Pull PA
AUDIO POWER AMPLIFIER:-
It is small signal power amplifier which is
used to raise power level of audio frequency
range(20Hz-20KHz).
It is small signal power amplifier which is
used to raise power level of audio frequency
range(20Hz-20KHz).
RADIO POWER AMPLIFIER:-
It is large signal power amplifiers raise
the power level of signals that have radio
frequency range from 20kHz to several MHz.
It is large signal power amplifiers raise
the power level of signals that have radio
frequency range from 20kHz to several MHz.
CLASS-A POWER AMPLIFIER:-
In this case transistor is so biased that output
current flows for the entire cycle (for 360 degree) of
the input signal thus the output wave is exactly same
as the input wave.
It’s collector efficiency is for-
(a) Direct coupling-25%
(b) Transformer coupling-50%
In this case transistor is so biased that output
current flows for the entire cycle (for 360 degree) of
the input signal thus the output wave is exactly same
as the input wave.
It’s collector efficiency is for-
(a) Direct coupling-25%
(b) Transformer coupling-50%
In this case the transistor bias and signal amplitude
are such that output current flows only during
positive half cycle(180 degree) of the input signal.
Output from Class-B operation is a rectified half
wave. Such a amplifiers are mostly used in Push-
Pull arrangements.
It’s collector efficiency is 78.5%.
CLASS-B POWER AMPLIFIER:-
are such that output current flows only during
positive half cycle(180 degree) of the input signal.
Output from Class-B operation is a rectified half
wave. Such a amplifiers are mostly used in Push-
Pull arrangements.
It’s collector efficiency is 78.5%.
CLASS-B POWER AMPLIFIER:-
CLASS-AB POWER AMPLIFIER:-
The characteristics of Class-AB amplifier lies
between Class-A and Class-B amplifiers.
It is so biased that it works for complete
positive half cycle and half of the negative
cycle.
Total conduction period is less than 360 degree but
more than 180 degree. Output signal obtained in
Class-AB operation is distorted.
The characteristics of Class-AB amplifier lies
between Class-A and Class-B amplifiers.
It is so biased that it works for complete
positive half cycle and half of the negative
cycle.
Total conduction period is less than 360 degree but
more than 180 degree. Output signal obtained in
Class-AB operation is distorted.
CLASS C POWER AMPLIFIER:-
A class C power amplifier is biased for operation for
less than 180 degree of the input signal cycle and
will operate only with a tuned or resonant circuit. In
its operation the output current flow for less than one
half cycle.
Collector efficiency of class C amplifier is 85-90%.
It is used in Tuned circuits for the purpose of Radio
or communication in RF range.
A class C power amplifier is biased for operation for
less than 180 degree of the input signal cycle and
will operate only with a tuned or resonant circuit. In
its operation the output current flow for less than one
half cycle.
Collector efficiency of class C amplifier is 85-90%.
It is used in Tuned circuits for the purpose of Radio
or communication in RF range.
CLASS D POWER AMPLIFIER:-
A Class D power amplifiers are designed to operate
with digital or pulse type signals and it’s overall
efficiency above 90 degree.
MOSFET is mainly used in Class-D amplifiers.
The efficiency of class D amplifier is above 90%.
A Class D power amplifiers are designed to operate
with digital or pulse type signals and it’s overall
efficiency above 90 degree.
MOSFET is mainly used in Class-D amplifiers.
The efficiency of class D amplifier is above 90%.
SINGLE ENDED POWER AMPLIFIER:-
It uses single transistor and derives output power
with one end permanently ground.
It uses single transistor and derives output power
with one end permanently ground.
DOUBLE ENDED POWER AMPLIFIER:-
Double ended uses two transistors in single stage. It
consists of two loops in which the transistor
collector current flows in opposite direction but add
in the load.
Double ended uses two transistors in single stage. It
consists of two loops in which the transistor
collector current flows in opposite direction but add
in the load.
PUSH PULL POWER AMPLIFIER:-
It uses two transistors having
complementary symmetry (one PNP and
NPN). They have symmetry as they are
made with the same material and technology.
It uses two transistors having
complementary symmetry (one PNP and
NPN). They have symmetry as they are
made with the same material and technology.
COMPLEMENTRAY & SYMMETRY
POWER AMPLIFIER
This is nothing but a Push Pull amplifier in which we
use a phase-splitter circuit to make phase shift of
180`
POWER AMPLIFIER
This is nothing but a Push Pull amplifier in which we
use a phase-splitter circuit to make phase shift of
180`
QUASI SYMMETRY POWER AMPLIFIER
This is amplifier in which we use four transistor of
two group as:
Group 1: Darlington pair ; Q1(NPN),Q3(NPN)
Group 2: Feedback pair ; Q2(PNP),Q3(NPN)
This is amplifier in which we use four transistor of
two group as:
Group 1: Darlington pair ; Q1(NPN),Q3(NPN)
Group 2: Feedback pair ; Q2(PNP),Q3(NPN)
CLASS-A AMPLIFIER:-
In this type, the transistor is so biased that the output
current flows for the full cycle of the input signal, as
shown in fig.
This means that the transistor remains forward biased
throughout the input cycle. From fig. it is seen that the
operating point Q is located at the centre of the load
line. So that the output current flows for complete
cycle of the input signal.
In this type, the transistor is so biased that the output
current flows for the full cycle of the input signal, as
shown in fig.
This means that the transistor remains forward biased
throughout the input cycle. From fig. it is seen that the
operating point Q is located at the centre of the load
line. So that the output current flows for complete
cycle of the input signal.
CLASS-A AMPLIFIER OPERATION:-
Class A power amplifier:-
INPUT
WAVEFORM
OUTPUT
WAVEFORM
INPUT
WAVEFORM
OUTPUT
WAVEFORM
CLASS-A AMPLIFIER:-
In class A operation, signal is faithfully reproduced at
the output without any distortion. This is an important
feature of class A operation. The efficiency of class-A
operation is very small.
As the collector current flows for 360degree (full cycle)
of the input signal. We can say that the angle of the
collector current flow is 360degree.
In class A operation, signal is faithfully reproduced at
the output without any distortion. This is an important
feature of class A operation. The efficiency of class-A
operation is very small.
As the collector current flows for 360degree (full cycle)
of the input signal. We can say that the angle of the
collector current flow is 360degree.
CLASS-A AMPLIFIER:-
DRAWING LOAD LINES :-
When the base current Ib is zero, the collector current Ic
is also equal to zero (neglecting reverse saturation
current Ico).
Therefore the voltage drop across load resistance is
also equal to zero and hence the collector-emitter
voltage Vce becomes equal to Vcc. Thus we get the
point 1 on fig which represents the condition of Ic =0
and Vce=Vcc.
DRAWING LOAD LINES :-
When the base current Ib is zero, the collector current Ic
is also equal to zero (neglecting reverse saturation
current Ico).
Therefore the voltage drop across load resistance is
also equal to zero and hence the collector-emitter
voltage Vce becomes equal to Vcc. Thus we get the
point 1 on fig which represents the condition of Ic =0
and Vce=Vcc.
CLASS-A AMPLIFIER:-
The Q point:-
For or class-A power amplifier, it is necessary that
transistor should conduct for the full 360 degree of
the input cycle.
Thus Q point (or operating point) selected
approximately is the mid of load line. It is shown in
the fig. as Q, here Icq represents zero signal
collector current and Vcq represents zero signal
voltage between collector and emitter respectively.
The Q point:-
For or class-A power amplifier, it is necessary that
transistor should conduct for the full 360 degree of
the input cycle.
Thus Q point (or operating point) selected
approximately is the mid of load line. It is shown in
the fig. as Q, here Icq represents zero signal
collector current and Vcq represents zero signal
voltage between collector and emitter respectively.
CLASS-A AMPLIFIER:-
operation :-
Let us assume that the input signal to the amplifier
to the amplifier is sinusoidal which results in a
sinusoidal variation of base current Ib. this in turn will
cause the collector current Ic and collector-emitter
voltage Vce to vary sinusoidally around the Q point.
operation :-
Let us assume that the input signal to the amplifier
to the amplifier is sinusoidal which results in a
sinusoidal variation of base current Ib. this in turn will
cause the collector current Ic and collector-emitter
voltage Vce to vary sinusoidally around the Q point.
CLASS-A AMPLIFIER:-
D.C. power input:-
The D.C. power input is provided by the supply Vcc
And with no input signal, the d.c. current drawn is
the collector bias current Icq. Hence d.c. power input
is
Pdc = Vcc.Icq
D.C. power input:-
The D.C. power input is provided by the supply Vcc
And with no input signal, the d.c. current drawn is
the collector bias current Icq. Hence d.c. power input
is
Pdc = Vcc.Icq
CLASS-A AMPLIFIER:-
A.C. power input:-
Let the peak value of collector current swing be
and that of collector voltage swing be
Vm.Hence there r.m.s. values are :
Im / Ö2 and Vm / Ö2
A.C. Power output using R.M.S. values :-
The a.c. power output is given by
Pac = Vrms Irms
= I
2
rms RL
= V
2
rms/RL
A.C. power input:-
Let the peak value of collector current swing be
and that of collector voltage swing be
Vm.Hence there r.m.s. values are :
Im / Ö2 and Vm / Ö2
A.C. Power output using R.M.S. values :-
The a.c. power output is given by
Pac = Vrms Irms
= I
2
rms RL
= V
2
rms/RL
CLASS-A AMPLIFIER:-
A.C. Power output using maximum values :-
The a.c. power output is given by :
Pac = Vrms.Irms
=(Vm/Ö2).Im/Ö2
=(Vm Im)/2
Since Im = Vm / RL the above equation can be rewritten :
Pac = V
2
m/2RL ----------(1)
Pac = I
2
mRL / 2 ----------(2)
Equation (1)and(2) represents a.c. power output using
peak
values.
A.C. Power output using maximum values :-
The a.c. power output is given by :
Pac = Vrms.Irms
=(Vm/Ö2).Im/Ö2
=(Vm Im)/2
Since Im = Vm / RL the above equation can be rewritten :
Pac = V
2
m/2RL ----------(1)
Pac = I
2
mRL / 2 ----------(2)
Equation (1)and(2) represents a.c. power output using
peak
values.
ANALYSIS OF CLASS-A AMPLIFIERS:-
The class –A amplifiers are further classified as :
(1) Series fed directly coupled class-A amplifiers
(2) Transformer coupled class–A amplifiers
The class –A amplifiers are further classified as :
(1) Series fed directly coupled class-A amplifiers
(2) Transformer coupled class–A amplifiers
SERIES FED DIRECTLY COUPLED
AMPLIFIER:-
In directly coupled type, the load is connected
directly in the collector. While in the transformer-
coupled type, the load is coupled to the collector
through a transformer.
AMPLIFIER:-
In directly coupled type, the load is connected
directly in the collector. While in the transformer-
coupled type, the load is coupled to the collector
through a transformer.
