Power amplifire analog electronics

POWER AMPLIFIER

Rakesh Mandiyahttps://in.linkedin.com/in/[email protected]

https://in.linkedin.com/in/rakeshmandiya

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 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 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.

Signal

pick up

transducer

Voltage

amplifier

Voltage

amplifier

power

amplifier Output

transducer

Non

Electrical

Energy inputEnergy

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.

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

The ratio of AC output power to DC powersupplied 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.

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:-

S.NO Voltageamplifier

Poweramplifier

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 mV Input 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

AUDIO POWER AMPLIFIER:-

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.

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 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:-

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.

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.

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%.

SINGLE ENDED POWER AMPLIFIER:-

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.

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.

COMPLEMENTRAY & SYMMETRY

POWER AMPLIFIERThis 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)

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.

CLASS-A AMPLIFIER OPERATION:-

Class A power amplifier:-

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.

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.

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.

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.

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

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 / 2 and Vm / 2

A.C. Power output using R.M.S. values :-

The a.c. power output is given by Pac = Vrms Irms = I2rms RL

= V2rms/RL

CLASS-A AMPLIFIER:-

A.C. Power output using maximum values :-

The a.c. power output is given by : Pac = Vrms.Irms =(Vm/2).Im/2 =(Vm Im)/2

Since Im = Vm / RL the above equation can be rewritten :

Pac = V2m/2RL ----------(1) Pac = I2mRL / 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

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.

CIRCUIT DIAGRAM OF SERIES FED CLASS A AMPLIFIER:-

DC OPERATION OF SERIES FED CLASS A AMPLIFIER:-

AC OPERATION OF SERIES CLASS A AMPLIFIER:-

o/p current

swing

o/p voltage swing

i/p signal

Ic

Vcc/Rc

Vcc Vce0

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.

A.C. POWER OUTPUT USING PEAK TO PEAK VALUES:-

wt

Imax

0Imin

/2

1

2

3

D.C. bias value

ic

A.C. POWER OUTPUT USING PEAK TO PEAK VALUES:-

We have from Equation Pac = Vm Im/2

but as Vm = Vpp/2 = (Vmax-Vmin)/2

Also Im = Ipp/2 = (Vmax-Vmin)/2

We get Pac = (Vpp/2.Ipp/2)/2 = Vpp.Ipp/8

= (Vmax–Vmin)(Imax-Imin)/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 = Pac/Pdc *100 %η = ( Vmax-Vmin) (Imax- Imin)/8VccIcq*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.

Vmax = Vcc and Vmin = 0 Imax = 2Icq and Imin =0 % η = (Vcc-0)(2Icq-0)/8Vcc.I cq.100 = (2VccIcq/8VccIcq).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%.

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 Pd = Pdc – Pac

Maximum power dissipation occurs when there is zero a.c. input signal, the a.c. power output (Pac) is also zero.

Pac = 0 because (Iac=0)

POWER DISSIPATION:-

Then Pd = Pdc

or Pd = Vcc Icq

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.

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.

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.

TRANSFORMER 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 ).


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 RL is actual load resistance

RL’ 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.

AC AND DC LOAD LINE OF TRANSFORMER

COUPLED CLASS A AMPLIFIER:-

Ic

Vce

Ac load line

Dc load lineQ

Ic max 2 Icq

Ic min=0

Vce min=0

Vce max=0

2Vcc

Icq

Vceq=Vcc

At center of ac

load line


D.C. POWER INPUT:- The d.c. power input is given by Pdc=VccIcq

This expression is same as for series fed directly 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.


AC OUTPUT POWER:- Generalized expression for a.c. output power for

transformer coupled amplifier is same as Pac = (Vpp/2.Ipp/2)/2 = Vpp.Ipp/8 = (Vmax –Vmin)(Imax-Imin)/8

EFFICIENCY:-%η = Pac/Pdc .100%η = (Vmax-Vmin) (Imax- Imin)/8VccIcq.100


MAXIMUM EFFICIENCY:- Vmax = 2Vcc and Vmin = 0 Imax = 2Icq and Imin =0 % η = (2Vcc-0)(2Icq-0)/8Vcc.Icq.100 = (4VccIcq/8VccIcq).100 = 50%


POWER DISSIPATION:-The power dissipation by the transistor is given by

: Pd=Pdc-Pac

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=VccIcq


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.

