ME2-HMTX - Lecture Notes 2012

Imperial College London

ME2-HMTX Mechatronics Aims: To demystify electronics and control — through hands-on experience of building electronic systems and of using the instrumentation needed to see how they behave. To develop understanding of, and an ability to use, some of the analogue electronic stages used in instrumentation and control systems — in particular: (a) ‘front ends’ which generate voltage signals from changes detected by sensors, (b) stages for processing (e.g. adding, differentiating or filtering) these signals, and (c) circuits for amplifying signals to drive output devices (e.g. a meter, motor, or switch). To introduce the concepts, advantages and potential problems of closed-loop control using analogue and digital controllers. To familiarise students with the basic vocabulary needed to understand the technical literature of mechatronic systems and to communicate with specialists in the field.

Ferdinando Rodriguez y Baena 2012 – 2013 Academic Edition

Table of Contents ME2-HMTX 2012-2013

©Copyright of Imperial College London (edited by Dr F. Rodriguez y Baena) 2

Table of Contents

CHAPTER 1: ELECTRICAL CIRCUITS ....................................................................................... 7

1.1 LEARNING OUTCOMES ........................................................................................................................ 7

1.2 INTRODUCTION ................................................................................................................................... 7

1.3 DEVICE CHARACTERISTICS ................................................................................................................... 7

1.3.1 Ports, Sources & Loads .......................................................................................................................... 7

1.3.2 “Volt-Amp” Characteristic of a One-Port Electrical Device ................................................................... 8

1.3.3 Ideal Sources ......................................................................................................................................... 8

1.3.4 Basic Electrical Elements ....................................................................................................................... 9

1.4 PARALLEL AND SERIES RESISTOR NETWORKS .................................................................................... 11

1.5 CIRCUITS AND NETWORKS ................................................................................................................ 12

1.5.1 Potential Divider ................................................................................................................................. 12

1.5.2 Superposition ...................................................................................................................................... 13

1.5.3 Kirchhoff’s Laws .................................................................................................................................. 13

1.6 EQUIVALENT NETWORKS .................................................................................................................. 14

1.6.1 Real Voltage and Current Sources ....................................................................................................... 14

1.7 OPERATING POINT OF NETWORKS: GRAPHICAL SOLUTION ............................................................... 16

1.8 REVIEW OF KEY CONCEPTS ................................................................................................................ 17

1.9 FURTHER READING ............................................................................................................................ 17

1.10 PROBLEMS: LINEAR NETWORKS AND COMPONENTS ....................................................................... 18

1.10.1 Resistor Conventions ...................................................................................................................... 18

1.10.2 Resistor Networks 1 ........................................................................................................................ 18





1.10.7 “Real-Life” Problem 1 ..................................................................................................................... 20

1.10.8 “Real-Life” Problem 2 ..................................................................................................................... 20

1.11 SOLUTIONS ........................................................................................................................................ 21

CHAPTER 2: ELECTRONIC CIRCUITS .................................................................................... 22

2.1 LEARNING OUTCOMES ...................................................................................................................... 22

2.2 INTRODUCTION ................................................................................................................................. 22

2.3 BASIC SEMICONDUCTOR PHYSICS ..................................................................................................... 22

2.3.1 “pn” Junctions ..................................................................................................................................... 24



2.4 IDEAL DIODES .................................................................................................................................... 25

2.5 EQUIVALENT NON-LINEAR DEVICES................................................................................................... 25

2.6 ZENER DIODE ..................................................................................................................................... 27

2.6.1 Zener Diode to Protect Input Ports ..................................................................................................... 28

2.6.2 Zener Voltage Regulator Stage ............................................................................................................ 29


2.8 FURTHER READING ............................................................................................................................ 32

2.9 PROBLEMS: NON-LINEAR NETWORKS AND COMPONENTS ............................................................... 32

2.9.1 Non Linear Series Networks ................................................................................................................ 32

2.9.2 Diode Characteristics .......................................................................................................................... 32

2.9.3 Zener Diodes ....................................................................................................................................... 33

2.9.4 Zener Diodes in Series ......................................................................................................................... 33

2.9.5 Non Linear Network 1 ......................................................................................................................... 33

2.9.6 Non Linear Network 2 ......................................................................................................................... 33

2.10 SOLUTIONS ........................................................................................................................................ 35

CHAPTER 3: ANALOGUE TO DIGITAL CONVERSION & BACK ....................................... 36


3.2 INTRODUCTION ................................................................................................................................. 36

3.3 ANALOGUE TO DIGITAL CONVERSION AT A GLANCE ......................................................................... 36

3.4 ANALOGUE TO DIGITAL CONVERSION ............................................................................................... 38

3.4.1 Quantization ........................................................................................................................................ 38

3.4.2 Coding ................................................................................................................................................. 39

3.4.3 Resolution ........................................................................................................................................... 39

3.4.4 Input Range vs. Converter Range ........................................................................................................ 40

3.4.5 Aliasing and the Nyquist Sampling Criterion ....................................................................................... 41

3.4.6 Information Channels .......................................................................................................................... 41

3.4.7 Information Content and Binary Coding ............................................................................................. 42

3.5 THE WAY BACK: DIGITAL TO ANALOGUE CONVERSION ..................................................................... 43

3.5.1 The R2R Ladder ................................................................................................................................... 43

3.5.2 Worked Solution for the Standard R2R Ladder ................................................................................... 44

3.5.3 A Symmetric R2R Ladder ..................................................................................................................... 51


3.7 FURTHER READING ............................................................................................................................ 52

3.8 PROBLEMS: ANALOGUE TO DIGITAL CONVERSION & BACK ............................................................... 53

3.8.1 Resolution and Sensitivity ................................................................................................................... 53

3.8.2 The Symmetric R2R Ladder ................................................................................................................. 53



3.8.3 Sampling Rate and A/D Conversion .................................................................................................... 53

3.9 SOLUTIONS ........................................................................................................................................ 53

CHAPTER 4: INTRODUCTION TO THE BASICATOM PRO™ & NI ELVIS II™ SYSTEM

54


4.2 THE BASICATOM PRO™ MICROCONTROLLER..................................................................................... 54

4.2.1 Main features ...................................................................................................................................... 55

4.2.2 Setup ................................................................................................................................................... 55

4.2.3 Programming in BASIC ........................................................................................................................ 56

4.2.4 Defining Variables ............................................................................................................................... 56

4.2.5 Programming Blocks ........................................................................................................................... 56

4.2.6 Debugging ........................................................................................................................................... 56

4.3 THE NI ELVIS II DATA ACQUISITION SYSTEM ...................................................................................... 57

4.4 FURTHER READING ............................................................................................................................ 57

CHAPTER 5: TRANSISTORS ..................................................................................................... 58


5.2 INTRODUCTION ................................................................................................................................. 58

5.3 THE BIPOLAR JUNCTION TRANSISTOR ............................................................................................... 59

5.3.1 Basic Operating Principle .................................................................................................................... 59

5.3.2 Active, Cut-Off and Saturated Conditions ........................................................................................... 60

5.4 COMMON-EMITTER CONFIGURATION .............................................................................................. 61

5.4.1 Configuration Characteristics .............................................................................................................. 61

5.4.2 The Transistor as a Switch ................................................................................................................... 62

5.4.3 The Transistor as an Amplifier ............................................................................................................ 62

5.5 COMMON-COLLECTOR CONFIGURATION .......................................................................................... 64

5.5.1 The Emitter-Follower .......................................................................................................................... 64

5.5.2 Darlington Pair .................................................................................................................................... 65

5.6 THE SOUND OF MUSIC ...................................................................................................................... 65

5.7 FIELD EFFECT TRANSISTORS (INFORMATION) .................................................................................... 66

5.8 PULSE WIDTH MODULATION ............................................................................................................. 66


5.10 FURTHER READING ............................................................................................................................ 67

5.11 PROBLEMS: TRANSISTORS ................................................................................................................. 68



5.11.1 Bipolar Junction Transistors 1 ......................................................................................................... 68

5.11.2 Bipolar Junction Transistors 2 ......................................................................................................... 68

5.11.3 “Real Life” Example 1 ...................................................................................................................... 68

5.11.4 “Real Life” Example 2 ...................................................................................................................... 69

5.12 SOLUTIONS ........................................................................................................................................ 69

CHAPTER 6: OPERATIONAL AMPLIFIERS .......................................................................... 70


6.2 INTRODUCTION ................................................................................................................................. 70

6.3 WHAT IS AN ELECTRONIC AMPLIFIER ................................................................................................ 70

6.3.1 An Ideal Electronic Amplifier ............................................................................................................... 70

6.3.2 A Real Electronic Amplifier .................................................................................................................. 71

6.4 OPERATIONAL AMPLIFIERS ............................................................................................................... 71

6.4.1 An Ideal Op-Amp: ................................................................................................................................ 72

6.4.2 A Real Op-Amp: ................................................................................................................................... 72

6.5 COMPARATORS ................................................................................................................................. 72

6.6 USING OP-AMPS WITH FEEDBACK ..................................................................................................... 73

6.6.1 Voltage Follower (Buffer Amp) ........................................................................................................... 74

6.6.2 Non-Inverting Voltage Amplifier ......................................................................................................... 74

6.6.3 Inverting Amplifier .............................................................................................................................. 75

6.6.4 “Sample and Hold” Op-Amp Stage ...................................................................................................... 76

6.6.5 Current-to-Voltage (Trans-Resistance) Amplifier ................................................................................ 76

6.7 ANALOGUE SIGNAL PROCESSING USING OP-AMP STAGES ................................................................ 77

6.7.1 Summing Amplifier (Adder or Mixer) .................................................................................................. 78

6.7.2 Difference Amplifier (Subtractor) ....................................................................................................... 78

6.8 SCHMITT TRIGGERS ........................................................................................................................... 80

6.9 CHOOSING COMPONENT VALUES FOR OP-AMP STAGES ................................................................... 82


6.11 FURTHER READING ............................................................................................................................ 83

6.12 PROBLEMS: OPERATIONAL AMPLIFIERS ........................................................................................... 84

6.12.1 Op-Amp Design ............................................................................................................................... 84

6.12.2 Difference Amplifier Proof .............................................................................................................. 84

6.12.3 Summing Amplifier ......................................................................................................................... 84

6.12.4 Signal Conditioning ......................................................................................................................... 85

6.12.5 Schmitt Triggers .............................................................................................................................. 85

6.13 SOLUTIONS ........................................................................................................................................ 86



CHAPTER 7: SIGNAL CONDITIONING - FILTERS ............................................................... 87


7.2 INTRODUCTION ................................................................................................................................. 87

7.3 A REVIEW OF AC AND COMPLEX NOTATION ..................................................................................... 88

7.3.1 Alternating Current at a Glance .......................................................................................................... 88

7.3.2 Complex Notation and Phasors ........................................................................................................... 89

7.3.3 Impendence in AC Circuits .................................................................................................................. 90

7.3.4 Gain and Phase Shift ........................................................................................................................... 90

7.3.5 Bode Diagrams .................................................................................................................................... 91

7.4 PASSIVE FILTERS ................................................................................................................................ 92

7.4.1 Passive Low-Pass Filter ........................................................................................................................ 93

7.4.2 Passive High-Pass Filter ....................................................................................................................... 94

7.5 ACTIVE TIME-DEPENDENT STAGES .................................................................................................... 97

7.5.1 Voltage Integrator ............................................................................................................................... 97

7.5.2 Voltage Differentiator ......................................................................................................................... 98

7.5.3 Active High-Pass Filter ....................................................................................................................... 101

7.5.4 Active Low-Pass Filter ....................................................................................................................... 103

7.5.5 Stages in Cascade .............................................................................................................................. 105

7.6 REVIEW OF KEY CONCEPTS .............................................................................................................. 106

7.7 FURTHER READING .......................................................................................................................... 106

7.8 PROBLEMS: SIGNAL CONDITIONING ................................................................................................ 107

7.8.1 Passive High-Pass Filter Proof ........................................................................................................... 107

7.8.2 Active Low-Pass Filter proof .............................................................................................................. 107

7.8.3 Voltage Integrator ............................................................................................................................. 107

7.8.4 “Real Life” Example 1 ........................................................................................................................ 107




7.9 SOLUTIONS ...................................................................................................................................... 109

UNIT 1 – DC Networks and Components ME2-HMTX 2012-2013


Chapter 1: Electrical Circuits

1.1 LEARNING OUTCOMES The aim of this UNIT is to provide a review of the basic concepts of DC electrical and electronic

circuits, which were covered extensively in the ME1-HMTX course. As such, most of the material

included in this handout should be revision. By the end of Chapter 1, which is the first of the two

chapters in UNIT 1, you should be able to:

1. Understand the difference between electrical and electronic circuits

2. Define resistance, capacitance and inductance

3. Apply Kirchhoff’s Voltage and Current laws to passive and active circuits

4. Understand the difference between ideal and real voltage and current sources

5. Be able to describe and apply the concept of “potential divider”

6. Be familiar with the concept of volt-amp characteristics to describe the inputs and outputs

of electronic stages

7. Be able to establish the operating point of a source/load network

8. Apply current superposition for networks with more than one source

1.2 INTRODUCTION The ME1-HMTX course introduced “network analysis” to determine voltages and currents in a DC

network. This UNIT revisits particular aspects of DC network analysis, which are needed to deal with

transducers, actuators and transistors (these are topics of subsequent UNITS). The aim is to describe

electronic stages simply in terms of the characteristics of their input and output ports, without the

need to worry about their internal workings (i.e. a “black box” approach). Specifically, this Chapter

focuses on DC networks composed of linear elements (e.g. resistors, capacitors and inductors),

which are elements that exhibit a linear relationship between current and voltage. Such networks,

also known as electrical circuits/networks, are associated with the flow of charge through a metallic

medium (e.g. copper). This is in contrast with electronic circuits/networks, which are composed of

non linear elements (e.g. diodes and transistors) and are associated with the flow of charge through

non-metal conductors (e.g. silicon, which is a semiconductor material). Electronic circuits will be

introduced in Chapter 2 with diodes and then again in Chapter 4, where transistors are covered.

A basic understanding of both electrical and electronic DC circuits is needed to grasp most of the

concepts described early in this course. AC network analysis follows similar concepts, but has the

added complexity of frequency dependency. AC, which has been covered extensively in ME1-HTMX,

will be revisited when filters are introduced in the Spring Term (UNIT 3 – Signal Conditioning).

1.3 DEVICE CHARACTERISTICS The easiest way of characterising the behaviour of a device using a “black-box” approach is to

describe the relationship between the potential difference (or voltage) across its terminals and the

current flowing through it. These are also known as “volt-amp” characteristics of a device.

1.3.1 Ports, Sources & Loads

In general, a port is a point at which energy enters or leaves a device. The simplest electrical device

has one port with two terminals. Subscript notation:



All voltages are relative to a reference value, known as the “0” voltage or ground (“ ”)

Unless otherwise stated, the voltage at a point “a” (i.e. va) actually means the potential

difference of a relative to ground

vab defines the potential difference of point a relative to b (i.e. va – vb)

Current, by convention, always travels in the direction of decreasing voltage potential

(i.e. from high potential to lower potential)

ia defines the current flowing into terminal a

An arrow between two points represents the potential of “the arrow head” relative to the

tail (i.e. vx = va – vb)

a

b

Vx = Vab

The rate of external work (i.e. Power) done by (or of heat dissipated from) a port with terminals a

and b is always:

P = vab x ia

1.3.2 “Volt-Amp” Characteristic of a One-Port Electrical Device

To record the volt-amp characteristics of a device (i.e. the relationship between current and voltage

at any given time), the current through the device needs to be rerecorded with respect to the

voltage applied across its terminals. A suitable circuit arrangement for this task is illustrated in Figure

1.1 (note that the ammeter and voltmeter here are ideal i.e. no current is flowing through the

voltmeter and no voltage is dropped across the ammeter).

Figure 1.1 – Typical “volt-amp” characteristics of a one-port device “x”.

A port is completely defined by the volt-amp characteristics relating v to i and, for resistors, this is

independent of time. This determines whether energy enters through, is stored in, or is dissipated

by the device.

Any volt-amp characteristic behaviour, however complicated, can be simulated using a network

constructed from one-port ‘ideal’ devices: sources, resistors, inductors and capacitors.

1.3.3 Ideal Sources

An electrical source is a device which supplies energy to a circuit. An ideal voltage source maintains

“v = E = const.” potential difference (i.e. voltage) between its two terminals for any current

requirement imposed by the network i.e. v is constant for any i.

VVariablevoltagesource

Characteristic



Figure 1.2 – Volt-amp characteristics of an ideal voltage source.

An ideal current source maintains a current “i = I = const.” whatever voltage rise it must develop in

order to do so (i.e. the current produced by the source is constant for any load attached to its

terminals).

Figure 1.3 – Volt-amp characteristics of an ideal current source.

Note: As the load resistance is increased, the voltage drop across the load terminals will increase

without limit (i.e. as R → ∞, VR → ∞).

It is also important to note that a voltage source can be treated as a short circuit if switched off,

while a current source will act as an open circuit.

1.3.4 Basic Electrical Elements

1.3.4.1 Resistor

Resistive devices oppose current flow: voltage always drops in the direction of current flow (this is

equivalent to the statement that current always flows from positive to negative potential). Due to

Ohm’s Law, a linear resistor has the characteristic VR = iR, the approximate hydraulic analogy of

which is a viscous restriction: a simple flow control valve (note that, if no current flows through R, VR

is zero).

Figure 1.4 – Volt-amp characteristics of a resistor element.

Resistors are colour-coded to identify the resistive value (Ohms, ) and tolerance band (i.e. the

accuracy of manufacture) of a particular element.

i

v

I I

Voltage

does not

depend on

current

R

Characteristic

+

E E

i

v



Figure 1.5 - Resistor colour bands.

Table 1 – Resitor colour reference values and computation methodology.

10 tolerance (%)CR ab

1.3.4.2 Reactive Elements

Reactive elements, as considered in this course, are inductors and capacitors. Inductors and

capacitors can store and release energy over time; thus, the voltage may rise or fall in the direction

of current flow, depending on what has happened previously.

A capacitor, C, opposes the forward movement of charge.

Figure 1.6 – Basic symbolic representation of a capacitor (left) and hydraulic equivalent (right).

The hydraulic analogy is a piston driving a spring of compliance C (stiffness 1/C). For equilibrium,

Equation 1 – Relationship between current and voltage in a capacitor.

This means that the current through a capacitor is proportional to the rate of change of the voltage

across its terminals with respect to time. Capacitors are often described as “storing” charge. In fact,

for any charge which enters one terminal, an equal charge leaves the other — but in doing so it

stores a voltage difference between the two. Thus a capacitor is a reservoir of potential energy (it

can be thought of as a small rechargeable battery), which can be “charged” with a constant current

a b c tolerance

Colour Value Colour Value

Black 0 Gold ±5%

Brown 1 Silver ±10%

Red 2 Nothing ±20%

Orange 3

Yellow 4

Green 5

Blue 6

Violet 7

Gray 8

White 9

Tolerance banda,b,c bands

Stiffness 1/C

C

Circuit symbol: Hydraulic analogue: unit

area

v

i q.

p



through its terminals, or discharged by connecting its two terminals with or without a load between

the two (if there is no load and the capacitor and circuit are ideal, shorting the terminals of a

charged capacitor will produce an infinite current through its terminals). Under DC conditions (i.e.

with a constant input voltage to the circuit), a capacitor acts as an open circuit: no current can flow

through it.

