3. Electrical Fundamental

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Module 3 Electrical Fundamentals for

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Preface Thank you for purchasing the Total Training Support Integrated Training System. We are sure you will need no other reference material to pass your EASA Part-66 exam in this Module. U

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These notes have been written by instructors of EASA Part-66 courses, specifically for practitioners of varying experience within the aircraft maintenance industry, and especially those who are self-studying to pass the EASA Part-66 exams. They are specifically designed to meet the EASA Part-66 syllabus and to answer the questions being asked by the UK CAA in their examinations. The EASA Part-66 syllabus for each sub-section is printed at the beginning of each of the chapters in these course notes and is used as the "Learning Objectives".

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1 U We suggest that you take each chapter in-turn, read the text of the chapter a couple of times, if only to familiarise yourself with the location of the information contained within. Then, using your club66pro.co.uk membership, attempt the questions within the respective sub-section, Li and continually refer back to these notes to read-up on the underpinning knowledge required to answer the respective question, and any similar question that you may encounter on your real Part-66 examination. Studying this way, with the help of the question practice and their Li explanations, you will be able to master the subject piece-by-piece, and become proficient in the subject matter, as well as proficient in answering the CAA style EASA part-66 multiple choice

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questions.

We regularly have a review of our training notes, and in order to improve the quality of the notes, and of the service we provide with our Integrated Training System, we would appreciate your feedback, whether positive or negative. So, if you discover within these course notes, any errors or typos, or any subject which is not particularly well, or adequately explained, please tell us, using the `contact-us' feedback page of the club66pro.co.uk website. We will be sure to review your feedback and incorporate any changes necessary. We look forward to hearing from you.

Finally, we appreciate that self-study students are usually also self-financing. We work very hard to cut the cost of our Integrated Training System to the bare minimum that we can provide, and in making your training resources as cost efficient as we can, using, for example, mono printing, but providing the diagrams which would be better provided in colour, on the club66pro.co.uk website. In order to do this, we request that you respect our copyright policy, and refrain from copying, scanning or reprinting these course notes in any way, even for sharing with friends and colleagues. Our survival as a service provider depends on it, and copyright abuse only devalues the service and products available to yourself and your colleagues in the

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Module 3 Chapters

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1. Electron Theory 2.Static Electricity and Conduction 3.Electrical Terminology 4.Generation of Electricity 5.DC Sources of Electricity

6.DC Circuits

7.Resistance/Resistor

8.Power

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9.Capacitance/capacitor 10.Magnetism 11.Inductance/inductor 12.DC Motor/Generator Theory 13. AC Theory 14. Resistive (R), Capacitive (C) and Inductive (L) Circuits 15. Transformers 16. Filters 17. AC Generators 18. AC Motors

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Knowledge Levels - Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, 131 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows:

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • •

A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents

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Module 3.1 Electron Theory Matter Elements and Compounds Molecules Atoms Energy Levels Shells and Sub-shells Valence Compounds Ionisation Conductors, Semiconductors, and Insulators

5 5 5 5 5 8 9 11 11 11 11

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Module 3.1 Enabling Objectives Objective

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Electron Theory Structure and distribution of electrical charges within: atoms, molecules, ions, compounds

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Module 3.1 Electron Theory Matter Matter is defined as anything that occupies space and has weight; that is, the weight and dimensions of matter can be measured. Examples of matter are air, water, automobiles, clothing, and even our own bodies. Thus, we can say that matter may be found in any one of three states: solid, liquid, and gaseous.

Elements and Compounds An ELEMENT is a substance which cannot be reduced to a simpler substance by chemical means. Examples of elements with which you are in everyday contact are iron, gold, silver, copper, and oxygen. There are now over 100 known elements. All the different substances we know about are composed of one or more of these elements.

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When two or more elements are chemically combined, the resulting substance is called a compound. A compound is a chemical combination of elements which can be separated by chemical but not by physical means. Examples of common compounds are water which consists of hydrogen and oxygen, and table salt, which consists of sodium and chlorine. A mixture, on the other hand, is a combination of elements and compounds, not chemically combined, that can be separated by physical means. Examples of mixtures are air, which is made up of nitrogen, oxygen, carbon dioxide, and small amounts of several rare gases, and sea water, which consists chiefly of salt and water.

Molecules A molecule is a chemical combination of two or more atoms, (atoms are described in the next paragraph). In a compound the molecule is the smallest particle that has all the characteristics of the compound. Consider water, for example. Water is matter, since it occupies space and has weight. Depending on the temperature, it may exist as a liquid (water), a solid (ice), or a gas (steam). Regardless of the temperature, it will still have the same composition. If we start with a quantity of water, divide this and pour out one half, and continue this process a sufficient number of times, we will eventually end up with a quantity of water which cannot be further divided without ceasing to be water. This quantity is called a molecule of water. If this molecule of water divided, instead of two parts of water, there will be one part of oxygen and two parts of hydrogen (H20).

