Chapter 2 POWER AMPLIFIER V2 [Compatibility Mode]

Chapter 2:Chapter 2:Power AmplifiersPower Amplifiers

SEE 3263: ELECTRONIC SYSTEMS

11

Power AmplifiersPower Amplifiers

SEE 3263 POWER AMPLIFIERS

INTRODUCTIONINTRODUCTION Amplifiers main characteristics:

Linearity Efficiency Output Power Signal Gain

2

Generally there are relationship between these characteristics. Increase the linearity will reduce the efficiency. Must plan well before designing an amplifier.For example, amplifier with higher output power must be used in transmitter while amplifier with higher linearity must be used in receiver.

2


AMPLIFIER SYSTEM AMPLIFIER SYSTEM APPLICATIONAPPLICATION

33


POWER AMPLIFIERPOWER AMPLIFIER Large signal amplifier. To deliver large power to the load. Final stage of an amplifier system. Able to dissipates large power. Widely used as audio components in radio, TV

4

Widely used as audio components in radio, TV receivers, CD/DVD players, PA System etc. The load in these applications is most often a loudspeaker which requires considerable power to convert electrical signals to sound waves.

Sometime used to drive a motor in control systems.

Used power transistor as the main device.4


AMPLIFIER AMPLIFIER CLASSIFICATIONCLASSIFICATION

Categorize according to the percentage input cycle when amplifier operate in linear region:

Class A

5

Class AClass BClass ABClass CClass D

5


6

Collector current waveform of a transistor when amplifier operating as (a) class A, (b) class B, (c) class AB, and (d) class C.

6


CLASS A POWER AMPLIFIERSCLASS A POWER AMPLIFIERS Class A amplifiers are biased so that the

entire input waveform is amplified without clipping.

In other words, the DC bias and level of

7

In other words, the DC bias and level of input signal are set so the output signal is unclipped and undistorted.

Can be divided into two configurations i.eRC-coupled and transformer- coupled.

7


SMALLSMALL--SIGNAL CLASS A SIGNAL CLASS A AMPLIFIERAMPLIFIER

Is the first stage of the amplifier system.

Should be biased this way because the

8

way because the signal voltage swings are not great until later stages.

Biasing in this way reduces the DC supply current and increases the efficiency of the entire system.

8


LARGELARGE--SIGNAL CLASS A SIGNAL CLASS A AMPLIFIERAMPLIFIER

9

Shows how the later stages of an amplifier system should be biased.

The midpoint bias is used to allow maximum signal swing.

The DC current requirement is higher because of the higher ICQ and a higher bias current.

9


Small-signal and large-signal class A amplifiers operate at 100% duty cycle.

This means the transistor is ON all the time and is dissipating power in

10

the time and is dissipating power in the form of heat.

For this reason, the class A amplifier is the least efficient of all classes of amplifiers.

10


OUTPUT SIGNAL SWING OF OUTPUT SIGNAL SWING OF CLASS A AMPLIFIERCLASS A AMPLIFIER

1111


COLLECTOR CURRENT COLLECTOR CURRENT COMPONENTSCOMPONENTS

1212

Comprises of AC and DC components


THE CLASS A QTHE CLASS A Q--POINTPOINT The Q-point should be in the middle of the AC load line.

For the DC load line, that is no problem because it is easily to design the class A amplifier to have a Q-point in the center

13

amplifier to have a Q-point in the center of the DC load line.

Adding a load will creates the AC load line superimposed over the DC Q-point.

This will limits the amount of undistorted output voltage swing.

13


Class A Power Amplifiers

A class A power amplifier class A power amplifier is a large signal amplifier that operates in the linear region. Ideally, a class A amplifier is

Ic(sat)

ICdesigned to operate in the center of the ac

1414

0

ICQ

Ic(sat)

QAC load line

VCEQ Vce(cutoff)VCE

DC load line

the center of the ac load line.