An inductor, L, opposes a change in current, i.e. a change in the flow rate of charge.

Figure 1.7 - symbolic representation of an inductor (left) and hydraulic equivalent (right).

Equation 2 - Relationship between current and voltage in an inductor.

This means that the voltage across an inductor is proportional to the rate of change of the current

flowing through its terminals with respect to time. An inductor “stores current” using a magnetic

field: it is a reservoir of “kinetic” energy. Under DC conditions an inductor acts as a short circuit:

there is no voltage drop across its terminals.

1.4 PARALLEL AND SERIES RESISTOR NETWORKS Resistors placed in series can simply be added up e.g. the total resistance of the resistor chain

below is:

250 750 200A B ∑

The equivalent resistor of a parallel network is the inverse of the sum of the entire parallel resistor

inversed e.g. the total resistance of the resistor chain below is:

[

]

[

]

250 750 500

(∑

)

L

Circuit symbol: Hydraulic analogue: unit

area

v

i

Mass

L

q.

p



Warm-Up Exercise

1.1 Calculate the total resistance, as measured between a and b, of the following resistor

network:

100 100 300

50 200

300

a

b

Figure 1.8 - Resistor network for warm-up exercise.

1.5 CIRCUITS AND NETWORKS

1.5.1 Potential Divider

The resistance of a sensor is usually detected using some form of a potential divider.

Figure 1.9 - Example of potential divider setup.

If and only if the output current is zero (i.e. no current is flowing out between R1 and R2):

In other words, the output voltage of a potential divider is proportional to the resistance across

which it appears. Potential dividers, among others, are also widely used to reduce the voltage from

some source to a precisely defined lower level.



1.5.2 Superposition

This technique is useful for calculating the total current through a network in which there is more

than one source. The resulting current at any point is the sum of the currents driven by each source

separately, so the procedure for analysis is as follows:

1. Label every independent current (i1, i2, i3…) and label every source (A, B, C…).

2. For each source in turn, analyse the network to determine currents (i1A, i2A, i3A…, i1B, i2B, i3B…)

(remember that switched off voltage sources act as short circuits and current sources can be

thought of as open circuits).

3. Sum the contribution of each source to each current (i1 = i1A + i1B +… etc.).

For instance, the following network with three voltage sources can be analysed as follows:

DCDC

(current)

DC R1

R2 R3

Figure 1.10 - Simple network with two voltage sources.

Due to the principle of superposition, the circuit can be treated as three circuits, as per below:

1.

DC R1

R2 R3

2.

DC

R1

R2 R3

3.

DC

(current)

R1

R2 R3

The current passing through each resistor in each of the three circuits above can be computed using

Ohm’s law, then summed together to produce the overall circuit characteristics.

1.5.3 Kirchhoff’s Laws

Systems of linear components can be simplified according to a set of two very powerful rules, known

as Kirchhoff’s Laws. Hydraulic circuits follow similar rules, with pressure and flow rate substituted

for voltage (potential) and current. Kirchhoff’s voltage and current laws are summarised briefly in

the following sections.

1.5.3.1 Kirchhoff’s Voltage Law (KVL)

The algebraic sum of potential differences around a closed loop at any instant is zero. KVL is

analogous to a mechanical force-equilibrium condition.



Note that, since “vx” is defined as the voltage difference between the “arrow head” and “arrow tail”

(Section 1.3.1), v1, v2 and v3 are negative, while v4 is a positive voltage.

1.5.3.2 Kirchhoff’s Current Law (KCL)

The algebraic sum of currents flowing into a node at any instant is zero. KCL is analogous to a

mechanical compatibility (or mass conservation) condition.

Not that, since current always flows from higher potential (or voltage) to lower potential, i1 and i3

are positive, while i2 is a negative current.

1.6 EQUIVALENT NETWORKS Ports of real devices always have more complicated characteristics than those of ideal sources,

resistors, capacitors and inductors. Such ports can be represented by equivalent networks,

assembled from ideal devices, with equivalent characteristics. Good examples of equivalent

networks are those used to describe real (as opposed to ideal) energy sources, which are briefly

summarised below.

1.6.1 Real Voltage and Current Sources

A real voltage source (e.g. a battery) cannot provide an infinite amount of current and a real current

source cannot develop an infinite voltage. In other words, the nominal voltage of a voltage source is

only available if the load connected to its terminals draws negligible current. Similarly, a current

source can only provided the nominal current if the load connected to its terminals has negligible

resistance. Thus, batteries are usually described in terms of a “reference voltage”, which is the

voltage which the battery can provide for a given current (i.e. the “reference current”).

Thus, real sources can be represented by a combination of an ideal source and a linear resistance.

Real voltage and current sources are known as “Thévenin source” and “Norton source” respectively

and can be found in Figure 1.11. These two circuits are given specific names because, apart from

serving as representation of real energy sources, any one-port (two-terminal) network containing

only voltage sources, current sources and resistors can be modelled by an equivalent Thévenin or

Norton source i.e. such networks can be analysed as “black boxes”, where the output voltage across

and current through its terminals can be measured by solving the equivalent Thévenin or Norton

circuit.

EV2

V1

V3

V4

+

E i1

i2i3

+



As such, these two networks are equivalent (i.e. they have identical volt-amp characteristics) for

corresponding values of current, voltage and resistance (i.e. if RTh = RN and ETh = IN RN).

Figure 1.11 – Real voltage and current sources, also known as “Thévenin” and “Norton” sources.

The resistance, also known as internal resistance Rs, of a voltage source is very small but finite; thus,

a small voltage will drop across the terminals of the source itself. Similarly, for a current source, Rs is

very large but finite and, as such, a small current is drawn by the source itself.

Example

Transform the circuit below, which captures the functionality of a potentiometer, into a

representative Thévenin source, the wiper (arrow) being ¾ of the way from the bottom position.

10 V 10kVout

Figure 1.12 – Potentiometer setup, where the output voltage can be varied by moving the wiper up and down.

Given a known wiper position (three quarters of the way up from the bottom), the circuit to be

modelled with a Thévenin source is illustrated below.

10 V

2.5k

Vout

7.5k

Figure 1.13 – Detailed circuit schematic to be analysed.

IN

Eth



First, compute the Thévenin voltage Eth by measuring Vout in the circuit represented in Figure 1.13.

Using potential divider:

Next, compute the Thévenin resistance RTh by switching off the original source, then computing the

equivalent resistance of the network (remember, a voltage source acts as a closed circuit when

switched off).

2.5kVout

7.5k

Vout

7.5k

2.5k

1.875k Vout

Figure 1.14 – Steps needed to reduce the original network to a single Thévenin resistance.

Finally, the Thévenin model for the circuit in this example can be found:

7.5 V

1.875kVout

V-

Figure 1.15 - Thévenin equivalent model of the circuit in Figure 1.13.

1.7 OPERATING POINT OF NETWORKS: GRAPHICAL SOLUTION By reducing any circuit to a source driving a load, the operating point of the load (i.e. the voltage

and corresponding current drawn by the load) can be found by solving the “source” and “load”

equations simultaneously.

For non-linear sources and loads, this is best done graphically — a powerful concept known

generally as an operating diagram, and in Electronics as a load line diagram.

Rs

E

+

Source:

X

Load:

Figure 1.16 – simple “black box” example of a source driving a load.



a)

Source characteristics

b)

Load characteristics

c)

Operating Point

Figure 1.17 - Graphical solution of the operating point of a simple source-load circuit. First, draw the volt-amp characteristics of the source (a); then, draw the volt-amp characteristics of the load (b); finally, superimpose the two:

the operating point is that at which the two volt-amp lines intersect (c).

1.8 REVIEW OF KEY CONCEPTS This Chapter reviewed the theory of linear elements and electrical DC circuits. The following

concepts, which were covered in the text, deserve particular attention:

1. Notation and conventions

2. “Volt-amp” characteristics of any one port, two-terminals device

3. Series and parallel resistor networks: ∑ , (∑

)

4. Ohm’s law:

5. The governing law of capacitors:

6. The governing law of inductors:

7. Potential divider:

8. KVL: the algebraic sum of potential differences around a closed loop at any instant is zero

9. KCL: the algebraic sum of currents flowing into a node at any instant is zero

10. “Thévenin” and “Norton” sources and equivalent networks

11. The “operating point” of networks

1.9 FURTHER READING 1. Introduction to MECHATRONICS and Measurement Systems, David Alciatore and Michael

Histand, McGraw Hill, 1999, ISBN: 0072963050, Chapter 2: “Electric Circuits and

Components”.

2. Hughes Electrical and Electronic Technology, Edward Hughes, Prentice Hall, 2008, ISBN:

0132060116, Chapters 3 and 4, pp. 30-79.



1.10 PROBLEMS: LINEAR NETWORKS AND COMPONENTS

1.10.1 Resistor Conventions

Calculate the resistor nominal value and upper and lower limits for each of the following

parameters:

a b c tol R (Ω) Rmin (Ω) Rmax (Ω)

Red Orange Red Nothing

Yellow Black Orange Gold

Green Blue Red Silver

1.10.2 Resistor Networks 1

Calculate the total resistance as measured between points a and b in the network below:

100k 100k 100k

100k 100k

100k100k

100k

a

b


By applying KCL to the circuit below, show that the total resistance of the circuit can be calculated

as:

Vs R1 R2




Calculate the voltage vout measured in the circuit below (remember that, unless otherwise stated,

vout means the voltage relative to ground). Also calculate the current flowing through the resistor R1.

10V

20V

R2=2k

R4=4k

R5=5k R6=6kR1=1k

R3=3k

Vout


In the circuit below, determine the current through and voltage drop across each resistor as well as

the power dissipated by each.

1 A

(current)1V

10

100


Calculate the total resistance as measured between points a and b of the network below. You will

have to apply Kirchhoff’s laws and set-up a set of linear equations to solve this. It is also

advantageous to apply a virtual voltage source across ab. This question is a bonus question for

strong students.



100k 100k 100k

100k 100k

100k100k

100k

a

b

1.10.7 “Real-Life” Problem 1

A car battery has an internal resistance of 0.05 Ω and a nominal voltage of 14.4 V. Calculate the drop

in voltage across the battery’s internal resistance when the electrical circuit is used to:

a. Power the interior car lights, which can be assumed to be simply resistive and which

dissipate 10W at 12 V.

b. In addition to the interior lights, power the headlights, which dissipate 150W at 12 V.

c. The engine is started by using a permanent magnet DC motor with 0.1 Ω armature

resistance. Calculate the drop in voltage when all lights (interior and headlights) are on at

the point in time when the engine is ignited.

d. What should you do to extend battery life?

1.10.8 “Real-Life” Problem 2

Unfortunately your mobile phone charger has broken. From a 1 kΩ potentiometer and a 9V battery

you need to build a power source in order to charge it. The phone has an internal resistance of 100 Ω

and requires a 5 V input.

a. Draw the circuit diagram of the potentiometer and battery setup, then add a resistor

representing the load (e.g. the internal resistance of the phone)

b. Calculate the Thévenin equivalent of your power supply for 5 V no load voltage and draw the

load and source characteristics. What is the operating voltage?

c. At what position does the potentiometer need to be in order to guarantee operation at 5V

under full load?



1.11 SOLUTIONS Warm-Up Exercises:

1.1: Rab=37.5 Ω.

Problems:

1.10.1:

a B c tol R Rmin Rmax

Red Orange Red Nothing 2,300 Ω 1840 Ω 2,760 Ω

Yellow Black Orange Gold 40,000 Ω 38000 Ω 42,000 Ω

Green Blue Red Silver 5,600 Ω 5040 Ω 6,160 Ω

1.10.2: 83.3 kΩ; 1.10.3: Q.E.D.; 1.10.4: 14.23V, 10mA; 1.10.5: iR1: -0.9A, VR2: 0.1A, VR1: -9V, VR2: 10V,

PR1: 8.1W, PR2: 1W; 1.10.6: 70kΩ; 1.10.7: (a) 0.05V, (b) 0.76V, (c) 5.14V, (d) start engine without

lights on.

1.10.8: (a)

9V 1k

100

, (b) ETh = 5V, RTh = 246.9Ω, Operating Voltage = 1.4V, (c) 927.8Ω.



Chapter 2: Electronic Circuits

2.1 LEARNING OUTCOMES Following the revision of simple DC electrical networks and linear components (Chapter 1), these

notes look at electronic networks i.e. networks composed of electric components which exhibit

nonlinear behaviour. Special focus will be given to diodes as the first semiconductor component

introduced in the course. Transistors will be introduced in Chapter 4. Following this section, you

should be able to:

1. Describe the basics of semiconductor theory

2. Analyse circuits that contain diodes

3. Design constant voltage supplies using Zener diodes

4. Apply the principle of superposition to the analysis of electronic networks

2.2 INTRODUCTION Electronic networks contain components that behave in a manner which changes nonlinearly with

respect to the current flowing through and voltage dropping across them. This section will look at

basic semiconductor devices known as diodes and Zener diodes and how they can be used in

practice to regulate voltage supplies. First however, we must investigate the basic concepts behind

semiconductor Physics.

2.3 BASIC SEMICONDUCTOR PHYSICS Extract from “Introduction to MECHATRONICS and Measurement Systems”, by David Alciatore:

“Metals have a large number of weakly bound electrons in what is called their conduction band.

When an electric field is applied to a metal, the electrons migrate freely, thus producing a current

through the metal. Because of the ease by which large currents can flow in metals, they are called

conductors (Figure 2.1). In contrast, other materials have atoms with valence electrons (i.e. the

electrons in the conduction band) which are tightly bound, and when an electric field is applied, the

electrons do not move easily. These materials are called insulators (e.g. rubber and wood). A very

useful class of materials, however, have properties somewhere between conductors and insulators.

They are called Semiconductors. Semiconductors such as Silicon and Germanium have current-

carrying characteristics that depend on temperature or the amount of light falling on them. As

illustrated in Figure 2.2, when a voltage is applied to a semiconductor, some of the valence electrons

easily jump to the conductance band and then move in the electric field to produce a current,

although smaller than that which would be produced in a conductor.

In a semiconductor crystal, a valence electron can jump to the conduction band, and its absence in

the valence band is called a hole. A valence electron from a nearby atom can move to the hole,

leaving another hole in its former place. This chain of events can continue, resulting in a current that

can be thought of as “the movement of holes in one direction” or “the movement of electrons in the

other” [...].The properties of pure semiconductor crystals can be significantly changed by inserting

small quantities of elements from group III or V of the periodic table into the crystal lattice of the

semiconductor. These elements, known as dopants, can be diffused or implanted into the

semiconductor [...].



6P6N

-e

Figure 2.1 - Metal structure with free valence electrons.

6P6N

-e

6P6N

-e

6P6N

-e

6P6N

-e

No excitation

6P6N

-e

6P6N

-e

6P6N

-e

6P6N

-e

Excitation applied

Figure 2.2 - Semiconductor material (without and with excitation).

Properties really get interesting when different amounts and different types of dopants are added to

the semiconductor. Consider what happens if dopants are embedded in the crystal lattice of silicon.

Silicon has four valence electrons that form symmetrical electron bonds in the crystal lattice.

However, if e.g. Arsenic or Phosphorus from group V is added to the crystal lattice, one of the five

valence electrons in each dopant atom remains freer to move around. In this case, the dopant is

called a donor element because it enhances the electron conductivity of the semiconductor. The

resulting semiconductor is called n-type silicon due to the electrons available in the crystal lattice as

charge carriers. Conversely, if the silicon is doped with Boron or Gallium from group III, holes form

due to missing electrons in the lattice where the so-called acceptor dopant atoms have replaced

silicon atoms. This is because the dopant atom only has three valence electrons. A hole can jump

from atom to atom, effectively producing a positive current. What really happens is that the

electrons move to occupy the holes, and this looks like holes moving. The resulting semiconductor is

called p-type silicon due to the holes, which are effectively positive charge carriers. In summary, the

purpose for doping a semiconductor such as silicon is to elevate and control the number of charge

carriers in the semiconductor. In an n-type semiconductor, the charge carriers are electrons, and in a

p-type semiconductor, they are holes.”



2.3.1 “pn” Junctions

The combination of “n-type” and “p-type” semiconductor materials forms a pn junction, which

offers very interesting properties. Here, the terminal connected to the p-type semiconductor is

known as the anode, while the other terminal is referred to as the cathode.

If a pn junction is not connected to a voltage source, a small depletion region forms where the holes

in the p section are filled by spare electrons in the n section (Figure 2.3, top). If a voltage source is

connected as in Figure 2.4a, then a current starts flowing, as electrons are able to move into the

depleted areas on the cathode side (remember that, while current flows from positive to negative by

convention, the flow of electrons, which are negatively charged, is in the opposite direction). This is

known as a forward biased configuration. If, on the other hand, the connections of the voltage

source are reversed (Figure 2.4b), the pn junction is known to be reversed biased and the depletion

region extends, preventing current flow. This configuration is also known as “cut-off”, since no

current is allowed to flow between the two terminals of the pn junction.

P N

holes electrons

cathodeanode

P N

holes electrons

cathodeanode

depletion layer

Figure 2.3 - Forming of depletion region in pn-junction.

a)

P N

Forward bias (conduction)

cathodeanode

+V

b) reversed bias (cutoff)

cathodeanode

+V

P N

depletion layer (expands)

Figure 2.4 – A pn junction in both the forward biased (a) and reversed biased (b) configurations.

From a practical point of view, we only need to be able to describe this behaviour in terms of volt-

amp characteristics, as will be expanded in the following sections.



2.4 IDEAL DIODES Many important electronic devices are time-independent but non-linear. The simplest and most

common example is a diode. A perfect diode has infinite resistance in one direction and zero

resistance in the other. Its ends must be labelled (cathode, k and anode, a) and the direction of

current and increasing voltage specified before plotting its characteristic volt-amp behaviour.

Remember that vak means voltage of a above k and ia means current into a.

a) b)

Figure 2.5 - The symbolic representation of a diode (a) and ideal, approximate and real diode volt-amp curves (b).

Figure 2.6 - Hydraulic analogy: a non-return valve.

2.5 EQUIVALENT NON-LINEAR DEVICES Non-linear devices (e.g. real diodes, which drop a finite voltage, typically 0.6V, before conducting in

the forward biased configuration) can always be represented by combinations of linear devices (e.g.

resistors, perfect sources etc.) and perfect diodes. Remember that it is always easier to begin by

considering zero-current and zero-voltage conditions and that the characteristics of linear networks

must consist of straight-line sections.

V

I

Idealdiode

0.6-0.7

breakdownregion

reversebiasregion

forwardbiasregion

Approximationof real diode

failure

failure

mA

µA

V

I

Idealzenerdiode

0.6-0.7

real zenerdiode

ZenerVoltage



Warm-Up Exercises: non-linear series networks

2.1 Sketch the volt-amp characteristics, ia vs. vab, for each of the ‘equivalent networks’ shown

below, which contain perfect diodes.

(a)

a

6 V+

3

b (b)

a

b

0.6 V+

3

(c)

a

+

3

0.6 V

b (d)

6 V

a+

3

b

To deal with parallel networks, remember that each branch containing a perfect diode will pass no

current if reverse biased and will dissipate no voltage if forward biases and that the characteristics

of parallel branches add.

Each branch can be analysed by placing the “black box” containing the branch itself across a-b in the

circuit illustrated in Figure 2.7. Then, by applying different voltage settings to the voltage source “E”

and measuring the current and voltage properties of the branch through the ammeter and

voltmeter, connected in series and in parallel with the branch respectively, the volt-amp

characteristic can be found.