Atoms Molecules are made up of smaller particles called atoms. An atom is the smallest particle of an element that retains the characteristics of that element. The atoms of one element, however, differ from the atoms of all other elements. Since there are over 100 known elements, there must be over 100 different atoms, or a different atom for each element. Just as thousands of

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words can be made by combining the proper letters of the alphabet, so thousands of different materials can be made by chemically combining the proper atoms.

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Any particle that is a chemical combination of two or more atoms is called a molecule. The oxygen molecule consists of two atoms of oxygen, and the hydrogen molecule consists of two atoms of hydrogen. Sugar, on the other hand, is a compound composed of atoms of carbon, hydrogen, and oxygen. These atoms are combined into sugar molecules. Since the sugar molecules can be broken down by chemical means into smaller and simpler units, we cannot have sugar atoms. The atoms of each element are made up of electrons, protons, and, in most cases, neutrons, which are collectively called subatomic particles. Furthermore, the electrons, protons, and neutrons of one element are identical to those of any other element. The reason that there are

different kinds of elements is that the number and the arrangement of electrons and protons within the atom are different for the different elements

The electron is considered to be a small negative charge of electricity. The proton has a positive charge of electricity equal and opposite to the charge of the electron. Scientists have measured the mass and size of the electron and proton, and they know how much charge each possesses. The electron and proton each have the same quantity of charge, although the mass of the proton is approximately 1837 times that of the electron. In some atoms there exists a neutral particle called a neutron. The neutron has a mass slightly greater than that of a proton, but it has no electrical charge. According to a popular theory, the electrons, protons, and neutrons of the atoms are thought to be arranged in a manner similar to a miniature solar system. The protons and neutrons form a heavy nucleus with a positive charge, around which the very light electrons revolve. Figure 1.1 shows one hydrogen and one helium atom. Each has a relatively simple structure. The hydrogen atom has only one proton in the nucleus with one electron rotating about it. The helium atom is a little more complex. It has a nucleus made up of two protons and two neutrons, with two electrons rotating about the nucleus. Elements are classified numerically according to the complexity of their atoms. The atomic number of an atom is determined by the number of protons in its nucleus.

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11

ELECTRON S c-) J I

HYDROGEN ' PROTONS

* NEUTRONS

Figure 1.1 - Structure of Hydrogen and Helium In a neutral state, an atom contains an equal number of protons and electrons. Therefore, an r atom of hydrogen - which contains one proton and one electron - has an atomic number of 1; L i and helium, with two protons and two electrons, has an atomic number of 2. The complexity of atomic structure increases with the number of protons and electrons.

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Energy Levels Since an electron in an atom has both mass and motion, it contains two types of energy. By virtue of its motion the electron contains kinetic energy. Due to its position it also contains potential energy. The total energy contained by an electron (kinetic plus potential) is the factor which determines the radius of the electron orbit. In order for an electron to remain in this orbit, it must neither GAIN nor LOSE energy. It is well known that light is a form of energy, but the physical form in which this energy exists is not known. One accepted theory proposes the existence of light as tiny packets of energy called photons. Photons can contain various quantities of energy. The amount depends upon the colour of the

light involved. Should a photon of sufficient energy collide with an orbital electron, the electron will absorb the photon's energy, as shown in figure 1.2. The electron, which now has a greater than normal amount of energy, will jump to a new orbit farther from the nucleus. The first new orbit to which the electron can jump has a radius four times as large as the radius of the original orbit. Had the electron received a greater amount of energy, the next possible orbit to which it could jump would have a radius nine times the original. Thus, each orbit may be considered to represent one of a large number of energy levels that the electron may attain. It must be emphasized that the electron cannot jump to just any orbit. The electron will remain in its lowest orbit until a sufficient amount of energy is available, at which time the electron will accept the energy and jump to one of a series of permissible orbits. An electron cannot exist in the space between energy levels. This indicates that the electron will not accept a photon of energy unless it contains enough energy to elevate itself to one of the higher energy levels. Heat energy and collisions with other particles can also cause the electron to jump orbits.

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Figure 1.2 -- Energy levels in an atom Once the electron has been elevated to an energy level higher than the lowest possible energy level, the atom is said to be in an excited state. The electron will not remain in this excited condition for more than a fraction of a second before it will radiate the excess energy and return to a lower energy orbit. To illustrate this principle, assume that a normal electron has just received a photon of energy sufficient to raise it from the first to the third energy level. In a short

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period of time the electron may jump back to the first level emitting a new photon identical to the one it received. A second alternative would be for the electron to return to the lower level in two jumps; from the third to the second, and then from the second to the first. In this case the electron would emit two photons, one for each jump. Each of these photons would have less energy than the original photon which excited the electron. This principle is used in the fluorescent light where ultraviolet light photons, which are not visible to the human eye, bombard a phosphor coating on the inside of a glass tube. The phosphor electrons, in returning to their normal orbits, emit photons of light that are visible. By using the proper chemicals for the phosphor coating, any colour of light may be obtained, including white.

