Additional Note Poweramp

8/3/2019 Additional Note Poweramp

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CHAPTER 2POWER AMPLIFIERS

2.1 INTRODUCTION

A power amplifier is designed to deliver a large amount of power to the load. To perform this function, a power amplifier must itself be capable of dissipating large

amount of power.

It must so designed that the heat generated when it is operated at high current andvoltage levels is release into surroundings at a rate fast enough to prevent destructive

temperature buildup.

Consequently, power amplifiers typically contain bulky component s having largesurface areas to enhance heat transfer to environment.

A power amplifier is often the last, or output, stage of an amplifier system. The preceding stages may be designed to provide amplification, to provide buffering

to a high impedance signal source, or to modify signal characteristics in some

predictable way.

Power amplifier are widely used in audio component such as radio and televisionreceivers, phonographs and tape players, stereo and high fidelity system, recording-

studio equipment, and public address system.

The load of these applications is most often a loudspeaker (speaker) which requiresconsiderable power to convert electrical signal to sound waves.

It also used in electromechanical control system to drive electric motors. Exampleinclude computer disk and tape drives, robotic manipulators, autopilots, antenna

rotator

One important aspect in the design power amplifiers is that it delivers a specifiedpower to the load efficiently.

Power amplifiers are classified according to the percent time the output transistors areconducting, or turn on.

The five principle classifications are class A, class B, class AB, class C and class D.Theseclassifications are illustrated in figure 2.1 for a sinusoidal signal.


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Figure 2.1 Collector current versus time characteristics (a) class A (b) class B (c) class

AB (d) class C

Amplifier circuits are classified as A, B, AB and C for analog designs, and class D andE for switching designs. For the analogue classes, each class defines what proportion

of the input signal cycle is used to actually switch on the amplifying device.

In class A operation, an output transistor is biased at a quiescent currentIQandconducts for the entire cycle of the input signal.

For class B operation, an output transistor conducts for only one-half of each sinewave input cycle.

In class AB operation, an output transistor is biased at small quiescent currentIQ andconduct slightly more than half a cycle.

In contrast, in class C operation an output transistor conducts for less than half acycle.

In class D, the operation is using pulse (digital) signals which are on for a shortinterval and off for a long interval.


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2.2 EFFICIENCY

The power efficiency of an amplifier, defined as the ratio of power output to powerinput. The other hands, the efficiency of an amplifier refers to the ratio of output-

signal power compared to the total input power.

An amplifier has two input power sources: one from the signal, and one from thepower supply.

Since every device takes power to operate, an amplifier that operates for 360 degreesof the input signal uses more power than if operated for 180 degrees of the input

signal.

By using more power, an amplifier has less power available for the output signal; thusthe efficiency of the amplifier is low. This is the case with the class A amplifier. It

operates for 360 degrees of the input signal and requires a relatively large input from

the power supply. Even with no input signal, the class A amplifier still uses power

from the power supply. Therefore, the output from the class A amplifier is relatively

small compared to the total input power.

This result in low efficiency, which is acceptable in class A amplifiers because theyare used where efficiency is not as important as fidelity.

Class AB amplifiers are biased so that collector current is cut off for a portion of onealternation of the input, which results in less total input power than the class A

amplifier. This leads to better efficiency.

Class B amplifiers are biased with little or no collector current at the dc operatingpoint. With no input signal, there is little wasted power. Therefore, the efficiency of

class B amplifiers is higher still.

The efficiency of class C is the highest of the four classes of amplifier operations. Table 2.1 shows the summarizes the operation of the various amplifier classes.


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Table 2.1 Comparison of Amplifier Classes

2.3 AMPLIFIER OPERATION

2.3.1 Class A Amplifier Operation

Class A amplifiers are biased so that variations in input signal polarities occur withinthe limits of CUTOFF and SATURATION.

In a PNP transistor, for example, if the base becomes positive with respect to theemitter, holes will be repelled at the PN junction and no current can flow in the

collector circuit. This condition is known as cutoff. Saturation occurs when the base

becomes so negative with respect to the emitter that changes in the signal are not

reflected in collector-current flow.