Figure 2.7 - Measurement setup to analyse a branch.

Example 1

Find the ia vs.vab characteristics of the network shown. What will be the power dissipated across ab

at an applied voltage of (a) 3V and (b) 6V. How is the power distributed between the left- and right-

hand branches in each case?

2

a

b

+

3

6 V

vab Left branch Right branch total

3V

I =

P =

I =

P =

I =

P =

6V

I =

P =

I =

P =

I =

P =

2.6 ZENER DIODE A Zener diode is a special kind of pn junction where the usual characteristics of a diode are only

applicable to a limited range of voltages. In the forward direction, a Zener diode behaves like a

normal diode, conducting at about 0.6V. In the reverse direction, however, it “breaks down” at a

fixed voltage, known as the Zener voltage vZ, after which it will start to conduct again. Ideal Zener

diodes can “pass” infinite currents, while real diodes, whether normal or Zener, will be damaged if a

current greater than the rated current for the device is allowed to flow though its terminals.



a) b)

Figure 2.8 – Symbolic representation of a Zener diode (a) and ideal, approximate and real Zener volt-amp curves (b).

Zener diodes are specified by:

vz (e.g. Zener diode ‘C6V3’ has a 6.3V Zener voltage) and

The maximum power which they can dissipate ( max Z k,maxP V i ).

2.6.1 Zener Diode to Protect Input Ports

Electronic measuring equipment (e.g. a data acquisition card) often has very sensitive input ports

which have limited voltage ranges (Figure 2.9). A typical application of a Zener diode is to protect

such equipment from experiencing too high a voltage across its terminals. This is done by simply

placing a Zener diode across the input terminals, as illustrated in Figure 2.10.

Figure 2.9 - Unprotected input signal.

Figure 2.9 shows an unprotected data acquisition system, which has a maximum acceptable input

voltage of 5V. If the range of input signals cannot be bound to within the range of the measuring

equipment, permanent hardware damage is a real possibility (and hopefully you will not find this out

for yourselves during the laboratory experiments!). Figure 2.10, on the other hand, shows a circuit

protected by a Zener diode with vz = 5V. As soon as the signal voltage exceeds 5V, the difference is

“clipped” by Zener (i.e. shorted back to ground), thus protecting the input to the equipment.

vz

i

vak

aiai

vak

ki

vka

Vak

Ia

Idealzenerdiode

0.6-0.7

real zenerdiode

Zener VoltageVz



Note however that this solution can give rise to very high loads on the source as it short-circuits the

signal voltage to the Zener voltage vz. The current entering the source will be determined by the

internal resistance of the source, which, if low, could be dangerously high.

Figure 2.10 - Protected data acquisition system.

2.6.2 Zener Voltage Regulator Stage

The previous section has shown how a Zener diode can be used to clip the output of a voltage source

down to the Zener voltage. A similar arrangement can be used to build a constant power supply

where the output voltage does not vary with load i.e. a real voltage supply that, within a predefined

range, will act as an ideal voltage supply. If, as illustrated in Figure 2.11, a voltage source is clipped

using a Zener diode, then, applying KVL (remember sign conventions):

( )

Now, applying KCL at the top node:

( )

( )

Note that a current iz is required from the supply even if no load is taken from the system (i.e. if io is

zero), which is not very efficient! The aim of a good regulator design is to minimize iz whilst ensuring

the required current range.

Vs+

Rs

Vz

Vo

io

iz

Figure 2.11- Zener clipped power supply.

The maximal possible current that the power supply can provide without a reduction in output

voltage Vo is the point at which iz = 0. Hence Io max is:



( )

If this current is exceeded, the Zener diode will no longer be active and the voltage Vo will fall below

the Zener voltage Vz.

The sequence below shows a regulated system with an external load. The thick lines indicate the

current path.

Vs+

Rs

Vz

Vo

io

iz

Figure 2.12 - No load (only internal currents) - point “A” in Figure 2.16.

Vs+

Rs

Vz

Vo

io

iz

Rl

Figure 2.13 - Part load (internal current and load current) - point “B” in Figure 2.16.

Vs+

Rs

Vz

Vo

io

iz

Rl

Figure 2.14 - Maximum current (i.e. smallest Rl) at required voltage (no internal currents iz) - point “C” in Figure 2.16.

Vs+

Rs

Vz

Vo

io

iz

Rl

Figure 2.15 - Short circuit - point “D” in Figure 2.16.



Figure 2.16 - Zener voltage regulator characteristic.

Note that, between points “C” and “D, a Zener regulated supply is identical to a normal supply with

the same source voltage and internal resistance i.e. the voltage can be maintained at a constant

level only for currents in the range of: ( )

.

Example 2

A Zener diode regulator stage is used to provide an output voltage of 9V using a C9V, 400mW diode.

The supply voltage varies between 12V and 13V. Calculate the minimum required series resistance.

Using this value, what is the maximum value of output current which the regulated supply can

deliver before the output voltage drops below 9V?

Maximum current through the Zener diode occurs when no output current (io) is drawn. This means:

The worst case occurs if Vs = 13V (as a higher output voltage will generate a higher current):

( )

( )

This time, the lower voltage Vs = 12V is selected, since a smaller supply will produce a smaller

maximum current. Then, the maximum available load current occurs when no current flows through

the Zener diode i.e. iz = 0:

Warm-Up Exercises: Zener Diodes and Series Resistors

2.2 What is the maximum allowable current for a C6V3, 1W Zener diode?



2.3 What is the maximum power dissipation in a C9V Zener diode connected to a constant 15V

supply through a 470Ω series resistor?

2.4 A 400mW C9V Zener diode is connected through a 1kΩ series resistor to a varying DC voltage

supply. What is the maximum permissible supply voltage?

2.5 A Zener diode regulator must supply up to 100mA at a constant 9V, from a DC supply which

may vary between 12 and 16V. What is the minimum possible series resistor value?

2.7 REVIEW OF KEY CONCEPTS This Chapter reviewed the theory of non-linear elements (e.g. semiconductors) and electronic DC

circuits. The following concepts, which were covered in the text, deserve particular attention:

1. Basic theory of semiconductor materials: doping, “p” and “n” type silicon, pn junctions

2. The definition and characteristics of an ideal and real diode

3. Solving simple non-linear networks using equivalent circuits

4. The definition and characteristics of a Zener diode, including use as a voltage regulator


Histand, McGraw Hill, 1999, ISBN: 0072963050, Chapter 2: “Semiconductor Electronics”.


0132060116, Chapters 3 and 4, Chapter 20-21: “Semiconductor Materials” and “Rectifiers”.

2.9 PROBLEMS: NON-LINEAR NETWORKS AND COMPONENTS

2.9.1 Non Linear Series Networks

Sketch the volt-amp characteristic, ia vs. vab, for each of the ‘equivalent networks’ shown below,

which contain PERFECT diodes.

(a) (b)

2.9.2 Diode Characteristics

The diagram below shows a circuit for testing components in electronic teaching kits. The signal

generator provides a -10V to 10V square wave.

0 V

20 V

meter

20 V

pp

+

V Component

under testvo

3Ω

Rs

3Ω 3Ω



Sketch the output voltage waveform, vo, showing maximum and minimum values, and indicate the

approximate DC voltmeter reading, when the component under test is:

a. An open circuit;

b. A real silicon diode with the polarity shown in the diagram above;

c. A real C6V3 Zener diode, with the polarity shown in the diagram above; and

d. An internally short-circuited (i.e. blown) diode.

2.9.3 Zener Diodes

Zener diodes with Vz = 5.0V and Pmax = 1W are used to protect the input of a Data Acquisition Card

with a maximum input range of +/- 5V. The DA card is linked to a sensor with a maximum output of

15V and a rated power of 1W. In order to protect the sensor, a resistor is placed in series with it.

Draw a circuit diagram of the system and select a suitable resistor value. Assume that no current can

flow into the DA card.

2.9.4 Zener Diodes in Series

A Zener diode voltage regulator is required to supply a constant 6V from a maximum voltage supply

of 9V DC. BZX 79 Zener diodes, rated at 500mW, are available for C2V4 and C3V6 but not for C6V0.

Calculate the required series resistor and the maximum regulated output current for this series

combination of Zener diodes. Note that Zener diodes can be used in series to achieve non-standard

voltages (i.e. their Zener voltages can be added, while the same current flows through each one).

2.9.5 Non Linear Network 1

What is the current that the source has to provide in the below circuit:

100k 100k 100k

100k 100k

100k100k

100k

a

b

5V DC

2.9.6 Non Linear Network 2

Calculate the power dissipated in all components of the circuit below.



R4 1k R2 2k

10V DC 10V DC

R1 2k R3 1k



2.10 SOLUTIONS Warm-up Exercises

2.1: (a) , (b) , (c) , (d) .

2.2: 159mA; 2.3: 115mW; 2.4: 53.4V; 2.5: 30Ω.

Problems

2.9.1a: ; 2.9.1b:

2.9.2: (a) -10V to 10V, VDC = 0V, (b) -10V to 0.6V, VDC = -4.7V, (c) -6.3V to 0.6V, VDC = -2.85V, (d) 0V

constant, VDC = 0V; 2.9.3: 150Ω; 2.9.4: 21.6Ω, 139mA; 2.9.5: 67µA; 2.9.6: PR1 = 72mW, PR2 = 0W, PR3 =

64mW, PR4 = 4mW.

UNIT 2 – Data Acquisition ME2-HMTX 2012-2013


Chapter 3: Analogue to Digital Conversion & Back

3.1 LEARNING OUTCOMES Many modern control systems (e.g. the BASIC Atom microcontroller used in the labs) employ digital

electronics, which means analogue signals, like voltages and currents, first need to be converted into

a digital format (i.e. 0s and 1s) before they can be processed. This section looks at some methods

and technologies which can be used to convert data from the analogue to the digital domain (and

vice versa). By the end of this unit you should be able to:

1. Be familiar with the basic terminology used in the analogue-to-digital conversion world, such as

sampling rate, resolution, aliasing, sensitivity, quantization, etc.

2. Understand and describe the concepts of digital data representation

3. Understand and describe the function of an analogue-to-digital converter

4. Understand and describe the function of a digital-to-analogue converter

5. Be able to design a simple digital-to-analogue converter known as the R2R ladder

3.2 INTRODUCTION Digital technology is ever more present in our daily lives in applications ranging from MP3 players to

digital televisions, mobile phones, cars, etc. Microprocessors and personal computers are also widely

used in mechatronic and engineering measurement systems and their ability to create ever more

accurate digital reconstructions of analogue signals is what drives the current surge in this

technology’s development. More accurate digital reconstruction requires faster processing power

(e.g. sampling points closer together) and larger storage (since more points stored need more

memory), both of which are apparent in the astonishing speed of progress of consumer electronics

today (think of the increasing size of hard drives or the increasing speed of computer processors).

One of the best and most accessible examples where such transition occurred is the evolution from

tapes and LPs to compact discs and digital music. Although some enthusiasts still use analogue

media (e.g. LP, tape), digital sound, whether stored on disk (e.g. a Compact Disk or the Hard Disk of a

computer) or on flash/solid state memory (e.g. the new portable IPods, memory sticks, memory

cards, etc.), is now widely considered as the medium of choice to enjoy and share music. With the

increased popularity of digital systems, it has become more important for engineers to be able

understand and use this technology. As such, this chapter will introduce some basic nomenclature,

as well as the underlying principles governing the conversion of analogue signals into a digital form

and vice versa.

3.3 ANALOGUE TO DIGITAL CONVERSION AT A GLANCE Taking digital music as an example, Figure 3.1 illustrates the lifecycle of a music signal, from the

moment it is created (e.g. musicians playing instruments or singing) to the moment when it is

delivered to the headphones of a digital music player, such as an iPod, or conventional speakers. As

one can see, the digital sound mixer (which is a Data Acquisition or simply “DAQ” System) is at the

heart of the conversion process. On the input side, sounds produced by vibrating cords (e.g. the

vocal cords or the strings of a violin) travel through the air and are captured by a microphone, which

causes small vibration of a thin membrane within. These vibrations are then converted into a

continuous voltage signal by means of suitable electronics.



Figure 3.1 - The lifecycle of sound within the digital world.

The microphone, which converts mechanical vibrations into an analogue voltage signal, is known as

a transducer. The mixer’s main role is to convert this continuous analogue signal into a discrete

digital readout, which can then be processed by a computer and/or stored onto a suitable medium

(e.g. a hard drive).

In order to reproduce the sound captured in this way, the mixer also needs to be able to translate

the digital sound data back into a continuous analogue voltage, which can then be used to vibrate

the thin membrane present inside conventional speakers (which are transducers too). Alternatively,

the sound data, in a digital format, can be transferred to a portable digital player, where the

conversion from digital to analogue takes place to reproduce the sound into the headphones (which,

you might have guessed, are transducers as well). Thus, to summarise, the mixer performs

analogue-to-digital conversion to process and store music in a digital medium, and digital-to-

analogue conversion to reproduce the stored sound through speakers, while the iPod performs

digital-to-analogue conversion to reproduce the stored sound through headphones.

Figure 3.2 shows a general analogue signal with corresponding digital representation, which is

illustrated here by evenly distributed points (connected by a dotted line). The following sections

focus on same basic techniques which can be used to convert between the two.

Voltage

Time

digitized signal

analogue signalsample point

Figure 3.2 - Analogue Signal and Sampled Equivalent.

inupt

transducer

sound

mixer

portable

digital mediaoutput

transducer

sound

"generator"

ANALOGUE TRANSFER

THROUGH AIR MEDIUM

D/A CONVERSION

speakers I II

note: 2 options for

sound reproduction



3.4 ANALOGUE TO DIGITAL CONVERSION As described earlier, the process required to convert an analogue signal into digital is called

analogue-to-digital conversion (ADC). Two main stages are required:

1. Quantization, which is defined as the transformation of a continuous analogue input into a

set of discrete outputs (known as states), which are evenly distributed in time.

2. Coding, which is the assignment of a unique digital code (usually a binary number) to each

output state.

Each of these stages, alongside a description of important parameters which govern this

transformation, is summarised in the following sections.

3.4.1 Quantization

During quantization, the analogue signal is sampled at discrete, evenly distributed points, according

to the sampling rate, which is the rate at which the analogue signal is discretised (note that a low

sampling rate means longer time spans between each of the sampled points).

Figure 3.3 - Quantization process (http://www.dspguide.com/graphics/F_3_1.gif).

http://www.dspguide.com/graphics/F_3_1.gif



“Sample-and-Hold” (Figure 3.3) refers to the process of measuring individual samples of the input

without the need to stop or pause the input signal itself i.e. “Sample and Hold” allows “timed

snapshots” to be taken of the analogue input signal without disruption. This process can easily be

performed with Operation Amplifiers (Chapter 6).

3.4.2 Coding

During the “coding” process, each discretised sample (after quantization) is assigned a unique code.

For instance, in Figure 3.4, a 3-bit device is used to convert an analogue voltage signal, then a binary

3-bit code is used to label each output state.

Figure 3.4 – The coding process for a simple 3-bit converter.

The device in Figure 3.4 has 23 or 8 output states, which are listed in the first column. The output

states are usually numbered consecutively from 0 to (N-1). The corresponding code word for each

output state is listed in the second column. Most A/D converters are 8-, 10-, or 12-bit devices that

resolve 256, 1,024, and 4,096 output states, respectively.

The number of analogue decision points that occur in the process of quantizing is (N-1). In Figure

3.4, decision points occur at 1.25V, 2.50V, ..., and 8.75V. The difference between an instantaneous

analogue sample and its closest decision point is known as quantization error (see Figure 3.4), since,

after conversion, this difference is lost forever i.e. the instantaneous sample value is “rounded

down” to the closest decision point. Clearly, a higher number of output states will reduce

quantization error for a given input range, since decision points will become increasingly closer to

each other. This concept is related to the “resolution” of the device, which is described next.

3.4.3 Resolution

The resolution of a transducer refers to the largest change in the measured value of a given quantity

(e.g. a voltage) which can occur without a detectable change in the output signal. The word “refers”

is used purposely here, since resolution can actually be described in several different ways, which

can be used interchangeably. This is most easily illustrated with the simple 4-bit converter example

in Figure 3.5.



Figure 3.5 – Resolution, range and quantization error for a simple 4-bit A/D converter.

For the case shown, resolution can be simply described as the number of bits in the converter – in

this example 4 (i.e. “4-bit resolution”). Additionally, resolution can be described in units of the

measured quantity, in this case Volts, as the ratio between the range and the number of output

states available in the converter (in this case 16, since the analogue input can vary within one of 24 =

16 intervals without the digital output changing) i.e. resolution = 15 / 16 = 0.93V. Thus, to

summarise, resolution can be defined:

1. As the number of bits in the ADC: 3-bits resolution

2. As a proportion of the converter range: (i.e. the units of the input)

3.4.4 Input Range vs. Converter Range

As described in Figure 3.5, input range is defined as the difference between the maximum and

minimum values of the input for which the sensor signal is meaningful (e.g., for a load cell, 0 to 1kN

or ±1 kN, for a strain gauge, 0-0.05V). The input range is determined by the physical capacity of the

sensor attached to the DAQ system, and, possibly, by saturation: the inability of its electrical

parameter to change any further.

Conversely, converter range is defined as the difference between the maximum and minimum

values of the input which the device can handle (e.g., 0-10 V (or just ‘10 V’); -24-24 V; 0-20mA, etc.).

The converter range is set by the manufacturer of a DAQ system, according to manufacturer

tolerances and the rating of the components within.

Since both these values are “ranges” and some times the words “converter” or “input” can be

omitted, it is important that the context of the sentence is used to identify the correct meaning.

Finally, the ratio, or log(ratio) in dB, of input range and resolution (described in the units of the input

measured) is defined as dynamic range.

maximum value (e.g. 15V)in

pu

t ra

qn

ge

b0 b3b2b1

analogue measurement

(e.g. 5.5V)

corresponding digital value

(e.g. 0 1 0 1)

bit is "low" bit is "high"

resolution is:

4 bits, or

0.93V

minimum value (e.g. 0V)

quantization error

for the example:

5.5V - 4.7V = 0.8V



Example

A 12-bit ADC has a range of 0-10V. Determine its resolution (in volts), and hence the effective dynamic range, in dB, if the ADC is used to encode a signal which varies only from 0.2 to 0.5 V.

3.4.5 Aliasing and the Nyquist Sampling Criterion

To avoid aliasing, the sampling frequency (Figure 3.6) must exceed twice the maximum frequency

present in the signal (i.e. no input frequency should exceed half of the sampling frequency). Note:

1 This criterion does not ensure that the signal is ‘well’ described by the samples. To do so, a much

higher frequency should be used.

2 The criterion refers to sine waves. Other periodic signals, e.g. the triangular wave shown,

contain sine waves of much higher frequencies [Maths (Fourier analysis)], which need to be

dealt with explicitly (i.e. Need to apply Nyquist theorem to the maximum frequency present in

the Fourier series, not the triangular wave frequency).

Figure 3.6 - The formation of Aliasing due to under sampling.

3.4.6 Information Channels

A channel is the medium through which a signal passes from its source to its destination. In the

analogue world, a channel is a physical path along which signals travel. For digital signals there are

two main types of channel:

1. Parallel The bits of each sample are represented simultaneously by the potentials of M

wires relative to ground (e.g., parallel printer cable, SCSI bus, PCI bus, etc.)

