Post on 29-Dec-2015
CIT 852 – Electronic Signals and Systems
Chapter 4: Analogue Amplifiers4.1 Characteristics of Analogue Amplifiers
4.2 Feedback: Gain Control and Frequency Response
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Lecture 8 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 or higher.
• Ideal power amplifier will deliver 100% of the power it draws from the supply to load. In practice, this can never occur.
• The reason for this is the fact that the components in the amplifier will all dissipate some of the power that is being drawn form the supply.
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Amplifier Power Dissipation
P1 = I12R1
P2 = I22R2
ICQRC
RE
R1
R2
VCC
I1
I2
ICC
PC = ICQ2 RC
PT = ITQ2 RT
PE = IEQ2 RE
IEQ
The total amount of power being dissipated by the amplifier, Ptot , is
Ptot = P1 + P2 + PC + PT + PE
The difference between this total value and the total power being drawn from the supply is the power that actually goes to the load – i.e. output power.
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Amplifier Efficiency • A figure of merit for the power amplifier is its efficiency, . • Efficiency ( 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 much higher efficiency ratings than others.
• This is primary consideration when deciding which type of power amplifier to use for a specific application.
• Amplifier Classifications
%100)(
)(%100
dcP
acP
powerinputdc
poweroutputac
i
o
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Amplifier Classifications • Power amplifiers are classified according to the percent of
time that collector current is nonzero. • The amount the output signal varies over one cycle of
operation for a full cycle of input signal.
vin voutAv Class-A
vin voutAv Class-B
vin voutAv 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 shape input waveform + phase shift.
• The collector current is nonzero 100% of the time. inefficient, since even with zero input signal, ICQ is nonzero(i.e. transistor dissipates power in the rest, or quiescent, condition)
vin voutAv
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Basic OperationCommon-emitter (voltage-divider) configuration (RC-coupled amplifier)
RC
+VCC
RE
R1
R2
RL
v in
ICQI1
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
v in
ICQI1
ICCThe total dc power, Pi(dc) , that an amplifier 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//R2
vcevin
vo
ic
RC//RLrC
AC output (or load) power, Po(ac)
Above equations can be used to calculate 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, 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+RE)
DC Load LineIC
VCE
IC(sat) = ICQ + (VCEQ/rC)
VCE(off) = VCEQ + ICQrC
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%10022
1
%100)(
)( CQCEQ
CQCEQ
dci
aco
IV
IV
P
P
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Limitation
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ExampleCalculate the input power [Pi(dc)], output power [Po(ac)], and efficiency [] of the amplifier circuit for an input voltage that results in a base current of 10mA peak.
RCRB
+VCC = 20V
IC
Vi
25
20k1
Vo
%5.6%100
6.9)48.0)(20(
625.0)20(2
10250
2
250)10(25
20
1100020
20
4.10)20)(48.0(20
48.05.482)3.19(25
3.191
7.020
)(
)(
)(
232)(
)(
)(
)()(
)(
dci
aco
CQCCdci
CpeakC
aco
C
CCsatc
B
P
P
WAVIVP
WA
RI
P
peakmApeakmAII
VVV
AmAV
R
VI
VAVRIVV
AmAmAII
mAk
VV
R
VVI
peakbpeakC
CCcutoffCE
CCCCCEQ
CQ
B
BECC
BQ
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Part 1
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Power Amplifier
• Small-signal approximation and models either are not applicable or must be used with care.
• Deliver the power to the load in efficient manner.
• Power dissipation is as low as possible.
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Classification of Power Amplifier
• Power amplifiers are classified according to the collector current waveform that results when an input signal is applied.
• Conducting angle.
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Classification of Power Amplifier
Collector current waveforms for transistors operating in (a) class A, (b) class B
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Classification of Power Amplifier
class AB class C
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Class B Output Stage
A class B output stage.
Complementary circuits.
