Large Signal Power Amp_intro_class A_PA

8/13/2019 Large Signal Power Amp_intro_class A_PA

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Class-A Power Amplifier


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Introduction Power amplifiers are used to deliver a relatively high amount of

power, usually toa low resistance load.

Typical load values range from 300W (for transmission antennas)to 8W (for audio speaker).

Typical output power rating of a power amplifier will be 1W orhigher.

Idealpower amplifier will deliver 100%of the power it drawsfrom the supply to load. In practice, this can never occur.

The reason for this is the fact that the components in theamplifier will all dissipatesome of the power that is beingdrawn form the supply.


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Amplifier Power Dissipation

P1= I

12R

1

P2= I

22R

2

ICQR

C

RE

R1

R2

VCC

I1

I2

ICC

PC= I

CQ2 R

C

PT= I

TQ2 R

T

PE= I

EQ2 R

E

IEQ

The totalamount of power

being dissipated by the

amplifier, Ptot,is

Ptot= P

1+ P

2+ P

C+ P

T+ P

E

The difference between this

total value and the total power

being drawn from the supply is

the power that actually goes tothe loadi.e. output power.


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Amplifier Efficiency

A figure of meritfor the power amplifier is its efficiency, .

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

By formula :

As we will see, certain amplifier configurations have muchhigher efficiency ratings than others.

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

%100)(

)(%100

dcP

acP

powerinputdc

poweroutputac

i

oh


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Amplifier Classifications

Power amplifiers are classified according to the percent oftime that collector current is nonzero.

The amount the outputsignal varies over one cycleof

operation for a full cycleof inputsignal.

vin

vout

Av Class-A

vin

vout

Av Class-B

vin

vout

Av Class-C


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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%


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Class A Amplifier

output waveform same shapeinput waveform + phase shift.

The collector current is nonzero100% of the time.

inefficient, since even with zero input signal, ICQisnonzero

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

vin

vout

Av


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Basic Operation

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

RC

+VCC

RE

R1

R2

RL

vin

ICQI

1

ICC


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Typical Characteristic Curves

for Class-A Operation


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DC Input Power

RC

+VCC

RE

R1

R2

RL

vin

ICQ

I1

ICCThe total dc power, Pi(dc) , that anamplifier draws from the power

supply :

CCCCi IVdcP )(

1III CQCC

CQCC II )(

1II

CQ

CQCCi IVdcP )(

Notethat this equation is valid for most amplifier power analyses.

We can rewrite for the above equation for the idealamplifier as

CQCEQi IVdcP 2)(


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AC Output Power

R1//R

2

vcevin

vo

ic

RC//R

LrC

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

Above equations can be used tocalculate the maximumpossible

value of ac load power. HOW??

L

rmso

rmsormsco

R

vviacP

2

)(

)()()(

Disadvantageof 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.


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IC

(mA)

VCE

VCE(off)

= VCC

IC(sat)

= VCC

/(RC+R

E)

DC Load Line

IC

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%100221

%100)(

)( CQCEQ

CQCEQ

dci

aco

IV

IV

P

Ph


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Limitation


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Transformer-Coupled Class-A Amplifier

Input

+VCC

RE

R1

R2

RL

N1:N2

Z2

= RL

Z1

A transformer-coupled class-A amplifier

uses a transformer to couple the outputsignal 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)


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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.


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DC Operating CharacteristicsThe 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 VCEQis designedto 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 VCEQwill be approximately equal to

VCCfor all the values of IC.

The nearly verticalload line of the transformer-coupled amplifier is caused by the extremelylow dc

resistanceof the transformer primary.

VCEQ= VCCICQ(RC+ RE)

The value of RLis ignoredin the dc analysis of the

transformer-coupled class-A amplifier.

The reason for this is the fact that transformer

provides dc isolationbetween the primary and

secondary. Since the load resistance is in the

secondary of the transformer it dose notaffect the dcanalysis of the primary circuitry.


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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 VCEcannot change by an amountgreater than (VCEQ0V), vce= VCEQ.

2.Determine the corresponding change inIC

Find the value ofZ1for the transformer:Z1=

(N1/N2)2Z2 and ic= vce/ Z1

3. Plot a line that passes through the Q-point and

the value ofIC(max).

IC(max) = ICQ+ ic

4. Locate the two points where the load line passes

through the lies representing the minimum and

maximum values ofIB. These two points are then

used to find the maximum and minimum values of

ICand VCE


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Input

+VCC

RE

R1

R2

RL

N1:N

2

Z2 = RL

Z1

R1//R

2

vcevin

ic

Z1

vo

VCE

IC

IB

= 0mA

DC load line

ac load line

IC(max)

= ??

~ VCEQ

~ VCC

~ 2VCC

Q-pointICQ


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There are several reasons for the differencebetween the practical and theoretical efficiencyratings for the amplifier:

The derivation of the h= 50% value assumes that VCEQ

= VCC. In practice, VCEQ will always be some value thatis less the VCC.

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 thederivation of the h= 50% value.


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One of theadvantagesof using the transformer-

coupled class-A amplifier is the increased

efficiencyover the RC-coupled class-A circuit.

Another advantageis the fact that the

transformer-coupled amplifier is easily convertedintoa type of amplifier that is used extensively in

communications :- thetuned amplifier (A tuned

amplifier has a circuit that is designed to have a

specific value of power gain over a specific rangeof frequency).

.

Advantages of Transformer Coupled Class-A Amplifier

Large Signal Power Amp_intro_class A_PA

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