2. Serial The bits of each sample are represented one after another in time by the

changing potential of one wire relative to ground (e.g., the USB bus, FireWire,

Network links, modem, serial port, etc.)

The most important property of analogue and digital channels is the rate at which they can carry

information —> the channel capacity. For digital signals, channel capacity is measured in units of 1

baud = 1 bit/second. For analogue channels it is measured in terms of bandwidth, which will be

covered later in the course.



3.4.7 Information Content and Binary Coding

The information content of a signal is defined as the total number of bits (which are two-state

elements) required to fully capture its contents. If the signal is continuous, an infinite number of bits

would be required to capture the information. Conversely, the information content of discrete

signals depends on the range and resolution which is required for a specific application. For instance,

if an input signal with a range of 0-10V needs to be measured to within 0.5V, the required

information content is log2(10V / 0.5V) = 4.3219 bits.

Not surprisingly, the information content of an M-bit binary number is M bits, where M is an integer.

In general, the information content “I” in a message (e.g. a measurement) is a real number (see

example above). Thus, to represent a measurement as a binary number without losing any

information, M must be greater than or equal to I (in bits), which requires I to be rounded up to the

closest integer value.

Warm-Up Exercises: Information Handling

3.1 Determine the information content (in bits) of a temperature measurement made using a 0-

100°C thermometer with a resolution of ±0.5 °C.

3.2 A digital voltmeter is described as having ‘4 digits’ if every digit on the display can vary from 0

to 9. Determine the information content (in bits) of a reading.

3.3 A digital voltmeter is described as having “3 and a half” digits if the first of four displayed

digits can only be 0 or 1. Determine the information content (in bits) of a reading.

3.4 What is the maximum frequency at which a signal can vary if it is to be correctly represented

by 8-bit samples transmitted through a 28,800 baud channel (remember that, for one channel,

information must be transferred serially)?



3.5 THE WAY BACK: DIGITAL TO ANALOGUE CONVERSION In the above section we explored the concept of converting an analogue signal into digital. In this

section we explore how the opposite can be accomplished e.g. convert the digital music stored in an

mp3 player back into a voltage, which can then be used to vibrate the thin membranes in stereo

headphones to produce sound. This process is known as digital-to-analogue conversion.

3.5.1 The R2R Ladder

The most common way of achieving this is by means of what it known as an R2R ladder. This is a

very simple network that has one output and as many input terminals as bits in the signal that needs

to be converted. In Error! Reference source not found., a 4 bit signal is fed into an R2R ladder with

the least significant bit at the bottom. This connection point can either be set to low (zero volts in

TTL) or high (5V in TTL logic). By solving the resistor network using Kirchhoff’s laws and current

superposition it can be shown that:

where “biti” is either “0” or “1”.

[Note: current superposition states that we can analyse a network containing more than one source

by calculating the current produced by each source separately and then adding them up.]

Microprocessor

2R

2R

2R

2R

R

R

R

2R

0V / 5V

23 (MSB)

0V / 5V

22

0V / 5V

21

0V / 5V

20 (LSB)

Vout

Figure 3.7 - 4Bit R2R ladder (LSB: Least Significant Bit, MSB: Most Significant Bit).



3.5.2 Worked Solution for the Standard R2R Ladder

Using the principle of superposition, Ohm’s law, potential divider and series and parallel

combinations of resistors only, this complex network can be solved as follows:















3.5.3 A Symmetric R2R Ladder

The circuit shown in Figure 3.7 is probably the simplest of its kind. This solution, however, has the

disadvantage that the load, as seen on each pin of the digital device, is different for different bits

(please refer to the document “ME2-HMTX - LAB 1 Primer Handout.pdf” for a detailed analysis of

this fact). This, in turn, might lead to output voltage fluctuations (remember source volt-amp

characteristic), which may reduce the accuracy of the conversion process. To overcome this

limitation, the ladder can be made “symmetric” by adding an additional link to ground as shown in

Figure 3.8, which shows a 2R resistor connecting Vout to 0V. By doing so, it can be shown that the

input impedance becomes the same for all input pins. Like the circuit in Figure 3.7, this one can be

extended indefinitely to give higher resolution. Also, by applying KCL appropriately, it can be shown

that:

Microprocessor

2R

2R

2R

2R

R

R

R

2R

0V / 5V

23 (MSB)

0V / 5V

22

0V / 5V

21

0V / 5V

20 (LSB)

Vout

2R

Figure 3.8 - R2R ladder with equal input resistance for each bit.



3.6 REVIEW OF KEY CONCEPTS This Chapter reviewed the theory of digital-to-analogue and analogue-to-digital conversion. The

following concepts, which were covered in the text, deserve particular attention:

1. A/D conversion

a. Quantization (also, link to op-amp sample-and-hold stage)

b. Coding

c. Sampling time and Nyquist theorem

d. Resolution

e. Information content and information channels

f. Input range, converter range, dynamic range

2. D/A conversion

a. Solve from first principles the R2R ladder

b. Solve from first principles the modified R2R ladder

3. How to specify the requirements of an A/D/A converter

4. How to use an A/D converter to serve a particular application


Histand, McGraw Hill, 1999, ISBN: 0072963050, Chapter 7: “Data Acquisition”.


0132060116, Chapter 6: “Signal conditioning”, pp. 224-243.



3.8 PROBLEMS: ANALOGUE TO DIGITAL CONVERSION & BACK

3.8.1 Resolution and Sensitivity

The diagram below illustrates the operation of a 4-bit optical encoder for angular position. As the

disc rotates in the direction shown, four optical sensors with switching outputs, arranged along the

line “ab”, generate a 4-bit natural binary signal which represents its angular position. The encoder is

attached to a lead screw of 0.8mm pitch, which controls the linear position of a slider.

a. What is the resolution (mm) of the linear position measurement?

b. If a computer is to be used to count the number of complete

rotations made by the encoder for a total linear range of

500mm, what is the exact information content for this task?

c. What is the smallest number of bits which is needed to store a

complete position measurement (rotation and translation) made

using this system?

3.8.2 The Symmetric R2R Ladder

For the digital-to-analogue converter in Figure 3.8:

a. Calculate the input resistance for each of the input pins as a function of R. Remember each

input is either high (e.g. 5 V) or low (0 V -> ground).

b. Derive the expression relating Vout to the state of the 4 digital lines, as found in Section 3.5.3.

3.8.3 Sampling Rate and A/D Conversion

Investigate and discuss with your peers and tutors:

a. What is the standard rate and resolution used to store music digitally on a CD?

b. Given that the human ear has problems resolving frequencies above 20kHz what do you

think the minimum sampling frequency should be?

c. Why does your answer in (b) vary from the findings in (a)?

d. When ripping a CD to disc, it is possible to resample the sound files. What effect on the

required file size has:

I. The reduction of the sampling rate from 128kS/s to 96kS/s

II. The reduction from 24bit to 16bit resolution

III. (I) and (II) together.

IV. What is the required file size in bytes (1byte = 8 bits) for 1 min of stereo sound recorded

with 16bit resolution at 96kS/s?

3.9 SOLUTIONS Warm-Up Exercises

3.1: 6.65 bits; 3.2: 13.3 bits; 3.3: 11 bits; 3.4: 1.8kHz.

Problems

3.8.1: (a) 0.05 mm, (b) 9.3 bits, (c) 14 bits (for 13.3 bits content); 3.8.2: (a) 3R, (b) Q.E.D.; 3.8.3: (a)

44.1kS/s (44.1kHz) with 16bit resolution, (b) 40kHz, (c) 40kHz only gets frequency right, but not

shape, (d) (I) 1/4, (II) 1/3, (III) 1/2, (IV) 23.04MBytes.

a b



Chapter 4: Introduction to the BasicATOM Pro™ &

NI ELVIS II™ System

4.1 LEARNING OUTCOMES Most of the laboratory exercises in the ME2-HMTX Mechatronics course use the BasicATOM Pro

microprocessor/microcontroller system to introduce you to digital control. Additionally the National

Instruments NI ELVIS II virtual instrumentation suite is used for computer based measurement and

data acquisition. While this Chapter offers a brief summary of the most common functions and

characteristics, a comprehensive description of each of these devices is available through the

“Online Help” function in the relative software, or offline, through the Atom Syntax Manual, which is

available through the Moodle website.

On completion of this Chapter you should be able to:

1. Understand the basic operation of the BasicATOM Pro microcontroller

2. Be able to program basic code in the BASIC programming language

3. Understand the basic operation of the National Instruments ELVIS II Instrumentation Suite

4.2 THE BASICATOM PRO™ MICROCONTROLLER The ATOM chip is a tiny computer, better known as a microcontroller, and it is designed for use in a

wide range of applications. As the microcontroller can be reprogrammed over and over, it can be

used to implement a number of different functions depending on the software it contains. The

ATOM chip (Figure 4.1a)is placed in the centre of a development board (Figure 4.1b) which contains

a number of extra elements which help to interface with the chip. A quick look at the ATOM chip pin

layout (Figure 4.1c) shows it has numerous input/output (I/O) pins, as well as other electronic

elements mounted on its surface (see Figure 1). The development board contains an input for the

power supply, a reset switch, a power LED which indicates when it is powered on, a number of LEDs

representing the input and output of the pins, a serial port for downloading software onto the chip

and a breadboard for creating basic electronic circuits.

a) b) c)

Figure 4.1 – The Atom chip (a), development board schematic (b) and pin layout (c).

Pins P0-P15 are general purpose I/O pins, so they can be used to receive or to output a signal.



It is important to remember that the maximum current that can be input into the ATOM chip is

25mA, and the maximum current output and input is 20mA. Anything greater than these values will

result in the destruction of the ATOM, which is expensive!

4.2.1 Main features

The BasicATOM Pro 24-M, which you will be using throughout the experiments, has the following

specifications:

RAM (memory that can be allocated whilst the chip runs): 2 kilobytes

Flash Memory to store program: 32 kilobytes

Clock speed (how many basic instructions per second): 16 MHz

The BasicATOM Pro also features built in 10bit analogue to digital (A/D) conversion on pins P0-P3

(these can be accessed through the ADIN command, which converts an analogue 0 – 5V signal to a

digital value from 0 – 1023). Also, Pins 10 and 11 can be used to output high frequency pulse width

modulated signals (PWM) to create variable voltage sources (these can be accessed through the

PWM or HPWM commands). See the BASIC Atom user manual on Moodle for further details.

4.2.2 Setup

The microcontroller is programmed using a PC running a custom Integrated Development

Environment (IDE) available at www.basicmicro.com, which connects to the chip development board

via a serial link. This is used to download the built code onto Flash Memory (i.e. similar to

downloading music to an MP3 player) so that it can be run on chip while the PC is powered off or

disconnected. The RS232 serial link uses pins Tx, Rx, Atn and Gnd as shown in Figure 4.2. Also shown

in this figure are the LEDs used to visualise if pins are active. Remember that all input pins are

limited to 5V. Any higher voltage applied will destroy the chip!!

Figure 4.2 - Atom chip as configured in the Lab.

http://www.basicmicro.com/



4.2.3 Programming in BASIC

The BasicATOM Pro is programmed using compiler, which needs to be installed on a PC with an

available serial port. The languages available are Assembler, C and BASIC. We will only use BASIC

although it should be noted that programs written in C or in Assembler will run considerably faster!

4.2.4 Defining Variables

Variables reserve and name a location in RAM to store data. Several types are available.

Type Bit Size Range

Bit 1 0 or 1

Nib 4 0 to 15

Byte 8 0 to 255

SByte 8 -127 to +128

Word 16 0 to 65,535

SWord 16 -32,767 to +32,768

Long 32 0 to 4,294,967,295

SLong 32 -2,147,483,647 to +2,147,483,648

Float 32 ±2-126 to ±2127

Examples: DOG Var Bit ‘ 0 or 1

POST Var Nib ‘ 0 to 15

LOG Var Byte ‘ 0 to 255

STICK Var Word ‘ 0 to 65535

TREE Var Long ‘ 0 to 4,294,967,295

Numerical Types: In the program, numbers can be written as Decimal, Hexadecimal or Binary. An

indicator is used to show which basis is used:

1. %1001: The % sign defines Binary

2. $1F2A: The $ sign defines Hexadecimal

3. If no precursor is used, the number is assumed to be Decimal.

4.2.5 Programming Blocks

Variables - to store values for a program

Text input and output via serial communication port

Interfacing input/output

Commands/operators - to manipulate variables

Jumps - to change path of program execution

Decisions - do different things depending on variables (e.g. IF – THEN – ELSE)

Repeating commands - fixed number of repeats or until some conditions is met (DO-WHILE)

The basic programming language includes all major programming commands, such IF-THEN-ELSE,

DO-WHILE and FOR-NEXT, and many floating point functions, such as sine, cosine, exponential, etc.

A detailed list of commands is included in the BasicATOM Pro Syntax Manual, available on Moodle.

4.2.6 Debugging

The current value stored in variables can be accessed at runtime through the use of the DEBUG

command. For instance:



' debugging information (position, signed position and setpoint ' will appear in the debug window if "debug" is pressed) CR CON 13

NUL CON 0

DEBUG [NUL, dec some_var, CR, dec some_other_var]

The above code will output the value of “some_var” and “some_other_var” in decimal form (“dec”)

to the debug screen on the PC, if the button “Debug” is pressed instead of the usual “Run” in the

chip IDE. More details are available on the BasicATOM Pro Syntax Manual on Moodle.

4.3 THE NI ELVIS II DATA ACQUISITION SYSTEM Computational Data Acquisition (DAQ) has been used for many decades now. In its standard from

this requires specialist knowledge and good programming skills. Modern DAQ systems have removed

this need and made the technology available to a much wider user base. The NI ELVIS II system goes

a step further by providing the user with Virtual Instruments which have the look and feel of their

analogue predecessors.

The NI ELVIS II system provides a number of instruments commonly used in the laboratory, all

compressed into one bench-top package. It interfaces with the PC via a conventional USB 2.0

connection, as can be seen in Figure 3.

Figure 4.3 - NI ELVIS II system components and layout (www.ni.com).

The control panel on the bench-top workstation provides a number of virtual “knobs” and

“switches” with which to control some of the functions the system offers, such as a digital

multimeter (DMM), a function generator, a variable power supply and an oscilloscope. The system

also provides a breadboard (or prototyping board) for building electronic circuitry that can connect

to the different functions of the ELVIS II.

4.4 FURTHER READING 1. www.basicmicro.com

2. www.ni.com

3. BasicATOM Pro Syntax Manual, available on the Moodle website.

http://www.basicmicro.com/

http://www.ni.com/

UNIT 3 – Signal Amplification and Processing ME2-HMTX 2012-2013


Chapter 5: Transistors

5.1 LEARNING OUTCOMES UNIT 1 looked at the basics of electrical and electronic systems, while UNIT 2 introduced the topics

of digital to analogue and analogue to digital conversion. UNIT 3, on the other hand, focuses on the

concept of “signal conditioning”, which is a necessary step in any mechatronic process. In particular,

Chapter 5 will look at semiconductor elements called transistors, Chapter 6 will introduce

operational amplifiers, while Chapter 7 will focus on the concept of filtering.

Specifically, these notes explain the theory and application of transistors, which are amongst the

most important electronic component developed in the last century. Following these notes you

should be able to:

1. Understand some of the underlying concepts behind “npn” and “pnp” bipolar junction

transistors, as well as the basic principles behind their operation.

2. Define the state of operation of a transistor, whether cut-off, active or saturated.

3. Analyse circuits containing transistors in “common-emitter” or “common-collector” modes.

4. Assemble a transistor circuit to act as a switch, as an amplifier or as a voltage follower.

5. Use multiple transistors to obtain higher current amplification (i.e. Darlington pairs)

6. Describe pulse width modulation and how it can be achieved with transistors.

5.2 INTRODUCTION We are often presented with very small signals that need to be amplified in order to become usable.

As engineers, we might need to control an electric motor or the heating in a room. A heat sensor

usually has an output of only a few mV, which needs to be processed and amplified in order for it to

be correctly interpreted. In addition, the concept of dynamic range was introduced in the previous

UNIT, which highlights the importance of “range matching” between sensors and acquisition

equipment: small signals should be amplified before e.g. they can be adequately processed by an

analogue to digital converter. Signal processing and amplification, otherwise known as signal

conditioning, is key to this process. However, this Chapter will focus on the amplification of signals

through transistors, while filtering will be covered in Chapter 7.

As described in the previous Chapter, one of the key tasks of a sound system (e.g. a digital sound

mixer or mp3 player) is to amplify low power signals, such that they can be used to drive the

actuators (e.g. a voice coil) in a speaker. Similarly, electric instruments, such as “rock guitars”, are

another example where signal conditioning is needed. The “pick-up” of an electric guitar has an

output amplitude range of about ±0.1V with a current of only a few micro amps. If such signal was to

be used in conjunction with a loudspeaker, nothing would happen. In order to produce a sound from

speakers, the raw signals generated by the pick-up need to be amplified up to a range of e.g ±12V

and tens of amps. Transistors, which are complex semiconductor devices which exploit some of the

unique properties of pn junctions, can be used to amplify such signals.

Pick-up

Signalcondi-tioning

Amplifier Speaker

mV,mAmV,mA V,A

Figure 5.1 - Block diagram of guitar amplification process.



5.3 THE BIPOLAR JUNCTION TRANSISTOR The Bipolar Junction Transistor, or BJT, is a sandwich of two diode-like bipolar junctions, usually Si.

Within the “filling” (i.e. the base), the two junctions interact to produce new behaviour. While the

physics behind BJTs is moderately taxing to visualise (and the able student is encouraged to expand

upon this topic further), the operation of these semiconductor devices is straightforward.

5.3.1 Basic Operating Principle

Two classes of BJTs exist, the “npn” and “pnp” types, which differ in terms of the ordering of the

semiconductor layers. We will deal mostly with npn transistors, although both configurations are

illustrated below.

n

n

p

collector (C)

emitter (E)

base (B)

VC

VE

VB

IC

IEIB

Physical equivalent Circuit representationPhysical Construction

Figure 5.2 – The “npn” bipolar junction transistor.

p

p

n

collector (C)

emitter (E)

base (B)

VC

VE

VB

IC

IEIB

Physical equivalent Circuit representationPhysical Construction

Figure 5.3 – The “pnp” bipolar junction transistor.

The relationships involving the transistor current and voltages are as follows:

; ;

Any pair of transistor terminals, when tested as a single port with the third terminal unconnected,

acts like its diode equivalent: be and bc junctions conduct only when forward biased, while the ce

junction hardly conducts at all in either direction (if the base is not connected!).

When connected into a suitable circuit, the characteristics of a BJT can be summarised as follows:

1. The input characteristics (i.e. the characteristics of the base port) remain similar to those of a

forward-biased diode: no base current ib flows until vbe > ~0.6V, then ib increases rapidly.

2. If vce is changed, the input characteristics are hardly affected.

3. Current flowing into the base flows out of the emitter, but ic >> ib, so that ie becomes

approximately equal to ic. Although ib should be considered when solving the output circuit, in



this course, where understanding rather than accuracy is given priority, you can assume that ic ≈

ie (i.e. that the effect of ib on the output current is negligible).

4. Since the base current is usually much smaller than the collector current, the power dissipated

by a transistor is approximately (ic x vce).

Warm-Up Exercises: Individual Transistor Junction Behaviour

5.1 One terminal of each transistor shown below is disconnected. Identify which applied voltage

is positive and state whether or not a current will flow (e.g. “vbe, yes”).