Notice that a class A amplifier dissipates dc power even with no signal. The dc power dissipated is the product of ICQ and VCEQ.


LOAD LINES OF THE CLASS A LOAD LINES OF THE CLASS A AMPLIFIERAMPLIFIER

1515


QQ--POINT AT THE MIDDLE POINT AT THE MIDDLE OF THE AC LOAD LINEOF THE AC LOAD LINE

1616


QQ--POINT NEAR CUTOFF POINT NEAR CUTOFF

1717


QQ--POINT NEAR SATURATION POINT NEAR SATURATION

1818


CLASS A POWER CONSIDERATIONSCLASS A POWER CONSIDERATIONS

19

For this amplifier, the Q-point is not in the center of the AC load line.This will limits the output maximum swing without distortion.PDQ = class A transistor power dissipates in the form

of heat= VCEQ x ICQ 19


This is the amount of power the class A biased transistor dissipates whether or not an AC input signal is present.

This means that the transistor will not dissipate any more power when the input signal is present and amplification is taking place. Why?

Because vce and ic increase and decrease by equal amounts with each alternation of the

2020

equal amounts with each alternation of the output signal.

The average AC current is equal to ICQ and the average AC collector-emitter voltage is equal to VCEQ.

From the circuit, PDQ(max) = 6V x 60mA = 360mW. The actual rating of the transistor should be at

least twice this value. A 1W transistor is best.


CLASS A OUTPUT POWER CLASS A OUTPUT POWER (P(POUTOUT = P= PLL))

PL power that is delivered to the load device (resistance) in rms.

The maximum rms load voltage is 0.707 or 2

v )p(O

v2

21

Notice that the AC load line limits the maximum output voltage swing to 3V.

( )( )

mW45)100(23P

R2v

R2

v

P

2

maxL

L

2)p(O

L

2)p(O

L

==

=

=

21


For any class A amplifier, the maximum AC output voltage is either:-

vo(pp) = 2 (vce(cutoff) VCEQ)where VCEQ VCC/2

22

where VCEQ VCC/2

vo(pp) = 2 (VCEQ VCE(SAT))where VCEQ VCE(SAT) < vce(cutoff) VCEQ

22


CLASS A DC POWER CLASS A DC POWER (P(PDCDC)) Is the amount of power delivered to the power amplifier from the DC supply (voltage-divider power excluded). Thus the DC power is:

P = V x I

23

PDC = VCC x ICQ= 12V x 60mA= 720mW

This is also the maximum power that is ever delivered to the amplifier, because when AC is applied, the average collector current is equal to ICQ.

23


CLASS A POWER CLASS A POWER EFFICIENCY (EFFICIENCY ( eta)eta)

Maximum efficiency occurs when the

%100PP%or

PP

DC

L

DC

L ==

24

Maximum efficiency occurs when the output voltage swing is maximum, which makes the load power maximum.

)lowvery(%3.6or063.0mW720

mW45==

24


Class A power amplifiers are not particularly efficient, so they are restricted to low power applications.

The maximum theoretical efficiency for a class A amplifier is 0.25 (or 25%) and usually they are considerably less.

If Q-point is at the middle of the AC load line, the maximum output voltage swing will be 6V.

25

If Q-point is at the middle of the AC load line, the maximum output voltage swing will be 6V.

Therefore:

%25or25.0mW720

W18.0P

P

W18.0)100(2)6(P

DC

(max)L(max)

2

(max)L

===

==

25


What is the efficiency of an amplifier that delivers 200 mW to a load if the power supply is 12 V at 400 mA? 6.7%

(a) If a 3 Vpp signal is applied to the input, what voltage do you expect to see at the speaker? (b) What power is delivered?