Push-pull operation
Maximum power-conversion efficiency is 78.5%
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Transfer Characteristic
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Crossover Distortion
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Power Dissipation
• The load power
• Maximum load power
L
oL R
VP
2ˆ
2
1
L
CC
VVL
oL R
V
R
VP
CCo
2
ˆ
2
1 2
ˆ
2
max
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Power Dissipation
• Total supply power
• Maximum total supply power
CCL
os V
R
VP
ˆ2
L
CC
VV
CCL
os R
VV
R
VP
CCo
2
ˆ
max
2ˆ2
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Power Dissipation
• Power-conversion efficiency
• Maximum power-conversion efficiency
CC
o
V
V̂
4
%5.78ˆ
4 ˆ
max CCo VVCC
o
V
V
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Power Dissipation
• Power dissipation
• Maximum Power dissipation
L
oCC
L
oD R
VV
R
VP
2ˆ
2
1ˆ2
max2
2
2ˆ
2
maxmax
2.02
ˆ
2
1ˆ2
LL
CC
VVL
oCC
L
oDPDN
PR
V
R
VV
R
VPP
CCo
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Class AB Output Stage
A bias voltage VBB is applied between the bases of QN and QP, giving rise to a bias current IQ . Thus, for small vI, both transistors conduct and crossover distortion is almost completely eliminated.
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A Class AB Output Stage Utilizing Diodes for Biasing
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A Class AB Output Stage Utilizing A VBE Multiplier for Biasing
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Part 2
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Control and Feedback• Introduction• Open-loop and Closed-loop Systems• Automatic Control Systems• Feedback Systems• Negative Feedback• The Effects of Negative Feedback• Negative Feedback – A Summary
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What is a Control System?
• System- a combination of components that act together and perform a certain objective.
• Control System- a system in which the objective is to control a process or a device or environment.
• Process- a progressively continuing operations/development marked by a series of gradual changes that succeed one another in a relatively fixed way and lead towards a particular result or end.
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Control Theory
• Branch of systems theory (study of interactions and behavior of a complex assemblage)
Control SystemManipulated Variable(s)
Control Variable(s)
Open Loop Control System
Control System
Manipulated Variable(s)
Control Variable(s)
Closed Loop Control System
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Introduction
• Earlier we identified control as one of the basic functions performed by many systems– often involves regulation or command
• Invariably, the goal is to determine the value or state of some physical quantity– and often to maintain it at that value, despite
variations in the system or the environment
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Open-loop and Closed-loop Systems
• Simple control is often open-loop– user has a goal and selects an input to a system to
try to achieve this
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• More sophisticated arrangements are closed-loop– user inputs the goal to the system
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Automatic Control Systems
• Examples of automatic control systems:– temperature control using a room heater
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• Examples of automatic control systems:– Cruise control in a car
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• Examples of automatic control systems:– Position control in a human limb
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• Examples of automatic control systems:– Level control in a dam
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Feedback Systems
• A generalised feedback system
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• By inspection of diagram we can add values
or rearrangingoi
o BXXAX
ABA
XX
i
o
1
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• Thus
• This the transfer function of the arrangement• Terminology:
• A is also known as the open-loop gain• G is the overall or closed-loop gain
ABA
XX
i
o
1G gain Overall
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• Effects of the product AB– If AB is negative
• If AB is negative and less than 1, (1 + AB) < 1• In this situation G > A and we have positive feedback
– If AB is positive• If AB is positive then (1 + AB) > 1• In this situation G < A and we have negative feedback• If AB is positive and AB >>1
- gain is independent of the gain of the forward path A
BABA
ABA
G1
1
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Negative Feedback
• Negative feedback can be applied in many ways– Xi and Xo could be temperatures, pressures, etc.– here we are mainly interested in voltages and currents
• Particularly important in overcoming variability– all active devices suffer from variability
• their gain and other characteristics vary with temperature and between devices
– we noted above that using negative feedback we can produce an arrangement where the gain is independent of the gain of the forward path
• this gives us a way of overcoming problems of variability
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• Consider the following example below:
• We will base our design on our standard feedback arrangement
Example: Design an arrangement with a stable voltage gain of 100 using a high-gain active amplifier. Determine the effect on the overall gain of the circuit if the voltage gain of the active amplifier varies from 100,000 to 200,000.