Biasing an amplifier in this manner places the dc operating point between cutoff andsaturation and allows collector current to flow during the complete cycle (360

degrees) of the input signal, thus providing an output which is a replica of the input.

The class A operated amplifier is used as an audio- and radio-frequency amplifier inradio, radar, and sound systems, just to mention a few examples.

For a comparison of output signals for the different amplifier classes of operation,refer to figure 2.2 during the following discussion.


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Figure 2.2 A comparison of output signals for the different amplifier classes of operation

2.3.2 Class AB Amplifier Operation

Class AB Amplifier Operation Amplifiers designed for class AB operation are biasedso that collector current is zero (cutoff) for a portion of one alternation of the input

signal.

This is accomplished by making the forward-bias voltage less than the peak value ofthe input signal. By doing this, the base-emitter junction will be reverse biased during

one alternation for the amount of time that the input signal voltage opposes and

exceeds the value of forward-bias voltage.

Therefore, collector current will flow for more than 180 degrees but less than 360degrees of the input signal, as shown in figure 2.2 view B.

As compared to the class A amplifier, the dc operating point for the class ABamplifier is closer to cutoff.

The class AB operated amplifier is commonly used as a push-pull amplifier toovercome a side effect of class B operation called crossover distortion.


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2.3.3 Class B Amplifier Operation

Amplifiers biased so that collector current is cut off during one-half of the inputsignal are classified class B.

The dc operating point for this class of amplifier is set up so that base current is zerowith no input signal.

When a signal is applied, one half cycle will forward bias the base-emitter junctionand IC will flow. The other half cycle will reverse bias the base-emitter junction and

IC will be cut off.

Thus, for class B operation, collector current will flow for approximately 180 degrees(half) of the input signal, as shown in figure 2.2 view C.

The class B operated amplifier is used extensively for audio amplifiers that requirehigh-power outputs. It is also used as the driver- and power-amplifier stages of

transmitters.

2.3.4 Class C Amplifier Operation

In class C operation, collector current flows for less than one half cycle of the inputsignal, as shown in figure 2.2 view D.

The class C operation is achieved by reverse biasing the emitter-base junction, whichsets the dc operating point below cutoff and allows only the portion of the input

signal that overcomes the reverse bias to cause collector current flow.

The class C operated amplifier is used as a radio-frequency amplifier in transmitters.


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2.5 CROSSOVER DISTORTION

Class B suffers from a fundamental problem in that the push-pull amplifier does not,in practice, move smoothly from one half of the waveform to the other half.

When the current outputs from the two valves are added together by the outputtransformer, a kink can be seen on class B amplifiers as shown in figure 2.3. Class

AB amplifiers can suffer from this also, if the bias current is too low.

The portion of the curve is called the dead band, and it produces a crossoverdistortion as illustrated in figure 2.3

In summary:Class B amplifiers generally introduce some crossover distortion.

Class AB amplifiers may introduce some crossover distortion.

Class A amplifiers introduce no crossover distortion, as both valves conduct

simultaneously.

Figure 2.3 Crossover Distortion Signal

2.5 SERIES-FED CLASS A AMPLIFIER

The simple fixed bias circuit connection as shown in figure 2.4 can be used to discussthe main features of a class A series fed.

The only difference between this circuit and small signal circuit are in range of volts,and transistor used is a power transistor that is capable of operating in the range of a

few to tens of watts.


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Figure 2.4 Series-fed class A large-signal Amplifier

DC Bias Operation

The dc bias set by Vcc and RB fixes the dc base-bias current at

B

CC

BR

VI

7.0= ------------------------------------------------(2.1)

BC II = -----------------------------------------------------(2.2)

CCCCCE RIVV = -----------------------------------------------(2.3)

Figure 2.5 shows the transistor characteristic showing load line and Q-point.


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Figure 2.5 Transistor characteristic showing load line and Q-point

AC Operation

When an input ac signal is applied to the amplifier of figure 2.4, the output will varyfrom its dc bias operating voltage and current.