CIRCUIT DIAGRAM OF SERIES FED
CLASS A AMPLIFIER:-
CLASS A AMPLIFIER:-
DC OPERATION OF SERIES FED CLASS
A AMPLIFIER:-
A AMPLIFIER:-
AC OPERATION OF SERIES
CLASS A AMPLIFIER:-
o/p current
swing
o/p voltage swing
i/p signal
Ic
Vcc/Rc
VccVce
0
CLASS A AMPLIFIER:-
o/p current
swing
o/p voltage swing
i/p signal
Ic
Vcc/Rc
VccVce
0
ANALYSIS OF SERIES FED DIRECT
COUPLED CLASS A AMPLIFIERS:-
The diagram shows a class of series fed amplifier. It
is so named because the load resistance RL is
connected directly in series with collector of the
transistor Q. Resistor R1,R2,Re and capacitor Ce
are used for biasing.
To understand the operation of the circuit, we take
help of graphical method.
COUPLED CLASS A AMPLIFIERS:-
The diagram shows a class of series fed amplifier. It
is so named because the load resistance RL is
connected directly in series with collector of the
transistor Q. Resistor R1,R2,Re and capacitor Ce
are used for biasing.
To understand the operation of the circuit, we take
help of graphical method.
A.C. POWER OUTPUT USING PEAK TO
PEAK VALUES:-
wt
Imax
0
Imin
p/2 p
1
2
3
D.C. bias value
ic
PEAK VALUES:-
wt
Imax
0
Imin
p/2 p
1
2
3
D.C. bias value
ic
A.C. POWER OUTPUT USING PEAK TO PEAK
VALUES:-
We have from Equation
P
ac
= V
m
I
m
/2
but as V
m
= V
pp
/2 = (V
max
-V
min
)/2
Also Im = I
pp
/2 = (V
max
-V
min
)/2
We get P
ac
= (V
pp
/2.I
pp
/2)/2 = V
pp
.I
pp
/8
= (V
max
–V
min
)(I
max
-I
min
)/8
VALUES:-
We have from Equation
P
ac
= V
m
I
m
/2
but as V
m
= V
pp
/2 = (V
max
-V
min
)/2
Also Im = I
pp
/2 = (V
max
-V
min
)/2
We get P
ac
= (V
pp
/2.I
pp
/2)/2 = V
pp
.I
pp
/8
= (V
max
–V
min
)(I
max
-I
min
)/8
EFFICIENCY:-
The efficiency of an amplifier represents the a.c.
power delivered to the load from the d.c. source. The
generalized expression for efficiency of an amplifier is
given as
%η = P
ac
/P
dc
*100
%η = ( V
max
-V
min
) (I
max
- I
min
)/8V
cc
I
cq
*100
This efficiency is also called as conversion efficiency of
an amplifier.
The efficiency of an amplifier represents the a.c.
power delivered to the load from the d.c. source. The
generalized expression for efficiency of an amplifier is
given as
%η = P
ac
/P
dc
*100
%η = ( V
max
-V
min
) (I
max
- I
min
)/8V
cc
I
cq
*100
This efficiency is also called as conversion efficiency of
an amplifier.
MAXIMUM EFFICIENCY:-
From dig. we can say that minimum voltage possible
is zero and maximum voltage possible is Vcc, for a
maximum swing. Similarly the minimum current is
zero and the maximum current possible is 2Icq, for a
maximum swing i.e.
V
max = V
cc and V
min = 0
I
max
= 2I
cq
and I
min
=0
% η = (Vcc-0)(2Icq-0)/8V
cc
.I
cq
.100
= (2V
cc
I
cq
/8V
cc
I
cq
).100 = 25%
From dig. we can say that minimum voltage possible
is zero and maximum voltage possible is Vcc, for a
maximum swing. Similarly the minimum current is
zero and the maximum current possible is 2Icq, for a
maximum swing i.e.
V
max = V
cc and V
min = 0
I
max
= 2I
cq
and I
min
=0
% η = (Vcc-0)(2Icq-0)/8V
cc
.I
cq
.100
= (2V
cc
I
cq
/8V
cc
I
cq
).100 = 25%
This the maximum efficiency possible in directly
coupled series fed class-A amplifier is just 25%
(ideally). In practical it is less than 25%.
coupled series fed class-A amplifier is just 25%
(ideally). In practical it is less than 25%.
POWER DISSIPATION:-
The amount of power that must be dissipated by the
transistor is the difference between the d.c. power
input Pdc and the a.c. power delivered to the load
Pac.
Power dissipation
P
d
= P
dc
– P
ac
Maximum power dissipation occurs when there is
zero a.c. input signal, the a.c. power output (P
ac
) is
also zero.
P
ac
= 0 because (Iac=0)
The amount of power that must be dissipated by the
transistor is the difference between the d.c. power
input Pdc and the a.c. power delivered to the load
Pac.
Power dissipation
P
d
= P
dc
– P
ac
Maximum power dissipation occurs when there is
zero a.c. input signal, the a.c. power output (P
ac
) is
also zero.
P
ac
= 0 because (Iac=0)
POWER DISSIPATION:-
Then
P
d
= P
dc
or P
d
= V
cc
I
cq
Equation gives the maximum power dissipation by
the transistor.
Then
P
d
= P
dc
or P
d
= V
cc
I
cq
Equation gives the maximum power dissipation by
the transistor.
ADVANTAGES:-
1. The circuit is simpler to design and
to implement.
2. The load is directly connected in the collector circuit
hence no output transformer is required.
3. It keeps DC power loss small because of small
resistance of primary winding of transformer.
4. It matching of load impedance with source
impedance is possible with transformer.
1. The circuit is simpler to design and
to implement.
2. The load is directly connected in the collector circuit
hence no output transformer is required.
3. It keeps DC power loss small because of small
resistance of primary winding of transformer.
4. It matching of load impedance with source
impedance is possible with transformer.
DISADVANTAGE:-
1.The load resistance is directly connected in the
collector, therefore the quiescent collector current
flows through it, there will be considerable power
loss in the load resistance.
2. Power dissipation is more, hence use of heat sink is
essential.
1.The load resistance is directly connected in the
collector, therefore the quiescent collector current
flows through it, there will be considerable power
loss in the load resistance.
2. Power dissipation is more, hence use of heat sink is
essential.
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
The above mentioned disadvantages of series fed
directly connected amplifier can be overcome can
be overcome by employing a transformer known as
output transformer.
The transformer is used to couple the load RL to the
collector circuit of the transistor.
AMPLIFIER:-
The above mentioned disadvantages of series fed
directly connected amplifier can be overcome can
be overcome by employing a transformer known as
output transformer.
The transformer is used to couple the load RL to the
collector circuit of the transistor.
TRANSFORMER CLASS –A AMPLIFIER:-
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
(A)CIRCUIT DIAGRAM:-
The basic circuit of a transformer-coupled amplifier
is shown in fig. the primary of the transformer,
having negligible d.c. resistance is connected in
the collector circuit. The secondary of transformer
is connected to load RL (loud speaker as load ).
AMPLIFIER:-
(A)CIRCUIT DIAGRAM:-
The basic circuit of a transformer-coupled amplifier
is shown in fig. the primary of the transformer,
having negligible d.c. resistance is connected in
the collector circuit. The secondary of transformer
is connected to load RL (loud speaker as load ).
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
Let us consider only the transformer and the load RL
as shown in fig.
Impedance RL is connected across transformer
secondary. This impedance is changed by the
transformer when viewed at the primary side(RL).
The current ratio between primary and secondary of
the transformer be N. Therefore, the voltage ratio
and current ratio can written as
V1/V2 = N
and I2/I1 = N
AMPLIFIER:-
Let us consider only the transformer and the load RL
as shown in fig.
Impedance RL is connected across transformer
secondary. This impedance is changed by the
transformer when viewed at the primary side(RL).
The current ratio between primary and secondary of
the transformer be N. Therefore, the voltage ratio
and current ratio can written as
V1/V2 = N
and I2/I1 = N
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
Where
N is equal to N1/N2
R
L
is actual load resistance
R
L
’ is equal to resistance presented by the
primary winding to the supply source.
The primary winding acts as a resistance equal to
load resistance across the secondary multiplied by
the square of turn ratio.
AMPLIFIER:-
Where
N is equal to N1/N2
R
L
is actual load resistance
R
L
’ is equal to resistance presented by the
primary winding to the supply source.
The primary winding acts as a resistance equal to
load resistance across the secondary multiplied by
the square of turn ratio.
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
D.C. OPERATION:-
There will be no voltage drop across the primary
winding of a transformer under quiescent condition.
The slop of d.c. load line is given as reciprocal of the
d.c. resistance in the collector circuit. The d.c.
resistance is zero hence slop will be infinite.
AMPLIFIER:-
D.C. OPERATION:-
There will be no voltage drop across the primary
winding of a transformer under quiescent condition.
The slop of d.c. load line is given as reciprocal of the
d.c. resistance in the collector circuit. The d.c.
resistance is zero hence slop will be infinite.
AC AND DC LOAD LINE OF TRANSFORMER
COUPLED CLASS A AMPLIFIER:-
Ic
Vce
Ac load line
Dc load line
Q
Ic max
2 Icq
Ic min=0
Vce min=0
Vce max=0
2Vcc
Icq
Vceq=Vcc
At center of ac
load line
COUPLED CLASS A AMPLIFIER:-
Ic
Vce
Ac load line
Dc load line
Q
Ic max
2 Icq
Ic min=0
Vce min=0
Vce max=0
2Vcc
Icq
Vceq=Vcc
At center of ac
load line
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
D.C. POWER INPUT:-
The d.c. power input is given by
P
dc
=V
cc
I
cq
This expression is same as for series fed directly
coupled class-A amplifier.
AMPLIFIER:-
D.C. POWER INPUT:-
The d.c. power input is given by
P
dc
=V
cc
I
cq
This expression is same as for series fed directly
coupled class-A amplifier.
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
AC OPERATION:-
In class-A operation, the Q-point is located at the
center of the load line.The a.c. load line is obtained
by drawing the line throw the operating point. Slope
of a.c. load line is equal to : -1/RL
When a.c. signal is applied collector current is varies
with input signal and accordingly operating point Q
is shifted its position up and down to the load line.
AMPLIFIER:-
AC OPERATION:-
In class-A operation, the Q-point is located at the
center of the load line.The a.c. load line is obtained
by drawing the line throw the operating point. Slope
of a.c. load line is equal to : -1/RL
When a.c. signal is applied collector current is varies
with input signal and accordingly operating point Q
is shifted its position up and down to the load line.