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.


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.


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 componentThird harmonic component

V V

V

V

V

0

0

0

0

0

Vm sin2wt

Vmcos2wt Vm sin3wt

Waveform due to

fundamental,2nd,3rd,harmonic

component

Vmsinwt

wt

wt wt

wt

wt

Vm cos2wt

Vm sin2wtVm sin3wt

Vmsinwt

SECOND HARMONIC DISTORTION:-

12

3

4

5

D.C. bias value

ImaxI1/2

Icq

I-1/2

Imin

wt

Ic


For linear amplification, the dynamic transfer characteristics relation b/w Ib and Ic 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).


Equation that describes distorted signal waveform is: ic= ICQ + I0 + I1cos(ωt) + I2cos(2ωt)

Where (ICQ+I0) =dc component independent of time.

I1= fundamental component of distorted ac signal. I2= second harmonic component , at twice the

fundamental frequency.

MEASUREMENT OF SECOND HARMONIC DISTORTION:-

At point 1,ωt=0 Ic=ICQ+I0+I1cos0+I2cos0

Ic=ICQ+I0+I1+I2 =IC(MAX)

At point 2,wt=/2 IC=ICQ+I0+I1cos/2+I2cos2/2 Ic=ICQ+I0 +I2

At point 3,wt=

Ic=ICQ+I0+I1cos+I2cos2 Ic=ICQ+I0-I1+I2 =IC(MIN)


At ωt=0,Ic=Imax

At ωt=/2,Ic=ICQ

At ωt=,Ic=Imin

Imax=ICQ+I0+I1+I2

Icq=ICQ+I0-I2

Imin=ICQ+I0-I1+I2, I0=I2

Imax-Imin=2I1


I1 =(Ic max-Ic min)/2I0 =I2

Imax+Imin=2ICQ+2(I0+I2)=2ICQ+4I2

I2=(Imax+Imin-2ICQ)/4

As the amplitude of the fundamental and second harmonic are known ,the % of second harmonic distortion can be calculated as

%D2=[ (I2/ I1)*100 ]

MEASUREMENT OF OVERALL HARMONIC DISTORTION:-

D2 =(Icmax +Icmin-2Icq ) *100/ 2(Icmax-Icmin)

D2 =(Vcemax +Vcemin-2Vceq ) *100/2(Vcemax-Vcemin)

5 POINT METHOD OF HARMONIC DISTORTION:- 1. Fundamental 2,3,4,-----------so on ,harmonic

distortion is given by D2=I2/I1

D3=I3/I1

D4 =I4/I1 and so on.

2. Total harmonic distortion is given by D2 = D2

2+D32+D4

2+---------

Measurement Of overall Harmonic Distortion:-3. When the distortion is negligible ,the power

delivered to the load given by Pac=I2

mRL/2Where Im=peak value of the output waveform =(Imax-Imin)/24. If B1=Fundamental frequency component

Pac =I21RL/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:

(Pac )D=I21RL/2+I2

2RL/2 =I2

1RL(1+I22/I2

1)/2 =Pac(1+D2

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 :

Ic=G1Ib+G2Ib2+G3Ib

3+……….

The input signal is given by : Ib = Ibmcosωt

Ic =G1Ibmcosωt+G2Ibmcos2 ωt+G3Ibmcos3ωt+……….