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This same principle is also used in lighting up the screen of a television picture tube. The basic principles just developed apply equally well to the atoms of more complex elements. In atoms containing two or more electrons, the electrons interact with each other and the exact path of any one electron is very difficult to predict. However, each electron lies in a specific energy band and the orbits will be considered as an average of the electron's position.

Shells and Sub-shells The difference between the atoms, insofar as their chemical activity and stability are concerned, is dependent upon the number and position of the electrons included within the atom. How are these electrons positioned within the atom? In general, the electrons reside in groups of orbits called shells. These shells are elliptically shaped and are assumed to be located at fixed intervals. Thus, the shells are arranged in steps that correspond to fixed energy levels. The shells, and the number of electrons required to fill them, may be predicted by the employment of Pauli's exclusion principle. Simply stated, this principle specifies that each shell will contain a maximum of 2n2 electrons, where n corresponds to the shell number starting with the one closest to the nucleus. By this principle, the second shell, for example, would contain 2(2 )2 or 8 electrons when full. In addition to being numbered, the shells are also given letter designations, as pictured in figure 1-3. Starting with the shell closest to the nucleus and progressing outward, the shells are labelled K, L, M, N, 0, P, and Q, respectively. The shells are considered to be full, or complete, when they contain the following quantities of electrons: two in the K shell, eight in the L shell, 18 in the M shell, and so on, in accordance with the exclusion principle. Each of these shells is a major shell and can be divided into sub-shells, of which there are four, labelled s, p, d, and f. Like the major shells, the sub-shells are also limited as to the number of electrons which they can contain. Thus, the "s" sub-shell is complete when it contains two electrons, the "p" sub-shell when it contains 6, the "d" sub-shell when it contains 10, and the "f" sub-shell when it contains 14 electrons.

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LETTER DESIGNA770N�

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Figure 1.3 - Shells in an atom In as much as the K shell can contain no more than two electrons, it must have only one subshell, the s sub-shell. The M shell is composed of three sub-shells: s, p, and d. If the electrons in the s, p, and d sub-shells are added, their total is found to be 18, the exact number required to fill the M shell. Notice the electron configuration for copper illustrated in figure 1.4. The copper atom contains 29 electrons, which completely fill the first three shells and sub-shells, leaving one electron in the "s" sub-shell of the N shell.

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Valence The number of electrons in the outermost shell determines the valence of an atom. For this reason, the outer shell of an atom is called the valence shell; and the electrons contained in this shell are called valence electrons. The valence of an atom determines its ability to gain or lose an electron, which in turn determines the chemical and electrical properties of the atom. An atom that is lacking only one or two electrons from its outer shell will easily gain electrons to complete its shell, but a large amount of energy is required to free any of its electrons. An atom having a relatively small number of electrons in its outer shell in comparison to the number of electrons required to fill the shell will easily lose these valence electrons. The valence shell always refers to the outermost shell.

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Pure substances made up more than 1 element which have been joined together by a chemical reaction therefore the atoms are difficult to separate. The properties of a compound are different from the atoms that make it up. Splitting of a compound is called chemical analysis.

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Note that a compound: • consists of atoms of two or more different elements bound together, • can be broken down into a simpler type of matter (elements) by chemical means (but not by physical means), • has properties that are different from its component elements, and • always contains the same ratio of its component atoms.

Ionisation r-,

When the atom loses electrons or gains electrons in this process of electron exchange, it is said to be ionized. For ionisation to take place, there must be a transfer of energy which results in a change in the internal energy of the atom. An atom having more than its normal amount of electrons acquires a negative charge, and is called a negative ion. The atom that gives up some of its normal electrons is left with less negative charges than positive charges and is called a positive ion. Thus, ionisation is the process by which an atom loses or gains electrons.

Conductors, Semiconductors, and Insulators

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In this study of electricity and electronics, the association of matter and electricity is important. Since every electronic device is constructed of parts made from ordinary matter, the effects of electricity on matter must be well understood. As a means of accomplishing this, all elements of which matter is made may be placed into one of three categories: conductors, semiconductors, and insulators, depending on their ability to conduct an electric current. conductors are elements which conduct electricity very readily, insulators have an extremely high resistance to the flow of electricity. All matter between these two extremes may be called semiconductors.

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The electron theory states that all matter is composed of atoms and the atoms are composed of smaller particles called protons, electrons, and neutrons. The electrons orbit the nucleus which contains the protons and neutrons. It is the valence electrons (the electrons in the outer shell) that we are most concerned with in electricity. These are the electrons which are easiest to break loose from their parent atom. Normally, conductors have three or less valence electrons; insulators have five or more valence electrons; and semiconductors usually have four valence electrons. The fewer the valence electrons, the better conductor of electricity it will be. Copper, for example, has just one valence electron.

