Transcript Lecture - Info Poster

CIT 852 – Electronic Signals and
Systems
Chapter 4: Analogue Amplifiers
4.1 Characteristics of Analogue Amplifiers
4.2 Feedback: Gain Control and Frequency Response
Powered by DeSiaMore
1
Lecture 8
Power Amplifier (Class A)
•
•
•
•
•
Induction of Power Amplifier
Power and Efficiency
Amplifier Classification
Basic Class A Amplifier
Transformer Coupled Class A Amplifier
Powered by DeSiaMore
2
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 lowresistance 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.
Powered by DeSiaMore
3
Amplifier Power Dissipation
VCC
The total amount of power
being dissipated by the
amplifier, Ptot , is
I CC
Ptot = P1 + P2 + PC + PT + PE
I1
I CQ
P =
I12R1
1
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.
2R
P
=
I
2
2
2
output power.
R1
RC
PC = I2CQR C
PT = I2TQ R T
I EQ
R2
RE
PE = I2EQ R E
I2
 Amplifier Efficiency h
Powered by DeSiaMore
4
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 :
ac output power
Po (ac)
h
 100% 
 100%
dc input power
Pi (dc)
• 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
Powered by DeSiaMore
5
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
Av
vout
Class-A
vin
Av
vout
Class-B
vin
Av
vout
Class-C
Powered by DeSiaMore
6
Efficiency Ratings
• The maximum theoretical efficiency ratings
of class-A, B, and C amplifiers are:
Amplifier
Maximum Theoretical
Efficiency, hmax
Class A
25%
Class B
78.5%
Class C
99%
Powered by DeSiaMore
7
Class A Amplifier
vin
Av
vout
• 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)
Powered by DeSiaMore
8
Basic Operation
Common-emitter (voltage-divider) configuration (RC-coupled amplifier)
+VCC
I CC
I1
I CQ
R1
RC
RL
v in
R2
RE
Powered by DeSiaMore
9
Typical Characteristic Curves for
Class-A Operation
Powered by DeSiaMore
10
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)
Powered by DeSiaMore
11
DC Input Power
+VCC
The total dc power, Pi(dc) , that an amplifier
draws from the power supply :
I CC
I1
I CQ
R1
RC
Pi (dc)  VCC I CC
RL
I CC  I CQ  I 1
I CC  I CQ
( I CQ  I 1 )
v in
R2
RE
Pi ( dc )  VCC I CQ
Note that this equation is valid for most amplifier power analyses. We can rewrite for the
above equation for the ideal amplifier as
Pi (dc)  2VCEQ I CQ
Powered by DeSiaMore
12
AC Output Power
AC output (or load) power, Po(ac)
ic
vo
2
Po (ac)  ic ( rms) vo ( rms)
vo ( rms)

RL
vce
vin
Above equations can be used to
calculate the maximum possible
value of ac load power. HOW??
rC
RC//RL
R1//R2
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
Powered by DeSiaMore
13
IC(sat) = VCC/(RC+RE)
IC(sat) = ICQ + (VCEQ/rC)
DC Load Line
ac load line
IC
IC
(mA)
VCE(off) = VCC
VCE(off) = VCEQ + ICQrC
VCE
VPP2
 VCEQ  I CQ  1
Po ( ac)  

  VCEQ I CQ 
8 RL
 2  2  2
ac load line
IC
VCE
Q - point
dc load line
h
Po ( ac )
Pi ( dc )
1
VCEQ I CQ
2
 100% 
 100%  25%
2VCEQ I CQ
VCE
Powered by DeSiaMore
14
Limitation
Powered by DeSiaMore
15
Example
+VCC = 20V
Calculate the input power [Pi(dc)], output power [Po(ac)],
and efficiency [h] of the amplifier circuit for an input
voltage that results in a base current of 10mA peak.
VCC  VBE 20V  0.7V

 19.3mA
RB
1k
ICQ  I B  25(19.3mA)  482.5mA  0.48 A
RB
1k
IC
RC
20
Vo
  25
IBQ 
Vi
VCEQ  VCC  ICRC  20V  (0.48 A)( 20)  10.4V
V
20V
I c ( sat)  CC 
 1000mA  1A
RC
20
VCE ( cu to ff )  VCC  20V
IC ( p ea k)  Ib ( p ea k)  25(10mA peak )  250mA peak
Po ( ac) 
Pi ( dc)
h
I C2 ( peak )
250  10 A)