V

26

(a) The CC amplifier has a gain of nearly 1. The output voltage is nearly equal to the input = 3 Vpp.(b) The power delivered to the speaker is:

( )22 1.06 V8

rmsVPR

= =

= 140 mW

VCC+12 V

Q1

R1C1

C2R2

RE

Q2

10 k

22 k100 F

0.22 F

22 2 W

Vou t

Vin

Speaker8


For the 3 Vpp input, what is the input power and what is the power gain? Assume the Darlington = 10,000.

VCC+12 V

( )k15.6

RRRRR LE21in(tot)=

=

27

The power gain is:

( )22( )

1.06 V6.15 k

rms

inin tot

VPR

= =

= 0.183 mW

Q1

R1C1

C2R2

RE

Q2

10 k

22 k100 F

0.22 F

22 2 W

Vou t

Vin

Speaker8

( )2 2 6.15 k18

in totp v

L

RA A

R

= =

= 769


28

Determine the power efficiency for the class A amplifier of figure shown above. Assume VCE(SAT) = 0.2 V.

28

{3.2%}{3.2%}


MATCHING THE LOAD FOR MATCHING THE LOAD FOR MAXIMUM POWER TRANSFERMAXIMUM POWER TRANSFER

The maximum power transfer states that: Maximum power is delivered to a load only when the load impedance (resistance) is equal to the output impedance of the

29

is equal to the output impedance of the amplifier.

Lets recalculate the power efficiency of the previous example if the load 600 is replace with a 100 resistance.

29

{9%}{9%}


TRANSFORMERTRANSFORMER--COUPLED COUPLED CLASS A POWER AMPLIFIERCLASS A POWER AMPLIFIER

The advantages of transformer-coupled over RC-coupled are:

transformer able to achieve impedance matching for maximum

30

transformer able to achieve impedance matching for maximum power transfer to the load.

blocks DC voltage/current from getting to the load resistor.

30

SEE 3263 POWER AMPLIFIERS+VCC

RLR1

R2

RSC1

VS

NP : NS

L

2

S

PL RN

Nr

=

+VO-

rL

IC

VCE

VCC(V )

QAC load line

)r

1slope(

L=

2VCC(v ( off))

0

0

ICQ

ic(sat) DC load line

31

Therefore the DC load line is vertical with slope

From the diagram,

(VCEQ) (vce(cut-off))

CCCEQ VV = 2i

I )sat(cCQ =

Question : Why DC load line vertical?Answer : At DC, frequency, f = 0 Hz.

Inductive reactance, =pi=pi= 0L)0(2fL2XL

==

DCR1

and

31


Question : Why VCEQ = VCC?

CEQLCQCC VXIV +=

CCCEQ VV =

When the Q-point is at the middle of the AC

= 0XLbecause

Answer : Since

32

When the Q-point is at the middle of the AC load line,

CCCEQ)offcut(ce V2V2v ==

CQ)sat(c I2i =

32


3333


TRANSFORMERTRANSFORMER--COUPLED MAXIMUM COUPLED MAXIMUM POWER EFFICIENCY, POWER EFFICIENCY, MAXMAX

( )L

2)p(o

L

NR2

vP

=

RNN

Vr

VIbut

IVIVP

L

2P

CC

L

CEQCQ

CQCCCCCCDC

==

==

34

( )L

2CC

2

P

S

L

CCP

S)p(O

R2

VNN

P

VNN

vbut

=

=

( )

%50or5.0

RNN

VR2

VNN

PP

N

L

2

S

P

CCL

2CC

2

P

S

DC

L

LS

=

==

34


(a) RE, R2 and R1.(b) power dissipation in RE, R2, R1

and Q when v = 0.

Given: VCC = 12 V, ICQ = 1 A, hFE = DC = 100, VCE(SAT) = 2 V, VE = 0.1 VCCDesign an amplifier with ic and vce output swing symmetry at Q-point. Assume DC current in R2 is 10 IBQ. Determine:

35

VE

and Q1 when vi = 0.(c) power dissipation in RLand Q1

when the amplitude of vi ismaximum.