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The electrical conductivity of matter is dependent upon the atomic structure of the material from

which the conductor is made. In any solid material, such as copper, the atoms which make up

the molecular structure are bound firmly together. At room temperature, copper will contain a considerable amount of heat energy. Since heat energy is one method of removing electrons from their orbits, copper will contain many free electrons that can move from atom to atom. When not under the influence of an external force, these electrons move in a haphazard manner within the conductor. This movement is equal in all directions so that electrons are not lost or gained by any part of the conductor. When controlled by an external force, the electrons move generally in the same direction. The effect of this movement is felt almost instantly from one end of the conductor to the other. This electron movement is called an electric current. Some metals are better conductors of electricity than others. Silver, copper, gold, and aluminium are materials with many free electrons and make good conductors. Silver is the best conductor, followed by copper, gold, and aluminium. Copper is used more often than silver because of cost. Aluminium is used where weight is a major consideration, such as in hightension power lines, with long spans between supports. Gold is used where oxidation or corrosion is a consideration and a good conductivity is required. The ability of a conductor to handle current also depends upon its physical dimensions. Conductors are usually found in the form of wire, but may be in the form of bars, tubes, or sheets. Non-conductors have few free electrons. These materials are called insulators. Some examples of these materials are rubber, plastic, enamel, glass, dry wood, and mica. Just as there is no perfect conductor, neither is there a perfect insulator. Some materials are neither good conductors nor good insulators, since their electrical characteristics fall between those of conductors and insulators. These in-between materials are classified as semiconductors. Germanium and silicon are two common semiconductors used in solid-state devices.

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Atomic

Element

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Ox en Fluorine Neon Sodium Magnesium Aluminium Silicon Phos horus Sul hur Chlorine Aron Potassium Calcium Scandium Titanium Vanadium Chromium Man anese Iron Cobalt Nickel Cer Zinc Gallium Germanium Arsenic Selenium Bromine K ton Rubidium Strontium Yttrium Zirconium Niobium Mol bdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium

No.

Electrons per Shell

Atomic

Element

Electrons per Shell KLMN

0

53 54 55 56 57 58 59

Iodine Xenon Cesium Barium Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium D s rosium Holmium Erbium Thulium Ytterbium Lutetium Halnium Tantalum Tun sten Rhenium Osmium Iridium Platinum Gold Mercu Thallium Lead Bismuth Polonium Asatine Radon Francium Radium Actinium Thorium Proactinium Uranium Ne tunium Plutonium Amerium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

7 8 8 8 9 9 9 9 9 9 9 9 9 9

No.

KLMNOPQ L

4 6 7

0 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52

I 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 2 3 4 5 -60 -

61 62 63 64 65 66 67 68 69

1

2 2

2 2 2 2 2 2 2

8 8 8 8

6 8 8 8 8 8 8 8 8 8 8 18 18 18 18 18 18 18 18 18 18 18 18 18 1

1 2 2 2 2 1 2 2 2 2 1 2 3 4 5 6 7 8 8 1 8 2 9 2 10 2 12 13 14 15 16 18 0 18 1 18 2 18 3 1B 4 18 5 18_1 18 6

71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103

8 8 8 8 8 8 8 8 8 8 S 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18

8 18 8 8 8 19 20 21 22 23 24 25 26 7

31

32

3

8 8 8 18 18 18 18 18 18 18 8 8 8 8 8 1B 18 18

3 3 32 32 32 32 32 32 32 32 32 32 32 32 32 32

9 9 9 10 11 12 13 14 15 16 18 18 18 18 18 18 18 18 18 18 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

PQ 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 3 4 5 6 7 8 8 8 9 9 9 9 9

9 9 9 9 9 9 9 9

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

Table 1.1 - Electrons per shell

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basic knowledge levels. The knowledge level indicators are defined as follows:

LEVEL 7 Objectives: • • examples. • The applicant should be able to use typical terms.

LEVEL 2 • An ability to apply that knowledge. Objectives: • • The applicant should be able to give a general description of the subject using, as appropriate, typicalexamples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing thesubject. • subject. •

The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • • A capacity to combine and apply the separate elements of knowledge in a logical and comprehensivemanner. Objectives: • and specific examples. • • The applicant should be able to read, understand and prepare sketches, simple drawings and schematicsdescribing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer'sinstructions. • The applicant should be able to interpret results from various sources and measurements and applycorrective action where appropriate.