(a) (b) (c) (d)

5.3.2 Active, Cut-Off and Saturated Conditions

A transistor is cut-off if the input current is zero and therefore the output current is zero.

As the input current increases, the transistor is:

Active, if the output current increases in response;

Saturated, if the output current is already at the maximum value which the output circuit

can supply, and therefore cannot increase any further.

The “be” junction contributes a small voltage drop to vce (since vce = vbe + vbc), so that even in the

saturated condition the transistor is not a perfect short circuit and usually drops a voltage of

approximately 0.2V (i.e. vce = ~0.2V when the transistor is saturated). However, we will neglect this

voltage drop in the context of this course (i.e. power consumption of a saturated transistor is 0W).

To determine whether the transistor is cut-off, active or saturated:

1. Focus on the base: Is the “be” junction conducting? If NO, then no collector current will flow

and the transistor is in a CUT OFF state.

2. Otherwise, calculate the base current, ib (remembering the ~0.6 V drop across be).

3. Calculate the collector current (hfe x ib) which the transistor should generate.

4. Calculate the maximum current ic which the collector circuit could generate if “ce” were a short.

5. If (4) exceeds (3), the transistor is ACTIVE; otherwise it is SATURATED.

Example 1

The transistor shown has an hfe of 100. Determine whether it is cut-off, active or saturated. Estimate the power dissipated by the transistor.

1 V

+1 V

+

1 V

++

1 V

100

10 k+

1 V

+

5 V



5.4 COMMON-EMITTER CONFIGURATION To treat the transistor as a two-port device, like a transducer, we must:

1. Select one of the three terminals and fix it to a constant value, and

2. Nominate one terminal as the output and the other as the input (control).

Most transistor stages are based on the common-emitter configuration, with the output measured

at the collector, the input at the base and the emitter fixed (in the case below, ve is defined as

ground). Remember that all voltages are relative to a reference voltage (i.e. ground), thus, when

speaking of e.g. “ve”, what is meant is the “voltage at the emitter relative to ground”.

5.4.1 Configuration Characteristics

Junction “be” is forward biased and, as expected, passes current.

Junction “ce” is reverse biased, but ‘unexpectedly’ conducts a current. The magnitude of this

current (ic) is proportional to that of the base current (ib), as will become clear in the

following sections.

For an “npn” transistor in common-emitter configuration, the volt-amp characteristics of the output

circuit vary with respect to the base current ib as follows:

Figure 5.4 – Volt-amp characteristics of the output port for a transistor in a “common-emitter” configuration.

Figure 5.4 illustrates how the collector current ic is proportional to the base current ib and is virtually

independent of the collector voltage vce. In other words, a transistor in this configuration acts like an

ideal current source, where the current through the load (which would be placed in series between

“c” and the rightmost voltage source) remains constant regardless of load size.

To summarise, in the common-emitter configuration, the transistor can be thought of a current

amplifier, since the change in base current ib is directly proportional to a “scaled up” change to the

collector current ic. Alternatively, since the base current is not actually scaled up, but rather acts as a

ib

ic

ie

vbe

v ce+

+

Variable DC supply

Base current,

mA i

b

Base voltage

vbe

10

1 V

Collector current, i

mA c

v ce

Collector voltage

10 100 µA

ib

80

60

40

20

10 V



control signal to enable current flow though ce, the transistor can be thought of as a throttle, where

ib controls the amount of current which can be allowed into the collector.

Specifically, the relationship between base and collector currents can be described mathematically

as follows:

where hfe is a constant scaling factor, which is generally about 100 and varies negligibly with vce.

Warm-Up Exercises: Saturation and Power Dissipation

An “npn” transistor in common-emitter mode has a 12 V supply and a 1.2kΩ collector resistor.

5.2 What is output saturation current?

5.3 What power does the transistor dissipate when it is saturated?

5.4 What power does the collector resistor dissipate when the transistor is saturated?

5.5 What power does the transistor dissipate when it is passing 50% of the saturation current?

5.4.2 The Transistor as a Switch

Imagine trying to control a transistor in common-emitter configuration without a base resistor, by

varying vbe. For all voltages up to Vbe = 0.6V nothing will happen. A very small further increase in

voltage, however, will cause a very large increase in base current and hfe times this large increase in

collector current, ic.

Thus an increasing base voltage applied directly or through a very small resistor switches “ce” from

cut-off (open circuit) to saturated (passing as much current as the collector circuit will allow).

This remains true for any base resistor small enough to ensure that:

,

Where imax is the maximum current which can be drawn by the load (i.e. the current ic when the

transistor is saturated and therefore vce is near zero). Any extra ib just adds to the emitter current.

Used like this, a transistor is smaller, cheaper, much faster and much more durable than a

mechanical switch. However it can only pass current in one direction!

5.4.3 The Transistor as an Amplifier

To act as an amplifier, a transistor must be controlled continuously through a range of active

conditions between cut-off and saturation. In a common-emitter configuration this can be achieved

by applying the input voltage through a relatively large base resistor (i.e. the base is “driven by a

current source”). As the base input voltage Vb is increased above 0.6V, the base current now

increases only very gently.

As the base current increases, the collector current will increase almost linearly with it. As the

collector current increases, the voltage across a load resistor Rc in series with the load increases.

This is the basis of the common-emitter voltage amplifier.



The output voltage might be measured across the collector resistor or across the transistor, however

note that these vary in opposite directions, since both change with input voltage but their sum

remains constant at Vcc.

Figure 5.5 – A transistor in common-emitter configuration, used as an amplifier.

Note: the relationship between ib and ic is independent of, but limited by Vcc i.e. the collector circuit

will behave as a “quasi-ideal” current source (providing a quasi-constant current independently of

the load Rc), as long as ic falls within the maximum capacity of the source, which is defined by the

diagonal line in Figure 5.5.

Example 2

The transistor circuit illustrated below is used to control the brightness of a light bulb. Calculate the setting for Rb to achieve 10%, 50% and 100% light intensity. Relate lamp power to the power dissipated by the transistor.

12V

Light bulb

(100W at 12V)

hfe=1000-1k?

5V

Collector current ic (mA)



5.5 COMMON-COLLECTOR CONFIGURATION Many practical applications with a single transistor use the simple circuit shown below, where it is

now the collector voltage which is fixed, while the output is measured at the emitter. One of the

classic common-collector circuits is the “emitter-follower”, where the load is connected in series

with the emitter, which enables a transistor to be used as a “near ideal voltage source”, as will be

explained in the following section.

5.5.1 The Emitter-Follower

Figure 5.6 – An “emitter-follower” transistor circuit, where the ports are in common-collector configuration.

The operation of an emitter-follower circuit can be visually described as follows:

1. Initially, Vin = 0 and no currents flow so VL=0.

2. As Vin increases just above ~0.6V, a current starts to flow into the base (ib)…

3. …which allows a current hfe times greater to flow from “c” to “e” and through RL (i.e. ic)…

4. …which causes the load voltage (VL) to rise…

5. …which reduces Vbe to 0.6V again.

This is an example of ‘internal’ feedback, since 2-4 above happen instantaneously, so that —

provided that the transistor remains active — whatever Vi is (> 0.6V), enough current will flow in the

load to maintain VL constant.

By applying DC circuit analysis to the input and output circuits (i.e. KVL and KCL) and eliminating ib,

the following governing equation for an emitter-follower circuit can be found:

Now, if hfe is large and Rb is much smaller than RL, the above equation reduces to

, which means:

Voltage gain

, but

Current gain

i.e. the stage output delivers up to hfe times the current delivered to the input, at the same voltage.

By keeping the transistor within its active region, an emitter-follower will behave like an ideal

voltage source, where the load voltage VL is maintained constant independently of how much

current it draws. For instance, if the resistance of a load varies with temperature, a voltage-follower



can be used to maintain a constant voltage across its terminals. Of course, for this to happen, the

transistor, once again, acts as a throttle, automatically adjusting ic to maintain this condition.

5.5.2 Darlington Pair

The current gain of an emitter-follower can be increased even further by connecting a second

emitter-follower between the emitter of the first transistor and the load.

Figure 5.7 – Two transistors in common-collector configuration form a Darlington pair.

This Darlington pair has a current gain of hfe1 x hfe2, but the base-emitter voltage drop vbe is ~1.2 V

instead of ~0.6 V.

5.6 THE SOUND OF MUSIC Normally we want an accurate representation of an input into the transistor to the loudspeaker.

Rock music (on the guitar side) is built on actually driving the transistor circuit into saturation. As

shown below:

t

Vbb

Vbe

t

Ic

t

Vbb

Vbe

t

Ic

t

Vbb

Vbe

t

IcV

cc/R

c

No output

Clean sound

Overdrive

Figure 5.8 – The concept of saturation as a “feature” in rock music generation.



5.7 FIELD EFFECT TRANSISTORS (INFORMATION) Field effect transistors (FETs) are in many ways simpler than BJTs, but there are more variants, so

they can be also more confusing.

The main difference between FETs and BJTs is that the gate (like the base of a BJT) is virtually

insulated, having almost infinite resistance to either the gate or drain (output) terminals. As a result,

FET output characteristics are similar to those of BJTs, but are controlled by input voltage rather

than current. This type of transistor is outside the scope of this course.

5.8 PULSE WIDTH MODULATION Linear amplifiers, using a common-emitter or common-collector output stage, are common for

driving medium-power loads, e.g. the 100W motor used in ME1-HMTX and ME2-HMTX laboratory

exercises. For both types of stage, all of the current which passes through the motor also passes

through the transistor. This means that a lot of power is dissipated in the output transistor, which

gets hot and can fail.

A transistor used as a switch, on the other hand, dissipates almost no power: when it is cut off,

there is no current through it. When it is saturated, there is almost no voltage across it. Thus, an

often used alternative to a variable voltage supply for DC power control is to switch the load on and

off at a very high frequency (usually >20 kHz). The frequency remains constant but the power is

controlled by varying the duty cycle: the proportion of the period for which the load is switched on.

Figure 5.9 – Typical example of pulse width modulation: the average voltage is proportional to the duty cycle of the signal.

This very effective method of power control is known as pulse width modulation, which you have

already encountered in ME1-HMTX. It is widely used to drive DC loads from digital systems without

the use of a digital-to-analogue convertor stage. The average power delivered is the product of the

maximum power and the duty cycle.



5.9 REVIEW OF KEY CONCEPTS This Chapter reviewed the theory semiconductor materials and bipolar junction transistors (or BJTs).

The following concepts, which were covered in the text, deserve particular attention:

1. Basic operation of the bipolar junction transistor (NPN)

a. Active, cut-off and saturated conditions

2. Transistors in stages

a. Common-emitter configuration

(applications: current amplifier, current regulator, switch)

b. Common-collector configuration

(applications: voltage regulator, Darlington pair)

3. Transistor operating states (active, cut-off, saturated)

4. 2 transistor configurations (common-emitter/collector)

5. How to use transistors in both configurations


Histand, McGraw Hill, 1999, ISBN: 0072963050, Chapter 2: “Semiconductor Electronics”.


0132060116, Chapters 22: “Junction Transistor Amplifiers”.



5.11 PROBLEMS: TRANSISTORS

5.11.1 Bipolar Junction Transistors 1

For each of the npn transistors shown below, for which hfe = 100:

(a) (b)

(c) (d)

1. Find the base current;

2. Classify operation as cut off, saturated or active and calculate the collector current, and

3. Calculate the power dissipation in the transistor.

5.11.2 Bipolar Junction Transistors 2

Find the emitter voltage and current in each transistor shown below, for which hfe is 100 and can be

considered to be very high (i.e. ic ~ ie). Remember vbe ~ 0.6V.

(a) (b)

Bonus question: if you cannot make the assumption that the base current is negligible (i.e. ie = ib + ic),

what would the voltage ve in the circuit in (a) be?

5.11.3 “Real Life” Example 1

An emitter-follower is to be used to control a resistive heating element with a power rating of 15W

at 5V. The emitter-follower is driven by a voltage source with maximum output voltage of 12V and

maximum output current of 20mA.

a. What is the minimum hfe of the required transistor for the load to operate at the rated

voltage and power?

100

10 k+

1 V

+

5 V

100

100 k+

0.5 V

+

10 V

1 k

100 k+

2 V

+

10 V

10 k

100 k+

2 V

+

5 V

10

k

10

k

10

0

12 V

0 V

4.7

k

1 k

22

0

12 V

0 V



Assuming that it is not possible to find a transistor with this specification and that two transistors

with hfe gains of 20 and 50 arranged as a Darlington pair are used instead, calculate:

b. A suitable value for the base resistor, and

c. The amount of current drawn from the input voltage source under the conditions in (b)


The circuit illustrated below applies a nearly constant armature voltage to the motor M. The

armature resistance of the motor is 3Ω, so that if it were stalled whilst switched on, this is the

minimum resistance which the transistor would need to drive. The transistor is a 2N 3055 type with

an hfe of 20.

a. What voltage will appear across the motor assuming that it draws negligible current?

If the motor voltage must not fall by more than 20% from this value under any circumstance:

b. Calculate the maximum possible value of the base resistor Rb.

c. Calculate the corresponding base current ib.

d. Assuming the value found in (c) to be the maximum current to be provided by the

voltage regulator circuit in figure, calculate the maximum Zener series resistor Rs.

e. What would the maximum power dissipation in the Zener diode be?


5.1: (a) vbc, yes, (b) vcb, yes, (c) vbe, no, (b) vbe, yes; 5.2: 10mA; 5.3: ~zero; 5.4: 120mW; 5.5: 30mW.

Problems

5.11.1: (a) ib = 40µA, ic = 4mA, active, Ptrans = 18.4 mW, (b) ib = 0A, ic = 0A, cut-off, Ptrans = 0W, (c) ib =

14µA, ic = 1.4mA, active, Ptrans = 12.0mW, (d) ib = 14 µA, ic = 0.5 mA, saturated, Ptrans ~ 0W; 5.11.2: (a)

ve = 3.6V, ie = 36mA, (b) ve = 1.45V, ie = 6.6mA, bonus question: ve = 3.61V; 5.11.3: (a) hfe = 281.25,

(b) Rb = 1333Ω, (c) ib = 3mA; 5.11.4: (a) ~5.7V, (b) 15Ω, (c) 76 mA, (d) 75Ω, (e) 479 mW.

Rs

0V

M

12V

Rb

C6V3



Chapter 6: Operational Amplifiers

6.1 LEARNING OUTCOMES Analogue electronic systems for instrumentation and control make extensive use of operational

amplifiers (op-amps): integrated amplifiers whose internal workings can be ignored for the purpose

of this course. Almost any computation (summing, integrating, differentiating, inverting, etc.) can be

performed on analogue signals using a few simple types of op-amp stage, each stage containing one

op-amp and a few resistors and/or capacitors. This unit considers what amplifiers are, then goes on

to describe how op-amps work and how stages, which perform a number of different operations,

can be assembled. Time-dependent stages (i.e. those used in the “AC world”), which form the heart

of dynamic control systems and active filtering, will be introduced in the third and final chapter of

this UNIT (i.e. Chapter 7). On completion of this chapter you should be able to:

1. Understand the basic concepts behind operational amplifiers

2. Describe the basic operating principles of operational amplifiers, with or without feedback

3. Outline the limitations of real op-amps, as compared to ideal op-amps

4. Design and use inverting amplifiers, non-inverting amplifiers, buffer amplifiers, summers,

difference amplifiers, as well as other typical op-amp stages

5. Design two-input comparator stages, with or without hysteresis (i.e. Schmitt triggers)

6.2 INTRODUCTION Since electric signals (e.g. the output of sensor or transducer) are often too low in voltage and

power for direct processing, it is important to be able to amplify the signals in order to match its

properties with those required by the processing device (e.g. ADC or speaker of your sound system).

Operational amplifiers are most often used to amplify and condition small signals. For high power

operation e.g. sound amplification or motor drives, transistors (Chapter 5) and MOSFETS are used.

6.3 WHAT IS AN ELECTRONIC AMPLIFIER An amplifier is a three-port device. Under the control of a signal detected at an input port, the

output port delivers power from a supply port to a load. An amplifier is effectively a control system,

the controlled output power of which is of the same kind as the input signal; the term is used for

mechanical and hydraulic systems, as well as electronics. For instance, think of transistors and how

they can be used to create voltage and current amplifiers (Chapter 5).

There are four basic amplifier types. From the most common to the least common, there exist

voltage amplifiers, current amplifiers, current-to-voltage (trans-resistance) amplifiers and voltage-to-

current (trans-conductance) amplifiers.

6.3.1 An Ideal Electronic Amplifier

The characteristics of an ideal amplifier are as follows:

THE INPUT: In order not to load the signal source while measuring the signal itself, an ideal

amplifier must have an input impedance of infinity (like an ideal voltmeter,

where voltage is measured without drawing any current) or zero (like an ideal

ammeter, where current is measured without “dropping” any voltage)



THE OUTPUT: In order to power any load, an ideal amplifier must act either as a perfect

voltage source or a perfect current source i.e. it must be able to produce

any voltage or current independently of the load size i.e. independently of

what is connected to the output

This means that an ideal amplifier will amplify a voltage or a current without affecting or being

affected by the surrounding circuit in an undetermined way.

6.3.2 A Real Electronic Amplifier

Unfortunately, ideal amplifiers do not exist because any electronic device is affected by limitations in

the material used, manufacturing and even the packaging processes used in their production.

As a result, real amplifier stages differ from ideal models in several ways, especially:

THE INPUT: The input impedance is large but finite (neither zero nor infinite)

THE OUTPUT: The output impedance is small but finite, so the output characteristics can

be modeled as a Thévenin or Norton source with a small output impedance

6.4 OPERATIONAL AMPLIFIERS These disadvantages can be overcome by:

Cascading well-designed transistor stages into an amplifier of very high voltage gain A and

Connecting external feedback components (next §) to define the function of the whole

stage: e.g. amplification (Section 6.3) and many other signal-processing operations (like

addition, subtraction, inversion, integration, differentiation, etc.)

These complex integrated circuits, which take the name of operational amplifiers, or op-amps in

short, have two input terminals, labeled v+ and v–, two supply terminals, labeled Vs+ and Vs–, and one

output terminal, labeled vo, which can sink current towards Vs- as well as source current from Vs+.

Disregarding any internal complexity, the simple operating principle of an op-amp is described by:

( )

Thus, as v+ increases, vo increases; as v- increases, vo decreases. The “missing” 0V output terminal,

which is not shown in the diagram, can be thought of being connected inside the op-amp to a

voltage midway between Vs+ and Vs–.

Figure 6.1 – Symbolic view of an op-amp (left) and chart describing the relationship between v+, v- and vo (right).

