Lecture # 16 &17
Complementary symmetry
& push-pull Amplifiers
Prepared by
Engr Sarfaraz Khan Turk
Lecturer at IBT LUMHS Jamshoro
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Class B Amplifier
In class B, the transistor is
biased just off. The AC signal
turns the transistor on.
The transistor only conducts
when it is turned on by onehalf of the AC cycle.
In order to get a full AC cycle
out of a class B amplifier, you
need two transistors:
There are two main types of types of class-B Amplifiers
1. The Basic or Standard Push-Pull Amplifiers.
2. The Complementary-symmetry(or quasi-complementary
amplifiers.
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The standard or Basic class-B
Push-Pull Amplifier
 The standard or Basic Push-Pull Amplifier contains two
transistors of the same type with the emitter tied together
 It uses a center-tapped transformer, or a transistor
phase splitter on the input and the center-tapped
transformer on the output.
 This Amplifier type is shown in the figure 1 .
 This configuration is less commonly used in the modern
electronics due to more expense spend on input and
output transformer which may occupy more area and
produce distortion in the nebouring electronic devices.
 Note the transistor types and the transformer. This is the
standard Push-pull Amplifier configuration.
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The Standard or Basic Class-B
push-pull Amplifier(contd)
•
During the positive halfcycle of the AC input,
transistor Q1 (npn) is
conducting and Q2 (npn) is
off.
•
During the negative halfcycle of the AC input,
transistor Q2 (npn) is
conducting and Q1 (npn) is
off.
Each transistor produces one-half of an AC cycle. The transformer combines the
two outputs to form a full AC cycle.
This circuit is less commonly used in modern circuits
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The Standard or Basic ClassB push-pull Amplifier
The two same npn transistors will be use in this configuration
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Complementary-symmetry
class-B Push-Pull amplifiers
 A class –B complementary-symmetry Push-Pull Amplifier
using complementary transistors (a pair of one npn and
one pnp transistors, with matched characteristics).
 A npn transistor that provides the positive half of the AC
cycle.
 An pnp transistor that provides the negative half of the
AC cycle.
 No required input and output Transformers.
 This configuration is most widely used in the modern
electronics due to less expensive as compared to the
standard push-pull amplifier .
 Two disadvantages of this circuit are:
1. Using two separate DC Power supplies for Vcc.
2. Cross over distortion.
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Complementary-symmetry
class-B Push-Pull amplifiers
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Why push-pull amplifier use?
Why the standard or Basic push-pull Amplifier
less commonly use in the modern electronics?
Why complementary symmetry amplifiers
preferred over the push-pull Amplifiers?
Why pnp transistor use in the complementary
symmetry push-pull Amplifier?
What is the current source and the current sink?
Which one transistor works as a current source?
Which one transistor works as a current sink?
Which one transistor works as a push?
Which one transistor works as a pull?
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Current sources and loads


when driving a reactive load we need to
supply current at some times (the output acts
as a current source)
at other times we need to absorb current (the
output acts as a current sink)
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Current sources and current sink


the circuit above is a good current source but
a poor current sink (stored charge must be
removed by RE)
an alternative circuit using pnp transistors
(below) is a good current sink but a poor
current source
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Complementary-symmetry
class-B Push-Pull amplifiers
Push-pull amplifiers


combining these
circuits can produce
an arrangement that
is both a good current
source and a good
current sink
this is termed a
push-pull amplifier
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Complementary-symmetry class-B
Push-Pull amplifiers
Since one part of the
circuit (T1) pushes the
signal high during the
positive +ve half-cycle
and the other part of the
circuit (T2) pulls the
signal low during the
negative -ve half cycle, of
the input AC signal the
circuit is referred to as a
push-pull circuit
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The Complementary-symmetry
Push-Pull Amplifier Stages
As Vin increases, Q1 is on and pushes a current into RL.(During
positive +ve half cycle of the input of AC signal.
As Vin decreases, Q2 is on and pulls a current out of RL.
.(During negative -ve half cycle of the input of AC signal. 13
Input DC power
The power supplied to the load by an amplifier is
drawn from the power supply
The amount of this DC power is calculated using
Pi ( dc )  VCC I dc
The DC current drawn from the source is the
average value of the current delivered to the
load
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Input DC power
The current drawn from a single DC supply has the
form of a full wave rectified signal, while that drawn
from two power supplies has the form of half-wave
rectified signal from each supply
On either case the average value for the current is
2
given by
I dc   I p

The input power can be written as
Pi ( dc ) 
2

VCC I p
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Output AC power
The power delivered to the load can be calculated
V
using the following equation P  V

L ( p p)
o ( ac )
8RL
L( p)
2RL
The efficiency of the amplifier is given by
Ip 
VL ( p )
Not that
R
Therefore the efficiency can be re-expressed as
L
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Output AC power
The maximum efficiency can be obtained if
The value of this maximum efficiency will be
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Power dissipated by the output
transistors
The power dissipated by the output transistors as
heat is given by
The power in each transistor is given by
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Example
Example 1: For class B amplifier providing a 20-V
peak signal to a 16-Ω speaker and a power supply
of VCC=30 V, determine the input power , output
power and the efficiency
Solution:
The input power is given by
Pi ( dc ) 
2

VCC I p
The peak collector load current can be found from
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Example
Solution:
The input power is
Pi ( dc ) 
2