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Table of Contents

Module 3.2 Static Electricity and Conduction

Introduction

Static Electricity Nature of Charges Charged Bodies Coulomb's Law of Charges Unit of Charge Electric Fields Conduction of Electricity in Solids, Liquids and a Vacuum

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5 6 7 7 8 8 8 9

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Module 3.2 Enabling Objectives Objective

EASA 66 Reference

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Static Electricity and Conduction Static electricity and distribution of electrostatic charges Electrostatic laws of attraction and repulsion Units of charge, Coulomb's Law Conduction of electricity in solids, liquids, gases and a

3.2

2

vacuum

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Module 3.2 Static- Electricity and Conduction Introduction

Electrostatics (electricity at rest) is a subject with which most persons entering the field of electricity and electronics are somewhat familiar. For example, the way a person's hair stands on end after a vigorous rubbing is an effect of electrostatics. While pursuing the study of electrostatics, you will gain a better understanding of this common occurrence. Of even greater significance, the study of electrostatics will provide you with the opportunity to gain important background knowledge and to develop concepts which are essential to the understanding of Ll electricity and electronics.

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Interest in the subject of static electricity can be traced back to the Greeks. Thales of Miletus, a Greek philosopher and mathematician, discovered that when an amber rod is rubbed with fur, the rod has the amazing characteristic of attracting some very light objects such as bits of paper and shavings of wood. About 1600, William Gilbert, an English scientist, made a study of other substances which had been found to possess qualities of attraction similar to amber. Among these were glass, when rubbed with silk, and ebonite, when rubbed with fur. Gilbert classified all the substances which possessed properties similar to those of amber as electrics, a word of Greek origin meaning amber. Because of Gilbert's work with electrics, a substance such as amber or glass when given a vigorous rubbing was recognized as being electrified, or charged with electricity. In the year 1733, Charles Dufay, a French scientist, made an important discovery about electrification. He found that when a glass was rubbed with fur, both the glass rod and the fur became electrified. This realization came when he systematically placed the glass rod and the fur near other electrified substances and found that certain substances which were attracted to the glass rod were repelled by the fur, and vice versa. From experiments such as this, he concluded that there must be two exactly opposite kinds of electricity. Benjamin Franklin, American statesman, inventor, and philosopher, is credited with first using the terms positive and negative to describe the two opposite kinds of electricity. The charge produced on a glass rod when it is rubbed with silk, Franklin labelled positive. He attached the term negative to the charge produced on the silk. Those bodies which were not electrified or charged, he called neutral.

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Static Electricity In a natural or neutral state, each atom in a body of matter will have the proper number of electrons in orbit around it. Consequently, the whole body of matter composed of the neutral atoms will also be electrically neutral. In this state, it is said to have a "zero charge." Electrons will neither leave nor enter the neutrally charged body should it come in contact with other neutral bodies. If, however, any number of electrons is removed from the atoms of a body of matter, there will remain more protons than electrons and the whole body of matter will become electrically positive. Should the positively charged body come in contact with another body having a normal charge, or having a negative (too many electrons) charge, an electric current will flow between them. Electrons will leave the more negative body and enter the positive body. This electron flow will continue until both bodies have equal charges. When two bodies of matter have unequal charges and are near one another, an electric force is exerted between them

because of their unequal charges. However, since they are not in contact, their charges

cannot equalize. The existence of such an electric force, where current cannot flow, is referred to as static electricity. ("Static" in this instance means "not moving.") It is also referred to as an electrostatic force. One of the easiest ways to create a static charge is by friction. When two pieces of matter are rubbed together, electrons can be "wiped off" one material onto the other. If the materials used are good conductors, it is quite difficult to obtain a detectable charge on either, since equalizing currents can flow easily between the conducting materials. These currents equalize the charges almost as fast as they are created. A static charge is more easily created between nonconducting materials. When a hard rubber rod is rubbed with fur, the rod will accumulate electrons given up by the fur, as shown in figure 2.1. Since both materials are poor conductors, very little equalizing current can flow, and an electrostatic charge builds up. When the charge becomes great enough, current will flow regardless of the poor conductivity of the materials. These currents will cause visible sparks and produce a crackling sound. +CHARGES AND ELECTRONS ARE PRESENT IN EQUAL QUANTITIES IN THE ROD AND FUR

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ELECTRONS ARE TRANSFERRED FROM THE FUR TO THE ROD

Figure 2.1 - Static charges

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Nature of Charges When in a natural or neutral state, an atom has an equal number of electrons and protons. Because of this balance, the net negative charge of the electrons in orbit is exactly balanced by the net positive charge of the protons in the nucleus, making the atom electrically neutral. An atom becomes a positive ion whenever it loses an electron, and has an overall positive charge. Conversely, whenever an atom acquires an extra electron, it becomes a negative ion and has a negative charge. Due to normal molecular activity, there are always ions present in any material. If the number of positive ions and negative ions is equal, the material is electrically neutral. When the number of positive ions exceeds the number of negative ions, the material is positively charged. The material is negatively charged whenever the negative ions outnumber the positive ions. Since ions are actually atoms without their normal number of electrons, it is the excess or the lack of electrons in a substance that determines its charge. In most solids, the transfer of charges is by movement of electrons rather than ions. The transfer of charges by ions will become more significant when we consider electrical activity in liquids and gases. At this time, we will discuss electrical behaviour in terms of electron movement.

