1 power amplifiers power amplifiers are used in the transmitter there are different classes for...

1

Power Amplifiers Power Amplifiers are used in the

transmitter There are different classes for the power

amplifier Class A, Class B, Class AB, Class C,

Class D and Class E

Those amplifier differs in their angle of conduction, efficiency, linearity and the in the amount of distortion they introduce to the system

2

Power Amplifiers

Class AClass A The amplifier conducts through the

full 360 of the input. The Q-point is set near the middle of the load line.

Class BClass B The amplifier conducts through

180 of the input. The Q-point is set at the cutoff point.

3

Power Amplifiers

Class ABClass ABThis is a compromise between the class A and B amplifiers. The amplifier conducts somewhere between 180 and 360 . The Q-point is located between the mid-point and cutoff.

Class CClass CThe amplifier conducts less than 180 of the input. The Q-point is located below the cutoff level.

4

Power Amplifiers

Class DClass DThis is an amplifier that is biased especially for digital signals.

5

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

6

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

7

Power Amplifiers Class C

8

Class C power amplifier

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

9

Class C power 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

10

Class C power amplifier According to the

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

Where ID is given by

11

Class C power amplifier 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(1t

t

DpCav dtItIT

I

1

1 t2

2 t

12

Class C power amplifier 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

13

Class C power amplifier 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

pCav

II

14

Class C power amplifier 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(

41212

121

pCC

CCO

IVIVRLIP

15

Class C power amplifier 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

16

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

17


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

18


Now the maximum collector current can be rewritten as

A normalized peak collector current is defined as

19


A plot of the normalized peak current versus the conduction angle is shown below

20


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

21


The power dissipated in the transistor is given by

Or PT can be rewritten as

2)

2sin2

cossin(2 11

IVIVP CC

CCT

22


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

23

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

OrVPRVCC OL 4.2255022

24

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 in slide 16

The peak collector current is given by

5.57

25


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

26


Now the transistor power equation can be modified as

The normalized transistor power dissipation is given by

Where

cos1

)2sin2()cos(sin4)(

f

27


28

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

29


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

30


Solution:Finally the value of the load resistance that

results in this output power is given by

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