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
AC OUTPUT POWER:-
Generalized expression for a.c. output power for
transformer coupled amplifier is same as
P
ac
= (V
pp
/2.I
pp
/2)/2 = V
pp
.I
pp
/8
= (V
max
–V
min
)(I
max
-I
min
)/8
EFFICIENCY:-
%η = P
ac
/P
dc
.100
%η = (V
max
-V
min
) (I
max
- I
min
)/8V
cc
I
cq
.100
AMPLIFIER:-
AC OUTPUT POWER:-
Generalized expression for a.c. output power for
transformer coupled amplifier is same as
P
ac
= (V
pp
/2.I
pp
/2)/2 = V
pp
.I
pp
/8
= (V
max
–V
min
)(I
max
-I
min
)/8
EFFICIENCY:-
%η = P
ac
/P
dc
.100
%η = (V
max
-V
min
) (I
max
- I
min
)/8V
cc
I
cq
.100
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
MAXIMUM EFFICIENCY :-
V
max
= 2V
cc
and V
min
= 0
I
max
= 2I
cq
and I
min
=0
% η = (2V
cc
-0)(2I
cq
-0)/8V
cc
.I
cq
.100
= (4V
ccI
cq/8V
ccI
cq).100 = 50%
AMPLIFIER:-
MAXIMUM EFFICIENCY :-
V
max
= 2V
cc
and V
min
= 0
I
max
= 2I
cq
and I
min
=0
% η = (2V
cc
-0)(2I
cq
-0)/8V
cc
.I
cq
.100
= (4V
ccI
cq/8V
ccI
cq).100 = 50%
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
POWER DISSIPATION:-
The power dissipation by the transistor is given
by : P
d
=P
dc
-P
ac
When there is no input signal the entire d.c.
power gets dissipation in the form of heat, which is
maximum power dissipation.
(Pd)
max
=V
cc
I
cq
AMPLIFIER:-
POWER DISSIPATION:-
The power dissipation by the transistor is given
by : P
d
=P
dc
-P
ac
When there is no input signal the entire d.c.
power gets dissipation in the form of heat, which is
maximum power dissipation.
(Pd)
max
=V
cc
I
cq
TRANSFORMER COUPLED CLASS-A
AMPLIFIER:-
Advantages:-
1 Efficiency of the operation is higher (50%)then
directly coupled series fed amplifier(25%).
2 The d.c. power bias current that flows through RL in
series fed amplifier and causes power loss in RL, is
stopped n case of transformer coupled. This is the
main reason for the increased efficiency over
series fed operation.
3 Impedance matching which is an essential
requirement for maximum power transfer is
possible.
AMPLIFIER:-
Advantages:-
1 Efficiency of the operation is higher (50%)then
directly coupled series fed amplifier(25%).
2 The d.c. power bias current that flows through RL in
series fed amplifier and causes power loss in RL, is
stopped n case of transformer coupled. This is the
main reason for the increased efficiency over
series fed operation.
3 Impedance matching which is an essential
requirement for maximum power transfer is
possible.
DISADVANTAGES:-
1. Due to transformer the circuit becomes bulkier
and costlier.
2. The circuit is complicated to design and
implement as compared to directly coupled series
fed amplifier.
3. Frequency response of the circuit is poor.
1. Due to transformer the circuit becomes bulkier
and costlier.
2. The circuit is complicated to design and
implement as compared to directly coupled series
fed amplifier.
3. Frequency response of the circuit is poor.
HARMONIC DISTORTION IN POWER
AMPLIFIER:-
Harmonic distortion means the presence of frequency
components in the output waveform which are not
present in the input signal.
Due to non linearity amplification of all the portion
of positive and negative half cycles is not same and it
causes the output waveform to be different from
input waveform.
When the input signal is applied to a transistor the
non-linear characteristics causes the positive half of
the signal to be amplified more than negative half
cycle.
AMPLIFIER:-
Harmonic distortion means the presence of frequency
components in the output waveform which are not
present in the input signal.
Due to non linearity amplification of all the portion
of positive and negative half cycles is not same and it
causes the output waveform to be different from
input waveform.
When the input signal is applied to a transistor the
non-linear characteristics causes the positive half of
the signal to be amplified more than negative half
cycle.
HARMONIC DISTORTION IN
POWER AMPLIFIER:-
Due to this output signal signal contains
fundamental frequency components and some
undesired frequency components,which are integral
multiple of input signal frequency.
These additional frequency components are called
harmonics. Hence the output is said to be distorted,
this is called harmonic distortion.
POWER AMPLIFIER:-
Due to this output signal signal contains
fundamental frequency components and some
undesired frequency components,which are integral
multiple of input signal frequency.
These additional frequency components are called
harmonics. Hence the output is said to be distorted,
this is called harmonic distortion.
HARMONIC DISTORTION IN
POWER AMPLIFIER:-
Out of all the harmonic components, the second
harmonics has the largest amplitude. As the order of
the harmonic increases. Since the second harmonic
amplitude is largest, the second harmonic distortion
is more important in the analysis of Audio Frequency
amplifier.
POWER AMPLIFIER:-
Out of all the harmonic components, the second
harmonics has the largest amplitude. As the order of
the harmonic increases. Since the second harmonic
amplitude is largest, the second harmonic distortion
is more important in the analysis of Audio Frequency
amplifier.
HARMONIC DISTORTION IN POWER AMPLIFIER:-
Distorted sinusoidal waveform
Fundamental sinusoidal waveform
Second harmonic component
Third harmonic component
V
V
V
V
V
0
0
0
0
0
Vm sin2wt
Vmcos2wt
Vm sin3wt
Waveform due to
fundamental,2
nd
,3
rd
,harmonic
component
Vmsinwt
wt
wt
wt
wt
wt
Vm cos2wt
Vm sin2wt
Vm sin3wt
Vmsinwt
Distorted sinusoidal waveform
Fundamental sinusoidal waveform
Second harmonic component
Third harmonic component
V
V
V
V
V
0
0
0
0
0
Vm sin2wt
Vmcos2wt
Vm sin3wt
Waveform due to
fundamental,2
nd
,3
rd
,harmonic
component
Vmsinwt
wt
wt
wt
wt
wt
Vm cos2wt
Vm sin2wt
Vm sin3wt
Vmsinwt
SECOND HARMONIC DISTORTION:-
1
2
3
4
5
D.C. bias value
Imax
I1/2
Icq
I-1/2
Imin
p/3p/22p/3 2p
p
wt
Ic
1
2
3
4
5
D.C. bias value
Imax
I1/2
Icq
I-1/2
Imin
p/3p/22p/3 2p
p
wt
Ic
SECOND HARMONIC DISTORTION:-
For linear amplification, the dynamic transfer
characteristics relation b/w I
b
and I
c
which is known
as transfer characteristic must be linear. To evaluate
the second harmonic distortion assume that the
dynamic transfer characteristics of the transistor is
parabolic (non-linear ) in nature rather than (linear).
For linear amplification, the dynamic transfer
characteristics relation b/w I
b
and I
c
which is known
as transfer characteristic must be linear. To evaluate
the second harmonic distortion assume that the
dynamic transfer characteristics of the transistor is
parabolic (non-linear ) in nature rather than (linear).
SECOND HARMONIC DISTORTION:-
Equation that describes distorted signal waveform is:
i
c
= I
CQ
+ I
0
+ I
1
cos(ωt) + I
2
cos(2ωt)
Where
(I
CQ
+I
0
) =dc component independent of time.
I
1= fundamental component of distorted ac signal. I
2=
second harmonic component , at twice the fundamental
frequency.
Equation that describes distorted signal waveform is:
i
c
= I
CQ
+ I
0
+ I
1
cos(ωt) + I
2
cos(2ωt)
Where
(I
CQ
+I
0
) =dc component independent of time.
I
1= fundamental component of distorted ac signal. I
2=
second harmonic component , at twice the fundamental
frequency.
MEASUREMENT OF SECOND HARMONIC
DISTORTION:-
At point 1,ωt=0
I
c
=I
CQ+I
0
+I
1cos0+I
2cos0
I
c
=I
CQ+I
0+I
1+I
2
=I
C(MAX)
At point 2,wt=P/2
I
C=I
CQ+I
0+I
1cosP/2+I
2cos2P/2
I
c=I
CQ+I
0 +I
2
At point 3,wt=P
I
c
=I
CQ+I
0+I
1cosP+I
2cos2P
I
c
=I
CQ+I
0-I
1+I
2 =I
C(MIN)
DISTORTION:-
At point 1,ωt=0
I
c
=I
CQ+I
0
+I
1cos0+I
2cos0
I
c
=I
CQ+I
0+I
1+I
2
=I
C(MAX)
At point 2,wt=P/2
I
C=I
CQ+I
0+I
1cosP/2+I
2cos2P/2
I
c=I
CQ+I
0 +I
2
At point 3,wt=P
I
c
=I
CQ+I
0+I
1cosP+I
2cos2P
I
c
=I
CQ+I
0-I
1+I
2 =I
C(MIN)
MEASUREMENT OF SECOND
HARMONIC DISTORTION:-
At ωt=0,I
c
=I
max
At ωt=P/2,I
c
=I
CQ
At ωt=P,I
c
=I
min
I
max
=I
CQ
+I
0
+I
1
+I
2
I
cq
=I
CQ
+I
0
-I
2
I
min
=I
CQ
+I
0
-I
1
+I
2
,
I
0
=I
2
I
max
-I
min
=2I
1
HARMONIC DISTORTION:-
At ωt=0,I
c
=I
max
At ωt=P/2,I
c
=I
CQ
At ωt=P,I
c
=I
min
I
max
=I
CQ
+I
0
+I
1
+I
2
I
cq
=I
CQ
+I
0
-I
2
I
min
=I
CQ
+I
0
-I
1
+I
2
,
I
0
=I
2
I
max
-I
min
=2I
1
MEASUREMENT OF SECOND HARMONIC
DISTORTION:-
I
1
=(I
c max
-I
c min
)/2
I
0
=I
2
I
max
+I
min
=2I
CQ
+2(I
0
+I
2
)=2I
CQ
+4I
2
I
2
=(I
max
+I
min
-2I
CQ
)/4
As the amplitude of the fundamental and second
harmonic are known ,the % of second harmonic
distortion can be calculated as
%D
2
=[ (I
2
/ I
1
)*100 ]
DISTORTION:-
I
1
=(I
c max
-I
c min
)/2
I
0
=I
2
I
max
+I
min
=2I
CQ
+2(I
0
+I
2
)=2I
CQ
+4I
2
I
2
=(I
max
+I
min
-2I
CQ
)/4
As the amplitude of the fundamental and second
harmonic are known ,the % of second harmonic
distortion can be calculated as
%D
2
=[ (I
2
/ I
1
)*100 ]
MEASUREMENT OF OVERALL HARMONIC
DISTORTION:-
D2 =(Ic
max
+Ic
min
-2Icq ) *100/ 2(Ic
max
-Ic
min
)
D2 =(Vce
max
+Vce
min
-2Vceq ) *100/2(Vce
max
-Vce
min
)
DISTORTION:-
D2 =(Ic
max
+Ic
min
-2Icq ) *100/ 2(Ic
max
-Ic
min
)
D2 =(Vce
max
+Vce
min
-2Vceq ) *100/2(Vce
max
-Vce
min
)
5 POINT METHOD OF HARMONIC
DISTORTION:-
1. Fundamental 2,3,4,-----------so on ,harmonic
distortion is given by
D
2
=I
2
/I
1
D
3
=I
3
/I
1
D
4
=I
4
/I
1
and so on.