(d) RL if the turn ratio of the transformer is 2

NN

2

1=

35


CLASS B OPERATIONCLASS B OPERATION

3636

A class B circuit provides an output signal varying over one-half of the input cycle.


This amplifier is biased at the cutoff point makes the transistor operates in its linear region for 180o of input signal or at 50% duty cycle.


37

duty cycle. For some input power, the class B amplifier is able to provide very large power to the load.

Therefore, class B power efficiency is very much better than the class A amplifier.

37


COMMON-COLLECTOR CLASS B AMPLIFIER

38



3939



4040


To produce the entire waveform, use two complementary symmetry transistors. Also known as push-pull class B amplifiers. These are a matching pair of npn/pnp BJTs using two emitter-followers and both positive and negative power supplies.

4141


CCCC

CC)OFF(CECEQ

CQ

VVI

VVV,0I

===

==

=

When transistor is OFF

42

CCCEQACCQCEQ)off(ce

L

CC

AC

CEQCQ)sat(c

CC

DC

CC)SAT(C

V0VRIVvRV0

RV

Ii

0RI

=+=+=

+=+=

===

42


AC AND DC LOAD LINES OF CLASS B AC AND DC LOAD LINES OF CLASS B PUSHPUSH--PULL AMPLIFIERPULL AMPLIFIER

L

CC

RV

4343


CLASS B PUSH AMPLIFIER CLASS B PUSH AMPLIFIER POWER EFFICIENCYPOWER EFFICIENCY

AC load power ( )

L

2CC

L

2)p(O

LAC R2V

R2V

PP ===

DC input power CCCCDC IV2P = (because of two VCC)

44

But ICC = ICQ + IP sin wt = L

CC

RVIppi

=

pi

Therefore L

2CC

DC RV2Ppi

=

Efficiency %5.78785.04

RV2R2

V

PP

L

2CC

L

2CC

DC

L==

pi=

pi

==

44


Determine the power efficiency and the power dissipation in each transistor when the peak output voltage, Vo(p) = 2 V.

+12 V

Q1

45

-12 V

Q2

Q1

80 Vi

t

vo(p)

45


( ) ( )mW25)80(2

V2R2

vP

2

L

2)p(o

L ===Load power,

)avg(CCCDC IV2P =Input power

L

)p(o)p(C)avg(C R

vII

pi=

pi=Average current,

)V2)(V12(2vV2

46

mW191)80()V2)(V12(2

RvV2

PL

)p(oCCDC =pi

=

pi=Thus,

%1.13%100mW191mW25%100

PP

DC

L===Power efficiency,

mW832

mA25mA1912

PPP LDCDQ =

=

=Power dissipation in each transistor,

46


AVERAGE POWER DISSIPATION IN AVERAGE POWER DISSIPATION IN EACH TRANSISTOR EACH TRANSISTOR vsvs PEAK OUTPUT PEAK OUTPUT

VOLTAGE OF CLASS B AMPLIFIER VOLTAGE OF CLASS B AMPLIFIER

47pi

CCV2

47


=

2

LDCDQ

VVV2

PPP

0VP

)p(O

DQ=

PDQ(maX) achieved when

48

pi=

=

pi=

pi=

CC)p(O

L

)p(O

L

CC

)p(O

DQ

L

)p(O

L

)p(OCC

V2V

0R2

VR

VVP

R4V

RVV

48


L

2CC

L

CCCC

(max)DQ R4

V2

R

V2VP

pi

pi

pi=

( ) ( ) ( )L

2

2CC

L2

2CC

L2

2CC

RV

RV

RV2

pi=

pi

pi=

49

%100PP

DC

L =

( )( ) %50%1002

1%100

RV2

RV

L2

2CC

L2

2CC

==

pi

pi=

49


If the output voltage of the circuit shown below is vo = 8 sin t V, determine the maximum power dissipation in each transistor.