3
2
RC
(20)  0.625W
2
2
 VCC I CQ  (20V )(0.48 A)  9.6W
Po ( ac)
Pi ( dc)
 100%  6.5%
Powered by DeSiaMore
16
Part 1
Powered by DeSiaMore
17
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.
Powered by DeSiaMore
18
Classification of Power Amplifier
• Power amplifiers are classified according to
the collector current waveform that results
when an input signal is applied.
• Conducting angle.
Powered by DeSiaMore
19
Classification of Power Amplifier
Collector current waveforms for transistors
operating in (a) class A, (b) class B
Powered by DeSiaMore
20
Classification of Power Amplifier
class AB
class C
Powered by DeSiaMore
21
Class B Output Stage
A class B output stage.
Complementary circuits.
Push-pull operation
Maximum power-conversion
efficiency is 78.5%
Powered by DeSiaMore
22
Transfer Characteristic
Powered by DeSiaMore
23
Crossover Distortion
Powered by DeSiaMore
24
Power Dissipation
• The load power
2
1 Vˆo
PL 
2 RL
• Maximum load power
PL max
2
1 Vˆo

2 RL
2
Vˆo VCC
VCC

2 RL
Powered by DeSiaMore
25
Power Dissipation
• Total supply power
2 Vˆo
Ps 
VCC
 RL
• Maximum total supply power
Ps max
2
2 Vˆo
2 VCC

VCC

 RL
 RL
Vˆ V
o
CC
Powered by DeSiaMore
26
Power Dissipation
• Power-conversion efficiency
h
 Vˆo
4 VCC
• Maximum power-conversion efficiency
h max 
 Vˆo
4 VCC
 78.5%
Vˆo VCC
Powered by DeSiaMore
27
Power Dissipation
• Power dissipation
2
2 Vˆo
1 Vˆo
PD 
VCC 
 RL
2 RL
• Maximum Power dissipation
PDN max  PDP max
2
2 Vˆo
1 Vˆo

VCC 
 RL
2 RL
2
Vˆo  VCC

2

2VCC
 0.2 PL max
2
 RL
Powered by DeSiaMore
28
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.
Powered by DeSiaMore
29
A Class AB Output Stage Utilizing Diodes
for Biasing
Powered by DeSiaMore
30
A Class AB Output Stage Utilizing A VBE
Multiplier for Biasing
Powered by DeSiaMore
31
Part 2
Powered by DeSiaMore
32
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
Powered by DeSiaMore
33
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.
Powered by DeSiaMore
34
Control Theory
• Branch of systems theory (study of interactions
and behavior of a complex assemblage)
Manipulated
Variable(s)
Manipulated
Variable(s)
Control System
Control System
Control
Variable(s)
Control
Variable(s)
Open Loop
Control
System
Closed Loop
Control
System
Feedback function
Powered by DeSiaMore
35
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
Powered by DeSiaMore
36
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
Powered by DeSiaMore
37
• More sophisticated arrangements are closedloop
– user inputs the goal to the system
Powered by DeSiaMore
38
Automatic Control Systems
• Examples of automatic control systems:
– temperature control using a room heater
Powered by DeSiaMore
39
• Examples of automatic control systems:
– Cruise control in a car
Powered by DeSiaMore
40
• Examples of automatic control systems:
– Position control in a human limb
Powered by DeSiaMore
41
• Examples of automatic control systems:
– Level control in a dam
Powered by DeSiaMore
42
Feedback Systems
• A generalised feedback system
Powered by DeSiaMore
43
• By inspection of diagram we can add values
Xo
 X i  BX o
A
or rearranging
Powered by DeSiaMore
Xo
A

X i 1  AB
44
• Thus
Overall gain G 
Xo
A

X i 1  AB
• 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
Powered by DeSiaMore
45
• 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
G
A
A
1


1  AB AB B
- gain is independent of the gain of the forward path A
Powered by DeSiaMore
46
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
Powered by DeSiaMore
47
• Consider the following example below:
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.
• We will base our design on our standard feedback arrangement
Powered by DeSiaMore
48
• We will use our active
amplifier for A and a stable
feedback arrangement for B
 Since we require an overall gain of 100
G
1
B
so we will use B = 1/100 or 0.01
Powered by DeSiaMore
49
• Now consider the gain of the
circuit when the gain of the
active amplifier A is 100,000
G
A
100 000