2. Total harmonic distortion is given by
D
2
= D
2
2
+D
3
2
+D
4
2
+---------
DISTORTION:-
1. Fundamental 2,3,4,-----------so on ,harmonic
distortion is given by
D
2
=I
2
/I
1
D
3
=I
3
/I
1
D
4
=I
4
/I
1
and so on.
2. Total harmonic distortion is given by
D
2
= D
2
2
+D
3
2
+D
4
2
+---------
Measurement Of overall Harmonic
Distortion:-
3. When the distortion is negligible ,the power
delivered to the load given by
P
ac
=I
2
m
R
L
/2
Where
I
m
=peak value of the output waveform
=(I
max
-I
min
)/2
4. If B
1
=Fundamental frequency component
P
ac
=I
2
1
R
L
/2
Distortion:-
3. When the distortion is negligible ,the power
delivered to the load given by
P
ac
=I
2
m
R
L
/2
Where
I
m
=peak value of the output waveform
=(I
max
-I
min
)/2
4. If B
1
=Fundamental frequency component
P
ac
=I
2
1
R
L
/2
MEASUREMENT OF OVERALL HARMONIC
DISTORTION:-
5) With distortion ,the power delivered to the load
increases proportional to the amplitude of the
harmonic component as:
(P
ac
)
D
=I
2
1
R
L
/2+I
2
2
R
L
/2
=I
2
1
R
L
(1+I
2
2
/I
2
1
)/2
=P
ac(1+D
2
2)
This is the power delivered to the load due the
second harmonic distortion.
DISTORTION:-
5) With distortion ,the power delivered to the load
increases proportional to the amplitude of the
harmonic component as:
(P
ac
)
D
=I
2
1
R
L
/2+I
2
2
R
L
/2
=I
2
1
R
L
(1+I
2
2
/I
2
1
)/2
=P
ac(1+D
2
2)
This is the power delivered to the load due the
second harmonic distortion.
HIGHER ORDER HARMONIC DISTORTION:-
The collector current due to higher order harmonic
be :
I
c
=G
1
I
b
+G
2
I
b
2
+G
3
I
b
3
+……….
The input signal is given by :
I
b
= I
bm
cosωt
I
c
=G
1
I
bm
cosωt+G
2
I
bm
cos
2
ωt+G
3I
bm
cos
3
ωt+……….
I
c
=B
0
+B
1
cosωt+B
2
cos2ωt+B
3
cos3ωt+………
The collector current due to higher order harmonic
be :
I
c
=G
1
I
b
+G
2
I
b
2
+G
3
I
b
3
+……….
The input signal is given by :
I
b
= I
bm
cosωt
I
c
=G
1
I
bm
cosωt+G
2
I
bm
cos
2
ωt+G
3I
bm
cos
3
ωt+……….
I
c
=B
0
+B
1
cosωt+B
2
cos2ωt+B
3
cos3ωt+………
HIGHER ORDER HARMONIC DISTORTION:-
At point 1, ωt=0,I
c
=I
max
I
max
=I
CQ
+B
0
+B
1
+B
2
+B
3
+…….
At point 2, ωt=P/3,I
c=I
1/2
I
1/2
=I
CQ
+B
0
+0.5B
1
-0.5B
2
….
At point 3, ωt=P/2,I
c
=I
CQ
At point 4, ωt=2P/3,Ic=I
-1/2
I
-1/2
=I
CQ
+B
0
-0.5B
1
-0.5B
2
+B
3
-0.5B
4
+…..
At point 5,wt=P ,Ic=I
min
I
min
=I
CQ
+B
0
-B
1
+B
2
-B
3
+B
4
……
At point 1, ωt=0,I
c
=I
max
I
max
=I
CQ
+B
0
+B
1
+B
2
+B
3
+…….
At point 2, ωt=P/3,I
c=I
1/2
I
1/2
=I
CQ
+B
0
+0.5B
1
-0.5B
2
….
At point 3, ωt=P/2,I
c
=I
CQ
At point 4, ωt=2P/3,Ic=I
-1/2
I
-1/2
=I
CQ
+B
0
-0.5B
1
-0.5B
2
+B
3
-0.5B
4
+…..
At point 5,wt=P ,Ic=I
min
I
min
=I
CQ
+B
0
-B
1
+B
2
-B
3
+B
4
……
Higher Order Harmonic distortion:-
Solving these equations
B
0
=(I
max
-2I
1/2
+2I
-1/2
+I
min
)/6-I
cq
B
1
=(I
max
+I
1/2
-I
-1/2
-I
min
)/3
B
2
=(I
max
-2I
cq
+I
min
)/4
B
3
=(Imax-2 I1/2+2I-1/2-Imin)/6
B
4
=(I
max
-4I
1/2
+6I
cq
-4I
-1/2
+I
min
)/12
Hence the harmonic distortion coefficients can be
obtained as
D
n
=|B
0
|/|B
1
|
Solving these equations
B
0
=(I
max
-2I
1/2
+2I
-1/2
+I
min
)/6-I
cq
B
1
=(I
max
+I
1/2
-I
-1/2
-I
min
)/3
B
2
=(I
max
-2I
cq
+I
min
)/4
B
3
=(Imax-2 I1/2+2I-1/2-Imin)/6
B
4
=(I
max
-4I
1/2
+6I
cq
-4I
-1/2
+I
min
)/12
Hence the harmonic distortion coefficients can be
obtained as
D
n
=|B
0
|/|B
1
|
POWER OUTPUT DUE TO DISTORTION:-
Now
P
ac
=B
2
1
R
L
/2
Hence the output power due to harmonic distortion is
(P
ac
)
D
=(B
2
1
R
l
+ B
2
2
R
l
+ B
2
3
R
l
+…..)/2
=B
2
1
R
l
(1+B
2
2
/B
2
1
+B
2
3
/B
2
1
+……)/2
(P
ac
)
D
=P
ac
(1+D
2
2
+D
2
3
+…….)
D
2
=D
2
2
+D
2
3
+……..
(P
ac
)
D
=P
ac
(1+D
2
)
Now
P
ac
=B
2
1
R
L
/2
Hence the output power due to harmonic distortion is
(P
ac
)
D
=(B
2
1
R
l
+ B
2
2
R
l
+ B
2
3
R
l
+…..)/2
=B
2
1
R
l
(1+B
2
2
/B
2
1
+B
2
3
/B
2
1
+……)/2
(P
ac
)
D
=P
ac
(1+D
2
2
+D
2
3
+…….)
D
2
=D
2
2
+D
2
3
+……..
(P
ac
)
D
=P
ac
(1+D
2
)
POWER OUTPUT DUE TO DISTORTION:-
If the total harmonic distortion is 15%
i.e.
D = 0.15
(P
ac
)
D
=P
ac
(1+0.15
2
)=1.0225P
ac
So,
There is 2.25%increases in power given to the load.
If the total harmonic distortion is 15%
i.e.
D = 0.15
(P
ac
)
D
=P
ac
(1+0.15
2
)=1.0225P
ac
So,
There is 2.25%increases in power given to the load.
CLASS-B AMPLIFIER:-
In class-B power amplifiers, the transistor is so
biased that the output current flows only for half
cycle of the input signal.
It means that the transistor is forward biased for half
of the input cycle. In the negative half cycle of the
input signals, the transistor enters in to cut-off region
and no signal is produced at the output.
In class-B power amplifiers, the transistor is so
biased that the output current flows only for half
cycle of the input signal.
It means that the transistor is forward biased for half
of the input cycle. In the negative half cycle of the
input signals, the transistor enters in to cut-off region
and no signal is produced at the output.
CIRCUIT DIAGRAM OF CLASS B
AMPLIFIER:-
AMPLIFIER:-
T1 !NPN
class A
Class B power amplifier:-
INPUT WAVEFORM
OUTPUT
WAVEFORM
class A
Class B power amplifier:-
INPUT WAVEFORM
OUTPUT
WAVEFORM
CLASS-B AMPLIFIER:-
As the collector current flows only for the 180
degree (half cycle) of the input signal. In this case
the transistor conduction angle is equal to
180degree as shown in fig.
As only a half cycle is obtained at the output, for full
input cycle, the output signal is said to be ‘distorted’
in class B operation. The efficiency of class B
amplifier is much higher than the class A amplifier.
As the collector current flows only for the 180
degree (half cycle) of the input signal. In this case
the transistor conduction angle is equal to
180degree as shown in fig.
As only a half cycle is obtained at the output, for full
input cycle, the output signal is said to be ‘distorted’
in class B operation. The efficiency of class B
amplifier is much higher than the class A amplifier.
POWER AND EFFICIENCY CALCULATIONS:-
P
in
dc
=V
cc
I
dc
Where,
I
dc
= the average or direct current taken
from the collector supply.
= I
cmax
/P
Thus
P
in dc
= V
cc
I
cmax
/ P
P
in
dc
=V
cc
I
dc
Where,
I
dc
= the average or direct current taken
from the collector supply.
= I
cmax
/P
Thus
P
in dc
= V
cc
I
cmax
/ P
POWER AND EFFICIENCY CALCULATIONS:-
P
dc
=(Ic
max
/P)
2
Rc
Thus
%η = P
ac
/P
dc
.100
%η =(¼* Ic
max
V
cc
/Ic
max
V
cc
/ P)*100
%η = P/4*100
%η=78.5%
P
dc
=(Ic
max
/P)
2
Rc
Thus
%η = P
ac
/P
dc
.100
%η =(¼* Ic
max
V
cc
/Ic
max
V
cc
/ P)*100
%η = P/4*100
%η=78.5%
CLASS-AB AMPLIFIERS:-
In this type, the transistor is so biased that the output
current flows for more than half, but less that the full
cycle. The transistor conduction angle is between
180degree and 360degree such a condition is
shown in fig. The Q point is very close to cut-off
value but well above X-axis.
In this type, the transistor is so biased that the output
current flows for more than half, but less that the full
cycle. The transistor conduction angle is between
180degree and 360degree such a condition is
shown in fig. The Q point is very close to cut-off
value but well above X-axis.
T1 !PNP
R
1
1
k
+
VG1
Vcc
Vin
Vout
CLASS AB PUSH PULL AMPLIFIER
R
1
1
k
+
VG1
Vcc
Vin
Vout
CLASS AB PUSH PULL AMPLIFIER
CLASS-AB AMPLIFIERS:-
The output signal obtained in the class AB operation
is distorted. The efficiency of class AB amplifier is
more than class A but less than class B operation.
The class AB operation is important in eliminating
the crossover distortion.
The output signal obtained in the class AB operation
is distorted. The efficiency of class AB amplifier is
more than class A but less than class B operation.
The class AB operation is important in eliminating
the crossover distortion.
Operation of Class AB amplifier:-
Ic
Vce
Q point
Vcc/RL
Load
line
Output
signal of
voltage
Imax
Ic
Vce
Q point
Vcc/RL
Load
line
Output
signal of
voltage
Imax
1. cross over distortion are not present.
2. efficiency is more than class A
configuration.
3. output per transistor is more than class A
operation.
4.non linear distortions are less than class B
amplifier.
Advantages:-
2. efficiency is more than class A
configuration.
