L08 Power Amplifier (Class A).ppt

7/27/2019 L08 Power Amplifier (Class A).ppt

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Power Amplifier (Class A)

Induction of Power Amplifier

Power and Efficiency

Amplifier Classification

Basic Class A Amplifier

Transformer Coupled Class A Amplifier


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

power, usually to a low resistance load.

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

Although these load values do not cover every possibility, they

do illustrate the fact that power amplifiers usually drive low-resistance loads.

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

Ideal power amplifier will deliver100% 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 dissipate some of the power that is beingdrawn form the supply.


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Amplifier Power Dissipation

P1

= I12R

1

P2

= I22R

2

ICQR

C

RE

R1

R2

VCC

I1

I2

ICC

PC

= ICQ2 R

C

PT

= ITQ2 R

T

PE

= IEQ2 R

E

IEQ

The total amount 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.

Amplifier Efficiency h


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Amplifier Efficiency h A figure of merit for the power amplifier is its efficiency, h. Efficiency ( h ) 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 muchhigher efficiency ratings than others.

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

Amplifier Classifications

%100)()(%100

dcPacP

powerinputdcpoweroutputac

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 output signal varies overone cycle of

operation for a full cycle ofinput signal.

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, hmaxClass A 25%

Class B 78.5%

Class C 99%


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Class A Amplifier

output waveformsame shapeinput waveform + phase shift.

The collector current is nonzero 100% of the time.

inefficient, since even with zero input signal, ICQ isnonzero

(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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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)


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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 :

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)(


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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 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, hmax

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


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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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ExampleCalculate the input power [Pi(dc)], output power [Po(ac)],

and efficiency [h] of the amplifier circuit for an inputvoltage that results in a base current of 10mA peak.R

C

RB

+VCC

= 20V

IC

Vi

25

20k1

Vo

)

%5.6%100

6.9)48.0)(20(

625.0)20(210250

2

250)10(25

20

1100020

204.10)20)(48.0(20

48.05.482)3.19(25

3.191

7.020

)(

)(

)(

232

)(

)(

)(

)()(

)(

dci

aco

CQCCdci

C

peakC

aco

C

CCsatc

B

P

P

WAVIVP

WARIP

peakmApeakmAII

VVV

AmAV

R

VI

VAVRIVV

AmAmAII

mAk

VV

R

VVI

peakbpeakC

CCcutoffCE

CCCCCEQ

CQ

B

BECC

BQ

h


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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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Transformer-Coupled Class-A Amplifier

An important characteristic of the transformeris the ability to produce a counter emf, orkickemf.

When an inductor experiences a rapid change insupply voltage, it will produce a voltage with apolarity that is opposite to the original voltage

polarity. The counter emf is caused by the

electromagnetic field that surrounds the inductor.


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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 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 extremelylow dc

resistance of the transformer primary.

VCEQ = VCCICQ(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.


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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 VCE cannot change by an amountgreater than (VCEQ0V), vce = VCEQ.

2.Determine the corresponding change inIC

Find the value ofZ1 for 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

Z1vo

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 thedifference between the practical and

theoretical efficiency ratings for theamplifier :

1. The derivation of the h= 50% value assumes

that VCEQ = VCC . In practice, VCEQ willalways be some value that is less the VCC .

2. The transformer is subject to various powerlosses. Among these losses are couple loss

and hysteresis loss. These transformer powerlosses are not considered in the derivation ofthe h= 50% value.


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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 convertedinto 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.

L08 Power Amplifier (Class A).ppt

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