A small input signal as shown in figure 2.6, will cause the base current to vary aboveand below the dc bias point, which will then cause the collector current (output) to

vary from the dc bias point set as well as the collector emitter voltage to vary aroundits dc bias value.


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Figure 2.6 Amplifier input and output signal variation

Power Consideration

The input power supplied to the class A series-fed amplifier is

QCCCiIVdcP =)( -----------------------------------------------------------------(2.4)

By using rms signals, the ac power delivered to the load (RC) can be expressed using

C

C

O

CCO

CCEO

R

rmsVacP

RrmsIacP

rmsIrmsVacP

)()(

)()(

)()()(

2

2

=

=

=

----------------------------------------------------(2.5)

By using peak signals, the ac power delivered to the load may be expressed using


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C

CO

C

C

O

CCE

O

RpVacP

RpI

acP

pIpVacP

2)()(

2

)()(

2

)()()(

2

2

=

=

=

--------------------------------------------------------(2.6)

By using peak-to-peak signals, the ac power delivered to the load may be expressedusing

C

CE

O

C

C

O

CCE

O

R

ppVacP

R

ppI

acP

ppIppVacP

8

)()(

8

)(

)(

8

)()()(

2

2

=

=

=

-----------------------------------------------(2.7)

The efficiency of a power amplifier is defined to be

%100)(

)(% =

dcP

acP

i

O ------------------------------------------------------(2.8)

The maximum efficiency for class A can be determined using the maximum voltageand current swings where

%25%100)(max

)(%

2)(

8

)(

max)(

max)(

2

2

==

=

=

=

=

dcP

acPMaxMax

R

VdcP

R

VacP

R

VppI

VppV

i

O

C

CC

i

C

CC

O

C

CC

C

CCCE


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Example 2.1

Calculate the input power, output power and efficiency of the amplifier circuit in figure

2.7 for an input voltage that results in a base current of 10 mA peak.

Figure 2.7

Solution:

mAkR

VI

B

CC

B 3.191

7.0207.0=

=

=

mAmII BC 5.482)3.19(25 ===

VmRIVV CCCCCE 4.10)20)(5.482(20 ===

Wm

RpI

acP

peakmAmpIpI

C

C

O

BC

625.0)20(2

)250(

2

)()(

250)10(25)()(

22

===

===

From figure 2.7 (b), mAICQ 480=


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WmIVdcPQCCCi

6.9)480(20)( ===

The amplifiers power efficiency can then be calculated using

%#5.6%1006.9

625.0%100

)(

)(% ===

dcP

acP

i

O

2.6 CLASS A COUPLING POWER AMPLIFIER

Class A amplifier having maximum efficiency of 50% uses a transformer to couple theoutput signal to the load as shown in figure 2.8.

Figure 2.8 Transformer-coupled audio power amplifier

2.6.1 Transformer Action

A transformer can increase or decrease voltage or current levels according to the turnsratio as shown in figure 2.9(a)

The voltage transformation is given by

1

2

1

2

N

N

V

V= ----------------------------------------------------------------------(2.9)

The current in the secondary winding is inversely proportional to the number of turnsin the winding. The current transformation is given by


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2

1

1

2

N

N

I

I= -----------------------------------------------------------------------(2.10)

This relationship is shown in figure 2.9(b). If the number of turns of wire on thesecondary is greater than that on the primary, the secondary will be less than the

current in the primary.

Figure 2.9 Transformer Operation (a) Voltage transformation (b) current transformation

(c) impedance transformation

The impedance transformation is given by2

2

2

1

2

1

'

aN

N

R

R

R

R

L

L =

== --------------------------------------------------(2.11)

where2

1

N

Na = and 'LR is the reflected impedance

We can express the load resistance reflected to the primary side as2

2

1 RaR = or LL RaR2'= --------------------------------------------(2.12)


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Example 2.2

Calculate the effective resistance seen looking into the primary of a 15:1 transformer

connected to an 8 load.

Solution

=== kRaR LL 8.1)8()15('22

Example 2.3

What transformer turns ratio is required to match a 16 speaker load so that the effective

load resistance seen at the primary is 10 k.