50

( )( )( )

( )( ) mW182804

V880

V8V12R4

v

RvV

P2

L

2)p(O

L

)p(OCC(max)DQ =

pi=

pi=

50


When VB = 0 V, both Q1 and Q2 is not conducting. The input signal must equal VBEfor Q1 or Q2 to conduct.

PROBLEMS IN CLASS B PROBLEMS IN CLASS B AMPLIFIERAMPLIFIER

(a) Crossover Distortion

51

for Q1 or Q2 to conduct.

There is a time lap between positive and negative alternation of input signal when Q1or Q2 conducting.

This time lap will cause a distortion at the output (crossover distortion).

51


5252


CLASS AB AMPLIFIERCLASS AB AMPLIFIER To reduce or eliminate crossover distortion, both transistors should be biased slightly into conduction at VBE (0.7V).

The amplifier now is no longer operate

53

The amplifier now is no longer operate as pure class B but instead operating as class AB.

Class AB push-pull amplifier with voltage- divider bias configuration is shown in the next slide.

53


+VCC

Q1

R1

VOLTAGEVOLTAGE--DIVIDER BIAS CLASS DIVIDER BIAS CLASS AB WITH TWO SUPPLIESAB WITH TWO SUPPLIES

54-VCC

Q

Q2

+

VO-

RL

R2

R1

Vi

54


Although class AB operation reduces crossover distortion in a push-pull amplifier, it has the disadvantage of reducing amplifier efficiency.

The fact that bias current is always present means that there is continuous power dissipation in both transistors.

5555


It is hard to maintain Q-point stable for class AB push-pull amplifier with voltage-divider bias because VBEchanges when temperature changes.

(b) Thermal Runaway

56

changes when temperature changes.

When temperature increases, VBEdecreases, will increase thus causing an increase in IC.

56


The increase in IC then causes further increases in temperature; VBE will further decreases and the transistor will finally burnt.

In other words, the class AB amplifier with

IC

T3 T1T2

57

In other words, the class AB amplifier with voltage-divider bias will create thermal runaway which can destroy the transistor.

To overcome this, use class AB amplifier with diode biasing.

VBE

0.7 V

57


VBE of diodes and transistors will decrease by 2.5 mV/oC when temperature increases. Thermal runaway is eliminated because the increase in temperature will not make IC to increase.

58

Cincrease.

The chosen diodes must match the characteristic values of VBE for the two transistors.

In a discrete circuit, the diodes and the transistors are installed on the same heat sink.

58


5959


R

VCC=+12V

If both transistors has VCE(SAT) = 2 V, determine:(i) Maximum peak output voltage, VO(p).(ii) Maximum output power, PL(max).(iii) Maximum power dissipation, PDQ(max) in both

transistors.(iv) Power efficiency, when PDQ = PDQ(max).

60

Q1

Q2

RL

10

R2

390

R1

390

D1

D2Vi

VCC=-12V

VO

60


REDUCING THE CROSSOVER REDUCING THE CROSSOVER DISTORTION WITH NEGATIVE DISTORTION WITH NEGATIVE

FEEDBACKFEEDBACK

vv =

6161

The diagram shows the buffered class B push-pull amplifier with negative feedback resulting in reduced crossover distortion.

io vv =


REDUCING THE CROSSOVER REDUCING THE CROSSOVER DISTORTION WITH NEGATIVE DISTORTION WITH NEGATIVE

FEEDBACKFEEDBACK

F vR1v +=

6262

The diagram shows the buffered class B push-pull amplifier with gain.

iF

o vRR1v

+=


Design a class B push-pull power amplifierwith an output power of 5W and an output signal swing of 20V(p-p), given that the input signal swing is 4V(p-p).

6363


PL = 5W, Vo(p) = 10V, Vi(p) = 2V

Pick RF = 12kand R = 3k.415R

RRR15

V2V10

v

vA

F

F

)p(i

)p(oV

==

+====

But V = (V + 2V) and V = (V + V )

6464

But VCC = (Vi(p) + 2V) and Vi(p) = (Vo(p) + VBE)Therefore VCC = Vo(p) + VBE + 2V = 10 + 0.7 + 2 = 12.7VPick VCC = 13V.

