Lecture -Power Amplifier

Lecture #10

IB >> 0

Three transistor operating modes:

IB = 0 IB > 0

Cutoff Linear (PC > 0)

Saturation

PC = 0 in both of these modes

• Amplifier efficiency

– an important consideration in the design of power amplifiers is efficiency

– efficiency determines the power dissipated in the amplifier itself

– power dissipation is important because it determines the amount of waste heat produced

• excess heat may require heat sinks, cooling fans, etc.

supplythefrom absorbed power

load the indissipatedpowerEfficiency

Concept Preview • Efficiency is most important in power amplifiers.

• Poor efficiency means that much of the input power is converted to heat.

• Class A amplifiers operate at the center of the load line and have a large quiescent current flow conducts for the entire signal cycle and has the lowest efficiency

• Class B amplifiers operate at cutoff and have no quiescent current flow. It conducts for only half of the signal cycle. They are usually operated in push-pull configurations. They have crossover distortion

• Class AB reduces crossover distortion.

• A class C amplifier conducts for less than half of the signal cycle.

• A class D amplifier switches between cutoff and saturation.

• Bridge amplifiers provide four times the output power and

eliminate the output coupling capacitor.

Power Amplifier

PIN Efficiency =

Input signal

POUT

POUT

PIN

Output signal

HEAT = PIN - POUT High efficiency means less heat.

Efficiency

• The dc power supplied to an amplifier is

PIN = VCC x IDC

• Efficiency = POUT/PIN x 100%

• The maximum efficiency for Class A

amplifiers with a dc collector resistance

and a separate load resistance is 25%.

• Class A is usually not acceptable when

watts of power are required.

0 2 4 6 8 10 12 14 16 18

0.2 0.4 0.6 0.8

1.0 1.2

1.4 A

B

C

AB

The class of an amplifier is determined by the bias

which establishes the Q-point.

Class C is established by reverse biasing the base-emitter junction.

Conduction Angles & theoretical max. efficiencies:

• Class A = 360o 50 %*

• Class B = 180o 78.5 %

• Class AB @ 200o (between A & B)

• Class C @ 90o 100 %

*Class A amplifiers are seldom driven to maximum output and typically provide much less efficiency.

t

IC

t

IC

t

IC

t

IC

ISAT

A B

C D

The major classes of amplifier operation

Class and efficiency quiz

If POUT = 100 W and PIN = 200 W, the efficiency is _________. 50%

The efficiency of an ideal amplifier is __________. 100%

When efficiency is poor, too much of the input is converted to ________. heat

An amplifier that conducts for the entire cycle is operating Class _______. A

An amplifier that conducts for half the cycle is operating Class _______. B

Class C power amplifiers

• Class C amplifier is obtained if the output current conduction angle is less than 180

• Class C have a greater efficiency than both class A and class B

• Class C also have larger distortion compared with the A and B

• Class C amplifier is used when there is no variation in signal amplitude and the output circuit contains a tuned circuit to filter out the harmonics

• Class C amplifier is used for applications similar to FM Modulation

Class C power amplifiers • The output of the class C conducts for less

than 180 of the AC cycle

• The Q-point is below cutoff as shown below

Power Amplifiers Class C

Class C power amplifier

The circuit shown to the left will be considered in analysis of the class C amplifier


The amplifier efficiency can be computed from the following equation

The output power is the AC power flowing in the load resistance

The input power is the supply power which can be determined from the multiplication of the supply voltage and the average collector current

The main step in the analysis is to compute the average collector current which is explained in the next slide

DC

AC

i

o

P

P

P

P


According to the figure shown to the left, the collector current is given by

Where ID is given by


The direct collector current is determined by its average value which is given by

It is desired to find the current as a function of the angles θ1 and θ2 rather than t1 and t2

This can be achieved by using the following mapping and

2

1

)sin(1

t

t

DpCav dtItIT

I

11 t

22 t


The average collector current can be rewritten as

By evaluating this equation we may have the following expression

To simplify the notation, the conduction angle will be defined as

2

1

)sin(1

dtItIT

I DpCav

ICav


By substituting the conduction angle expression into the average collector current equation we get