Solution:

1:25625

62516

10'

2

1

2

2

1

==

===

N

N

k

k

R

R

N

N

L

L

2.6.2 Operation of Amplifier Stage

The transformer (dc) winding resistance determines the dc load line for the circuit ofthe figure 2.8.

Figure 2.10 shows the load lines for class A transformer-coupled amplifier.


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Figure 2.10 Load lines for class A transformer-coupled amplifier

Figure 2.11 shows the voltage and current swings resulting in the circuit of figure 2.8.


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Figure 2.11 Graphical operation of transformer-coupled class A amplifier

From the signal variations shown in figure 2.11 the values of the peak-to-peak signalswings are

MINMAX CECECEVVppV = )(

minmax)( CCC IIppI =

The ac power developed across the transformer primary can then be calculated using( )( )

8)( minmaxminmax

CCCECE

O

IIVVacP

= --------------------------------(2.13)

For the ideal transformer the voltage delivered to the load can be calculated usingequation 2.9.

1

1

2

2 VN

NVVL ==

The power across load can then be expressed as

L

LL

R

rmsVP

)(2= -----------------------------------------------------------------(2.14)


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The load current is given byCL I

N

NII

2

1

2 == ----------------------------------------------------------------(2.15)

The output ac power then calculated usingLLL RrmsIP )(

2= ------------------------------------------------------------------(2.16)

The input (dc) power obtained from the supply is calculated from the supply dcvoltage and the average power drawn form the supply;

QCCCiIVdcP =)( ----------------------------------------------------------------(2.17)

For the transformer-coupled amplifier the power dissipated by the transformer is small(due to the small dc resistance of a coil) and will be ignored in the present calculation.

Thus the only power loss considered here is that dissipated by the power transistor andcalculated using

)()( acPdcPP oiQ = ---------------------------------------------(2.18)

wherePQis the power dissipated as heat.

The maximum theoretical efficiency for a class A transformer-coupled amplifier goesup 50%.

The efficiency can be expressed as%50%

2

minmax

minmax

+

=

CECE

CECE

VV

VV --------------------------------(2.19)

The larger the value ofmaxCE

V and the smaller the value ofminCE

V , the closer the

efficiency approaches the theoretical limit of 50%.


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Example 2.4

Calculate the efficiency of a transformer-coupled class A amplifier for a supply of 12 V

and outputs of

(a) VpV 12)( =

(b) VpV 6)( =

(c) VpV 2)( =

Solution

Since ,12VVV CCCEQ == the maximum and minimum of the voltage swing are

(a) VpVVV CEQCE 241212)(max =+=+=

VpVVV CEQCE 01212)(min ===

%50024

02450%

2

=

+

=

(b) VpVVV CEQCE 18612)(max =+=+=


%5.12618

61850%

2

=

+

=

(c) VpVVV CEQCE 14212)(max =+=+=


%39.11014

101450%

2

=

+

=

The amplifier efficiency drops dramatically from a maximum of 50% for V(p) = VCC to

slightly over 1% for V(p) =2 V.


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2.7 CLASS B AMPLIFIER OPERATION

Class B operation is provided when the dc bias leaves the transistor biased just off, thetransistor turning on when the ac signal is applied.

This is essentially no bias and the transistor conduct current for only one-half of thesignal cycle.

To obtain output for the full cycle of signal, it is necessary to use two transistor andhave each conduct on opposite half cycle.

Then the combined operation providing a full cycle of output signal. Since one part of the circuit pushes the signal high during one half-cycle and the other

parts pulls the signal low during the other half-cycles, the circuit is referred to as a

push-pull circuit.

Figure 2.12 shows a diagram for push pull operation.

Figure 2.12 Block representation of push pull operation.

An ac input signal is applied to the push-pull circuit, with each half operating onalternate half-cycles, the load then receiving a signal for the full ac cycle.

The power transistor used in the push-pull circuit are capable of delivering the desiredpower to the load, and the class B operation of these transistor provides greaterefficiency than was possible using a single transistor in class A operation.

The power supplied to the load by an amplifier is drawn from the power supply whichprovides the input or dc power as shown in figure 2.13.