1  AB 1  (100 000  0.01)
100 000

1  1 000
 99.90
1

B
Powered by DeSiaMore
50
• Now consider the gain of the
circuit when the gain of the
active amplifier A is 200,000
G
A
200 000

1  AB 1  (200 000  0.01)
200 000

1  2 000
 99.95
1

B
Powered by DeSiaMore
51
• Note that a change in the gain
of the active amplifier of 100%
causes a change in the overall
gain 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
Powered by DeSiaMore
52
• 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
Powered by DeSiaMore
53
• Thus we can implement our feedback
arrangement using an active amplifier and a
passive feedback network to produce a stable
amplifier
• The arrangement on
the right has a gain
of 100 …
… but how do we
implement the
subtractor?
Powered by DeSiaMore
54
• A differential amplifier is effectively an active
amplifier combined with a subtractor. A
common form is the operational amplifier or
op-amp
• The arrangement on
the right has a gain
of 100.
Powered by DeSiaMore
55
• In this circuit the gain is determined
by the passive components and we
do not need to know the gain of the
op-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
Powered by DeSiaMore
56
The Effects of Negative Feedback
7.6
• Effects on Gain
– negative feedback produces a gain given by
G
A
1  AB
– there, feedback reduces the gain by a factor of 1 +
AB
– this is the price we pay for the beneficial effects of
negative feedback
Powered by DeSiaMore
57
• 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
Powered by DeSiaMore
58
– therefore the bandwidth increases as the gain is
reduced with feedback
– in some cases the gain x bandwidth = constant
Powered by DeSiaMore
59
• 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
Powered by DeSiaMore
60
• 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
Powered by DeSiaMore
61
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
Powered by DeSiaMore
62
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
G
A
1  AB
• 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.
Powered by DeSiaMore
63
Classification of Systems
Classes of Systems
Distributed Parameter (Partial
Differential Equations,
Transmission line example)
Lumped Parameter
Deterministic
Continuous Time
Linear
Time Varying
Stochastic
Discrete Time
Nonlinear
Constant Coefficient
Non-homogeneous
Homogeneous (No External Input;
system behavior depends on
Powered by DeSiaMore
initial conditions)
64
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
Powered by DeSiaMore
65
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
Powered by DeSiaMore
66
Laplace Transform

• Definition L[ f (t )]  F (s)  0 f (t )e dt
• Condition for Existence
• Laplace Transforms of exponential,
step, ramp,
  0  Limit e | f (t ) | 0
sinusoidal, pulse, and impulse functions
• Translation of
and multiplication by
• Effect of Change of
time
scale
t
f (t )
e
• Real and complex differentiations, initial and final
value theorems, real integration, product theorem
• Inverse Laplace Transform
 st
t
suchthat
t 
Powered by DeSiaMore
67
Inverse Laplace Transform
• Definition L [F (s)]  f (t)  1  F (s)e ds
2j
• Formula is seldom or never used; instead,
Heaviside partial fraction expansion is used.
• Illustration with a problem: d y  4 dy  3 y  2r (t )
dt
dt
Initial conditions: y(0) = 1, y’(0) = 0, and
r(t) = 1, t >= 0. Find the steady state response
• Multiple pole case with s  2s  3



F ( s) 
L 

• Use the ideas to find
(s  1) and
 ( s  a)   
c  j
1
st
c i
2
2
2
1
3


sa
L1 
2
2
 ( s  a)   Powered by DeSiaMore
2
2
68
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
Powered by DeSiaMore
69
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
R(s)
+
G1
-
+
H2
G2
+
+
H3
Y(s)
G3
G4
H1
Powered by DeSiaMore
70
Signal Flow Graphs
• Mason’s Gain Formula
a11x1  a12 x2  r1  x1
a11
r1
a21x1  a22 x2  r2  x2
x1
a21
r2
Solve these two equations and generalize to
get Mason’s Gain Formula
H2
G1
R(s)
G5
Gij 
H3
G2
G3
G6
H8
G7
H7
G4
F
ijk
a12
x2
a22
 ijk
k

Y(s)
G8
Find Y(s)/R(s) using
the
formula
Powered by
DeSiaMore
71
Another Signal Flow Graph Problem
G7
1
G1
G2
G3
G4
R(s)
-H4
G8
G5
G6
-H1
C(s)
-H2
-H3
Powered by DeSiaMore
72