Vs+

v o

Vs-

v - -v+

Linear

(active)

region

Vs+

Vs-

v o

v -

v +



6.4.1 An Ideal Op-Amp:

The characteristics of an ideal op-amp can be summarized as follows:

Both inputs have infinite impedance (i.e. they draw no current)

The gain A is constant and infinite

The output impedance is zero (i.e. current flowing in or out of the output drops no voltage)

The output voltage can vary in any range between - to + V

The output current can vary in any range between - to + A

Vs+ and Vs- can be neglected, since ( ) should hold true under all

circumstances

6.4.2 A Real Op-Amp:

Conversely, for a real op-amp (which is almost always an integrated circuit in a single package):

The input impedance is high (usually > 10 MΩ) but finite

The gain A is very high (~104 – 106) but varies greatly from sample to sample

The output impedance is very low (~1 – 10Ω) but finite

The output current is limited to a few tens of mA

The output voltage can only vary between the two supply voltages Vs+ and Vs–

There is often an input offset voltage error (i.e. a ‘misreading’ of v+ – v–), which is amplified

by the large gain A to give an unpredictable output voltage when v+ = v–

Despite these limitations, the following operating principles still hold:

1. No current can flow into the op-amp through v+ or v-

2. Current can flow into or out of vo

3. A small positive difference between v+ and v- will cause the output vo to saturate at the

maximum available voltage the op-amp can produce i.e. Vs+

4. A small negative difference between v+ and v- will cause the output vo to saturate at the

minimum available voltage the op-amp can produce i.e. Vs-

6.5 COMPARATORS The simplest use of an op-amp is in a stage where two voltage signals need to be compared. In fact,

due to its high gain, an op-amp without feedback (i.e. in “open-loop” configuration) acts as a

comparator where:

if v+ > v–, the output switches to its positive limit Vs+;

if v+ < v–, the output switches to its negative limit Vs–.

[Remember: as v+ , vo ; as v–, vo .]

For instance, the circuit described in Figure 6.2 below switches on a red or a green led according to

whether the input voltage (vi) is greater or less than a pre-set reference level (vref). In this case, the

reverence voltage vref can simply be computed using potential divider between the two stage

resistors, R1 and R2, to give:

2

1 2

9ref

Rv v x V

R R



Figure 6.2 - A comparator setup (left) and the graphical relationship between vi and vo (right).

Purpose-designed comparators, which have the same circuit symbol as op-amps, are less linear than

op-amps; much faster; have higher gain; and provide a higher current (25-50mA) and may be

designed to drive logic gate inputs.

6.6 USING OP-AMPS WITH FEEDBACK Most of the problems associated with real op-amps are virtually eliminated by the use of

“feedback”, which also extends the capabilities of an op-amp stage significantly. In fact, op-amps are

almost always used with negative feedback, where the output is connected back (directly or

indirectly) to the inverting input, so that the op-amp action reduces the difference between v+ and v-

to zero (this fact is difficult to visualize, but can be derived mathematically for any op-amp stage

with feedback, as will become clear in the following sections). As will become clear in the following

sections, op-amps with feedback can be used for a variety of purposes, including amplification and

voltage/current regulation (like for transistors), but also more complex mathematical operations,

such as addition, subtraction, integration and differentiation.

For all of the stages described next (which employ feedback), the op-amp needs a ‘high’ gain, but

neither the actual value of the gain nor the input offset is very important. In fact, as soon as an op-

amp is used with feedback, the following assumption can be made:

v+ = v-

which will be derived in Section 6.6.1. Thus, there are two alternative ways to derive the relationship

between inputs and output of an op-amp-stage:

Write down the circuit equations and solve them as

Use the simple shortcut that, if the amplifier is active (i.e. ), then

Again, in both cases one can assume that the input currents are zero.

[Note: To understand the role of negative feedback in each case, consider what happens if the stage

output varies slightly from the value shown, whilst inputs remain constant.]



6.6.1 Voltage Follower (Buffer Amp)

Like an emitter follower stage (Chapter 5), this circuit can be used to produce an ideal voltage

source, where vo = vi independently of how much current is drawn by a load attached to the output.

In fact, if vo increases to a value which is greater than vi, “(v+ – v–)” becomes negative and the op-

amp then drives the output back down to reinstate the equality.

In addition, since the input impendence of the op-amp is very high, while the output impendence is

very low, large loads can be driven from a high-impedance source without affecting its output.

, but

Rearranging:

and, as 1

(q.e.d.)

o o

o o

o

o

v A v v v v

v A v v

Avv A

A

v v v

Figure 6.3 – “Voltage-follower” stage (left) and analytical solution describing the relationship between vo and vi.

For a voltage follower, the closed loop voltage gain of the circuit (i.e. vo / vi = 1) does not depend on

the open-loop gain “A” of the op-amp. It can also be shown that the linearity of the stage (i.e. the

dependence of gain on amplitude) is unaffected by that of the op-amp.

Warm-Up Exercises: Op-amps with and without Feedback

6.1 In ALL exercises, assume op-amps to be ideal, running from supplies Vs+ = –Vs– = 15V. Nothing

is connected to their outputs unless shown. Determine all voltages (“”) and currents (“→”).

(a) (b)

(c) (d) (e)

6.6.2 Non-Inverting Voltage Amplifier

Using feedback, op-amps can be used to make very linear amplifiers of all types (as listed in 6.3),

using only external resistors (which can be very linear, stable and accurate) to set gain parameters.

One of the most commonly used op-amp stages is the “non-inverting voltage amplifier”, illustrated

below, where the relationship between input and output voltages is:

1 2v

2

R RA

R

, where Av is the voltage gain between vi and vo.

+10 V

0 V

a

b

a

+10 V

0 V

1 k

b

0 V

a

b

0 V

a10 V

b

0 V

a10 V

b

1 k

c



2

1 2

2i

21 2

1 2

1 2i

2

( ), but and

v

1

Now, as

v

o o i

o i o

o

Rv A v v v v v v

R R

R Av A v v

RR R AR R

A

R Rv

R

Figure 6.4 – Non-inverting amplifier stage (left) and analytical solution describing the relationship between vo and vi.

Note that, by using the simple shortcut described in Section 6.6, the equation if Figure 6.4 can be

derived by simply using potential divider:

2

1 2

1 2i

2

v , but:

v

o

i o

Rv

R R

R Rv v v v

R

Thus, this stage can be used to amplify an input voltage vi by a set amount only by choosing an

appropriate combination of resistors e.g. if R1 = R2, then vo = 2 x vi.

6.6.3 Inverting Amplifier

This is probably the most commonly used op-amp stage, since, while the non-inverting amplifier can

only increase the magnitude of an input voltage, the inverting amplifier can also reduce the value of

vi (if R1 > R2). In this stage, the relationship between input and output voltages is defined by:

2v

1

RA

R , where Av is the voltage gain between vi and vo.

1

22 2

1 1

Using the shortcut:

(virtual earth) and i

i or

i

io o i

v v

vv ground

R

v Rv R R v v

R R

Figure 6.5 - inverting amplifier stage (left) and analytical solution describing the relationship between vo and vi.

Since v+ = 0V, the op-amp maintains v- = 0V, although it is physically not connected to ground. This

phenomenon is referred to as “virtual earth”, since v- can be completely fixed to a set voltage (in

this case ground) without it being physically connected to a hard wire carrying that same voltage.



6.6.4 “Sample and Hold” Op-Amp Stage

A sample and hold circuit is used extensively in analogue-to-digital conversion (Chapter 3), where a

signal value must be stabilized while it is converted to a digital representation. The sample and hold

stage illustrated in Figure 6.6 consists of a “voltage-holding” capacitor and a voltage follower.

Figure 6.6 - Sample and hold op-amp stage.

With switch “S” closed, the output vo will follow the input vi. Also, while the two voltages track each

other, the capacitor “C” charges and discharges, such that vc vi at all times. When S opens, vo

(which is equal to v+ and consequently to v-) will retain the last voltage stored in the capacitor i.e. the

value of vi just before the switch was opened. As a result, this stage can be used to temporarily

pause a changing input voltage without affecting the signal itself (remember that virtually no current

is drawn by the op-amp input).

The type of capacitor used for this application is of course important, as a low-leakage capacitor,

such as a polystyrene or polypropylene type, is needed to ensure that the capacitor voltage vc does

not drop during the “holding” period.

6.6.5 Current-to-Voltage (Trans-Resistance) Amplifier

Acting on negative feedback, this op-amp stage maintains v– = 0. Thus, as current arrives at the

inverting input point, the op-amp output voltage goes negative enough to pull it forwards through

the feedback resistor. Again, the inverting input point becomes a “virtual earth”: v– is forced to

remain at zero volts, but no current can actually flow from this point to the 0V line. Consequently, as

the magnitude of the input current increases, the magnitude of the output voltage increases.

i

(as i increases, increases proportionally)o in m

in out

v R

v

Figure 6.7 – Trans-resistance stage (left) and analytical solution describing the relationship between iin and vo.

A simple resistor could also be regarded as a trans-resistance amplifier. However, the op-amp stage

differs from a resistor since it can be thought of a short circuit when “seen from” the input (i.e. like



an ammeter), whilst the output is a low impedance voltage source i.e. the input and output are

virtually disconnected, which means the stage performs as an “ideal” device.

Trans-resistance amplifiers have numerous applications, including detector stages for photodiodes,

phototransistors etc., as will be seen in the tutorial problems at the end of this chapter.

Warm-Up Exercises: Trans-Resistance (Input and Output Currents)



(a) (b)

(c) (d)

6.7 ANALOGUE SIGNAL PROCESSING USING OP-AMP STAGES It is often necessary to process information by performing precise mathematical operations on

analogue voltage signals: e.g. a voltage representing position can be differentiated to give a voltage

representing velocity. Op-amp stages are ideal for this. Such stages are sometimes shown as:

Figure 6.8 – Schematic representation of (from left to right): a multiplier, an integrator and a differentiator op-amp stage.

These symbols take for granted that:

The input impedance is high,

The output impedance is low, and

The voltage supplies are high enough (maybe ±15V) to accommodate any required change in

output voltage.

0 V

a

b

10 mA

1 k

c

d

0 V

a

b

10 mA

1 k

c

d

1 k

e

dtddt

5



Signal processing stages involving constant voltages and currents (i.e. summers and difference

amplifiers) will be described in the following sections, while time-varying stages (i.e. differentiators,

integrators and filters) are the topic of Chapter 7, where AC is (re)introduced (see ME1-HMTX).

6.7.1 Summing Amplifier (Adder or Mixer)

The current which flows into the input of an inverting amp depends only on the input voltage, since

the op-amp input should remain at 0V (virtual earth). Thus, in general:

The input impedance of a “virtual earth circuit” is simply the impedance between the stage

input terminal and the inverting op-amp input terminal.

This remains true however many input resistors are connected to the virtual earth point, thus each

input voltage will be unaffected by the others. Currents which flow in through the separate input

resistors meet at the virtual earth point and the op-amp pulls the total forward through the

feedback resistor. Hence, it can be shown that the voltage output in this stage is given by:

io f

1,2

,N

n

n n

vv R

R

where Rf and R1-Rn can be chosen to provide individual amplification for each input signal.

Figure 6.9 - A summing amplifier stage.

In summary, the op-amp stage above sums and, if necessary, amplifies the input voltage signals. If R1

= R2 =… Rn = Rf, the output voltage is just the negative sum of the input voltages.

A practical detail: to “balance” the op-amp inputs, the positive op-amp input should be connected

to 0V through a resistor R0 equivalent to all of the inverting input resistors in parallel.

6.7.2 Difference Amplifier (Subtractor)

A “diff amp” combines inverting and non-inverting amps to generate an output which is the

deference between two input signals. In fact, it can be shown that the voltage output is given by:

o 2 1( ),i iv k v v

where “k” is a scaling factor which can be used to amplify the difference between the two inputs.



Figure 6.10 - A difference amplifier op-amp stage.

Subtraction of a reference voltage from a signal is often used in instrumentation and control, where,

for instance, the difference between a desired and actual position (or velocity) is amplified and used

to drive a closed-loop position (or speed) controller (Chapter 9).

A precision diff amp, the inputs of which are buffered to give extremely high input impedance, is

known as an instrumentation amplifier; it can be constructed using 3 op-amps but is also available

as a separate package. While its description is beyond the scope of this chapter, further details

about the architecture and operation of instrumentation amplifiers can be found in “Alciatore,

Introduction to Mechatronics and Measurement systems, Section 5.9”.

Warm-Up Exercises: Operational Stages



(a) (b)

(c) (d)

0 V

a b

10 mA

1 k

c1 k

d

0 V

a b

10 mA

500

c1 k d

50

0e



e) f)

Example

The output from a Wheatstone Bridge detector stage, for a range of zero to maximum input, runs from +2V to -2V. Design an op-amp stage which can convert this range to 0-10V. Note that the stage must run from ±15V supplies and have an input resistance of 100kΩ.

6.8 SCHMITT TRIGGERS A comparator circuit (Section 6.5) will work well under some conditions, but it is not always ideal. If

there is a slow waveform, or one with some noise on it, then there is the possibility that the output

will switch back and forth several times during the switch over phase as only small levels of noise on

the input will cause the output to change. This may not be a problem in some circumstances, but if



the output from the operational amplifier comparator is being fed into fast logic circuitry, then it can

often give rise to problems as the transients generated by the noise in the signals cannot be

differentiated from the signal itself.

Under these circumstances, circuits that help mitigate this problem are required. One such system is

known as the Schmitt trigger, which was originally invented by an American scientist named Otto

Schmitt. A Schmitt trigger is a comparator with hysteresis. As its input voltage increases through a

certain value, the output switches from positive to negative. However, the input voltage must fall to

a lower voltage to switch the output back to positive. A Schmitt trigger can be easily implemented

by taking the comparator circuit in Figure 6.2 and adding a resistor between the output and non-

inverting input, as illustrated in Figure 6.11, which represents “positive feedback”.

Figure 6.11 - A comparator with hysteresis, otherwise known as Schmitt trigger (left) and schematic representation of the relationship between vi and vo (vi1 – vi2 is called dead-band).

As in a regular comparator stage, in the op-amp stage above the output voltage vo will either be vs+

or vs-. However, in order to compute when the op-amp will switch, some circuit analysis is required.

To start with, let us assume that vo is at the negative op-amp source voltage vs- (i.e. the red led is on).

The switching point when vo becomes vs+ occurs when the voltage at node “a” is equal to vi (i.e. va =

v+ = vi), which we will call vi1. By applying KCL at node “a”, vi1 can be computed:

1 111 1 2 3

1 2 3 1 3

or v ( || || )cc i i s cc sii

v v v v v VvR R R

R R R R R

This means that the Schmitt trigger will switch as the input signal drops below this value.

Once the output is set to vs+, the same method can be used to find the upper limit, vi2:



2 1 2 3

1 3

v ( || || )cc si

v VR R R

R R

In other words, the Schmitt trigger produces a “dead-band” equal to the difference between the two

limiting conditions described above, where the output does not change as a result of a change in

input. An equation for the dead-band can be found by subtracting the two values of vi found above:

1 2 3 1 2 3 1 2 3

1 3 1 3 3

( || || ) ( || || ) ( || || )cc s cc s s sv V v V v vR R R R R R R R R

R R R R R

The reference voltage vref, which is simply the point around which the dead-band is cantered, is

equal to the midpoint between vi1 and vi2. This value can be chosen to represent e.g. the switching

point in the opening of a valve, or any other trigger around which “some hysteresis” is needed. Also

note that, as an inverting amplifier, the relationship between input and out voltages is inverted: as

vi drops below vi1, vo switches to its positive maximum value vs+ and, when vi increases above vi2, vo

becomes vs- (for the interested reader, there are plenty of interactive examples on the web).

Schmitt triggers of one kind or another have many uses in instrumentation and control, for instance:

1. Switching transducers. Here they prevent repeated indecisive changes of output as the

physical input makes small variations around the switching point

2. Comparator stages like that described above, to avoid the same effect

3. Analogue to digital convertors, where they are used to filter out noise from the input signal

4. ‘Bang-bang’ control systems (especially for temperature). Here they reduce the frequency

of switching, by allowing the controlled variable to vary within the deadband without

causing a change of state

6.9 CHOOSING COMPONENT VALUES FOR OP-AMP STAGES All of the stages described in this chapter only specify resistor values as ratios or products. In

practice, in order to make stages predictable and self-contained (therefore easily tested),

components around op-amps should have:

Much higher impedance than the impedance of the op-amp output (~10Ω), such that they

do not inadvertently load the op-amp input; and

Much lower impedance than the impedance of the op-amp input (~10MΩ), which will vary

from sample to sample.

Resistor values used within op-amp stages are therefore usually between 1kΩ and 100kΩ.

6.10 REVIEW OF KEY CONCEPTS This Chapter reviewed the theory of operational amplifiers, used with or without feedback. The

following concepts, which were covered in the text, deserve particular attention:

5. The basic operating principles of ideal and real amplifiers

6. The basic operating principles of ideal and real operational amplifiers (or “op-amps)

7. How to use op-amps in simple comparator stages



8. The effect of feedback on the output of an op-amp stage

9. The behaviour and operation behind the following feedback op-amp stages:

a. Voltage follower

b. Inverting and non-inverting amplifier

c. Trans-resistance amplifier

d. Summing amplifier

e. Difference amplifier

f. Schmitt trigger


Histand, McGraw Hill, 1999, ISBN: 0072963050, Chapter 5: “Analog Signal Processing Using

Operational Amplifiers”.



6.12 PROBLEMS: OPERATIONAL AMPLIFIERS

6.12.1 Op-Amp Design

The most readily available resistor values are the ‘E12 series’: 1, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7,

5.6, 6.8, 8.2 and 10Ω and factors of 10 larger or smaller (e.g. 180 Ω, 18 kΩ, etc.).

Design a non-inverting amplifier with a gain of 263, using fixed E12 resistors. A deviation of ± 1% of

the nominal gain is acceptable. Remember that resistors can be connected in parallel.

6.12.2 Difference Amplifier Proof

Derive an expression for the output of a difference amplifier with inverting input resistor R1,

negative feedback resistor R2, non-inverting resistor R3 and ground resistor R4. Show that this

expression reduces to the correct form, as given in 6.7.2, for a difference amplifier in which:

2 4

1 3

R R

kR R

6.12.3 Summing Amplifier

The operational amplifier in shown below runs from power supplies of ±15V. Square wave inputs of

100Hz and 50Hz respectively, from sources of very low input impedance, are applied at points A and

B as shown.

Figure 6.12 - Summing amplifier setup.

a. From first principles, derive the expression for a summing amplifier described in

Section 6.7.1.

b. What DC voltage should be applied at point C for the DC (i.e. time-average) output

voltage to be 0V?

c. Sketch the output waveform at point P, showing voltage and time scales, whilst this

voltage is applied.

d. What is the voltage at point “X” under these conditions?

e. What is the input impedance of this circuit at point “A”?

f. Point “C” is now disconnected from its DC input. Sketch the output waveform at

“P”, showing peak values and time intervals.



6.12.4 Signal Conditioning

The circuit shown below is used in an automatic camera to measure light intensity. The small

current passed by a reverse-biased ‘large area’ photodiode (having an area of 100mm2 and a

sensitivity of 0.5AW–1) is measured by an op-amp current to voltage amplifier stage.

Figure 6.13 - Special purpose op-amp setup.

a. What will be the voltage at the op-amp’s inverting input?

b. The maximum incident light intensity is estimated to be 1 Wm–2. What will be the

magnitude and direction of the maximum current in the feedback resistor?

c. Select a feedback resistor to maximise sensitivity whilst avoiding op-amp ‘limiting’.

d. Hence determine the sensitivity of the detector in V m2 W–1.

6.12.5 Schmitt Triggers

The diagram below schematically illustrates a constant pressure chamber setup, where a pressure

sensor, which produces a pressure-proportional output voltage between 0V and 9V, is used to

power an electric motor driving the compressor. The op-amp stage used here is responsible for

ensuring that the compressor is driven if the pressure sensor voltage drops below 5V by some

tolerance. A simple diode is used to ensure that no current flows through the driving motor if Vo

becomes negative.