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• We will use our active amplifier for A and a stable feedback arrangement for B
Since we require an overall gain of 100
so we will use B = 1/100 or 0.01
BG
1
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• Now consider the gain of the circuit when the gain of the active amplifier A is 100,000
B
ABA
G
1
90.99000 11000 100
)01.0000 100(1000 100
1
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• Now consider the gain of the circuit when the gain of the active amplifier A is 200,000
B
ABA
G
1
95.99000 21000 200
)01.0000 200(1000 200
1
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• Note that a change in the gainof the active amplifier of 100%causes a change in the overallgain of just 0.05 %
• Thus the use of negative feedback makes the gain largely independent of the gain of the active amplifier
• However, it does require that B is stable– fortunately, B can be based on stable passive
components
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• Implementing the passive feedback path– to get an overall gain of
greater than 1 requires a feedback gain B of less than 1
– in the previous example the value of B is 0.01
– this can be achieved using a simple potential divider
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• Thus we can implement our feedback arrangement using an active amplifier and a passive feedback network to produce a stable amplifier
• The arrangement onthe right has a gain of 100 …
… but how do weimplement thesubtractor?
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• A differential amplifier is effectively an active amplifier combined with a subtractor. A common form is the operational amplifier or op-amp
• The arrangement onthe right has a gain of 100.
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• In this circuit the gain is determinedby the passive components and wedo not need to know the gain of theop-amp– however, earlier we assumed
that AB >> 1– that is, that A >> 1/B– that is, open-loop gain >> closed-loop gain– therefore, the gain of the circuit must be much less
than the gain of the op-amp– see Example 7.2 in the course text
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The Effects of Negative Feedback
• Effects on Gain– negative feedback produces a gain given by
– there, feedback reduces the gain by a factor of 1 + AB
– this is the price we pay for the beneficial effects of negative feedback
7.6
ABA
1G
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• Effects on frequency response– from earlier lectures we know that all amplifiers have
a limited frequency response and bandwidth– with feedback we make the overall gain largely
independent of the gain of the active amplifier– this has the effect of increasing the bandwidth, since
the gain of the feedback amplifier remains constant as the gain of the active amplifier falls
– however, when the open-loop gain is no longer much greater than the closed-loop gain the overall gain falls
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– therefore the bandwidth increases as the gain is reduced with feedback
– in some cases the gain x bandwidth = constant
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• Effects on input and output resistance– negative feedback can either increase or decrease the
input or output resistance depending on how it is used.• if the output voltage is fed back this tends to make the output
voltage more stable by decreasing the output resistance• if the output current is fed back this tends to make the output
current more stable by increasing the output resistance• if a voltage related to the output is subtracted from the input
voltage this increases the input resistance• if a current related to the output is subtracted from the input
current this decreases the input resistance• the factor by which the resistance changes is (1 + AB)• we will apply this to op-amps in a later lecture
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• Effects on distortion and noise– many forms of distortion are caused by a non-linear
amplitude response• that is, the gain varies with the amplitude of the signal
– since feedback tends to stabilise the gain it also tends to reduce distortion - often by a factor of (1 + AB)
– noise produced within an amplifier is also reduced by negative feedback – again by a factor of (1 + AB)
• note that noise already corrupting the input signal is not reduced in this way – this is amplified along with the signal
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Negative Feedback – A Summary
• All negative feedback systems share some properties
1. They tend to maintain their output independent of variations in the forward path or in the environment
2. They require a forward path gain that is greater than that which would be necessary to achieve the required output in the absence of feedback
3. The overall behavior of the system is determined by the nature of the feedback path
Unfortunately, negative feedback does have implications for the stability of circuits – this is discussed in later lectures
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Key Points
• Feedback is used in almost all automatic control systems• Feedback can be either negative or positive• If the gain of the forward path is A, the gain of the feedback
path is B and the feedback is subtracted from the input then
• If AB is positive and much greater than 1, then G 1/B• Negative feedback can be used to overcome problems of
variability within active amplifiers• Negative feedback can be used to increase bandwidth, and to
improve other circuit characteristics.