The output power is 2IV

P )p(o)p(oL =

A1V10W52

VP2I

)p(o

L)p(o =

== === 10A1

V10IV

R)p(O

)p(O(min)L


The average DC current drawn from the powersupply is:

A318.0A1I

I )p(oDC =pi

=

pi=

The DC supply input power is given by:

W268.8A318.0V132

IV2P DCCCDC

=

=

=

6565

The efficiency is:

%47.60

%100W268.8

W5

%100PP

DC

L

=

=

=

W268.8=


DARLINGTON CLASS AB DARLINGTON CLASS AB AMPLIFIERAMPLIFIER

6666


67

Referring to figure above, if VBE = VD = 0.7 V, R1 = R2 = 1 k, VCC=12Vand RL = 8,(a) Name the function of the diodes D1 , D2 , D3 and D4(b) Sketch and label the DC and AC load lines.(c) Calculate the maximum load power.(d) Determine the DC input power, PDC, output power, PL and power

efficiency, if the input signal is vi = 4 sin t V.67


CLASS AB PUSHCLASS AB PUSH--PULL PULL COMPLIMENTARYCOMPLIMENTARY--SYMMETRY SYMMETRY

WITH TWO SUPPLIESWITH TWO SUPPLIES

6868


CLASS AB PUSHCLASS AB PUSH--PULL PULL MOSFET WITH TWO SUPPLIESMOSFET WITH TWO SUPPLIES

6969


Troubleshooting Assume a newly constructed push-pull amplifier shows only the lower part of the ac signal at the output. How should you find the problem? Let start with dc measurements.

Q1

VCC

R1

+15 V

510

+15 V

70

Q1

VSRL

VCC

D2

D1

Q2

15 V

30 R2510

Checking the dc voltages:0.0 V

3.0 V This is not the expected reading!

Can you figure out a likely problem?Diode D2 is likely to be open. Remove it an test it.

15 V


CLASS C POWER AMPLIFIERCLASS C POWER AMPLIFIER Class C amplifiers are biased into conduction much less than 180o.

Never use in any application that need higher linearity like in audio amplifier because of bad distortion in the output waveform.Gives higher efficiency than class A, class B

71

Gives higher efficiency than class A, class B and class AB because less power dissipated by the transistor.

They are usually used in RF applications, such as RF oscillators and modulators or tuned amplifier at the transmitter.

71


ADVANTAGES OF A CLASS C ADVANTAGES OF A CLASS C POWER AMPLIFIERPOWER AMPLIFIER

High power efficiency ranging from about 80% to nearly 100%.

Circuit simplicity. Not requiring a

72

Circuit simplicity. Not requiring a specified bias circuit.

Excellent temperature stability.

No danger of thermal runaway.72


BASIC CLASS BASIC CLASS C C AMPLIFIER AMPLIFIER (NON INVERTING)(NON INVERTING)

7373


OUTPUT CURRENT IN OUTPUT CURRENT IN CLASS C AMPLIFIERCLASS C AMPLIFIER

7474


CLASS C AMPLIFIER BASIC CLASS C AMPLIFIER BASIC OPERATIONOPERATION

The transistor is on when the input signal exceeds |VBB| + VBE. Because class C amplifiers are biased on for a small percentage of time, they can be very

7575

time, they can be very efficient.


CLASS C WAVEFORMCLASS C WAVEFORM

7676


CLASS C POWER CLASS C POWER DISSIPATIONDISSIPATION

The power dissipation of the transistor in a class C amplifier is low because it is on for only a small percentage of the input cycles.