Now the supplied input power can be written as

If the output RLC circuit is a narrow band filter tuned to the fundamental frequency of the current pulses, then the output power will be

)cos(sin

pCC

CavCCi IV

IVP

)cos(sin

p

Cav

II


Where I1 is the amplitude of the fundamental current components which is determined by the trigonometric Fourier series as shown below

By solving the previous equation we may have

The output power now can be written as

)2sin2(4

1212

121

pCC

CCO

IVIVRLIP


The efficiency of the amplifier can be written as

A plot of the efficiency as a a function of the conduction angle is shown below

)cos(sin4

2sin2121

CavCC

CC

i

O

IV

IV

P

P

Class C power amplifier design

• There are four important design parameters are of great importance for PA design in general

• These parameters are

– The output power

– Transistor power dissipation

– Maximum collector to emitter voltage VCEmax

– The maximum transistor output current Ip

Class C power amplifier design • The maximum collector current is given by

• Since • The collector current can be rewritten as

• The maximum current in terms of the output

current can be written as

• Note that the value of the collector voltage VCC can be written as

Class C power amplifier design • Now the maximum collector current can be

rewritten as

• A normalized peak collector current is defined as

Class C power amplifier design • A plot of the normalized peak current

versus the conduction angle is shown below

Class C power amplifier design • The power dissipated in the transistor is given by

• Note the value of Ip can be expressed as

• From we can conclude that

• If the value of Ip is substituted in the PT equation then

2)cos( 1IV

SinIV

PPP CCPCCOiT

cos1 M

p

II

2sin2

2 1

II p

27

Class C power amplifier design • The power dissipated in the transistor is

given by

• Or PT can be rewritten as

2)

2sin2

cossin(2 1

1

IVIVP CC

CCT


• A normalized plot of PT/PO versus theta is shown below

Class C power amplifier design example

Example: Design a class C amplifier that will deliver 5-W average power to a 50 Ω load at a frequency of 1 MHz using a transistor with a safe power dissipation rating of 0.5 W

Solution:

The average output power is given by

Or

VPRVCC OL 4.2255022

Class C power amplifier design example

Solution: Since the allowable power dissipation is The maximum conduction angle can be found

from the graph shown in slide 19 or by solving the PT/PO equation

The value of the normalized current corresponds to this angle is refer to the figure

The peak collector current is given by

5.57


• An alternate design procedure for class C amplifiers is – Select the power supply

– Select the transistor

– Determine the maximum output power without exceeding the transistor ratings

– The transistor then can be driven to its maximum allowed value of output current

– Determine the value of the load resistance that twill result in the maximum current according to

32


• Now the transistor power equation can be modified as

• The normalized transistor power dissipation is given by

• Where

cos1

)2sin2()cos(sin4)(

f

33


34

Class C power amplifier design example 2

• Example 2: Determine the maximum output power and the conduction angle of a class C amplifier using a transistor with maximum power dissipation rating of 4 W and a maximum output current of 1.5 A. The supply voltage is 48 V

Solution:

The normalized maximum transistor dissipation is given by

35


Solution:

The conduction angle for maximum normalized transistor power P΄T is found to be as

If we refer back (PT/PO vs θ ) plot we find that the value of PT/PO which corresponds to this angle is

The output power now can be found as

36


Solution:

Finally the value of the load resistance that results in this output power is given by

Class C • Class C amps are never used for audio circuits.

• They are commonly used in RF circuits.

• Class C amplifiers operate the output transistor in a state that results in tremendous distortion (it would be totally unsuitable for audio reproduction).

Basic class C amplifier operation (non inverting).

Basic class C operation.

Class C waveforms.

Tuned class C amplifier.

Tuned class C amplifier with clamper bias.

Class C

• However, the RF circuits where Class C amps are used, employ filtering so that the final signal is completely acceptable.

• Class C amps are quite efficient.

Class B

• A class B output stage can be far more efficient than a class A stage (78.5 % maximum efficiency compared with 25 %).

• It also requires twice as many output transistors…

• …and it isn’t very linear; cross-over distortion can be significant.

Class B • Class B amplifiers are used in low cost designs or

designs where sound quality is not that important.

• Class B amplifiers are significantly more efficient than class A amps.