3. output per transistor is more than class A
operation.
4.non linear distortions are less than class B
amplifier.
Advantages:-
DISADVANTAGES:-
1. Efficiency is less than class B amplifier.
2. Lower output per transistor in comparison to
class B configuration.
3. Non linear distortions are more than class A
configuration.
1. Efficiency is less than class B amplifier.
2. Lower output per transistor in comparison to
class B configuration.
3. Non linear distortions are more than class A
configuration.
CLASS-C AMPLIFIER:-
In class C amplifier, the transistor bias and signal
amplitude are such that the output current flows for
appreciably less than half cycle of the input signal.
Hence its conduction angle is up to 120 degree and
150 degree that is the transistor remains forward
biased for less than half the cycle.
Class C amplifier circuit consist of tuned circuit (L
and C tank circuit) in the output which is tuned to
desired RF frequency.
It select the fundamental and rejects the rest of the
harmonics.
In class C amplifier, the transistor bias and signal
amplitude are such that the output current flows for
appreciably less than half cycle of the input signal.
Hence its conduction angle is up to 120 degree and
150 degree that is the transistor remains forward
biased for less than half the cycle.
Class C amplifier circuit consist of tuned circuit (L
and C tank circuit) in the output which is tuned to
desired RF frequency.
It select the fundamental and rejects the rest of the
harmonics.
CLASS-C AMPLIFIER:-
CLASS-C AMPLIFIER:-
A class C power is biased to operate for less then
180° of the input signal cycle.
As shown in fig. the tuned circuit in the output
however, will provide a full cycle of output signal for
the fundamental or resonant frequency of tuned
circuit (L and C tank circuit) of the output.
A class C power is biased to operate for less then
180° of the input signal cycle.
As shown in fig. the tuned circuit in the output
however, will provide a full cycle of output signal for
the fundamental or resonant frequency of tuned
circuit (L and C tank circuit) of the output.
CLASS-C AMPLIFIER:-
The use of such amplifiers is limited for a fixed
frequency, as occurs in communication circuits.
Operation of class C circuit is not intended primarily
for large signal or power amplifiers.
The use of such amplifiers is limited for a fixed
frequency, as occurs in communication circuits.
Operation of class C circuit is not intended primarily
for large signal or power amplifiers.
CLASS –C AMPLIFIER OPERATION:-
Ic
VceQ point
Vcc/RL
Input cycle of
current
Load
line
Output
signal of
votage
Vcc=Vceq
Imax
Ic
VceQ point
Vcc/RL
Input cycle of
current
Load
line
Output
signal of
votage
Vcc=Vceq
Imax
CLASS-C AMPLIFIER:-
In class C operation as the collector current flows for
less than 180 degree, the output is much more
distorted.
Due to this reason class C amplifiers are never used
for audio frequency amplifiers. But the efficiency of
class C operation is much higher and can reach
very close to 100%.
In class C operation as the collector current flows for
less than 180 degree, the output is much more
distorted.
Due to this reason class C amplifiers are never used
for audio frequency amplifiers. But the efficiency of
class C operation is much higher and can reach
very close to 100%.
ADVANTAGES:-
1.The collector efficiency is very high (more than
80%). It is more than class A ,class B and class AB
operations.
2. Output delivered to load is free from harmonics
sience circuit is tuned to fundamental and harmonic
are rejected.
1.The collector efficiency is very high (more than
80%). It is more than class A ,class B and class AB
operations.
2. Output delivered to load is free from harmonics
sience circuit is tuned to fundamental and harmonic
are rejected.
DISADVANTAGES:-
1.The output is not complete waveform. It is a
distorted waveform and distortions are more than
class A, class B and class AB configurations.
2. Use of transformer in output make the circuit
heavy, expensive and large in size.
1.The output is not complete waveform. It is a
distorted waveform and distortions are more than
class A, class B and class AB configurations.
2. Use of transformer in output make the circuit
heavy, expensive and large in size.
CLASS-D AMPLIFIER:-CLASS-D AMPLIFIER:-
Since the transistor devices use to provide the
output are basically either OFF or ON, there will be
very little power loss due to their low ON voltage.
Hence most of the power supplied to the amplifier is
transferred to the load, the efficiency of the circuit is
typically very high. Power MOSFET devices have
been quite popular as the driver device for the class
D amplifier.
Since the transistor devices use to provide the
output are basically either OFF or ON, there will be
very little power loss due to their low ON voltage.
Hence most of the power supplied to the amplifier is
transferred to the load, the efficiency of the circuit is
typically very high. Power MOSFET devices have
been quite popular as the driver device for the class
D amplifier.
Converts digital
signal
back to
sinusoidal signal
Saw tooth
generator
comparator amplifier
Low pass
filter
feedback
Vin
Vout
BLOCK DIAGRAM OF
CLASS-D POWER:-
signal
back to
sinusoidal signal
Saw tooth
generator
comparator amplifier
Low pass
filter
feedback
Vin
Vout
BLOCK DIAGRAM OF
CLASS-D POWER:-
CLASS-D AMPLIFIER:-CLASS-D AMPLIFIER:-
The class D efficiency is largely determined by the
ratio of the load resistance to the total D.C. loop
resistance which is the sum of the rDS(ON) of the
MOSFET, wire resistance (including the output filter)
and the total resistance.
For highest efficiency, the MOSFET r
DS
(ON)
resistances shunt and filter resistances should be
small compared to the load resistance.
The class D efficiency is largely determined by the
ratio of the load resistance to the total D.C. loop
resistance which is the sum of the rDS(ON) of the
MOSFET, wire resistance (including the output filter)
and the total resistance.
For highest efficiency, the MOSFET r
DS
(ON)
resistances shunt and filter resistances should be
small compared to the load resistance.
CLASS-D AMPLIFIER:-CLASS-D AMPLIFIER:-
In audio class D application, MOSFETS are
employed instead of 1GBTs and BJTs because the
switching frequencies required to keep distortion low
at 20KHz signal frequencies can exceed 160KHz
and neither the bipolar power transistors or 1GBts
switch efficiency at such high frequencies.
In audio class D application, MOSFETS are
employed instead of 1GBTs and BJTs because the
switching frequencies required to keep distortion low
at 20KHz signal frequencies can exceed 160KHz
and neither the bipolar power transistors or 1GBts
switch efficiency at such high frequencies.
CLASS-D AMPLIFIER:-CLASS-D AMPLIFIER:-
Class D amplifiers are popular because of their very
high efficiency, an efficiency of over 90% is
achieved using this type of circuit.
A Class D amplifier is designed to operate with
digital or pulse type signals.
It is necessary to convert any input signal into plus
type waveform before using it to drive a large power
load and to convert the signal back to a sinusoidal
type signal to recover the original signal
Class D amplifiers are popular because of their very
high efficiency, an efficiency of over 90% is
achieved using this type of circuit.
A Class D amplifier is designed to operate with
digital or pulse type signals.
It is necessary to convert any input signal into plus
type waveform before using it to drive a large power
load and to convert the signal back to a sinusoidal
type signal to recover the original signal
CLASS-D AMPLIFIER:-CLASS-D AMPLIFIER:-
While the letter ‘D’ is used to describe the next type
of bias operation after class C, the D could also be
considered to stand ‘Digital’ since that is the nature
of signals provided to the class D amplifier.
Convert back to the sinusoidal-type signal
employing a low pass filter.
While the letter ‘D’ is used to describe the next type
of bias operation after class C, the D could also be
considered to stand ‘Digital’ since that is the nature
of signals provided to the class D amplifier.
Convert back to the sinusoidal-type signal
employing a low pass filter.
CLASS-A PUSH PULL AMPLIFIER:-
DEFINITION:-
The distortion introduced by non linearity discussed
earlier can be minimized by the circuit known as
push pull configuration and this amplifier is known
as push pull amplifier.
CONSTRUCTION:-
Two transistor are used (T
1
and T
2
),both of these
transistor are identical. Their emitters are Connected
together .But bases and collector are connected are
in opposite ends of input and output transformer
(I.e.T
r1
,T
r2
).
DEFINITION:-
The distortion introduced by non linearity discussed
earlier can be minimized by the circuit known as
push pull configuration and this amplifier is known
as push pull amplifier.
CONSTRUCTION:-
Two transistor are used (T
1
and T
2
),both of these
transistor are identical. Their emitters are Connected
together .But bases and collector are connected are
in opposite ends of input and output transformer
(I.e.T
r1
,T
r2
).
N1 N2
TR1
T1 !NPN
T2 !NPN
N1 N2
TR2
V
1
5
R1 1k
R2 1k
input
load
CLASS-A PUSH PULL AMPLIFIER:-
TR1
T1 !NPN
T2 !NPN
N1 N2
TR2
V
1
5
R1 1k
R2 1k
input
load
CLASS-A PUSH PULL AMPLIFIER:-
CLASS-A PUSH PULL AMPLIFIER:-
Both the transformer Tr1 and Tr2 are center-tapped
transformers. Both Resistors R
1
and R
2
provides
biasing arrangement.
Load is connected across secondary of T
r2
.To
ensure that maximum power is delivered to the load
from amplifier ,turns ratio of T
r2 is so chosen that
output impendence of transistor matches to that of
load impedance.
Both the transformer Tr1 and Tr2 are center-tapped
transformers. Both Resistors R
1
and R
2
provides
biasing arrangement.
Load is connected across secondary of T
r2
.To
ensure that maximum power is delivered to the load
from amplifier ,turns ratio of T
r2 is so chosen that
output impendence of transistor matches to that of
load impedance.
CLASS-A PUSH PULL AMPLIFIER:-
Operation:-
As shown in circuit diagram, I
c1
and I
c2
flow in opposite
direction through the primary of transformer(T
r2
). In
addition, Ic1 and Ic2 are equal in magnitude. So there
is no net d.c. component of collector current in primary
of transformer Tr2.
Hence there is no d.c. saturation in the transformer
core .This result increases in A.C. power output
compared with single transistor operation.
Operation:-
As shown in circuit diagram, I
c1
and I
c2
flow in opposite
direction through the primary of transformer(T
r2
). In
addition, Ic1 and Ic2 are equal in magnitude. So there
is no net d.c. component of collector current in primary
of transformer Tr2.
Hence there is no d.c. saturation in the transformer
core .This result increases in A.C. power output
compared with single transistor operation.
CLASS-A PUSH PULL AMPLIFIER:-
When a.c. signal is applied to the input, during
positive half cycle of the input a.c. signal, the base
of transistor T1 is more positive than the base of
transistor T2.
Collector currents are always in phase with base
currents. Hence the collector current Ic1 increases
while Ic2 of transistor T2 decreases.
When a.c. signal is applied to the input, during
positive half cycle of the input a.c. signal, the base
of transistor T1 is more positive than the base of
transistor T2.
Collector currents are always in phase with base
currents. Hence the collector current Ic1 increases
while Ic2 of transistor T2 decreases.
CLASS-A PUSH PULL AMPLIFIER:-
These currents(I
c1
and I
c2
) flow in opposite direction in
the two halves of the center-tapped output
transformer(T
r2
).Due to this,the magnitude of voltage
induced in the load will be proportional to the
difference of collector currents(I
c1
and I
c2
).