7777

)on(Don

)avg(D PTtP

=

)SAT(CE)SAT(C)on(D VIP =


A class C amplifier is driven by a 200kHz signal. The transistor is on for 1 s and the amplifier is operating over 100% of its load line. If IC(SAT) = 100mA and VCE(SAT) = 0.2V, what is the average power dissipation of the transistor?

1

78

mW4

)V2.0)(mA100(s5s1

VITt

P Therefore

s5kHz2001

T is, period The

)SAT(CE)SAT(Con

D(avg)

=

=

=

=

=


CLASS C MAXIMUM OUTPUT CLASS C MAXIMUM OUTPUT POWERPOWER

R2V

PL

2)p(O

L =

79

%100PPPPPP

R2

)avg(DL

L

)avg(DLDC

L

+

=

+=


Suppose the class C amplifier has a VCCequal to 24V, RL is 100 and PD(avg) = 4mW. Determine the efficiency.

W88.2)100(2)24(

R2V

P power, oadL2

L

2)P(O

L ===

80

%86.99

%100mW4W88.2

W88.2

%100PP

P ,Therefore

)100(2R2

D(avg)L

L

L

=

+

=

+

=


TUNED CLASS C AMPLIFIERTUNED CLASS C AMPLIFIER

8181

Class C operation is useful in oscillators. The collector circuit has a parallel resonant circuit (tank) and oscillations are sustained by the short pulse of collector current on each cycle.


RESONANCE CIRCUIT RESONANCE CIRCUIT OPERATIONOPERATION

8282


TANK CIRCUIT OSCILLATIONTANK CIRCUIT OSCILLATION

8383


CLASS C AMPLIFIER WITH CLASS C AMPLIFIER WITH RESISTIVE LOADRESISTIVE LOAD

8484


CLASS C AMPLIFIER WITH LC CLASS C AMPLIFIER WITH LC LOADLOAD

8585


A CLASS C AMPLITUDE A CLASS C AMPLITUDE MODULATORMODULATOR

8686


CLASS D POWER AMPLIFIERCLASS D POWER AMPLIFIER Class D is designed to operate with pulse or digital signals.

Because of its higher efficiency, class D is suitable to be used as a power amplifier.

Similar to class C amplifier, this higher efficiency is achieved because the transistor only in

87

Similar to class C amplifier, this higher efficiency is achieved because the transistor only in conduction for a short period of time thus dissipates minimum power.

For class D amplifiers, the input signal, vi is converted into square pulses, then filtered to obtain the output signal, vo that is identical to the shape of the input waveform.

87


This output signal, vo is then used to drive the load at high power.

Below is the block diagram of the class D power amplifier.

8888


The low pass filter is use to change the amplified pulse signal, vc to sinusoidal output signal, vo.

The basic component of the class D amplifier is the Pulse-Width

89

The basic component of the class D amplifier is the Pulse-Width Modulator(PWM) that produced the square pulse trains, vc proportional to vi.

89


vmg

90

vi

vc

90


The diagram shows how PWM produce the pulse trains, vc from sawtooth generator and voltage comparator.

Note that the peak to peak voltage, vmgmust be higher than the peak to peak voltage of vi.

91

voltage of vi. Also the frequency for vmg must be at least 10 times the input frequency vi.

Because almost all power is transferred to the load, class D amplifier is said to have the highest power efficiency of all classes.

91


POWER AMPLIFIER CIRCUIT BOARD POWER AMPLIFIER CIRCUIT BOARD AND TRANSISTOR PIN CONFIGURATIONAND TRANSISTOR PIN CONFIGURATION

9292


Power transistors are rated to handle (dissipate) a specified amount of power with a case temperature, TC of 25oC.

If the case temperature increases during operation, which of course it will, transistor must be derated.

POWER RATING AND DERATINGPOWER RATING AND DERATING

93

transistor must be derated. The transistor can be derated according to the derating factor specified by the manufacturer.

Derating factor is specified as watts per degree centigrade (above 25oC).