• They suffer from bad distortion when the signal level is low (the distortion in this region of operation is called "crossover distortion").

Class B • Class B is used most often where economy of design

is needed.

• Before the advent of IC amplifiers, class B amplifiers were common in clock radio circuits, pocket transistor radios, or other applications where quality of sound is not that critical.

Power Amplifiers

Crossover

distortion

Complementary Symmetry Power Amplifier (Class-B)

Power Amplifiers

Biasing the Push-Pull Amplifier (Class-AB) (OCL)

To overcome crossover distortion, the biasing is adjusted to just overcome the

VBE of the transistors; this results in a modified form of operation called class

AB. In class AB operation, the push-pull stages are biased into slight

conduction, even when no input signal is present.

Power Calculation is the same as class-B

}VCC

}VCC

Power Amplifiers

Single-Supply Push-Pull Amplifier (OTL)

The circuit operation is the same as that described previously, except the

bias is set to force the output emitter voltage to be VCC/2 instead of zero

volts used with two supplies. Because the output is not biased at zero

volts, capacitive coupling for the input and output is necessary to block

the bias voltage from the source and the load resistor.

Power dissipation of the class B output stage

versus amplitude of the output sinusoid.

Class AB • Class AB is probably the most common amplifier

class currently used in home stereo and similar amplifiers.

• Class AB amps combine the good points of class A and B amps.

• They have the improved efficiency of class B amps and distortion performance that is a lot closer to that of a class A amp.

Eliminating crossover distortion in a transformer-coupled push-pull amplifier. The diode compensates for the base-emitter drop of the transistors and produces

class AB operation.

Load lines for a complementary symmetry push-pull amplifier. Only the load lines for the npn transistor are shown.

Single-ended push-pull amplifier.

A Darlington class AB push-pull amplifier.

A Class AB push-pull amplifier with correct output voltage.

Incorrect output waveforms for the amplifier in previous Figure.

Class AB

• With such amplifiers, distortion is worst when the signal is low, and generally lowest when the signal is just reaching the point of clipping.

• Class AB amps use pairs of transistors, both of them being biased slightly ON so that the crossover distortion (associated with Class B amps) is largely eliminated.

• Distortion in push-pull amplifiers

• Improved push-pull output stage arrangements

– BJT POWER TRANSISTOR

EXAMPLE

Determine the required ratings

(current, voltage and power) of

the BJT.


EXAMPLE – Solution

For the maximum

collector current; 0@CEV

A 38

24max

L

CCC

R

VI

For the maximum collector-

emitter voltage; 0CI

V 24max CCCE VV



The load line equation

is;

The load line must lie

within the SOA

LCCCCE RIVV

The transistor power

dissipation;

LCCCCCLCCCCCET RIIVIRIVIVP 2



The maximum power occurs when

02 LCCC RIV

0C

T

dI

dP

i.e. when

or when A 5.1CI

At this point; V 12 LCCCCE RIVV

and; W18 CCET IVP

Differentiating



Thus the transistor ratings are;

W18

V 24

A 3

max

max

T

CE

C

P

V

I

In practice, to find a suitable transistor for a given

application, safety factors are normally used. The

transistor with

will be required.

W18 V, 24 A, 3 maxmax TCEC PVI

A large-signal amplifier can also be called a power amplifier.

This class A amplifier has a large quiescent collector current.

C

B E

VCC = 18 V

RL = 12 W RB = 1.2 kW

CC b = 60

IB = VCC

RB

18 V

1.2 kW = = 15 mA

IC = b x IB = 60 x 15 mA = 0.9 A

0 2 4 6 8 10 12 14 16 18

0.2 0.4 0.6 0.8

1.0 1.2

1.4

VCE in Volts

IC in A

5 mA

0 mA

25 mA

20 mA

15 mA

10 mA

ISAT = VCC

RL

18 V

12 W = = 1.5 A

Q

This is a Class A amplifier.

PC = VCE x IC = 7.2 V x 0.9 A = 6.48 W

0 2 4 6 8 10 12 14 16 18

0.2 0.4 0.6 0.8

1.0 1.2

1.4

VCE in Volts

IC in A

5 mA

0 mA

25 mA

20 mA

15 mA

10 mA Q

This is a Class B amplifier.