Charged Bodies ri

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One of the fundamental laws of electricity is that like charges repel each other and unlike charges attract each other. A positive charge and negative charge, being unlike, tend to move toward each other. In the atom, the negative electrons are drawn toward the positive protons in the nucleus. This attractive force is balanced by the electron's centrifugal force caused by its rotation about the nucleus. As a result, the electrons remain in orbit and are not drawn into the nucleus. Electrons repel each other because of their like negative charges, and protons repel each other because of their like positive charges.

-, The law of charged bodies may be demonstrated by a simple experiment. Two pith (paper pulp) Li balls are suspended near one another by threads, as shown in figure 2.2.

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(B)

(C)

Figure 2.2 - Repulsion and attraction of charged bodies

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If a hard rubber rod is rubbed with fur to give it a negative charge and is then held against the right-hand ball in part (A), the rod will give off a negative charge to the ball. The right-hand ball will have a negative charge with respect to the left-hand ball. When released, the two balls will be drawn together, as shown in figure 2.2 (A). They will touch and remain in contact until the left-hand ball gains a portion of the negative charge of the right-hand ball, at which time they will swing apart as shown in figure 2.2 (C). If a positive or a negative charge is placed on both balls (figure 2-2 (B)), the balls will repel each other.

Coulomb's Law of Charges The relationship between attracting or repelling charged bodies was first discovered and written about by a French scientist named Charles A. Coulomb. Coulomb's Law states that

Charged bodies attract or repel each other with a force that is directly proportional to the product of their individual charges, and is inversely proportional to the square of the distance between them. The amount of attracting or repelling force which acts between two electrically charged bodies in free space depends on two things - (1) their charges and (2) the distance between them.

Unit of Charge The process of electrons arriving or leaving is exactly what happens when certain combinations of materials are rubbed together: electrons from the atoms of one material are forced by the rubbing to leave their respective atoms and transfer over to the atoms of the other material. In other words, electrons comprise the "fluid" hypothesized by Benjamin Franklin. The operational definition of a coulomb as the unit of electrical charge (in terms of force generated between point charges) was found to be equal to an excess or deficiency of about 6,280,000,000,000,000,000 electrons. Or, stated in reverse terms, one electron has a charge of about 0.00000000000000000016 coulombs. Being that one electron is the smallest known carrier of electric charge, this last figure of charge for the electron is defined as the elementary charge. 1 coulomb = 6,280,000,000,000,000,000 electrons

Electric Fields The space between and around charged bodies in which their influence is felt is called an electric field of force. It can exist in air, glass, paper, or a vacuum. electrostatic fields and dielectric fields are other names used to refer to this region of force. Fields of force spread out in the space surrounding their point of origin and, in general, diminish in proportion to the square of the distance from their source. The field about a charged body is generally represented by lines which are referred to as electrostatic lines of force. These lines are imaginary and are used merely to represent the direction and strength of the field. To avoid confusion, the lines of force exerted by a positive charge are always shown leaving the charge, and for a negative charge they are shown entering. Figure 2.3 illustrates the use of lines to represent the field about charged bodies.

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(A)

(B)

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Figure 2.3 - Electrostatic lines of force Figure 2.3 (A) represents the repulsion of like-charged bodies and their associated fields. Part (B) represents the attraction of unlike-charged bodies and their associated fields.

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Conduction of Electricity in Solids, Liquids and a Vacuum

1

Solids Electric current is the movement of valence electrons. Conduction is the name of this process. It is more fully described in Chapter 1 of this Module. Generally, only metals conduct electricity. Some conduct better than others.

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The exception to this is graphite (one of the forms of the element Carbon). Carbon is a nonmetal which exhibits some electrical conductivity.

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Liquids The only liquid elements which conduct are the liquid metals. At room temperature liquid L i mercury is a conductor. Other metals continue to conduct electricity when they are melted. Non-metals such as water, alcohol, ethanoic acid, propanone, hexane and so on, are all non conductors of electricity. However, it is possible to make some non-conducting liquids conduct electricity, by a process called ionization. Ionized substances are called ionic substances. Ionic substances are made of charged particles - positive and negative ions. In the solid state they are held very firmly in place in a lattice structure. In the solid state the ions cannot move about at all. When the ionic solid is melted, the bonds holding the ions in place in the lattice are broken. The ions can then move around freely.

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When an electric current is applied to an ionic melt the electricity is carried by the ions that are now able to move. In an ionic melt the electric current is a flow of ions.