Ic =B0+B1cosωt+B2cos2ωt+B3cos3ωt+………

HIGHER ORDER HARMONIC DISTORTION:-

At point 1, ωt=0,Ic=Imax

Imax=ICQ+B0+B1+B2+B3+…….At point 2, ωt=/3,Ic=I1/2

I1/2=ICQ+B0+0.5B1-0.5B2….At point 3, ωt=/2,Ic=ICQ

At point 4, ωt=2/3,Ic=I -1/2

I-1/2=ICQ+B0-0.5B1-0.5B2+B3-0.5B4+…..At point 5,wt= ,Ic=Imin

Imin=ICQ+B0-B1+B2-B3+B4……

Higher Order Harmonic distortion:-

Solving these equations B0=(Imax-2I 1/2+2I-1/2+Imin)/6-Icq

B1=(Imax+I 1/2-I-1/2-Imin)/3B2=(Imax-2Icq+Imin)/4

B3=(Imax-2 I1/2+2I-1/2-Imin)/6 B4=(Imax-4I 1/2+6Icq-4I-1/2+Imin)/12

Hence the harmonic distortion coefficients can be obtained as

Dn=|B0|/|B1|

POWER OUTPUT DUE TO DISTORTION:-

Now Pac=B2

1R L /2Hence the output power due to harmonic distortion is

(Pac)D =(B21Rl+ B2

2Rl+ B23Rl+…..)/2

=B21Rl(1+B2

2/B21+B2

3/B21+……)/2

(Pac)D=Pac(1+D22+D2

3+…….) D2=D2

2+D23+……..

(Pac)D=Pac(1+D2)

POWER OUTPUT DUE TO DISTORTION:-

If the total harmonic distortion is 15% i.e. D = 0.15 (Pac)D=Pac(1+0.152)=1.0225Pac

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.

CIRCUIT DIAGRAM OF CLASS B AMPLIFIER:-

T1 !NPN

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.

POWER AND EFFICIENCY CALCULATIONS:-

Pin dc =Vcc Idc

Where, Idc = the average or direct current taken from the collector supply. = Icmax /Thus

P in dc= Vcc Icmax /

POWER AND EFFICIENCY CALCULATIONS:-

Pdc=(Icmax/)2RcThus

%η = Pac/Pdc .100

%η =(¼* Icmax Vcc /Icmax Vcc/ )*100%η = /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.

T1 !PNP

R1

1k

+

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.

Operation of Class AB amplifier:-

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:-

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.

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.

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.

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.

CLASS –C AMPLIFIER OPERATION:-

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%.

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.

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.

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.

Converts digital

signal

back to

sinusoidal signal

Saw tooth

generator comparator amplifierLow pass

filter

feedback

Vin

Vout

BLOCK DIAGRAM OF CLASS-D POWER:-


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 rDS(ON) resistances shunt and filter resistances should be small compared to the load resistance.


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 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


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 (T1 and T2),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.Tr1,Tr2).

N1 N2

TR1

T1 !NPN

T2 !NPN

N1 N2

TR2

V1

5

R1 1k

R2 1k

input

load



Both the transformer Tr1 and Tr2 are center-tapped transformers. Both Resistors R1and R2 provides biasing arrangement. Load is connected across secondary of Tr2.To ensure that maximum power is delivered to the load from amplifier ,turns ratio of Tr2 is so chosen that output impendence of transistor matches to that of load impedance.


Operation:- As shown in circuit diagram, Ic1 and Ic2 flow in opposite

direction through the primary of transformer(Tr2). 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.


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.


These currents(Ic1 and Ic2) flow in opposite direction in the two halves of the center-tapped output transformer(Tr2).Due to this,the magnitude of voltage induced in the load will be proportional to the difference of collector currents(Ic1 and Ic2).

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 (Ic1 and Ic2).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.


From the above discussion, it is seen that when Ic1 increases, Ic2 decreases and when Ic2 increases, Ic1 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 3600. Hence name class-A push-pull amplifier.

DISTORTION IN CLASS-A PUSH-PULL AMPLIFIER:-

Ib1=Ib sin wt Ib2=Ib sin (wt+) The collector current is represented as : Ic=I0+I1sin wt+I2sin 2wt+I3sin 3wt As the collector collector current of second

transistor is 180 out of phase. It can be represented as :


Ic2=I0+I1sin(wt+)+I2sin2(wt+)+I3sin3(wt+)

Ic2=I0-I1sinwt+I2sin2wt-I3sin3wt

In the push-pull amplifier, the output voltage induced in the secondary of the output transformer is proportional to the two collector currents (Ic1-Ic2).

DISTORTION IN CLASS-A PUSH-PULL AMPLIFIER:-Hence V0=K (Ic1-Ic2)

or V0=K (2I1sin2wt+2I3sin3wt+…….)

or V0=2K (I1sinwt+I3sin3wt+I5sin5wt+……)

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.