PC = VCE x IC = 18 V x 0 A = 0 W

Its quiescent power dissipation is zero.

0 2 4 6 8 10 12 14 16

0.2 0.4 0.6 0.8

1.0 1.2

1.4

5 mA

0 mA

25 mA

20 mA

15 mA

10 mA

The collector signal is too distorted for linear applications.

C

B

C

B E

E

+VCC

The complementary-symmetry Class B push-pull amplifier has acceptable

linearity for some applications.

NPN

PNP

NP

N

PN

P

Class B

C

B

C

B E

E

+VCC

Since the base-emitter junction potential is 0.7 V, there is some crossover distortion.

NPN

PNP

C

B

C

B E

E

+VCC

Crossover distortion is eliminated by applying some forward bias

to the transistors (class AB).

NPN

PNP

1.4 V

0 2 4 6 8 10 12 14 16 18

0.2 0.4 0.6 0.8

1.0 1.2

1.4

VCE in Volts

IC in A Q

The quiescent power dissipation is moderate for class AB.

The efficiency is much better than class A.

82

Push-pull amplifiers

Push-pull operation helps to increase values of input and output impedances and to additionally suppress even harmonics

2 , 0

0, sin c

c1

Ii

2 , sin

0, 0

c

c2I

i

first transistor collector current

second transistor collector current

RL

T1 T2

Vb Vcc

ic1

ic2

iL

icc

n2

n1

n1

Ic

2

Ic

2

Ic

2

Ic

2

Ic0

ic1

ic2

icL icc

For 50% duty cycle of each device (ideal Class B) with driving sinusoidal voltage:

Being transformed through output transformer T2, total collector current:

sin cc2c1L Iiii

Current flowing in center tap of primary winding of transformer T2:

sin cc2c1cc Iiii

83

Push-pull amplifiers

sin sin L L cL VRIv

c

2

0

ccco 2

2

1 IdiI

ccc0

2VIP

cccout2

1VIP

%5.78 4

0

out @

P

P

Ideally, even-order harmonics are canceled as they are in-phase and combined in center tap of primary winding of output transformer

RL

T1 T2

Vb Vcc

ic1

ic2

iL

icc

n2

n1

n1 To eliminate losses, it is necessary to connect bypass capacitance to this center point

As for 50% duty cycle, third- and higher-order odd harmonics do not exist, ideally sinusoidal signal will appear in load

Total DC collector current

For zero saturation resistance when collector voltage amplitude Vc = Vcc and equal turns of winding when VL = Vc, DC and fundamental output powers

Maximum theoretical collector efficiency that can be achieved in Class B operation

POWER TRANSISTOR

Transistor limitations

• Maximum rated current,

• Maximum rated voltage,

• Maximum rated power.

The maximum rated power is related to the maximum

allowable temperature of the transistor.

– BJT

Large-area devices – the geometry and doping

concentration are different from those of small-signal

transistors

Examples of BJT rating:

Parameter

Small-signal

BJT

(2N2222A)

Power BJT

(2N3055)

Power BJT

(2N6078)

VCE (max) (V) 40 60 250

IC (max) (A) 0.8 15 7

PD (max) (W) 1.2 115 45

b 35 – 100 5 – 20 12 – 70

fT (MHz) 300 0.8 1

POWER TRANSISTOR

Summary & Revision and some extra Material

Amplifier Power Dissipation

P1 = I

12R

1

P2 = I

22R

2

ICQ

RC

RE

R1

R2

VCC

I1

I2

ICC

PC = I

CQ2 R

C

PT = I

TQ2 R

T

PE = I

EQ2 R

E

IEQ

The total amount of power being dissipated by the amplifier, Ptot , is

Ptot = P1 + P2 + PC + PT + PE

The difference between this total value and the total power being drawn from the supply is the power that actually goes to the load – i.e. output power.

Amplifier Efficiency

Amplifier Efficiency • A figure of merit for the power amplifier is its efficiency, .

• Efficiency ( of an amplifier is defined as the ratio of ac output power (power delivered to load) to dc input power .

• By formula :

• As we will see, certain amplifier configurations have much higher efficiency ratings than others.