Similarly, for the negative half of the input a.c.signal,
the magnitude of voltage induced in the load will be
proportional to the difference (I
c1
and I
c2
).Thus, for the
complete a.c signal, there is an induced a.c. voltage
in the secondary of output transformer Tr2 and a.c.
power is delivered to the load.
These currents(I
c1
and I
c2
) flow in opposite direction in
the two halves of the center-tapped output
transformer(T
r2
).Due to this,the magnitude of voltage
induced in the load will be proportional to the
difference of collector currents(I
c1
and I
c2
).
Similarly, for the negative half of the input a.c.signal,
the magnitude of voltage induced in the load will be
proportional to the difference (I
c1
and I
c2
).Thus, for the
complete a.c signal, there is an induced a.c. voltage
in the secondary of output transformer Tr2 and a.c.
power is delivered to the load.
CLASS-A PUSH PULL AMPLIFIER:-
From the above discussion, it is seen that when I
c1
increases, I
c2
decreases and when I
c2
increases, I
c1
decreases.this means that one transistor is pushed
into conduction and other is pulled out of
conduction. Hence the name push-pull amplifier.
Biasing arrangement is so that the collector current
flows through 360
0. Hence name class-A push-pull
amplifier.
From the above discussion, it is seen that when I
c1
increases, I
c2
decreases and when I
c2
increases, I
c1
decreases.this means that one transistor is pushed
into conduction and other is pulled out of
conduction. Hence the name push-pull amplifier.
Biasing arrangement is so that the collector current
flows through 360
0. Hence name class-A push-pull
amplifier.
DISTORTION IN CLASS-A PUSH-PULL
AMPLIFIER:-
I
b1
=I
b
sin wt
I
b2
=I
b
sin (wt+P)
The collector current is represented as :
I
c
=I
0
+I
1
sin wt+I
2
sin 2wt+I
3
sin 3wt
As the collector collector current of second
transistor is 180 out of phase. It can be
represented as :
AMPLIFIER:-
I
b1
=I
b
sin wt
I
b2
=I
b
sin (wt+P)
The collector current is represented as :
I
c
=I
0
+I
1
sin wt+I
2
sin 2wt+I
3
sin 3wt
As the collector collector current of second
transistor is 180 out of phase. It can be
represented as :
DISTORTION IN CLASS-A PUSH-PULL
AMPLIFIER:-
I
c2
=I
0
+I
1
sin(wt+P)+I
2
sin2(wt+P)+I
3
sin3(wt+P)
I
c2
=I
0
-I
1
sinwt+I
2
sin2wt-I
3
sin3wt
In the push-pull amplifier, the output voltage induced
in the secondary of the output transformer is
proportional to the two collector currents (I
c1
-I
c2
).
AMPLIFIER:-
I
c2
=I
0
+I
1
sin(wt+P)+I
2
sin2(wt+P)+I
3
sin3(wt+P)
I
c2
=I
0
-I
1
sinwt+I
2
sin2wt-I
3
sin3wt
In the push-pull amplifier, the output voltage induced
in the secondary of the output transformer is
proportional to the two collector currents (I
c1
-I
c2
).
DISTORTION IN CLASS-A PUSH-PULL
AMPLIFIER:-
Hence
V
0
=K (I
c1
-I
c2
)
or V
0
=K (2I
1
sin2wt+2I
3
sin3wt+…….)
or V
0
=2K (I
1
sinwt+I
3
sin3wt+I
5
sin5wt+……)
where K is the constant of proportionality.
AMPLIFIER:-
Hence
V
0
=K (I
c1
-I
c2
)
or V
0
=K (2I
1
sin2wt+2I
3
sin3wt+…….)
or V
0
=2K (I
1
sinwt+I
3
sin3wt+I
5
sin5wt+……)
where K is the constant of proportionality.
DISTORTION IN CLASS-A PUSH-PULL
AMPLIFIER:-
It is important to note that no even harmonic term is
appears in the equation. Hence all even harmonics
are eliminated from the output.
In harmonic distortion, the magnitude of second
harmonic contribute to most of distortion. As the
order of harmonic increases (i.e. third, fourth
harmonic and so on) distortion is less objectionable.
AMPLIFIER:-
It is important to note that no even harmonic term is
appears in the equation. Hence all even harmonics
are eliminated from the output.
In harmonic distortion, the magnitude of second
harmonic contribute to most of distortion. As the
order of harmonic increases (i.e. third, fourth
harmonic and so on) distortion is less objectionable.
DISTORTION IN CLASS-A PUSH-PULL
AMPLIFIER:-
Second harmonic is considered to be most
objectionable. In push-pull amplifiers, all even
harmonics are cancelled.
Therefore, in the absence of second harmonic we
say that distortion in the output of a push-pull
amplifier is almost negligible in comparison to a
single ended power amplifier.
AMPLIFIER:-
Second harmonic is considered to be most
objectionable. In push-pull amplifiers, all even
harmonics are cancelled.
Therefore, in the absence of second harmonic we
say that distortion in the output of a push-pull
amplifier is almost negligible in comparison to a
single ended power amplifier.
CLASS-A PUSH PULL AMPLIFIER:-
Advantages:-
1.Less distortion due to cancellation of even
harmonics.
2.We get more output per transistor for a given
amount of distortion.
3.The d.c. components of the collector current of
transistor T
1
and T
2
flows in opposite direction
hence there is no net d.c. magnetization or core
saturation.
Advantages:-
1.Less distortion due to cancellation of even
harmonics.
2.We get more output per transistor for a given
amount of distortion.
3.The d.c. components of the collector current of
transistor T
1
and T
2
flows in opposite direction
hence there is no net d.c. magnetization or core
saturation.
CLASS-A PUSH PULL AMPLIFIER:-
Disadvantages:-
1.Two transformers used makes the circuit bulkier
and costlier.
2.Two hundred percentage identical transistors are
required, which is not possible in practical.
3.Since this is class-A, therefore, individually each
transistor has maximum efficiency of 50% and
hence the over-all efficiency of two transistors
together is also maximum 50%.
Disadvantages:-
1.Two transformers used makes the circuit bulkier
and costlier.
2.Two hundred percentage identical transistors are
required, which is not possible in practical.
3.Since this is class-A, therefore, individually each
transistor has maximum efficiency of 50% and
hence the over-all efficiency of two transistors
together is also maximum 50%.
CLASS-B PUSH PULL AMPLIFIER:-
We have studied that class A amplifier removes some
of the drawbacks of single ended transistor coupled
amplifier, but efficiency is only 50%.Class-B amplifier
helps in getting higher efficiency and higher output
power.
Circuit is similar to class-A amplifier except that the
biasing resistance R1&R2 are absent so no biasing is
provided to both transistors because of transistors are
to work in class-B operation. in this operating point is
set at the cutoff region for which no biasing is required.
We have studied that class A amplifier removes some
of the drawbacks of single ended transistor coupled
amplifier, but efficiency is only 50%.Class-B amplifier
helps in getting higher efficiency and higher output
power.
Circuit is similar to class-A amplifier except that the
biasing resistance R1&R2 are absent so no biasing is
provided to both transistors because of transistors are
to work in class-B operation. in this operating point is
set at the cutoff region for which no biasing is required.
CLASS-B PUSH PULL AMPLIFIER:-
CLASS-B PUSH PULL AMPLIFIER:-
OPERATION:-
The input transformer is a phase splitter providing
two signals 180º out of phase ,one going to each of
transistors.
When there is no signal, both the transistors are cut-
off, hence no current is drawn by either of them.
Thus there is no power wasted during this condition.
OPERATION:-
The input transformer is a phase splitter providing
two signals 180º out of phase ,one going to each of
transistors.
When there is no signal, both the transistors are cut-
off, hence no current is drawn by either of them.
Thus there is no power wasted during this condition.
CLASS-B PUSH-PULL AMPLIFIER:-
Now consider that an a.c. signal (Vs=Vosinωt) is
applied at input .
During positive half cycle of input signal i.e. Vs goes
positive ,the induced voltage on the secondary of
input transformer becomes positive for the base of T1
and is negative for T2 thus T1 conducts during
positive half cycle and during this T2 does not
conducts.
Now consider that an a.c. signal (Vs=Vosinωt) is
applied at input .
During positive half cycle of input signal i.e. Vs goes
positive ,the induced voltage on the secondary of
input transformer becomes positive for the base of T1
and is negative for T2 thus T1 conducts during
positive half cycle and during this T2 does not
conducts.
CLASS-B PUSH-PULL AMPLIFIER:-
In the negative half cycle when Vs goes negative T1
does not conducts but T2 conduct . In figure the
wave shapes shown for I1 and I2 each remaining
zero for 180º and conducts for next 180º (similar to
rectified half waves).
Due to transformer action the current induced in the
secondary is a full wave.
In the negative half cycle when Vs goes negative T1
does not conducts but T2 conduct . In figure the
wave shapes shown for I1 and I2 each remaining
zero for 180º and conducts for next 180º (similar to
rectified half waves).
Due to transformer action the current induced in the
secondary is a full wave.
CLASS -B PUSH PULL AMPLIFIER:-
D.C. power input:-
D.c. power output of the two transistors is
P(d.c.)=2×power input to the transistor
=2[I(d.c)×V cc]
we know
Idc=Im / π (for half wave)
therefore
Pdc.=(2Im/π) × Vcc
D.C. power input:-
D.c. power output of the two transistors is
P(d.c.)=2×power input to the transistor
=2[I(d.c)×V cc]
we know
Idc=Im / π (for half wave)
therefore
Pdc.=(2Im/π) × Vcc
CLASS-B PUSH PULL AMPLIFIER:-
A.C. power outputA.C. power output :- :-
The a.c power output using peak value is given asThe a.c power output using peak value is given as
P ac= Vm Im/2P ac= Vm Im/2
P ac=(V cc-V min)× Im/2 P ac=(V cc-V min)× Im/2
Efficiency:-Efficiency:-
This can be calculated using basic equationhis can be calculated using basic equation
%η = ( Pac/Pdc )×100 %η = ( Pac/Pdc )×100
= { [Im/2×(Vcc-Vmin) ]/ [2/π× ImVcc] }×100 = { [Im/2×(Vcc-Vmin) ]/ [2/π× ImVcc] }×100
= { [ π(Vcc -Vmin)] / [4Vcc] }×100= { [ π(Vcc -Vmin)] / [4Vcc] }×100
A.C. power outputA.C. power output :- :-
The a.c power output using peak value is given asThe a.c power output using peak value is given as
P ac= Vm Im/2P ac= Vm Im/2
P ac=(V cc-V min)× Im/2 P ac=(V cc-V min)× Im/2
Efficiency:-Efficiency:-
This can be calculated using basic equationhis can be calculated using basic equation
%η = ( Pac/Pdc )×100 %η = ( Pac/Pdc )×100
= { [Im/2×(Vcc-Vmin) ]/ [2/π× ImVcc] }×100 = { [Im/2×(Vcc-Vmin) ]/ [2/π× ImVcc] }×100
= { [ π(Vcc -Vmin)] / [4Vcc] }×100= { [ π(Vcc -Vmin)] / [4Vcc] }×100
CLASS-B PUSH PULL AMPLIFIER:-
Maximum Efficiency:-
The maximum efficiency arises when Vmin = 0.