+5 V

-5 V

0 V



a. Assuming that vo starts at the negative op-amp source voltage vs-, the switching

point when vo becomes vs+ occurs when the voltage at node “a” is equal to vi (i.e. va

= v+ = vi), which we will call vi1. Prove that: 1

1

1 3 1 2 3

1 1 1v cc s

i

v V

R R R R R

b. Using a similar approach to (b), show that the reference voltage Vref around which

the op-amp stage will “switch states” occurs at: 1

1 1 2 3

1 1 1v cc

ref

v

R R R R

c. If Vcc = Vs+ = 9V, Vs- = -9V and R1 = 9Ω, find the values for R2 and R3 which are needed

to ensure that the compressor is switched on (i.e. 9V are applied across its two

ports) when the pressure sensor voltage drops below 5V, while being turned off

otherwise. Allow for a ±1V “dead-band” around this value to avoid inadvertent

switch-on due to low amplitude noise in the sensor signal.

d. Draw a graph illustrating the output voltage Vo, as a function of the input voltage Vi.

Make sure to include the lower trigger voltage, Vi1, the upper trigger voltage Vi2, the

source voltages Vs+ and Vs- and numerical values for all x and y-axes intersects.


6.1: (a) a = 0, b = 15V, (b) a = 0, b = 15mA, (c) a = 0, b = 0, (d) a = 10V, b = 0, (e) a = 10V, b = 0, c =

10mA; 6.2: (a) a = 0, b = 10mA, c = 10mA, d = 10V, (b) a = 10mA, b = 20mA, c = 10V, d = 10mA,

e = 0, (c) a = 10mA, b = 30mA, c = 10V, d = 20mA, (d) a = 10mA, b = 20mA, c = 10V, d = 0, 6.3:

(a) a = 10V, b = 10mA, c = 10mA, d = 10V, (b) a = 10V, b = 10mA, c = 20 mA, d = 5V, (c) a = 0, b =

12V, c = 6V, d = 6mA, (d) a = 12mA, b = 9V, c = 6V, d = 3mA, (e) a = 10V, b = 10V, c = 5mA, d = 5V, (f)

a = 10V, b = 10V, c = 5mA, d = 5V.

Problems

6.12.1: e.g. 1kΩ, 27kΩ in parallel with 390kΩ; 6.12.2: Q.E.D.; 6.12.3: (a) Q.E.D., (b) -10V (to shift

output waveform up by 10V), (c) with period 20ms, (d) 0V, (e) 1kΩ, (f) the output

waveform drops by 10V and is clipped (limited) at -15V; 6.12.4: (a) 0V, (b) 50µA reverse current, (c)

to be conservative, assume vout should not exceed 3.75 V; then, the resistor should be 75kΩ, (d)

3.75Vm2W1; 6.12.5: (a) Q.E.D., (b) Q.E.D., (c) R2 = 15, R3 = 45, (d) refer to Figure 6.11 (right).



Chapter 7: Signal Conditioning - Filters

7.1 LEARNING OUTCOMES Chapter 7 concludes the topic of “Signal Amplification and Processing”, which was covered in UNIT 3.

Here, the specific subject of “signal conditioning” will be introduced, after a brief review of

Alternating Current (AC) and complex notation. Both passive (i.e. involving passive elements, such as

resistors and capacitors) and active (i.e. involving active elements, such as op-amps) circuits will be

considered, alongside a few other useful active time-dependant stages, which may be useful to you

in the future (e.g. integrator and differentiator stages). By the end of this chapter you should be able

to:

1. Solve AC circuits using complex notation, impendence and phasor diagrams.

2. Describe and utilise the concepts of “gain” (H) and “phase shift” (φ) in AC circuit analysis.

3. Set-up the complex equations describing the relationship between input and output in

simple 1st order passive low- and high-pass filters.

4. Describe and use Bode diagrams to represent filter characteristics graphically, including cut-

off frequency, bandwidth and Decibel notation.

5. Understand how op-amps can be used to create simple active time-dependant stages, such

as integrators and differentiators.

6. Set-up the complex equations describing the relationship between input and output in

simple 1st order active low- and high-pass filters.

7. Cascade a number of active stages to achieve more complex behaviour (e.g. band-pass and

band-reject filters).

8. Select a circuit’s resistances and capacitances to give a desired filter performance.

7.2 INTRODUCTION Filters are used vary widely in everyday applications. By turning up the “base” on your stereo, one is

changing the parameters of a low-pass filter enabling the low-frequency signals to pass through

whilst inhibiting higher frequencies. “Micro-filters” used in an ADSL internet connections (think of

home broadband) split the low-frequency voice signal of the telephone line and the high-frequency

internet signals. Similarly, while measuring an electrical signal (such as the voltage across a sensor),

one might need to filter out high frequency noise before the signal itself can be processed. Thus, it

is important to know the basic principles behind filters. However, while the majority of signals are

non-periodic and usually quite “erratic”, in the following sections we will only focus on simple

periodic signals, which will only vary in amplitude and phase. While this simplification may appear

unjustified, the way in which instrumentation responds to sinusoidal signals is important for two

reasons:

1. Many systems involve rotation or vibration only.

2. Fourier analysis [ME2 Maths] shows that the variation of any quantity with time can be

expressed as a sum of many sine waves at different frequencies and phase angles.

Therefore, the response of a system to sine waves says all there is to know about it, since its

response to any other input can be reconstructed by adding the responses to its constituent sine

waves. This realisation (i.e. that the performance of a system in time can be analysed simply by using



sinusoidal inputs) leads to the concept of frequency response analysis: a study of the effect that the

frequency of a sinusoidal input has on the performance of a system. More specifically, the following

sections will consider the effect which various time-dependent stages have on sinusoidal inputs of

different amplitudes and frequencies. Finally, please recall that, in systems of linear elements (and

in all stages considered here), all signals share the same frequency. Thus, the effect of a stage on a

sinusoidal input will be measured with respect to changes in amplitude and phase between input

and output.

The following sections will review the concept of Alternating Current (AC), complex notation, and

phasor diagrams and Bode plots, covered in ME1-HMTX Mechatronics. Then, passive high- and low-

pass filters will be introduced, followed by a number of time-dependant active stages based on

Operational Amplifiers (Chapter 6).

7.3 A REVIEW OF AC AND COMPLEX NOTATION

7.3.1 Alternating Current at a Glance

An Alternating Current (AC) signal, whether a voltage or a current, can be expressed in mathematical

notation as follow:

( ) sin( )dc mV t V V t (1)

Where:

Vm amplitude

radian frequency

phase angle

+ve leading

-ve lagging

VDC DC offset

T period = 2/

f frequency = 1/T

The general signal expressed in Equation (1) is graphically illustrated in Figure 7.1, with respect to a

reference sinusoidal signal.

Figure 7.1 - Graphical representation of a general AC signal.



7.3.2 Complex Notation and Phasors

The analysis of AC circuits is simplified by the use of phasors, which use complex numbers to

represent sinusoidal signals. Euler’s formula forms the basis for this analysis:

( ) cos( ) sin( )j te t j t (2)

where 1j is being used instead of “i” to avoid confusion with the symbol for current. This

implies that sinusoidal signals can be expressed as real and imaginary components of complex

exponentials. Because of the mathematical ease of manipulating exponential expressions (think of

integration and differentiation), this form of analysis is convenient for making and interpreting

calculations and will therefore be used throughout the remainder of this Chapter.

In networks of linear elements, once any transients (i.e. temporary disturbances) have dissipated in

an AC circuit, the voltage across and current through each element will oscillate with the same

frequency as the input. The amplitude of the voltage and current for each element will be

constant, but may differ in phase (lagging or leading the input) from the input. This fact lets us treat

circuit variables V and I as complex exponentials of the form:

( ) cos( ) sin( )j t

m mV V e V t j t (3)

Once again, in systems of linear elements, the frequency of all currents and voltages is the same

(i.e. only amplitude and phase change). Thus, a “phasor” is the vector representation of an AC signal

at a particular point in time i.e. when t = 0. Thus, the typical voltage in complex exponential form

described in Equation (3) can be written as V = Vm[cos() + jsin()]. This quantity is graphically

illustrated in Figure 7.2, where:

2 2r x y

is the phasor angle, measured from the “t” reference

The x coordinate corresponds to the real component

The y coordinate corresponds to the imaginary component

The phasor can be expressed in the condensed form: Vm

Figure 7.2 - Phasor representation of an AC voltage.

A phasor representation is:

Independent of frequency (frequency is represented by the speed of rotation of the phasor

diagram about the “z” axis i.e. the axis coming out of the page)



Can be thought of as the “snapshot” of a current or voltage at time t = 0

Can be manipulated using complex arithmetic

A few useful rules for manipulating phasors:

1. 1

jj

(4)

2. 1tany

x

(5)

3. 2 2r x y (6)

4. (r1 1) (r2 2) = r1 r2 (1 + 2) (7)

5. (r1 1) / (r2 2) = r1 / r2 (1 - 2) (8)

7.3.3 Impendence in AC Circuits

In the DC world:

A resistor has a resistance “R” = V / I (Ohm’s Law)

A capacitor has “infinite” resistance (open circuit)

An inductor has “zero” resistance (short circuit)

In the AC world, things become more complicated, since not only the magnitude, but also the

frequency and phase of a signal need to be considered. Thus, the concept of “impedance” is used to

extend that of “resistance” to include frequency dependency. Impendence, which can be thought

of as “frequency dependent” resistance, can be expressed in various ways (recall your ME1-HMTX

notes), but here we will only use complex notation, as described in Section 7.3.2.

Since Ohm’s Law still holds (“V = IR” in the DC world and “V = IZ” in the AC world), for resistors:

V = ZRI ZR = V/I = R = (R +0j)Ω = tan-1(0/R) = 0o

(This means that the voltage across and current through a resistor are in phase)

For inductors, assume a current I = Im e j(t + ):

V = L(dI/dt) = LjIme j(t + ) = (Lj)I

ZL = V/I = (Lj) = (0 + Lj)Ω = tan-1(L/0) = 90o

(This means that the voltage across an inductor leads the current by 90o)

For capacitors, assume a voltage V = Vme j(t + ):

I = C(dV/dt) = CjVme j(t + ) = (Cj)V

Zc = V/I = 1/(Cj) = (0 – j/(C))Ω = tan-1(-1/(Cx0)) = -90o

(This means that the voltage across a capacitor lags the current by 90o)

7.3.4 Gain and Phase Shift

Input and output voltages, both sinusoidal and of equal frequency, can be represented as phasors on

a diagram, as illustrated in Figure 7.3. Their relationship can then be simply described by defining

two important parameters, gain and phase shift, which will be used throughout the remainder of

this chapter. The gain is defined as the ratio between output and input magnitudes (weather



voltages or currents). Similarly, the phase shift describes by how much the output leads/lags the

input.

Using complex notation, both of these quantities can be expressed as:

| || |

| |o o o

i i i

A AH H

A A

(9)

where “A” can be a current or a voltage, the subscript “o” defines the output and “i” the input.

This ratio is thus another complex number H, which defines how the output phasor is related to the

input phasor. Using this notation, the gain is the magnitude of the complex ratio H, or |H|, which,

using Equation (8), is simply the ratio of the magnitudes of the output and input signals:

| || |

| |o

i

AH

A (10)

Similarly, the phase shift is the angle between the output and input phasors, or , which, using

Equation 8, is simply the difference between the phases of the output and input signals:

o i (11)

Figure 7.3 - Phasor diagram illustrating two phasor quantities describing the relationship between the input and output

of a system. The gain is |H| = |Ao| / |Ai|, while the phase shift is described by the angle H.

A “negative gain”, like that of an inverting amp, will from now on be represented by a positive gain

and a 180° phase shift. The magnitude of H (i.e. |H|) is the same whether peak, peak-to-peak or

Root Mean Square (RMS) voltages are used, and is generally expressed in Decibel (db) units:

|H|dB = 20log10(|H|).

7.3.5 Bode Diagrams

A Bode diagram or Bode plot displays the frequency response of an AC circuit as two separate

curves, which represent the gain and phase shift between the output and input signals:

The gain plot: |H|dB vs. log(frequency); and

The phase-shift plot: vs. log(frequency), where is usually expressed in degrees



The axis normally runs from –180° to +180° since e.g. a 270° phase lag is equivalent to a 90° phase

lead. Frequency may be angular (, in rad/sec) or cyclic (2

f

, in Hz); the shape of the plot is

clearly the same.

It is important to note a few important properties of log-log plots, which may explain why they are

used in this instance:

1. Multiplication by a constant “C” shifts all points rightwards (or upwards)

along the axis by log(C).

2. Division by a constant “C” shifts all points leftwards (or downwards)

along the axis by log(C).

3. A factor of 10 is referred to as a decade: log10(10) = 1.

4. A factor of 2 is referred to (especially for frequency) as an octave: log10(2) 0.3.

5. “0” on a log scale represents unity.

6. A function y = Cxn appears as a straight line log y = n log(x) + log(C).

Of particular importance is the frequency at which the power of the output signal is half that of the

input signal: the “corner” or “cut-off” frequency c. For this reason, c is also referred to as the

“half power point”. Since the power of a sinusoidal signal is proportional to the square of the

signal’s amplitude, at the cut-off value:

10.707

2o o

i i

A P

A P , or 30% amplitude attenuation (Ao 70% Ai)

This value can be expressed in Decibel as:

10

120log 3

2dB dB

Which means that c corresponds to the frequency at which the output signal is attenuated by 3dB.

The term bandwidth is used to quantify the range of frequencies which a system can adequately

reproduce. This parameter can be expressed analytically as the range of frequencies where the input

of the system is not attenuated by more than 3dB i.e. the frequency range between corner

frequencies. A system usually has two corner frequencies at which the attenuation is -3dB, which

are defined as the low and high corner or cut-off frequencies L and H. In practice, the bandwidth

of most amplifiers and servomechanisms is approximately equal to the upper cut-off frequency,

because the lower cut-off frequency is several decades (i.e. orders of magnitude) smaller.

The rest of this chapter will use Bode plots to describe the performance of several signal

conditioning stages, both passive and active.

7.4 PASSIVE FILTERS Filters affect different ranges of frequency in different ways. Typical uses include the removal of

low-frequency mains hum from high-frequency signals or avoidance of aliasing by removing high-

frequency signals before converting the output into digital form.



Passive filters contain only passive components: resistors, capacitors or (more rarely) inductors.

Here we will consider two, low-pass and high-pass filters, which will be solved both graphically

(through Bode diagrams) and analytically (using complex notation) in the following sections.

7.4.1 Passive Low-Pass Filter

A standard first-order low pass filter is illustrated in Figure 7.4. The input is represented by an AC

voltage V = Vme j(t + ) and the output Vo is measured across a capacitor of capacitance C.

2

1

1

1 ( )

tan ( )

HRC

RC

Figure 7.4 - Conventional 1st order low-pass filter (left) and gain and phase shift relationships (right).

Using complex notation, the relationship between input and output voltages can be described using

the potential divider rule as:

Co i

R C

ZV V

Z Z

(12)

Now, since:

1. The impedance of the Capacitor is: c

j 1Z

C j C

2. The impedance of the resistor stays: RZ R

and defining the complex relationship between input and output voltages as “H”:

o C

i R C

1

V Z 1j CH

1V Z Z 1 j RCRj C

(13)

Using Equation (13) and Equation (8), deriving equations for the gain and phase shift relating the

input and output voltages in a low-pass filter is straightforward:

2

1 1Gain H

1 j RC 1 ( RC)

(14)

1 1 1o i

0 RCPhase Shift tan tan tan RC

1 1

(15)

By inspecting Equation (13) and Equation (14), several important conclusions can be drawn:

1. As 0:

|H| 1 the magnitude of low frequency signals remains untouched.

0o the phase of low frequency signals remains untouched.

Cvi o

v

R



2. As :

|H| 0 the magnitude of high frequency signals is highly attenuated.

-90o high frequency signals lag the input by 90o.

3. The circuit has a cut-off frequency, c, when:

c2

1 1 1H

2 RC1 ( RC)

(16)

“RC” is known as the time constant of the system, sometimes referred to as .

4. At = c = -1:

|H| = 0.707 or -3dB (half power point)

= -tan-1(1) = -45o

Zc = ZR

Thus, a low-pass filter will suppress all high frequency components of a signal while leaving the low

frequencies unaffected. Any cut-off point can be defined by selecting appropriate values for the

resistor R and capacitor C.

The Mechanical Engineering Data and Formulae Book shows the exact Bode plot for this function,

which is also illustrated in Figure 7.5 below. Note that the phase shift line has straight-line

asymptotes for 0 and . Also, for convenience, the “x axis” in the phase plot is described

as a ratio between the frequency and c, which means the graph can be used as a reference to

identify suitable capacitor and resistor values for a given corner frequency.

Figure 7.5 - Bode diagram for a first order low-pass filter.

7.4.2 Passive High-Pass Filter

A standard first-order high-pass filter is illustrated in Figure 7.6. The input is represented by an AC

voltage V = Vme j(t + ) and the output Vo is measured across a resistor of resistance R.

First Order Lowpass Response

-100

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

0.1 1.0 10.0 100.0 1000.0 10000.0 100000.0

/ c

Phase (deg)

Gain (dB)



2

1

1 ( )

90 tan ( )o

RCH

RC

RC

Figure 7.6 - Conventional 1st

order high-pass filter (left) and gain and phase shift relationships (right).

Using similar logics as outlined in Section 7.4.1, it can be shown that the relationship between input

and output voltages is:

2

RCH

1 ( RC)

(17)

o 190 tan RC (18)

By inspecting Equation (17) and Equation (18), several important conclusions can be drawn:

1. As 0:

|H| 0 the magnitude of low frequency signals is highly attenuated.

90o high frequency signals lead the input by 90o.

2. As :

|H| 1 the magnitude of high frequency signals remains untouched.

0o the phase of high frequency signals remains untouched.


c2

RC 1 1 1H

2 RC1 ( RC)

(19)

4. At = c = -1:

|H| = 0.707 or -3dB (half power point)

= 90o - tan-1(1) = 45o

Zc = ZR

Thus, a high-pass filter will suppress all low frequency components of a signal while leaving the high

frequencies unaffected. Again, any cut-off point can be defined by selecting appropriate values for

the resistor R and capacitor C.

The Mechanical Engineering Data and Formulae Book shows the exact Bode plot for this function,

which is also illustrated in Figure 7.7 for convenience. Again, please note that the phase shift line has

straight-line asymptotes for 0 and .

R

Cv

iov



Figure 7.7 - Bode diagram for a first order high-pass filter.

Warm-up Exercises: Passive Filters

In the following exercises, please try to use both a graphical (read the numbers off Figure 7.5 and

Figure 7.7) and analytical solution (by solving Equations (14), (15), (17) and (18)), then compare your

answers to make sure they agree with each other.

7.1 Determine gain and phase shift of a signal transmitted by a passive first order low-pass filter

with a frequency 20 times the cut-off frequency c.

7.2 Determine gain and phase shift of a signal transmitted by a passive first order low-pass filter


7.3 Determine gain and phase shift of a signal transmitted by a passive first order high-pass filter


7.4 Determine gain and phase shift of a signal transmitted by a passive first order high-pass filter

with a frequency 1/20 times the cut-off frequency c.

7.5 A passive low-pass filter with R=1,000Ω and C=36F is being used to filter a signal of

2,800rad/sec. What is the reduction in signal in dB and as an absolute ratio of amplitudes?

7.6 A passive high-pass filter with R=500Ω and C=36F is being used to filter a signal of 20rad/sec.

What is the reduction in signal in dB and as an absolute ratio?