ABA
1G
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Classification of Systems
Classes of Systems
Lumped ParameterDistributed Parameter (Partial Differential Equations, Transmission line example) Deterministic
Discrete TimeContinuous Time
NonlinearLinear
Time Varying
Stochastic
Constant Coefficient
Non-homogeneous Homogeneous (No External Input; system behavior depends on initial conditions)
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Example Control Systems• Mechanical and Electo-mechanical (e.g. Turntable) Control Systems• Thermal (e.g. Temperature) Control System• Pneumatic Control System• Fluid (Hydraulic) Control Systems• Complex Control Systems • Industrial Controllers
– On-off Controllers– Proportional Controllers– Integral Controllers– Proportional-plus-Integral Controllers– Proportional-plus-Derivative Controllers– Proportional-plus-Integral-plus-Derivative Controllers
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Mathematical Background
• Why needed? (A system with differentials, integrals etc.)• Complex variables (Cauchy-Reimann Conditions, Euler
Theorem)• Laplace Transformation
– Definition– Standard Transforms – Inverse Laplace Transforms
• Z-Transforms• Matrix algebra
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Laplace Transform
• Definition• Condition for Existence• Laplace Transforms of exponential, step, ramp,
sinusoidal, pulse, and impulse functions• Translation of and multiplication by• Effect of Change of time scale• Real and complex differentiations, initial and final
value theorems, real integration, product theorem• Inverse Laplace Transform
dtetfsFtf st
0)()()]([L
0|)(|0
tfe t
t
suchthat Limit
te)(tf
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Inverse Laplace Transform
• Definition• Formula is seldom or never used; instead,
Heaviside partial fraction expansion is used.• Illustration with a problem: Initial conditions: y(0) = 1, y’(0) = 0, and r(t) = 1, t >= 0. Find the steady state response• Multiple pole case with • Use the ideas to find and
dsesFj
tfsFjc
ic
st
)(2
1)()]([1
L
)(2342
2
trydt
dy
dt
yd
22
1
)( as
L
22
1
)( as
asL
3
2
)1(
32)(
s
sssF
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Applications
• Spring-mass-damper- Coulomb and viscous damper cases
• RLC circuit, and concept of analogous variables• Solution of spring-mass-damper (viscous case)• DC motor- Field current and armature current
controlled cases• Block diagrams of the above DC-motor problems• Feedback System Transfer functions and Signal flow graphs
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Block Diagram Reduction
• Combining blocks in a cascade• Moving a summing point ahead of a block• Moving summing point behind a block• Moving splitting point ahead of a block• Moving splitting point behind a block• Elimination of a feedback loop
G1 G2 G3 G4
H2
H1
H3
R(s)Y(s)
+ +
+ -
-+
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Signal Flow Graphs
• Mason’s Gain Formula
Solve these two equations and generalize toget Mason’s Gain Formula
r1
r2
x1
x2
a21 a12
a22
a1111212111 xrxaxa
22222121 xrxaxa
k
ijkijk
ij
FG
G1 G2 G3 G4
G5G6 G7 G8
H2 H3
H7H8
R(s) Y(s)
Find Y(s)/R(s) using the formula 71Powered by DeSiaMore
Another Signal Flow Graph Problem
R(s) C(s)
1 G1 G2 G3 G4 G5 G6
G8G7
-H4-H1
-H2
-H3
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