30(1.25)  23.9 W
The output power is given by
The efficiency is
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Maximum power dissipated by the
output transistors
The maximum power dissipated by the two
transistors occurs when the output voltage across
the load is given by
The maximum power dissipation is given by
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Example
Example 2: For class B amplifier using a supply of
VCC=30 V and driving a load of 16-Ω, determine
the input power , output power and the efficiency
Solution:
The maximum output power is given by
The maximum input power drawn from the supply
is
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Example
Solution:
The efficiency is given by
The maximum power dissipated by each transistor
is
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Class B Amplifier circuits
A number of circuit arrangements can be used
to realize class B amplifier
We will consider in this course two
arrangements in particular
1.
2.
The first arrangement uses a single input signal fed
to the input of two complementary transistors
(complementary symmetry circuits)
The second arrangement uses two out of phase
input signals of equal amplitudes feeded to the input
of two similar NPN or PNP transistors (quasicomplementary push-pull amplifier)
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Complementary symmetry circuits
first arrangement
This circuit uses
both npn and pnp
transistor to
construct class B
amplifier as shown
to the left
One disadvantage
of this circuit is the
need for two
separate voltage
supplies
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Complementary symmetry circuits
another disadvantage of this circuit is the
resulting cross over distortion
Cross over distortion can be eliminated the by
biasing the transistors in class AB operation
where the transistors are biased to be on for
slightly more than half a cycle
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Class AB biasing to solve
crossover distortion
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Complementary symmetry circuits
A more practical version
of a push-pull circuit
using
complementary
transistors is shown to
the right
This circuit uses to
complementary
Darlington
pair
transistors to achieve
larger current driving
and
lower
output
impedance
Complementary-symmetry push-pull circuit using Darlingtion transistors
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Complementary symmetry circuits
Second arrangement
As stated previously the second arrangement
which uses two equal input signals of opposite
phase has to be preceded by a phase inverting
network as shown below
Phase splitter circuit
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Quasi-complementary push pull
amplifier second arrangement
In practical power amplifier circuits it is
preferable to uses npn for both transistors
Since the push pull connection requires
complementary devices, a pnp high power
transistor must be used.
This can be achieved by using the circuit shown
Quasi-complementary push-pull transformer less power amplifier
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Example
Example: For the circuit shown, calculate the input
power, output power and the power handled by
each transistor and the efficiency if the input
signal is 12 Vrms
Solution:
The peak input voltage is
The output power is
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Example
Solution:
The peak load current is
The dc current can be found from the peak as
The input power is given by
The power dissipated by each transistor is given by
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Crossover Distortion
There is a slight problem associated with the push pull amplifier
arrangement. Transistors require a 0.7V difference between the base
and the emitter in order for them to start conducting.
As we can see from the two
graphs the Output voltage does
not perfectly reflect the input
voltage. There is an area called
cross over distortion where neither
transistor is conducting due to
them requiring 0.7V to switch them
on.
Input Voltage
Output voltage
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Crossover Distortion
If the transistors Q1 and Q2
do not turn on and off at
exactly the same time, then
there is a gap in the output
voltage. For large Vin, the
output follows the input with
a fixed DC offset, however
as Vin becomes small the
output drops to zero and
causes “Crossover
Distortion.”
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Improved Push-Pull Stage
VB=VBE1+|VBE2|
With a battery of VB inserted between the
bases of Q1 and Q2, the dead zone is
eliminated.
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Amplifier Distortion
If the output of an amplifier is not a complete
AC sine wave, then it is distorting the output.
The amplifier is non-linear.
This distortion can be analyzed using Fourier
analysis. In Fourier analysis, any distorted
periodic waveform can be broken down into
frequency components. These components
are harmonics of the fundamental frequency.
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Harmonics
Harmonics are integer multiples of a fundamental frequency.
If the fundamental frequency is 5kHz:
1st harmonic
2nd harmonic
3rd harmonic
4th harmonic
etc.
1 x 5kHz
2 x 5kHz
3 x 5kHz
4 x 5kHz
Note that the 1st and 3rd harmonics are called odd harmonics and the
2nd and 4th are called even harmonics
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Harmonic Distortion
According to
Fourier analysis,
if a signal is not
purely
sinusoidal, then
it contains
harmonics.
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Harmonic Distortion Calculations
Harmonic distortion (D) can be calculated:
% nthharmonicdistortion %D n 
An
 100
A1
where
A1 is the amplitude of the fundamental frequency
An is the amplitude of the highest harmonic
The total harmonic distortion (THD) is determined by:
% THD  D 22  D 23  D 23    100
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Power Transistor Derating Curve
Power transistors dissipate
a lot of power in heat. This
can be destructive to the
amplifier as well as to
surrounding components.
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For Student References (for Complementary
symmetry & push-pull Amplifiers) Read:
Chapter 12 Power Amplifiers from the book Electronic devices,
circuit and systems (Micheal M cirovic) topics 12.6 sub topic
from (12.6.1 to 12.6.2)and 12.7 to 12.8.
Chapter 11 Power Amplifiers from the book introductory
electronic devices and circuits by (Robert T
.paynter)conventional flow version. topic 11.4 class B amplifiers
complementary-symmetry or push pull amplifiers.
chapter 8 Introduction to Amplifiers from (Robert T
.paynter)topic 8.3 (see only class B amplifiers)Wikipedia &
world wide web.
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