Therefore,
%ηmax = [(πVcc) / (4Vcc)]×100
%ηmax = 78.5%
Thus maximum efficiency possible for class-B push
pull amplifier is 78.5%.this is more than class-
A(50%).This increase is due to standing current
being zero and hence no loss of power during the
cut off half cycle.
Maximum Efficiency:-
The maximum efficiency arises when Vmin = 0.
Therefore,
%ηmax = [(πVcc) / (4Vcc)]×100
%ηmax = 78.5%
Thus maximum efficiency possible for class-B push
pull amplifier is 78.5%.this is more than class-
A(50%).This increase is due to standing current
being zero and hence no loss of power during the
cut off half cycle.
CLASS-B PUSH PULL AMPLIFIER:-
Maximum power DissipationMaximum power Dissipation:-:-
( Pd )max = [4/π² (Pac) max ]1/2 ( Pd )max = [4/π² (Pac) max ]1/2
= 2/π²(Pac) max = 2/π²(Pac) max
Maximum power DissipationMaximum power Dissipation:-:-
( Pd )max = [4/π² (Pac) max ]1/2 ( Pd )max = [4/π² (Pac) max ]1/2
= 2/π²(Pac) max = 2/π²(Pac) max
CLASS-B PUSH PULL AMPLIFIER:-
Advantages:-
• Efficiency is much higher than class-A .
• When no input signal power dissipation is zero .
• Even harmonics gets cancelled, this reduces
harmonic distortions .
• As the d.c component of current flows in
opposite direction through the primary windings , there
is no d.c saturation of the core.
• Due to transformer , impedance matching is possible
.
• Ripple present in supply voltage also gets
eliminated .
Advantages:-
• Efficiency is much higher than class-A .
• When no input signal power dissipation is zero .
• Even harmonics gets cancelled, this reduces
harmonic distortions .
• As the d.c component of current flows in
opposite direction through the primary windings , there
is no d.c saturation of the core.
• Due to transformer , impedance matching is possible
.
• Ripple present in supply voltage also gets
eliminated .
CROSSOVER DISTORTION:-
Crossover distortion in the output signal refers to the
fact that during the time when input signal crossover
from positive to negative (or negative to positive)
,there is some nonlinearity in the output signal.
This is due the fact that as long as the magnitude of
input signal is less than cut-in voltage of base
emitter junction of transistor (0.7v for Si & 0.2 for
Ge), the collector current remain zero and transistor
remain in cut-off region.
Crossover distortion in the output signal refers to the
fact that during the time when input signal crossover
from positive to negative (or negative to positive)
,there is some nonlinearity in the output signal.
This is due the fact that as long as the magnitude of
input signal is less than cut-in voltage of base
emitter junction of transistor (0.7v for Si & 0.2 for
Ge), the collector current remain zero and transistor
remain in cut-off region.
CROSSOVER DISTORTION:-
Hence there is a period between the cross over of
the half cycles of the input signal,for which none of
the transistor is active & the output is zero.
Thus the output signal does not follow the input Thus the output signal does not follow the input
signal and hence get distorted. Such distortion is signal and hence get distorted. Such distortion is
crossover distortion.crossover distortion.
Crossover distortion is eliminated in class-AB Crossover distortion is eliminated in class-AB
amplifier. amplifier.
Hence there is a period between the cross over of
the half cycles of the input signal,for which none of
the transistor is active & the output is zero.
Thus the output signal does not follow the input Thus the output signal does not follow the input
signal and hence get distorted. Such distortion is signal and hence get distorted. Such distortion is
crossover distortion.crossover distortion.
Crossover distortion is eliminated in class-AB Crossover distortion is eliminated in class-AB
amplifier. amplifier.
Output
signal
Both transistor
OFF
WT
WT
signal
Both transistor
OFF
WT
WT
PHASE SPLITTER CIRCUIT:-
It is circuit to produce the two output ;both are out of
phase with 180 degree.
It is circuit to produce the two output ;both are out of
phase with 180 degree.
PHASE SPLITTER CIRCUIT:-
T1 !NPN
T2 !PNP
Vcc1
Vcc2
C
RL
T1 !NPN
T2 !PNP
Vcc1
Vcc2
C
RL
COMPLEMENTRY SYMMETRY CLASS –B
PUSH PULL AMPLIFIER:-
Circuit is as shown in diagram. This does not use a
input or output transformer .Input transformer has
the function of providing two inputs 180º out of
phase, which makes only one transistor (T1 or T2)
to conduct at a time.
PUSH PULL AMPLIFIER:-
Circuit is as shown in diagram. This does not use a
input or output transformer .Input transformer has
the function of providing two inputs 180º out of
phase, which makes only one transistor (T1 or T2)
to conduct at a time.
COMPLEMENTRY SYMMETRY CLASS –B
PUSH PULL AMPLIFIER:-
However using complementary transistor (one NPN
other PNP) if we introduce same input to these
transistor, two collector currents of transistors would
be 180 º out of phase .
Thus avoiding transformer in the push pull
amplifier circuit . Also the output transformer is
avoided using two separate but equal power
supplies (Vcc1 and Vcc2) .
PUSH PULL AMPLIFIER:-
However using complementary transistor (one NPN
other PNP) if we introduce same input to these
transistor, two collector currents of transistors would
be 180 º out of phase .
Thus avoiding transformer in the push pull
amplifier circuit . Also the output transformer is
avoided using two separate but equal power
supplies (Vcc1 and Vcc2) .
COMPLEMENTARY SYMMETRY CLASS –B
PUSH PULL AMPLIFIER:-
Fig:-circuit diagram
PUSH PULL AMPLIFIER:-
Fig:-circuit diagram
COMPLEMENTARY SYMMETRY CLASS –B
PUSH PULL AMPLIFIER:-
Simplified circuit diagram
PUSH PULL AMPLIFIER:-
Simplified circuit diagram
COMPLEMENTARY SYMMETRY
CLASS-B PUSH PULL AMPLIFIER:-
Operation:-
In positive half cycle of input T1 is forward biased
hence conducts ,T2 is reversed biased hence does
not conducts . This results into positive half cycle
across load RL as in figure .
Just opposite in negative half cycle of input i.e.
T2conducts and T1 being reverse biased does not
conduct.
CLASS-B PUSH PULL AMPLIFIER:-
Operation:-
In positive half cycle of input T1 is forward biased
hence conducts ,T2 is reversed biased hence does
not conducts . This results into positive half cycle
across load RL as in figure .
Just opposite in negative half cycle of input i.e.
T2conducts and T1 being reverse biased does not
conduct.
COMPLEMENTARY SYMMETRY CLASS –B
PUSH PULL AMPLIFIER:-
This results into negative half cycle across load R
L
.
Thus I
1
and I
2
flow in one half cycle each but flow
through R
L
. Hence total current through R
L
is in both
half cycles, a complete cycle of output signal comes
across the load .
As in figure
PUSH PULL AMPLIFIER:-
This results into negative half cycle across load R
L
.
Thus I
1
and I
2
flow in one half cycle each but flow
through R
L
. Hence total current through R
L
is in both
half cycles, a complete cycle of output signal comes
across the load .
As in figure
COMPLEMENTARY SYMMETRY CLASS –
B PUSH PULL AMPLIFIER:-
Advantages:-
1.As the circuit is without transformer so its weight ,
size and cost are less.
2.Frequency response improves due to transformer
less circuit .
B PUSH PULL AMPLIFIER:-
Advantages:-
1.As the circuit is without transformer so its weight ,
size and cost are less.
2.Frequency response improves due to transformer
less circuit .
DISADVANTAGE:-
1.Circuit needs two separate supply voltages (Vcc1
and Vcc2) .
2. Output is distorted due to cross over distortion.
1.Circuit needs two separate supply voltages (Vcc1
and Vcc2) .
2. Output is distorted due to cross over distortion.
THERMAL RESISTANCE:-
It is the resistance in which heat flow
between two temperature point.
Pt= T1-T2
Q
Where
Q = Thermal resistance
It is the resistance in which heat flow
between two temperature point.
Pt= T1-T2
Q
Where
Q = Thermal resistance
VARIATION OF PD VERSUS
TEMPERATURE:-
Q
Temperature
P.D
(watts)
TEMPERATURE:-
Q
Temperature
P.D
(watts)
QUASI COMPLIMENTARY SYMMETRY PUSH QUASI COMPLIMENTARY SYMMETRY PUSH
PULL AMPLIFIER:-PULL AMPLIFIER:-
It is similar to complimentary symmetry
except it uses transistor pair in each NPN and
PNP transistor.
Above pair in known to be “Darlington pair”
and below pair is know to be “feedback pair”.
PULL AMPLIFIER:-PULL AMPLIFIER:-
It is similar to complimentary symmetry
except it uses transistor pair in each NPN and
PNP transistor.
Above pair in known to be “Darlington pair”
and below pair is know to be “feedback pair”.
Q3
Q4 !NPN
Q1
Q2 !PNP
R
1
R
2
1
k
R
L
CL
Cin1
Cin2
R
3
1
k
QUASI COMPLIMENTARY SYMMETRYQUASI COMPLIMENTARY SYMMETRY
PUSH PULL AMPLIFIER:-PUSH PULL AMPLIFIER:-
Q4 !NPN
Q1
Q2 !PNP
R
1
R
2
1
k
R
L
CL
Cin1
Cin2
R
3
1
k
QUASI COMPLIMENTARY SYMMETRYQUASI COMPLIMENTARY SYMMETRY
PUSH PULL AMPLIFIER:-PUSH PULL AMPLIFIER:-
OPERATION:-OPERATION:-
AB operation is to be performed ,base potential of`
T2 must be higher than potential of point k ,while the
base potential of T3 must be lowered than that of
point K .this is accomplished by adjusting of biasing
of T1.
When no signal is applied at ten base of transistor
t1.capacitor “C” charges to potential of point K i.e.
Vcc/2.
AB operation is to be performed ,base potential of`
T2 must be higher than potential of point k ,while the
base potential of T3 must be lowered than that of
point K .this is accomplished by adjusting of biasing
of T1.
When no signal is applied at ten base of transistor
t1.capacitor “C” charges to potential of point K i.e.
Vcc/2.
OPERATION:-OPERATION:-
When signal is applied to the amplifier and positive
half cycle appears at the base of T1,T1 conducts
base potential of T3 is lowered , because point B3 is
grounded due to conduction of T1. Thus T1 and T2
are ON and T3 is in cutoff .Capacitor C charges
additionally via transistor T2 and load. Current is
delivered to load.
During negative half cycle ,T1 and T2 becomes During negative half cycle ,T1 and T2 becomes
OFF and T3 becomes ON therefore , capacitor OFF and T3 becomes ON therefore , capacitor
charges through T3 and load .Thus ,signal power is charges through T3 and load .Thus ,signal power is
delivered to the load.delivered to the load.