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Taking water as an example. Remember firstly, that water is considered to be a non-conductor of electricity. It can allow some electricity through it if a high voltage is applied to it. This is due to the presence of a minute concentration of Hfi and OH- ions in the water. However, electrons cannot flow through water. Covalent substances do not conduct at all in solution. Ionic substances are able to conduct electricity when they are dissolved in water. The reason lies again in the fact that ionic substances are made of charged particles - ions. When the ionic solid is dissolved in water the ionic lattice breaks up and the ions become free to move around in the water. When you pass electricity through the ionic solution, the ions are able to carry the electric current because of their ability to move freely. A solution conducts by means of freely moving ions. An electrolyte is a liquid which can carry an electric current through it. Ionic solutions and ionic melts are all electrolytes. Electrolysis describes the process which takes place when an ionic solution or melt has electricity passed through it. Gases A gas in its normal state is one of the best insulators known. However, in a similar way as liquid, it can be forced to conduct electricity by ionisation of the gas molecules. Ionisation of the gas molecules can be effected by extremely high voltages. For example, lightning, is electric current flowing through an ionised path through air due to the huge electrical potential difference between the storm cloud and the ground. In air, and other ordinary gases, the dominant source of electrical conduction is via a relatively small number of mobile ions produced by radioactive gases, ultraviolet light, or cosmic rays. Since the electrical conductivity is extremely low, gases are dielectrics or insulators. However, once the applied electric field approaches the breakdown value, free electrons become sufficiently accelerated by the electric field to create additional free electrons by colliding, and ionizing, neutral gas atoms or molecules in a process called avalanche breakdown. The breakdown process forms a plasma that contains a significant number of mobile electrons and positive ions, causing it to behave as an electrical conductor. In the process, it forms a light emitting conductive path, such as a spark, arc or lightning.

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Figure 2.4 - Lightning is electric current flowing through an ionized plasma of its own making

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Plasma is the state of matter where some of the electrons in a gas are stripped or "ionized" from their molecules or atoms. A plasma can be formed by high temperature, or by application of a high electric or alternating magnetic field as noted above. Due to their lower mass, the electrons in a plasma accelerate more quickly in response to an electric field than the heavier positive ions, and hence carry the bulk of the current. Vacuum It is a common belief that electricity cannot flow through a vacuum. This is however incorrect. Remember that a conductor is "something through which electricity can flow," rather than

"something which contains movable electricity." A vacuum offers no blockage to moving

charges. Should electrons be injected into a vacuum, the electrons will flow uninhibited and unretarded. As such, a vacuum is an ideal conductor.

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This fact is taken advantage of in many situations, from televisions to vacuum valves. A vacuum arc can arise when the surfaces of metal electrodes in contact with a good vacuum begin to emit electrons either through heating (thermionic emission) or via an electric field that is sufficient to cause field emission. Once initiated, a vacuum arc can persist since the freed particles gain kinetic energy from the electric field, heating the metal surfaces through high speed particle collisions. This process can create an incandescent cathode spot which frees more particles, thereby sustaining the arc. At sufficiently high currents an incandescent anode spot may also be formed. Electrical discharge in vacuum is important for certain types of vacuum tubes and for high voltage vacuum switches.

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Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Ltd.

Knowledge Levels - Category A, B1, B2 and C Aircraft Maintenance Licence ._i

Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels.

The knowledge level indicators are defined as follows:

7 �. J

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and

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examples. • The applicant should be able to use typical terms.

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • A detailed knowledge of the theoretical and practical aspects of the subject. • A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents u UModule 3.3 Electrical Terminology Electrical Energy Electrical Charges Electric Current Electrical Resistance iL �i

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LI LI

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Conductance

Electrical Laws

5 5 6 8 13

13 14

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Module 3.3 Enabling Objectives Objective Electrical Terminology The following terms, their units and factors affecting them: potential difference, electromotive force, voltage, current, resistance, conductance, charge, conventional current flow, electron flow

EASA 66 Reference

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3.3

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Module 3.3 Electrical Terminology Electrical Energy In the field of physical science, work must be defined as the product of force and displacement. That is, the force applied to move an object and the distance the object is moved are the factors of work performed. U It is important to notice that no work is accomplished unless the force applied causes a change in the position of a stationary object, or a change in the velocity of a moving object. A worker may tire by pushing against a heavy wooden crate, but unless the crate moves, no work will be accomplished. i

U Energy In our study of energy and work, we must define energy as the ability to do work. In order to perform any kind of work, energy must be expended (converted from one form to another). } U Energy supplies the required force, or power, whenever any work is accomplished. One form of energy is that which is contained by an object in motion. When a hammer is set in motion in the direction of a nail, it possesses energy of motion. As the hammer strikes the nail, the energy of motion is converted into work as the nail is driven into the wood. The distance the nail is driven into the wood depends on the velocity of the hammer at the time it strikes the nail. 1 Energy contained by an object due to its motion is called kinetic energy. Assume that the hammer is suspended by a string in a position one meter above a nail. As a result of gravitational attraction, the hammer will experience a force pulling it downward. If the string is suddenly cut, the force of gravity will pull the hammer downward against the nail, driving it into the wood. While the hammer is suspended above the nail it has ability to do work because of its elevated position in the earth's gravitational field. Since energy is the ability to do work, the

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hammer contains energy.