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Figure 2.13 Connection of push-pull amplifier to load: (a) using two voltage supplies, (b)

using one voltage supply.

The amount of this input power cab be calculated usingdcCCi IVdcP =)( ------------------------------------------------------(2.20)

where Idc is the average or dc current drawn from the power supplies.

The average current drawn can be expressed as)(

2pIIdc

= -----------------------------------------------------------(2.21)

Thus, using equation 2.21 in the power input equation 2.20 results

= )(

2)( pIVdcP CCi

----------------------------------------------(2.22)

The output power can be calculated as

L

L

OR

rmsVP

)(2= -------------------------------------------------------(2.23)

L

L

L

LO

R

pV

R

ppVacP

2

)(

8

)()(

22

=

= -------------------------------------(2.24)

The larger the rms or peak output voltage, the larger the power delivered to the load. The efficiency of the class B amplifier can be calculated using the basic equation;


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%100)(

)(% =

dcP

acP

i

O

( )[ ]

%100)(

4%

%100)(/2

2/)(%

2

=

=

CC

L

CC

LL

V

pV

pIV

RpV

-------------------------------(2.25)

using LL RpVpI /)()( = . Equation 2.25 shows that the larger the peak voltage, the higher

the circuit efficiency, up to a maximum value when CCL VpV =)( , this maximum

efficiency then being

max efficiency %5.78%1004

==

The power dissipated ( as heat) by the output power transistors is the differencebetween the input power delivered by the supplies and the output power delivered to

the load.

)()(2 acPdcPP oiQ = --------------------------------------------(2.26)

whereP2Q is the power dissipated by the two output power transistor.

The dissipated power handled by each transistor is then

2

2Q

QPP = --------------------------------------------------------(2.27)


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Example 2.5

For class B operation the maximum output power is delivered to the load whenCCL VpV =)( .

MaximumL

CC

O

R

VacP

2

)(2

= ------------------------------------(2.28)

The maximum value of average current from the power supply isMaximum

L

CC

dcR

VpII

2)(

2==


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WhereL

CC

R

VpI =)(

Then, the maximum value of input power isMaximum

L

CC

L

CCCCdcCCiR

V

R

VVIimumVdcP

222)(max)( =

== ------------(2.29)

The maximum power dissipated by the two transistor accurs when the output voltageacross the load is

CCL VpV 636.0)( =

For a maximum transistor power dissipation ofMaximum

L

CC

QR

VP

2

2

2

2

= -----------------------------------------(2.30)

Example 2.6

For a class B amplifier using a supply of VCC = 30V and driving a load 0f 16 ,

determine the maximum input power, output power , and transistor dissipation.

Solution

Maximum WR

V

acPL

CC

O 125.28)16(2

30

2)(

22

===

Maximum WR

VdcP

L

CC

i 81.35)16(

)30(22)(

22

===

Maximum %54.78%10081.35

125.28%100

)(

)(% ===

dcP

acP

i

O

The maximum power dissipated by each transistor is

Maximum( )

WR

VPimumP

L

CCQ

Q7.5

16

3025.0

25.0

2

max

2

2

2

22

=

=

==


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The maximum efficiency of a class B amplifier can also be expressed as follows:

L

L

L

LO

R

pV

R

ppVacP

2

)(

8

)()(

22

=

=

==

L

LCCdcCCiR

pVVIVdcP

)(2)(

( )( )[ ]

%)(

54.78%

%100/)(/2

2/)(%100

)(

)(%

2

CC

L

LLCC

LL

i

O

V

pV

RpVV

RpV

dcP

acP

=

==

-------------(2.31)

Example 2.7

Calculate the efficiency of a class B amplifier for a supply voltage of VCC = 24 V with

peak output voltage of

(a) VpVL 22)( =

(b) VpVL 6)( =

Solution:

(a) %72

24

2254.78%

)(54.78% =

==

CC

L

V

pV

(b) %6.1924

654.78%

)(54.78% =

==

CC

L

V

pV

Notice that a voltage near maximum to VCC results in efficiency near the maximum,while a small voltage swing still provides poor efficiency.