• This is primary consideration when deciding which type of power amplifier to use for a specific application.

• Amplifier Classifications

%100)(

)(%100

dcP

acP

powerinputdc

poweroutputac

i

o

Amplifier Classifications • Power amplifiers are classified according to the percent of

time that collector current is nonzero.

• The amount the output signal varies over one cycle of operation for a full cycle of input signal.

vin

vout

Av Class-A

vin

vout

Av Class-B

vin

vout

Av Class-C

Efficiency Ratings

• The maximum theoretical efficiency ratings of class-A, B, and C amplifiers are:

Amplifier Maximum Theoretical

Efficiency, max

Class A 25%

Class B 78.5%

Class C 99%

Class A Amplifier

• output waveform same shape input waveform + phase shift.

• The collector current is nonzero 100% of the time.

inefficient, since even with zero input signal, ICQ is nonzero

(i.e. transistor dissipates power in the rest, or quiescent, condition)

vin

vout

Av

Basic Operation

Common-emitter (voltage-divider) configuration (RC-coupled amplifier)

RC

+VCC

RE

R1

R2

RL

vin

ICQ

I1

ICC

Typical Characteristic Curves for Class-A Operation

Typical Characteristic

• Previous figure shows an example of a sinusoidal input and the resulting collector current at the output.

• The current, ICQ , is usually set to be in the center of the ac load line. Why?

(DC and AC analyses discussed in previous sessions)

DC Input Power

RC

+VCC

RE

R1

R2

RL

vin

ICQ

I1

ICC

The total dc power, Pi(dc) , that an amplifier draws from the power supply :

CCCCiIVdcP )(

1III

CQCC

CQCCII )(

1II

CQ

CQCCiIVdcP )(

Note that this equation is valid for most amplifier power analyses. We can rewrite for the above equation for the ideal amplifier as

CQCEQiIVdcP 2)(

AC Output Power

R1//R

2

vcev

in

vo

ic

RC//R

LrC

AC output (or load) power, Po(ac)

Above equations can be used to calculate the maximum possible value of ac load power. HOW??

L

rmso

rmsormsco

R

vviacP

2

)(

)()()(

Disadvantage of using class-A amplifiers is the fact that their efficiency ratings are so low, max 25% .

Why?? A majority of the power that is drawn from the supply by a class-A amplifier is used up by the amplifier itself.

Class-B Amplifier

IC

(mA)

VCE

VCE(off)

= VCC

IC(sat)

= VCC

/(RC+R

E)

DC Load LineIC

VCE

IC(sat)

= ICQ

+ (VCEQ

/rC)

VCE(off)

= VCEQ

+ ICQ

rC

ac load line

IC

VCE

Q - point

ac load line

dc load line

L

PP

CQCEQ

CQCEQ

o

R

VIV

IVacP

82

1

22)(

2

%25%10022

1

%100)(

)( CQCEQ

CQCEQ

dci

aco

IV

IV

P

P

Limitation

Example Calculate the input power [Pi(dc)], output power [Po(ac)], and efficiency [] of the amplifier circuit for an input voltage that results in a base current of 10mA peak.

RC

RB

+VCC

= 20V

IC

Vi

25b

W20Wk1

Vo

%5.6%100

6.9)48.0)(20(

625.0)20(2

10250

2

250)10(25

20

1100020

20

4.10)20)(48.0(20

48.05.482)3.19(25

3.191

7.020

)(

)(

)(

232

)(

)(

)(

)()(

)(

W

W

W

@

W

dci

aco

CQCCdci

C

peakC

aco

C

CCsatc

B

P

P

WAVIVP

WA

RI

P

peakmApeakmAII

VVV

AmAV

R

VI

VAVRIVV

AmAmAII

mAk

VV

R

VVI

peakbpeakC

CCcutoffCE

CCCCCEQ

CQ

B

BECC

BQ

b

b

Transformer-Coupled Class-A Amplifier

Input

+VCC

RE

R1

R2

RL

N1:N

2

Z2

= RL

Z1

A transformer-coupled class-A amplifier uses a transformer to couple the output signal from the amplifier to the load.

