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
SEE 3263 POWER AMPLIFIERS
AMPLIFIER SYSTEM AMPLIFIER SYSTEM APPLICATIONAPPLICATION
33
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
OUTPUT SIGNAL SWING OF OUTPUT SIGNAL SWING OF CLASS A AMPLIFIERCLASS A AMPLIFIER
1111
SEE 3263 POWER AMPLIFIERS
COLLECTOR CURRENT COLLECTOR CURRENT COMPONENTSCOMPONENTS
1212
Comprises of AC and DC components
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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.
SEE 3263 POWER AMPLIFIERS
LOAD LINES OF THE CLASS A LOAD LINES OF THE CLASS A AMPLIFIERAMPLIFIER
1515
SEE 3263 POWER AMPLIFIERS
QQ--POINT AT THE MIDDLE POINT AT THE MIDDLE OF THE AC LOAD LINEOF THE AC LOAD LINE
1616
SEE 3263 POWER AMPLIFIERS
QQ--POINT NEAR CUTOFF POINT NEAR CUTOFF
1717
SEE 3263 POWER AMPLIFIERS
QQ--POINT NEAR SATURATION POINT NEAR SATURATION
1818
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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.
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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%}
SEE 3263 POWER AMPLIFIERS
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%}
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
3333
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
(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
SEE 3263 POWER AMPLIFIERS
CLASS B OPERATIONCLASS B OPERATION
3636
A class B circuit provides an output signal varying over one-half of the input cycle.
SEE 3263 POWER AMPLIFIERS
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.
CLASS B OPERATIONCLASS B OPERATION
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
SEE 3263 POWER AMPLIFIERS
COMMON-COLLECTOR CLASS B AMPLIFIER
38
SEE 3263 POWER AMPLIFIERS
CLASS B OPERATIONCLASS B OPERATION
3939
SEE 3263 POWER AMPLIFIERS
CLASS B OPERATIONCLASS B OPERATION
4040
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
AC AND DC LOAD LINES OF CLASS B AC AND DC LOAD LINES OF CLASS B PUSHPUSH--PULL AMPLIFIERPULL AMPLIFIER
L
CC
RV
4343
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
( ) ( )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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
=
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
5252
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
+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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
5959
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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 =
SEE 3263 POWER AMPLIFIERS
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
+=
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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=
SEE 3263 POWER AMPLIFIERS
DARLINGTON CLASS AB DARLINGTON CLASS AB AMPLIFIERAMPLIFIER
6666
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
CLASS AB PUSHCLASS AB PUSH--PULL PULL COMPLIMENTARYCOMPLIMENTARY--SYMMETRY SYMMETRY
WITH TWO SUPPLIESWITH TWO SUPPLIES
6868
SEE 3263 POWER AMPLIFIERS
CLASS AB PUSHCLASS AB PUSH--PULL PULL MOSFET WITH TWO SUPPLIESMOSFET WITH TWO SUPPLIES
6969
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
BASIC CLASS BASIC CLASS C C AMPLIFIER AMPLIFIER (NON INVERTING)(NON INVERTING)
7373
SEE 3263 POWER AMPLIFIERS
OUTPUT CURRENT IN OUTPUT CURRENT IN CLASS C AMPLIFIERCLASS C AMPLIFIER
7474
SEE 3263 POWER AMPLIFIERS
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.
SEE 3263 POWER AMPLIFIERS
CLASS C WAVEFORMCLASS C WAVEFORM
7676
SEE 3263 POWER AMPLIFIERS
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 =
SEE 3263 POWER AMPLIFIERS
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)
=
=
=
=
=
SEE 3263 POWER AMPLIFIERS
CLASS C MAXIMUM OUTPUT CLASS C MAXIMUM OUTPUT POWERPOWER
R2V
PL
2)p(O
L =
79
%100PPPPPP
R2
)avg(DL
L
)avg(DLDC
L
+
=
+=
SEE 3263 POWER AMPLIFIERS
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
=
+
=
+
=
SEE 3263 POWER AMPLIFIERS
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.
SEE 3263 POWER AMPLIFIERS
RESONANCE CIRCUIT RESONANCE CIRCUIT OPERATIONOPERATION
8282
SEE 3263 POWER AMPLIFIERS
TANK CIRCUIT OSCILLATIONTANK CIRCUIT OSCILLATION
8383
SEE 3263 POWER AMPLIFIERS
CLASS C AMPLIFIER WITH CLASS C AMPLIFIER WITH RESISTIVE LOADRESISTIVE LOAD
8484
SEE 3263 POWER AMPLIFIERS
CLASS C AMPLIFIER WITH LC CLASS C AMPLIFIER WITH LC LOADLOAD
8585
SEE 3263 POWER AMPLIFIERS
A CLASS C AMPLITUDE A CLASS C AMPLITUDE MODULATORMODULATOR
8686
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
vmg
90
vi
vc
90
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
POWER AMPLIFIER CIRCUIT BOARD POWER AMPLIFIER CIRCUIT BOARD AND TRANSISTOR PIN CONFIGURATIONAND TRANSISTOR PIN CONFIGURATION
9292
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
POWER DERATING FOR POWER DERATING FOR 2N3055 (NPN), MJ2955 (PNP)2N3055 (NPN), MJ2955 (PNP)
94
DERATING FACTOR DERATING FACTOR 0.657W/0.657W/ooCC
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
HEAT SINKHEAT SINK
9696
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
OHMS LAW OF THERMODYNAMICS OHMS LAW OF THERMODYNAMICS FOR POWER TRANSISTORSFOR POWER TRANSISTORS
100
JA
AJD
TTP
=
RVVI 21 =
100
SEE 3263 POWER AMPLIFIERS
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
SEE 3263 POWER AMPLIFIERS
D
CJJC P
TT +=
Transistors thermal resistance,
Case temperature,
102
JCDJC PTT =
C65)W/C8)(W2(C81 ooo ==
Case temperature,
102
SEE 3263 POWER AMPLIFIERS
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.
SEE 3263 POWER AMPLIFIERS
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.