Lecture # 16 &17 Complementary symmetry & push-pull Amplifiers

Prepared by Engr Sarfaraz Khan Turk Lecturer at IBT LUMHS Jamshoro 1

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 one half 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.

2

The standard or Basic class-B Push-Pull Ampl

ifier

  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 .

3

•

The Standard or Basic Class-B push-pull Amplifier(contd

)

During the positive half cycle of the AC input, transistor Q 1 (

npn

) is conducting and Q 2 (

npn

) is off.

•

During the negative half cycle of the AC input, transistor Q 2 (

npn

) is conducting and Q 1 (

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

4

The Standard or Basic Class B push-pull Amplifier

The two same npn transistors will be use in this configuration 5

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.

6

Complementary-symmetry class-B Push-Pull amplifiers

7

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? 8

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

output acts as a

current sink

) current (the 9

Current sources and current sink

  the circuit above is a

good

current source but a

poor

current sink (stored charge must be removed by

R E

) an alternative circuit using

pnp

(below) is a

good

transistors current sink but a

poor

current source 10

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

11

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 12

The Complementary-symmetry Push-Pull Amplifier Stages

As V in increases, Q 1 is on and pushes a current into R L .(During positive +ve half cycle of the input of AC signal. As V in decreases, Q 2 is on and pulls a current out of R L . .(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

P i

(

dc

) 

V CC I dc

The DC current drawn from the source is the average value of the current delivered to the load 14

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 given by

I dc

 2  

I p

The input power can be written as

P i

(

dc

)  2 

V CC I p

15

Output AC power

The power delivered to the load can be calculated using the following equation

P o

(

ac

) 

V L

(

p



p

) 8

R L



V L

(

p

) 2

R L

The efficiency of the amplifier is given by Not that

I p



V L

(

p

)

R L

Therefore the efficiency can be re-expressed as 16

Output AC power

The maximum efficiency can be obtained if The value of this maximum efficiency will be 17

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 18

Example

Example 1:

For class B amplifier providing a 20-V peak signal to a 16 Ω speaker and a power supply of

V CC =30 V

, determine the input power , output power and the efficiency

Solution:

The input power is given by

P i

(

dc

)  2 

V CC I p

The peak collector load current can be found from 19

Example

Solution:

The input power is

P i

(

dc

)  2  30 ( 1 .

25 )  23 .

9

W

The output power is given by The efficiency is 20

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 21

Example

Example 2:

For class B amplifier using a supply of

V CC =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 22

Example

Solution:

The efficiency is given by The maximum power dissipated by each transistor is 23

Class B Amplifier circuits A number of circuit arrangements can be used to realize class B amplifier 1.

2.

We will consider in arrangements in particular this course two 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 (quasi complementary push-pull amplifier) 24

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 25

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 26

Class AB biasing to solve crossover distortion 27

Complementary symmetry circuits A more practical version of a push-pull circuit using transistors is shown to the right complementary This circuit complementary Darlington transistors larger and to current lower impedance uses to pair achieve driving output Complementary-symmetry push-pull circuit using Darlingtion transistors 28

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 29 Phase splitter circuit

Quasi-complementary push pull amplifier second arrangement In practical power amplifier circuits preferable to uses npn for both transistors it is Since the push pull transistor must be used.

connection requires complementary devices, a pnp high power This can be achieved by using the circuit shown Quasi-complementary push-pull transformer less power amplifier 30

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

V rms Solution:

The peak input voltage is The output power is 31

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 32

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

33

Crossover Distortion

If the transistors Q 1 and Q 2 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.” 34

Improved Push-Pull Stage

V B =V BE1 +|V BE2 | With a battery of V B bases of Q 1 and Q 2 inserted between the , the dead zone is eliminated.

35

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

.

36

Harmonics

Harmonics are integer multiples of a fundamental frequency.

If the fundamental frequency is 5kHz: 1 st 2 nd harmonic harmonic 3 rd harmonic 4 th harmonic etc.

1 2 3 4 x x x x 5kHz 5kHz 5kHz 5kHz Note that the 1 st 2 nd and 4 th and 3 rd harmonics are called odd harmonics and the are called even harmonics

37

Harmonic Distortion

According to Fourier analysis, if a signal is not purely sinusoidal, then it contains harmonics.

38

Harmonic Distortion Calculations

Harmonic distortion (D) can be calculated: % nth harmonic distortion



%D n



A n A 1



100 where A 1 A n is the amplitude of the fundamental frequency is the amplitude of the highest harmonic The total harmonic distortion (THD) is determined by: % THD



D 2 2



D 2 3



D 2 3

  

100

39

40

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.

41

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

.paynter)topic 8.3

from (Robert T

(see only class B amplifiers)

Wikipedia & world wide web.

42