PD(max at operating temp) = PD(max at 25C)-[(TC-25oC).D]

93


POWER DERATING FOR POWER DERATING FOR 2N3055 (NPN), MJ2955 (PNP)2N3055 (NPN), MJ2955 (PNP)

94

DERATING FACTOR DERATING FACTOR 0.657W/0.657W/ooCC


The 2N3055 is a very popular high-power NPN transistor. It is rated for a maximum power dissipation of 115W at a case temperature of 25oC. The specified derating factor is 0.657W/oC. If the transistors case temperature reaches 65oC during operation, what is the transistor power rating?

95

PD(max at operating temp) = PD(max at 25C)-[(TC- 25oC).D]= 115W [(65oC-25oC)x0.657W/oC]= 115W 26.3W= 88.7W


HEAT SINKHEAT SINK

9696


To remove dissipated power in the form of heat from the transistor case quickly as possible.

When power dissipated by the transistor, the heat produced at the collector-base junction must be

HEAT SINKHEAT SINK

97

produced at the collector-base junction must be quickly removed into the air.

For small signal amplifier, the surface area of the transistor is large enough to dissipate the heat but not for the large signal amplifier.

97


Large power dissipation in the transistor will increase the collector junction temperature, TJ and if greater than TJ(max), the transistor will damage.TJ(max) for silicon 150oC -200oCT for germanium 100oC -

98

200 CTJ(max) for germanium 100oC -110oC

Generally, the power transistor is cooled by three mechanism of heat transfer.

First the heat is transferred to the collector junction through the semiconductor material, then through the transistor case and finally into the air.

98


To increase the flow of heat transfer quickly from the transistor case to air, add a heat sink (to increase the surface area).

Average power dissipated by the transistor is given by:

99

transistor is given by: TJ - TA = JAPD

99


OHMS LAW OF THERMODYNAMICS OHMS LAW OF THERMODYNAMICS FOR POWER TRANSISTORSFOR POWER TRANSISTORS

100

JA

AJD

TTP

=

RVVI 21 =

100


A transistor can dissipate power of 2W at its base-collector junction. Junction to case thermal resistance is 8oC/W and case to ambient thermal resistance is 20oC/W. Ambient temperature is 25oC. Determine (a) junction temperature (b) case temperature.

101

AJADJ TPT +=

CAJCJA +=W/C28W/C20W/C8 ooo =+=

Junction to ambient thermal resistance,

Junction temperature, C81C25)W/C28)(W2( ooo =+=

101


D

CJJC P

TT +=

Transistors thermal resistance,

Case temperature,

102

JCDJC PTT =

C65)W/C8)(W2(C81 ooo ==

Case temperature,

102


A 2N3055 power transistor dissipates 20W during operation. The amplifier circuit is designed to operate over an ambient temperature range of 0oC to 80oC. The worst case condition exists when the ambient temperature is 80oC. The temperature case to heat sink thermal resistance is 0.5oC/W and the heat sink is rated for a thermal resistance of 3oC/W. Calculate the case temperature of the transistor for worst case operating conditions.

103

TC = (PD.CA) + TA = (20W x 3.5oC/W) + 80oC= 70oC + 80oC= 150oC

PD(max at operating temp) = PD(max at 25C)-[(TC - 25oC).D]PD(max at 80C) = 115W [(150oC-25oC)x0.657W/oC]

= 115W 82W= 33W

The transistor is actually dissipating 20W, so the maximum of 33W is a safe margin.


Class A Class A

Power gainPower gain

A type of amplifier that operates entirely in its linear (active) region.

The ratio of output power to the input power of an amplifier.

Selected Key Terms

EfficiencyEfficiency

Class BClass B

of an amplifier.

The ratio of the signal power delivered to a load to the power from the power supply of an amplifier.

A type of amplifier that operates in the linear region for 180o of the input cycle because it is biased at cutoff.

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Chapter 2 POWER AMPLIFIER V2 [Compatibility Mode]

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