The relationship between the primary and secondary values of voltage, current and impedance are summarized as:

LR

Z

Z

Z

N

N

I

I

V

V

N

N

1

2

1

2

2

1

1

2

2

1

2

1

N1, N2 = the number of turns in the primary and secondary V1, V2 = the primary and secondary voltages I1, I2 = the primary and secondary currents Z1, Z2 = the primary and seconadary impedance ( Z2 = RL )

Transformer-Coupled Class-A Amplifier

• An important characteristic of the transformer is the ability to produce a counter emf, or kick emf.

• When an inductor experiences a rapid change in supply voltage, it will produce a voltage with a polarity that is opposite to the original voltage polarity.

• The counter emf is caused by the electromagnetic field that surrounds the inductor.

Counter emf

10V

+

-

+

-

10V

SW1

10V

+

-

+

-

10V

This counter emf will be present only for an instant.

As the field collapses into the inductor the voltage decreases in value until it eventually reaches 0V.

DC Operating Characteristics The dc biasing of a transformer-coupled class-A amplifier is very similar to any other class-A amplifier with one important exception : the value of VCEQ is designed to be as close as possible to VCC.

Input

+VCC

RE

R1

R2

RL

N1:N

2

Z2

= RL

Z1

VCE

IC

IB

= 0mA

DC load line

The dc load line is very close to being a vertical line indicating that VCEQ will be approximately equal to VCC for all the values of IC.

The nearly vertical load line of the transformer-coupled amplifier is caused by the extremely low dc resistance of the transformer primary.

VCEQ = VCC – ICQ(RC + RE)

The value of RL is ignored in the dc analysis of the transformer-coupled class-A amplifier. The reason for this is the fact that transformer provides dc isolation between the primary and secondary. Since the load resistance is in the secondary of the transformer it dose not affect the dc analysis of the primary circuitry.

AC Operating Characteristics

Input

+VCC

RE

R1

R2

RL

N1:N

2

Z2

= RL

Z1

VCE

IC

IB

= 0mA

DC load line

ac load line

IC(max)

= ??

~ VCEQ

~ VCC

~ 2VCC

Q-point

1. Determine the maximum possible change in VCE

•Since VCE cannot change by an amount greater than (VCEQ – 0V), vce = VCEQ.

2. Determine the corresponding change in IC

•Find the value of Z1 for the transformer: Z1 = (N1/N2)2Z2 and ic = vce / Z1

3. Plot a line that passes through the Q-point and the value of IC(max).

•IC(max) = ICQ + ic

4. Locate the two points where the load line passes through the lies representing the minimum and maximum values of IB. These two points are then used to find the maximum and minimum values of IC and VCE

Input

+VCC

RE

R1

R2

RL

N1:N

2

Z2

= RL

Z1

R1//R

2

vcev

in

ic

Z1

vo

VCE

IC

IB

= 0mA

DC load line

ac load line

IC(max)

= ??

~ VCEQ

~ VCC

~ 2VCC

Q-pointICQ

Maximum load power and efficiency

The Power Supply for the amplifier : PS = VCCICC

Maximum peak-to-peak voltage across the primary of the transformer is approximately equal to the difference between the values of VCE(max) and VCE(min) : VPP = VCE(max) – VCE(min)

Maximum possible peak-to-peak load voltage is found by V(P-P)max = (N2 / N1)V PP

The actual efficiency rating of a transformer-coupled class-A amplifier will generally be less than 40%.

N1 : N

2

RL

VPP V

(P-P) max

There are several reasons for the difference between the practical and theoretical efficiency ratings for the amplifier :

1. The derivation of the = 50% value assumes that VCEQ = VCC . In practice, VCEQ will always be some value that is less the VCC .

2. The transformer is subject to various power losses. Among these losses are couple loss and hysteresis loss. These transformer power losses are not considered in the derivation of the = 50% value.

• One of the primary advantages of using the transformer-coupled class-A amplifier is the increased efficiency over the RC-coupled class-A circuit.

• Another advantage is the fact that the transformer-coupled amplifier is easily converted into a type of amplifier that is used extensively in communications :- the tuned amplifier.

• A tuned amplifier is a circuit that is designed to have a specific value of power gain over a specific range of frequency.