7.7 Two passive filters, separated by a buffer amplifier (Section 6.6.1) to avoid loading of the first

filter, are used in series to from a band-pass filter. The high-pass filter has fcHP=40Hz and the

low-pass fcLP = 4,000Hz. Signals at 2Hz, 400Hz, and 80,000Hz are sent through the filter.

Determine gain in dB and phase shift for each case.

7.8 Investigate what impact a low-pass pass filter will have on a square wave signal.

First Order Highpass Response

-40

-20

0

20

40

60

80

100

0.01 0.10 1.00 10.00 100.00 1000.00 10000.00

/ c

Phase (deg)

Gain (dB)



7.5 ACTIVE TIME-DEPENDENT STAGES Operational amplifiers were introduced in Chapter 6, alongside a number of useful stages that

perform mathematical manipulations on input voltages and currents (e.g. amplifiers, inverters,

summers and difference op-amp stages). These “signal conditioning” circuits only dealt with fixed

signals i.e. voltages and currents which did not change in time. This section will introduce a few op-

amp-based stages which become useful in the “AC world”: differentiators, integrators and filters.

The main differences between passive (i.e. composed of resistors, capacitors, inductors and

resistors) and active stages (i.e. employing op-amps) are that:

1. An op-amp stage does not load the signal source when connected, which is an important

advantage, since multiple stages can then be cascaded to produce complex behaviour

(Section 7.5.5).

2. An op-amp stage can have a gain which is greater than “1” i.e. the amplitude of the output

signal can be larger than the magnitude of the input signal.

3. Generally, an op-amp stage inverts the output with respect of the input. In the AC world,

inversion can be thought of as a phase shift of 180˚, which means many active time-

dependant stages introduce a fixed additional phase shift of ±180° for all due to inversion.

What follows is a brief description of four of the most common active time-dependant stages for

signal conditioning, which are often found in modern electronic systems: integrators, differentiators,

and active low- and high-pass filters.

7.5.1 Voltage Integrator

By replacing the feedback resistor of an inverting amplifier (Chapter 6) with a capacitor, a voltage

integrator is produced.

1

90o

HRC

Figure 7.8 - Voltage integrator stage (left) and gain and phase shift relationships (right).

A value for the complex ratio “H” between the output and input voltages of an integrator stage can

be obtained by first analyising the relationship between input and output voltages in the time

domain. Assuming the shorcut v+ = v-:

virtual earth and I 0 oio

dVV dV C V C

R dt dt

1

or o io i

dV VV Vdt

dt RC RC (20)



Now, assuming Vi = Vme j(t + ), equation (20) can be rewritten as:

( ) ( )1 1 1j t j t oo i m m

i

V jV Vdt V e dt V e H

RC RC j RC V RC

(21)

From (21), values for the gain and phase shift introduced by an integrator stage can be found:

j 1H

RC RC

(22)

1 1 o o oo i

1 0tan tan 90 0 90

0 RC

(23)

Thus, while in the time domain the output of an integrator stage will be proportional to the integral

of the input, in the frequency domain the magnitude of the output signal is attenuated linearly as a

function of the input frequency “”. To summarise:

1. As 0:

|H| the integral of a DC signal will grow without limits.

2. As :

|H| 0 the integral of a high frequency signal will tend to 0.

Note that an integrator stage will introduce a constant 90˚ phase shift between the output and the

input and the frequency at which the gain becomes unity is called unity gain frequency (0 or f0

depending if the frequency is expressed in rad/sec or Hz), which occurs when 0 = 1/RC.

Figure 7.9 - Bode diagram for a voltage integrator stage.

7.5.2 Voltage Differentiator

This is another virtual earth circuit. As current flows up to the inverting input through C, the op-amp

pulls it out through the feedback resistor, thus maintaining v- = 0V.



90o

H RC

Figure 7.10 - Voltage differentiator stage (left) and gain and phase shift relationships (right).

By simply analyising the mathemathical derivation for an op-amp differentiator stage using complex

notiation, values for the gain and phase shift can be found. In the time domain:

and 0

i

o

dVI C

dt

V IR

i

o

dVV RC

dt (24)

Now, assuming Vi = Vme j(t + ), equation (24) can be rewritten as:

( ) ( )j t j t oio m m

i

VdV dV RC RC V e j RC V e H j RC

dt dt V (25)

From (25), values for the gain and phase shift introduced by a differentiator stage can be found:

H j RC RC (26)

1 oRCtan 90

0

(27)

Thus, while in the time domain the output of a differentiator stage will be proportional to the rate of

change of the input, in the frequency domain the magnitude of the output signal will be amplified

linearly as a function of the input frequency “”. To summarise:

1. As 0:

|H| 0 the rate of change of a DC signal is zero.

2. As :

|H| the derivative of a high frequency signal will tend to .

Note that a differentiator stage will introduce a constant -90˚ phase shift between the output and

the input and unity gain frequency occurs, once again, when 0 = 1/RC.



Figure 7.11 - Bode diagram for a voltage differentiator stage.

Warm-up Exercises: Time-Dependant Active Filters

7.9 All op-amps shown are ideal, running from supplies Vs+ = Vs = 15V. Nothing is connected to

their outputs unless shown. Determine voltages ( ) and currents ().

(a)

(b)

(c)

(d)

(e)

-100

-80

-60

-40

-20

0

20

40

60

80

100

0.1 1.0 10.0 100.0 1000.0 10000.0 100000.0

/ 0

First Order Lowpass Response

Phase (deg)

Gain (dB)

0 V

a b

e

d

1 µ

1 V/s

(constantly)

c1M

0 V

a

b

1 µ

10 µA

since t = 0 c

0 V

a

b

1 µ

10 V

c

1 M



7.5.3 Active High-Pass Filter

In a differentiator, the gain tends to infinity as id

d

v

t tends to infinity, which can sometimes cause

problems in fast changing circuits. The gain can be limited by connecting a capacitor C2 in parallel

with the feedback resistor, as illustrated in Figure 7.12.

1 2

22 2

12

1 ( )

90 tan ( )o

C RCH

C RC

RC

Figure 7.12 – Active high-pass filter (left) and gain and phase shift relationships (right).

If the input voltage varies slowly, so does the input current and most of it is conducted forward by

the feedback resistor. The feedback capacitor can therefore be neglected and the stage remains a

differentiator (Section 7.5.2).

If the input voltage varies rapidly, so does the input current, and most of it is conducted forward by

the feedback capacitor. The feedback resistor can therefore be neglected. Because both capacitors

pass the same current, their voltages vary with the ratio of their capacitances. The stage just

becomes an inverting amplifier whose “high frequency” voltage gain is:

1

2

| |hf

CH

C

A more analytical solution for this op-amp circuit can be obtained by using complex notation. Since

the stage illustrated in Figure 7.12 is equivalent to that of an inverting amplifier, where the feedback

impendence is simply the parallel combination of the capacitor “C2” and resistor “R”, the

relationship between input and output voltages can be described as:

2

2F C

2

2

1R

j C RZ Z ||R

1 1 j C RRj C

Thus, according to Section 6.6.3 in Chapter 6:

1 1

F 1o i i i

C 2 C 2

Z Rj CR 1V V V V

Z 1 j C R Z 1 j C R

(28)



Now, defining the complex relationship between input and output voltages as “H” and multiplying

the top and bottom of (28) by “C2” leads to the following complex ratio:

o 1 2 1 2

2i 2 2 2 2

V Rj C C C Rj CH

V C j C R C 1 j C R

(29)

Using (29) and (8), deriving equations for the gain and phase shift relating the input and output

voltages in an active high-pass filter is straightforward:

1 2 1 2

22 2 2 2

C Rj C C RCGain H

C 1 j C R C 1 ( RC )

(30)

2

1 1 o 11 2 2o i 2

2

RC C RCtan tan 90 tan RC

0 C

(31)

By comparing equations (30) and (31) with equations (17) and (18) respectively, it becomes clear

that the stage illustrated in Figure 7.12 is just a high-pass filter with the following additional

features:

While for a passive high-pass filter the maximum gain is “1” (i.e. input and output signal

magnitudes are the same), an active high-pass filter can have a gain which is > 1

The high-frequency gain for an active high-pass filter is |H|inf = C1/C2

The phase shift plot is shifted by 180˚ (i.e. it is inverted)

This means that:

1. As 0:

|H| 0 the magnitude of low frequency signals is highly attenuated.

-90o low frequency signals lag the input by 90o.

2. As :

|H| C1/C2 the magnitude of high frequency signals is constant.

-180o the phase of high frequency signals is shifted by 180˚ (i.e. it is inverted).


2c2

22

RC 1 1 1H

2 RC1 ( RC )

(32)

In this case, the fixed gain of “C1/C2” is not considered...think of it as an amplifier in series

with a filter.

4. At = c = -1:

|H| = 0.707 or -3dB (half power point) + |H|hf

= -90o - tan-1(1) = -135o

The Bode diagram for an active high-pass filter, where a fixed gain (|H|inf) equal to 10dB is used for

illustration purposes, is shown in Figure 7.13. In this example, the output amplitude of high

frequency signals is amplified by 10(10/20) or 3.1 times with respect to the input (and C1 = 3.1 x C2).



Figure 7.13 - Bode diagram of an active high-pass filter.

7.5.4 Active Low-Pass Filter

Any stray DC voltage at the input of an integrator stage causes a continuous change in output

voltage, so that an op-amp integrator quickly limits and stops working. This problem can be

eliminated by connecting a resistor “R2” in parallel with the feedback capacitor, as illustrated in

Figure 7.14.

2

21 2

12

1

1 ( )

180 tan ( )o

RH

R R C

R C

Figure 7.14 – Active low-pass filter (left) and gain and phase shift relationships (right).

As usual, all of the input current must pass through the feedback loop. Now, if the input voltage

varies rapidly, so does the input current and most of it is conducted through the feedback loop

through the capacitor. The feedback resistor can therefore be neglected and the stage remains an

integrator (Section 7.5.1).

If the input voltage varies slowly, so does the input current, and most of it is conducted through the

feedback loop through the resistor. The capacitor can therefore be neglected and the stage just

becomes an inverting amp whose “DC gain” is:

2

1

| | DC

RH

R



Using complex notation and the analytical procedure adopted in Section 7.5.3 for an active high-pass

filter, it can be shown that the gain and phase shift of an active low-pass filter are:

2

21 2

R 1H

R 1 ( R C)

(33)

12180 tan ( ) o R C (34)

Again, by comparing equations (33) and (34) with equations (14) and (15) respectively, it becomes

clear that the stage illustrated in Figure 7.14 is just a low-pass filter with the following additional

features:

An active low-pass filter can have a gain which is > 1

The DC gain for an active low-pass filter is |H|DC = R2/R1

The phase shift plot is shifted by 180˚

This means that:

1. As 0:

|H| R2/R1 the magnitude of low frequency signals is constant.

180o the phase of DC signals is shifted by 180˚ (i.e. it is inverted).

2. As :

|H| 0 the magnitude of high frequency signals is highly attenuated.

90o high frequency signals lead the input by 90o


c222

RC 1 1 1H

2 R C1 ( R C)

(32)

In this case, the fixed gain of “R2/R1” is not considered.

4. At = c = -1:

|H| = 0.707 or -3dB (half power point) + |H|DC

= 180o - tan-1(1) = 135o

The Bode diagram for an active low-pass filter, where a fixed gain (|H|DC) equal to 10dB is used for

illustration purposes, is shown in Figure 7.14.



Figure 7.15 - Bode diagram of an active low-pass filter.

7.5.5 Stages in Cascade

For active stages connected in cascade, where the input of each stage is not loaded (think of virtual

earth), it can be shown that:

1. The total gain is the product of the individual gains, therefore log (total gain) is the sum of

the individual log (gain)s.

2. Since the phase shift for each stage does not depend on the phase angle of the input signal,

the total phase shift is simply the sum of the individual phase shifts.

Thus, the Bode plot for a cascade system can be built up in stages by simple graphical addition.

Please note that the phase axis “wraps round” at 180° = –180°, as if drawn on a cylinder. For

instance, 120° + 90° becomes –150°. A simple example describing how filters in cascade can be

useful is the band-pass filter, which can be constructed by cascading a low- and high-pass filter

together, as illustrated in Figure 7.16, and only affects frequencies outside a predefined range.

Figure 7.16 - Band-pass filter, composed of an active low-pass filter in cascade with an active high-pass filter.



Example

Sketch the Bode plot for an inverting integrator with f0 = 100Hz, in cascade with an active low-pass

filter with a low-frequency gain of 12dB and fc = 200Hz, for frequencies between 10 and 1,000Hz.

Figure 7.17 - The sketch shows gain (G) and phase shift () plots for the integrator (I) and low-pass (L). The plots for the

cascade combination (C) are derived by addition.

7.6 REVIEW OF KEY CONCEPTS This Chapter reviewed the theory of AC circuit analysis using complex notation, then introduced

passive and active signal conditioning filters. The following concepts, which were covered in the text,

deserve particular attention:

1. AC circuit analysis using complex notation.

2. The concept of impedance, with expressions for resistors, capacitors and inductors.

3. The concepts of “gain” and “phase shift”, which represent the relationship between input

and output in a signal conditioning stage.

4. Passive low- and high-pass filters: how to derive expressions for their gain and phase shift

plots and how to represent them on a Bode diagram.

5. Cut-off frequency, bandwidth and Decibel notation.

6. the use op-amps to create integrator and differentiator stages, including how they can be

represented on a Bode diagram.

7. Active low- and high-pass filters: how to derive expressions for their gain and phase shift

plots and how to represent them on a Bode diagram. Need particular attention to the

concepts of the “fixed gain” component (i.e. ||DC and |H|hf) and inversion (180˚ phase shift).

8. Cascading a number of active stages to achieve more complex behaviour (e.g. band-pass).

9. Select resistances and capacitances to give a desired filter performance.


Histand, McGraw Hill, 1999, ISBN: 0072963050, Section 2.6 – Alternating Current Circuit

Analysis” and Chapter 4 – System Response.

2. ME1-HMTX – Mechatronics lecture notes - Notes 12: AC Networks 2.



7.8 PROBLEMS: SIGNAL CONDITIONING

7.8.1 Passive High-Pass Filter Proof

Using complex impedances develop the gain and phase shift relationships between the input and

output voltages of the passive high-pass filter below and draw a Bode diagram of the filter.

C

R

Vo

0V0V

Vi

7.8.2 Active Low-Pass Filter proof

Using complex impedances develop the gain and phase shift relationships between the input and

output voltages of the active low-pass filter below and draw a Bode diagram for the filter when the

ratio of R2/R1 is 2 (i.e. the amplitude of the output is twice that of the input for DC signals).

7.8.3 Voltage Integrator

Using two op-amps, three E12 standard resistors and a 1µF capacitor, design a system which simply

integrates its input voltage:

o i dv v t

The stage should have a positive gain of 1, with a resistive input impedance of 10kΩ.


A data acquisition card has an input range of 0 to 10V with a resolution of 10 bits. The card can

sample at 10kHz.

a. What is the highest frequency the card can sample without the danger of aliasing?

b. Anti aliasing filters are used to remove signals that could cause aliasing. What would a filter

for the DA card need to achieve?

c. Determine the corner frequency of a filter that would eliminate the chance of aliasing.

d. What impact would this filter have on a signal with a frequency of 50Hz, 500Hz and at 5kHz?


An accelerometer with an output of 0 to 5V is used to monitor the vibrations of a frame of a train.

The signal is recorded with 12-bit ADC card that has a sampling frequency of 100,000 samples per

second and an input range of ±5V. A low pass filter is to be designed in order to reduce the possible

aliasing to 10% of the ADC range. The filter should have an input resistance of 1kΩ.




Rotational speed can be measured using a small generator producing a voltage which is proportional

the rotational speed (as seen in ME2-HMTX Lab4). Due to the characteristics of the generator, there

will always be a “ripple” on the voltage signal caused by construction of the coils (stator and rotor

rubbing against each other cause a high frequency noise which disrupts the main voltage). This

ripple can be removed using a low pass filter. The speed sensor used in the lab generates 15% ripple

voltage (peak to peak) at 4 times the frequency of the speed measured. For a signal at 1,000rpm, the

output of the sensor is 4V plus ripple (i.e. 4V DC plus ripple voltage).

a. Select a resistor which, in combination with a 36F capacitor, will reduce the ripple by 20dB.

What will the peak-to-peak amplitude of the ripple voltage be after filtering?

b. Select a resistor which, in combination with a 36F capacitor, will reduce the ripple by 40dB.

c. Select a resistor which, in combination with a 36F capacitor, will reduce the ripple by 60dB.

d. What impact could this filter have on the control of the motor speed?


An equalizer is to be used to process a sound signal with a range of 0-250mV to boost spoken

language while suppressing background noise i.e. the equalizer should amplify those frequencies

within the normal range of speech (300Hz – 3,400Hz), while suppressing lower and higher

frequencies which are outside this range.

Given that the sound signal is weak, all frequencies within the range of interest should be amplified

by a factor of 2 (or 6dB), while the phase should remain untouched.

a. Design a suitable cascade of active filters for an equalizer which will produce the

desired effect. Note that an amplification of 3dB at 300Hz and 3,400Hz would be

acceptable, thus design your filter accordingly.

b. Draw the compound stage and give suitable values to all elements in the circuit.



7.9 SOLUTIONS Warm-up:

7.1: -26dB, -87o; 7.2: -31dB, -88 o; 7.3: 0dB, 2.8o; 7.4: -26dB, 87o; 7.5: -40dB, 0.01; 7.6: -9dB, 0.36;

7.7: (-26, 0, -26)dB, (87, 0, -87)o; 7.8: Lookup the Fourier decomposition of a square wave: the first

harmonic has the lowest frequency, with a period equal to that of the square wave i.e. a very low-

low pass filter will turn the square wave into a sine wave of the same period; 7.9: (a) a = 1µA, b = 0V,

c = 1µA, d = -1V, e = -1µA, (b) a = 0V, b = -1mA, c = 2mA, d = 1V, (c) a = 0V, b = -10µA, c = -10tV, (d)

a = 0V, b = -10(1+t)mA, c = -10tV, (e) a = 10µA, b = -10µA, c = -10tV.

Problems:

7.8.1: Q.E.D. Section 7.4.2; 7.8.2: Q.E.D. Section 7.5.4; 7.8.3: Two stages in series: an inverting amp

with R1 = R2 = 10kΩ and an integrator with R = 1MΩ, C = 1µF; 7.8.4: (a) 5kHz, (b) need to reduce the

amplitude of any signal higher than 5kHz to a value which is smaller than the ADC resolution, (c) 5Hz,

(d) -20, -40, -60dB signal reduction respectively; 7.8.5: Resolution: 2.4mV, fmax = 50kHz due to

aliasing gain at 50kHz: (10% of 10V)/5V = 0.2 or -14dB from Bode plot c = (50,000*2*)/5

=62.8krad/s, C=16nF; 7.8.6: (a) 663.14, 60mV, (b) 6,631.4, (c) 66,314, (d) Might make control

unstable due to phase shift introduced by the filter; 7.8.7: (a) band-pass filter with CL =

21.4krad/sec and CH = 1.9krad/sec, (b) e.g. choose L-P: C = 46.7nF and R2 = 1kΩ and H-P: C2 = 0.53F

and R = 1kΩ; since gain plots add, |H|DC = |H|hf = 3db or 10(3/20) = 1.4 C1 = 1.4C2 and R1 = R2 / 1.4.

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ME2-HMTX - Lecture Notes 2012

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