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2.8 CLASS B AMPLIFIER CIRCUIT

Figure 2.14 shows different way to obtain phase-inverted signal from a single inputsignal.

Figure 2.14 Phase-splitter circuits

Figure 2.14(a) shows a center-tapped transformer to provide opposite phase signals. If the transformer is exactly center-tapped, the two signals are exactly opposite in

phase and of the magnitude(no crossover distortion).

Figure 2.14(b) shows a circuit uses a BJT stage with in-phase output from the emitterand opposite phase output from the collector.


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2.8.1 Transformer-Coupled Push-Pull Circuits

Figure 6.15 shows a circuit uses a center tapped input transformer to produce oppositepolarity signals to the two transistor inputs

An output transformer to drive the load in a push-pull mode.

Figure 6.15 Push-pull circuit.

During the first half-cycle, Q1 conduct whereas Q2 off. The current i1 through transformer results in the first half-cycle of signal to the load. During the second half-cycle, Q2 conduct whereas Q1 off. The current i2 through transformer results in the second half-cycle of signal to the

load.

The overall signal developed across the load then varies over the full cycle of signaloperation.

2.8.2 Complementary-Symmetry Circuits

Figure 6.16(a) shows a circuit using complementary transistors (npn andpnp) toobtain a full cycle output across a load using half-cycles of operation from each

transistor.


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Figure 6.16 Complementary-symmetry push-pull circuit


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The npn transistor will be biased into conduction by the positive half-cycle of signal,with a resulting half-cycle signal across the load as shown in figure 6.16(b).

During the negative half-cycle of signal thepnp transistor is biased into conductionwhen the input goes negative as shown in figure 6.16(c).

During a complete cycle of the input a complete cycle of output signal is developedacross the load.

One disadvantage of this circuit is the need for two separate voltage supplies. Another, less obvious disadvantage with the complementary circuit is shown in the

resulting crossover distortion in the output signal.

Crossover distortion refers to the fact that during the signal crossover from positive tonegative (or vice versa) there is some nonlinearity in the output signal.

A more practical version of a push-pull circuit using complementary transistors isshown in figure 6.17.

Figure 6.17 Complementary-symmetry push-pull circuit using Darlington transistor.


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The load is driven as the output of an emitter-follower so that the load resistance of theload is matched by the low output resistance of the driving source.

This circuit connection to provide higher output current and lower output resistance.

Example 2.8

For the circuit of figure 6.18 calculate the input power, output power and power handled

by each output transistor and the circuit efficiency for an input of 12 V rms.

Figure 6.18 Class B power amplifier for example 2.8

Solution:

VrmsVpV ii 17)12(2)(2)( ===

Since the resulting voltage across the load ideally the same as the input signal,

VpVL 17)( =

The output power across the load is

WR

pVacP

L

LO 125.36

)4(2

17

2

)()(

22

===


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The peak load current is

AR

pVI

L

L

L 25.44

17)(===

dc current from supplies is calculated to be

( )ApII Ldc 71.2

25.42)(

2===

Power supplied to the circuit is

WIVdcP dcCCi 75.67)71.2(25)( ===

Power dissipated by each output transistor is

WPPP

P OiQ

Q 8.152

125.3675.67

22

2=

=

==

The circuit efficiency is then

%3.53%10075.67

125.36%100

)(

)(% ===

dcP

acP

i

O

2.9 POWER TRANSISTOR HEAT SINK

The definition of power is the rate at which energy is consumed or dissipated ( 1 W =1 J/s).

If the rate at which heat energy dissipated in device is less than the rate at which it isgenerated, the temperature of the device must rise.

The power dissipated in a transistor increases its internal temperature above theambient temperature.

If the device or junction temperature Tj becomes too high, the transistor may sufferpermanent damage.

Special precautions must be taken in packaging power transistors and in providingheat sinks so that heat can be conducted from the transistor.

Figure 6.18(a) and (b) show two packaging scheme, and figure 6.18(c) shows atypical heat sink.


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Figure 6.19 Two packaging schemes: (a) and (b) for power transistor and (c) typical heat

sink

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Additional Note Poweramp