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:

• An npn transistor that provides the

negative half of the AC cycle

• A pnp transistor that provides the

positive half.

Class B Amplifier

Class B Amplifier

• Since one part of the circuit pushes the signal high during one half-cycle and the other part pulls the signal low during the other half cycle, the circuit is referred to as a push-pull circuit

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

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

dcCCdci IVP )(

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

• The input power can be written as

pdc II

2

pCCdci IVP

2)(

Output AC power

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

• The efficiency of the amplifier is given by

• Not that

• Therefore the efficiency can be re-expressed as

L

pL

L

ppL

acoR

V

R

VP

28

)()(

)(

L

pL

pR

VI

)(

Output AC power

• The maximum efficiency can be obtained if

• The value of this maximum efficiency will be

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

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

The peak collector load current can be found from

pCCdci IVP

2)(

Example

Solution:

The input power is

The output power is given by

The efficiency is

WP dci 9.23)25.1(302

)(

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

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

Example

Solution:

The efficiency is given by

The maximum power dissipated by each transistor is

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. The first arrangement uses a single input signal fed to the input of two complementary transistors (complementary symmetry circuits)

2. 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)

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

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

Class B Output Stage

A class B output stage.

Complementary circuits.

Push-pull operation

Maximum power-conversion efficiency is 78.5%

Transfer Characteristic

Crossover Distortion

Class AB biasing to solve crossover distortion

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

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

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

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

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

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.

Class B Amplifier Push-Pull Operation

• During the positive

half-cycle of the AC

input, transistor Q1

(npn) is conducting

and Q2 (pnp) is off.

• During the negative

half-cycle of the AC

input, transistor Q2

(pnp) 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

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.

Harmonics

Harmonics are integer multiples of a fundamental frequency.

If the fundamental frequency is 5kHz:

1st harmonic 1 x 5kHz

2nd harmonic 2 x 5kHz

3rd harmonic 3 x 5kHz

4th harmonic 4 x 5kHz

etc.

Note that the 1st and 3rd harmonics are called odd harmonics and the

2nd and 4th are called even harmonics

Harmonic Distortion

According to Fourier

analysis, if a signal is not

purely sinusoidal, then it

contains harmonics.

138

Harmonic Distortion Calculations

The total harmonic distortion (THD) is determined by:

100A

A%Ddistortion harmonic nth %

1

nn

100DDDTHD %23

23

22

Harmonic distortion (D) can be calculated:

where

A1 is the amplitude of the fundamental frequency

An is the amplitude of the highest harmonic

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.

Power Dissipation

• The load power

• Maximum load power

L

oL

R

VP

2ˆ

2

1

L

CC

VVL

oL

R

V

R

VP

CCo

2

ˆ

2

12

ˆ

2

max

Power Dissipation

• Total supply power

• Maximum total supply power

CC

L

os V

R

VP

ˆ2

L

CC

VV

CC

L

os

R

VV

R

VP

CCo

2

ˆ

max

2ˆ2

Power Dissipation

• Power-conversion efficiency

• Maximum power-conversion efficiency

CC

o

V

V̂

4

%5.78ˆ

4ˆ

max

CCo VVCC

o

V

V

Power Dissipation

• Power dissipation

• Maximum Power dissipation

L

oCC

L

oD

R

VV

R

VP

2ˆ

2

1ˆ2

max2

2

2ˆ

2

maxmax

2.02

ˆ

2

1ˆ2

L

L

CC

VVL

oCC

L

oDPDN

PR

V

R

VV

R

VPP

CCo

Class AB Output Stage

A bias voltage VBB is applied between the bases of QN and QP, giving rise to a bias current IQ . Thus, for small vI, both transistors conduct and crossover distortion is almost completely eliminated.

A Class AB Output Stage Utilizing Diodes for Biasing

A Class AB Output Stage Utilizing A VBE Multiplier for Biasing

Amateur Radio Extra Class

Amps & Power Supplies

• E1XXX… A push-pull type amplifier reduces or eliminates even-order harmonics.

• E7B17… A grounded-grid amplifier has low input impedance.

• E7B07… A vacuum-tube power amplifier can be neutralized by feeding back an out-of-phase component of the output to the input.

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Lecture -Power Amplifier

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