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.


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 T1 and T2 flows in opposite direction hence there is no net d.c. magnetization or core saturation.


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.



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.

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.

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


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/2P 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× = [Im/2×(Vcc-Vmin) ]/ [2/π× ImVcc] ×100ImVcc] ×100

= [ π(Vcc -Vmin)] / [4Vcc] ×100= [ π(Vcc -Vmin)] / [4Vcc] ×100


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 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


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:-

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

PHASE SPLITTER CIRCUIT:-

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

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.

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) .

COMPLEMENTARY SYMMETRY CLASS –B PUSH PULL AMPLIFIER:-

Fig:-circuit diagram


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.


This results into negative half cycle across load RL. Thus I1 and I2 flow in one half cycle each but flow through RL . Hence total current through RL is in both half cycles, a complete cycle of output signal comes across the load .As in figure


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.

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)

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”.

Q3

Q4 !NPN

Q1

Q2 !PNP

R1

R2

1k

RL

CL

Cin1

Cin2

R3

1k

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.

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.

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 = ΔIcRL = 100mA*5ohm = 500mV

CASE 2: When loudspeaker is transformer coupledOutput impedance of the transformer =ΔVce/ΔIc =12/100mA=120ohm

For maximum power transfer the load resistancereferred to primary side (i.e.RL) must be equal to output

impedance of the transistorSo RL=120

We know that turns ratio is given by Turns ratio (N) = (R’

L/RL)1/2

= (120/5)1/2

= 4.898Now secondary voltage i.e. voltage across the speaker Vs=VP/N=VCC/N=12/4.898=2.45V

Load current= Vs/RL=2.45/5=0.489

So power transferred to speaker =ILVL

= ILVL=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(N1/N2) is 10,determine the maximum power delivered to the load. Take a zero signal collector current of 500mA.

ANS:-Given that RL=8 ohm, N=10 and ICQ=500mA

The load resistance seen by the primary of the transformer R’L = N2RL =10*10*8= 800 ohm

So maximum power delivered to the 8 ohm loud speaker Po= 0.5*I2

CQR’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

ic=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 D2=(B2/B1)*100=(1.5/15)*100=10%

D3=(B3/B1)*100=(1.2/15)*100=8%

D4=(B4/B1)*100=(0.5/15)*100=3.33%

Distortion factor:- D=(D2

2+D23+D2

4)1/2=0.132=13.2% (2) The net power output in percentage of distortion P=(1+D2)P1=(1+.132*.132)P1 =1.0174P1

(3) Hence percentage increase in power due to distortion = (1.0174 P1- P1)/P1*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 Ipeak=Vpeak/RL=(24/2)/100=0.12A

DC power drawn from battery Pdc=Vcc*Idc=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%.

Therefore a.c. power delivered to load= 1/2( V2

peak/RL)=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, Vcc=20 volt, VCEQ=10 Volt, ICQ=600mA and collector load resistor RL=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 (Pin)dc=VCCICQ

= 20*600*0.001=12 Watts(2)d.c. power consumed by the load resistor PRc=I2

CQ*RC=(600*0.001)2*16=5.76 Watts

(3)d.c. power delivered to the transistor Ptr=(Pin)dc-PRc=6.24 Watts

(4) A.C. power developed across load resistor (Po)ac=I2

rmsRc=(Im/1.404)2Rc

=(.3/1.404)2*16=0.72 Watts(5) Collector efficiency (Po)ac/Ptr*100=(0.72/6.24)*100=11.5%

(6) Overall efficiency ((Po)ac/Pin)*100=0.72/12*100=6%

SUBMITTED TO:

MRS. SARITA CHOUHANd

SUBMITTED BY:

PRATEEK LAMORIA

RAKESH YADAV

RAMESH BISHNOI

RAM CHANDRA

PRATIBHA SONI

PRINYANKA JANGIR

SHAMBHU LAL VYAS

RAM NIWAS

ROHITASH

RAMBHAROSE NAGAR

Power amplifire analog electronics

Engineering