When signal is applied to the amplifier and positive
half cycle appears at the base of T1,T1 conducts
base potential of T3 is lowered , because point B3 is
grounded due to conduction of T1. Thus T1 and T2
are ON and T3 is in cutoff .Capacitor C charges
additionally via transistor T2 and load. Current is
delivered to load.
During negative half cycle ,T1 and T2 becomes During negative half cycle ,T1 and T2 becomes
OFF and T3 becomes ON therefore , capacitor OFF and T3 becomes ON therefore , capacitor
charges through T3 and load .Thus ,signal power is charges through T3 and load .Thus ,signal power is
delivered to the load.delivered to the load.
NUMERICALS
Q.1 :- A power transistor working in a class-A operation is
supplied from a 12-volt battery if the maximum collector current
change is 100mA, find the power transferred to a 5ohm
loudspeaker if it is:
(1)Directly connected in the collector:
(2)Transformer coupled for maximum power transfer.
Also find the turn ration of the transformer in (2)case.
ANS:-
CASE 1: When loudspeaker is directly connected in the
collector
Maximum voltage across loudspeaker
= ΔI
c
R
L
= 100mA*5ohm = 500mV
supplied from a 12-volt battery if the maximum collector current
change is 100mA, find the power transferred to a 5ohm
loudspeaker if it is:
(1)Directly connected in the collector:
(2)Transformer coupled for maximum power transfer.
Also find the turn ration of the transformer in (2)case.
ANS:-
CASE 1: When loudspeaker is directly connected in the
collector
Maximum voltage across loudspeaker
= ΔI
c
R
L
= 100mA*5ohm = 500mV
CASE 2: When loudspeaker is transformer coupled
Output impedance of the transformer
=ΔV
ce
/ΔI
c
=12/100mA=120ohm
For maximum power transfer the load resistance
referred to primary side (i.e.R
L
) must be equal to output
impedance of the transistor
So R
L
=120
Output impedance of the transformer
=ΔV
ce
/ΔI
c
=12/100mA=120ohm
For maximum power transfer the load resistance
referred to primary side (i.e.R
L
) must be equal to output
impedance of the transistor
So R
L
=120
We know that turns ratio is given by
Turns ratio (N) = (R
’
L
/R
L
)
1/2
= (120/5)
1/2
= 4.898
Now secondary voltage i.e. voltage across the speaker
V
s
=V
P
/N=V
CC
/N=12/4.898=2.45V
Load current= V
s
/R
L
=2.45/5=0.489
So power transferred to speaker =I
L
V
L
= I
LV
L=2.5*0.49=1.2005W
=1200mW
Turns ratio (N) = (R
’
L
/R
L
)
1/2
= (120/5)
1/2
= 4.898
Now secondary voltage i.e. voltage across the speaker
V
s
=V
P
/N=V
CC
/N=12/4.898=2.45V
Load current= V
s
/R
L
=2.45/5=0.489
So power transferred to speaker =I
L
V
L
= I
LV
L=2.5*0.49=1.2005W
=1200mW
Q 2:- A single ended class-A amplifier has a transformer
coupled load of 8 ohm. If the transformer turns-ratio(N
1
/N
2
) is
10,determine the maximum power delivered to the load. Take a
zero signal collector current of 500mA.
ANS:-
Given that R
L
=8 ohm, N=10 and I
CQ
=500mA
The load resistance seen by the primary of the transformer
R’
L
= N
2
R
L
=10*10*8= 800 ohm
So maximum power delivered to the 8 ohm loud speaker
P
o
= 0.5*I
2
CQ
R’
L
=(0.5)*(500*.001)(800) = 200 W
coupled load of 8 ohm. If the transformer turns-ratio(N
1
/N
2
) is
10,determine the maximum power delivered to the load. Take a
zero signal collector current of 500mA.
ANS:-
Given that R
L
=8 ohm, N=10 and I
CQ
=500mA
The load resistance seen by the primary of the transformer
R’
L
= N
2
R
L
=10*10*8= 800 ohm
So maximum power delivered to the 8 ohm loud speaker
P
o
= 0.5*I
2
CQ
R’
L
=(0.5)*(500*.001)(800) = 200 W
Q 3:- When a sinusoidal signal is fed to an amplifier, the output
current is given by
i
c
=15 sin 400t + 1.5 sin 800t + 1.2 sin 1200t + 0.5 sin 1600t ,
calculate
(1) second third and fourth percentage harmonic distortion.
(2) percentage increase in power due to distortion.
ANS:-
(1)Percentage harmonic distortions of various components are
given by
D
2
=(B
2
/B
1
)*100=(1.5/15)*100=10%
D
3
=(B
3
/B
1
)*100=(1.2/15)*100=8%
D
4=(B
4/B
1)*100=(0.5/15)*100=3.33%
current is given by
i
c
=15 sin 400t + 1.5 sin 800t + 1.2 sin 1200t + 0.5 sin 1600t ,
calculate
(1) second third and fourth percentage harmonic distortion.
(2) percentage increase in power due to distortion.
ANS:-
(1)Percentage harmonic distortions of various components are
given by
D
2
=(B
2
/B
1
)*100=(1.5/15)*100=10%
D
3
=(B
3
/B
1
)*100=(1.2/15)*100=8%
D
4=(B
4/B
1)*100=(0.5/15)*100=3.33%
Distortion factor:-
D=(D
2
2
+D
2
3
+D
2
4
)
1/2
=0.132=13.2%
(2) The net power output in percentage of distortion
P=(1+D
2
)P
1
=(1+.132*.132)P
1
=1.0174P
1
(3) Hence percentage increase in power due to distortion
= (1.0174 P
1
- P
1
)/P
1
*100= 1.74%
D=(D
2
2
+D
2
3
+D
2
4
)
1/2
=0.132=13.2%
(2) The net power output in percentage of distortion
P=(1+D
2
)P
1
=(1+.132*.132)P
1
=1.0174P
1
(3) Hence percentage increase in power due to distortion
= (1.0174 P
1
- P
1
)/P
1
*100= 1.74%
Q 4:- A complementary class B power amplifier uses a 15 volt
d.c. supply with a sinusoidal input, a maximum peak to peak of
24 volt is desired across a load of 100 ohm find the power
dissipated by each transistor.
ANS:- Peak value of current is
I
peak=V
peak/R
L=(24/2)/100=0.12A
DC power drawn from battery
P
dc
=V
cc
*I
dc
=15*(2/π*0.12)=1.146 Watts
d.c. supply with a sinusoidal input, a maximum peak to peak of
24 volt is desired across a load of 100 ohm find the power
dissipated by each transistor.
ANS:- Peak value of current is
I
peak=V
peak/R
L=(24/2)/100=0.12A
DC power drawn from battery
P
dc
=V
cc
*I
dc
=15*(2/π*0.12)=1.146 Watts
Thus the maximum efficiency possible in directly
coupled series fed class-A amplifier is just 25%
(ideally). In practical it is less than 25%.
coupled series fed class-A amplifier is just 25%
(ideally). In practical it is less than 25%.
Therefore a.c. power delivered to load
= 1/2( V
2
peak
/R
L
)=1/2(12*12/100)=0.72 Watts
Hence power dissipated in both the transistors
= 1.146-0.72=0.426 watts
Therefore, power dissipated by each transistor
=0.426/2=0.213 Watts
= 213 mW
= 1/2( V
2
peak
/R
L
)=1/2(12*12/100)=0.72 Watts
Hence power dissipated in both the transistors
= 1.146-0.72=0.426 watts
Therefore, power dissipated by each transistor
=0.426/2=0.213 Watts
= 213 mW
Q 5:- For a single stage class A amplifier, V
cc
=20 volt, V
CEQ
=10
Volt, I
CQ
=600mA and collector load resistor R
L
=16 ohm ,The ac
output current various by (+-)300 mA, with the ac input
signal.
Determine
(1)Power supplied by the d.c. source to amplifier or d.c. power
input to amplifier.
(2)D.C. power consumed by the load resistor.
(3)D.C. power delivered to transistor.
(4) Output power ac or ac power developed across the load resistor.
(5)Collector efficiency.
(6)Overall efficiency.
cc
=20 volt, V
CEQ
=10
Volt, I
CQ
=600mA and collector load resistor R
L
=16 ohm ,The ac
output current various by (+-)300 mA, with the ac input
signal.
Determine
(1)Power supplied by the d.c. source to amplifier or d.c. power
input to amplifier.
(2)D.C. power consumed by the load resistor.
(3)D.C. power delivered to transistor.
(4) Output power ac or ac power developed across the load resistor.
(5)Collector efficiency.
(6)Overall efficiency.
ANS:-
(1)Power supplied by the dc source to amplifier circuit
(P
in
)
dc
=V
CC
I
CQ
= 20*600*0.001=12 Watts
(2)d.c. power consumed by the load resistor
P
Rc
=I
2
CQ*
R
C
=(600*0.001)
2
*16=5.76 Watts
(3)d.c. power delivered to the transistor
P
tr
=(P
in
)
dc
-P
Rc
=6.24 Watts
(1)Power supplied by the dc source to amplifier circuit
(P
in
)
dc
=V
CC
I
CQ
= 20*600*0.001=12 Watts
(2)d.c. power consumed by the load resistor
P
Rc
=I
2
CQ*
R
C
=(600*0.001)
2
*16=5.76 Watts
(3)d.c. power delivered to the transistor
P
tr
=(P
in
)
dc
-P
Rc
=6.24 Watts
(4) A.C. power developed across load resistor
(P
o
)
ac
=I
2
rms
R
c
=(I
m
/1.404)
2
R
c
=(.3/1.404)
2
*16=0.72 Watts
(5) Collector efficiency
(P
o
)
ac
/P
tr
*100=(0.72/6.24)*100=11.5%
(6) Overall efficiency
((P
o)
ac/P
in)*100=0.72/12*100=6%
(P
o
)
ac
=I
2
rms
R
c
=(I
m
/1.404)
2
R
c
=(.3/1.404)
2
*16=0.72 Watts
(5) Collector efficiency
(P
o
)
ac
/P
tr
*100=(0.72/6.24)*100=11.5%
(6) Overall efficiency
((P
o)
ac/P
in)*100=0.72/12*100=6%
SUBMITTED TO:
MRS. SARITA CHOUHAN d
SUBMITTED BY:
PRATEEK LAMORIA
RAKESH YADAV
RAMESH BISHNOI
RAM CHANDRA
PRATIBHA SONI
PRINYANKA JANGIR
SHAMBHU LAL VYAS
RAM NIWAS
ROHITASH
RAMBHAROSE NAGAR
MRS. SARITA CHOUHAN d
SUBMITTED BY:
PRATEEK LAMORIA
RAKESH YADAV
RAMESH BISHNOI
RAM CHANDRA
PRATIBHA SONI
PRINYANKA JANGIR
SHAMBHU LAL VYAS
RAM NIWAS
ROHITASH
RAMBHAROSE NAGAR
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