Energy contained by an object due to its position is called potential energy. The amount of potential energy available is equal to the product of the force required to elevate the hammer and the height to which it is elevated. L11

Another example of potential energy is that contained in a tightly coiled spring. The amount of energy released when the spring unwinds depends on the amount of force required to wind the spring initially.

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Electrical Charges From the previous study of electrostatics, you learned that a field of force exists in the space surrounding any electrical charge. The strength of the field is directly dependent on the force of the charge. The charge of one electron might be used as a unit of electrical charge, since charges are created by displacement of electrons; but the charge of one electron is so small that it is impractical to use. The practical unit adopted for measuring charges is the coulomb, named after the scientist Charles Coulomb. One coulomb is equal to the charge of 6,280,000,000,000,000,000 (six quintillion two hundred and eighty quadrillion) or 6.28 x 1018 electrons. When a charge of one coulomb exists between two bodies, one unit of electrical potential energy exists, which is called the difference of potential between the two bodies. This is referred to as electromotive force, or voltage, and the unit of measure is the volt. Electrical charges are created by the displacement of electrons, so that there exists an excess of electrons at one point, and a deficiency at another point. Consequently, a charge must always have either a negative or positive polarity. A body with an excess of electrons is considered to be negative, whereas a body with a deficiency of electrons is positive. A difference of potential can exist between two points, or bodies, only if they have different charges. In other words, there is no difference in potential between two bodies if both have a deficiency of electrons to the same degree. If, however, one body is deficient of 6 coulombs (representing 6 volts), and the other is deficient by 12 coulombs (representing 12 volts), there is a difference of potential of 6 volts. The body with the greater deficiency is positive with respect to the other. In most electrical circuits only the difference of potential between two points is of importance and the absolute potentials of the points are of little concern. Very often it is convenient to use one standard reference for all of the various potentials throughout a piece of equipment. For this reason, the potentials at various points in a circuit are generally measured with respect to the metal chassis on which all parts of the circuit are mounted. The chassis is considered to be at zero potential and all other potentials are either positive or negative with respect to the chassis. When used as the reference point, the chassis is said to be at ground potential. Occasionally, rather large values of voltage may be encountered, in which case the volt becomes too small a unit for convenience. In a situation of this nature, the kilovolt (kV), meaning 1,000 volts, is frequently used. As an example, 20,000 volts would be written as 20 kV. In other cases, the volt may be too large a unit, as when dealing with very small voltages. For this purpose the millivolt (mV), meaning one-thousandth of a volt, and the microvolt (µV), meaning one-millionth of a volt, are used. For example, 0.001 volt would be written as 1 mV, and 0.000025 volt would be written as 25 µV. When a difference in potential exists between two charged bodies that are connected by a conductor, electrons will flow along the conductor. This flow is from the negatively charged body to the positively charged body, until the two charges are equalized and the potential difference no longer exists. 3-6

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An analogy of this action is shown in the two water tanks connected by a pipe and valve in figure 3.1. At first the valve is closed and all the water is in tank A. Thus, the water pressure across the valve is at maximum. When the valve is opened, the water flows through the pipe from A to B until the water level becomes the same in both tanks. The water then stops flowing in the pipe, because there is no longer a difference in water pressure between the two tanks.

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TAN K B

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Electron movement through an electric circuit is directly proportional to the difference in potential or electromotive force (EMF), across the circuit, just as the flow of water through the pipe in figure 3.1 is directly proportional to the difference in water level in the two tanks. A fundamental law of electricity is that the electron flow is directly proportional to the applied voltage. If the voltage is increased, the flow is increased. If the voltage is decreased, the flow is decreased.

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Electric Current Electron flow It has been proven that electrons (negative charges) move through a conductor in response to an electric field. Electron current flow will be used throughout this explanation. Electron current is defined as the directed flow of electrons. The direction of electron movement is from a region of negative potential to a region of positive potential. Therefore electron flow can be said to flow from negative to positive. The direction of current flow in a material is determined by the polarity of the applied voltage. Conventional Current Flow In the UK and Europe, conventional current flow is said to be from positive to negative potential the opposite way to the actual flow of electrons. Conventional current was defined early in the history of electrical science as a flow of positive charge. In solid metals, like wires, the positive charge carriers are immobile, and only the negatively charged electrons flow. Because the electron carries negative charge, the electron current is in the direction opposite to that of conventional (or electric) current.

Flow of positive charge L..J

Electric charge moves from the positive side of the power source to the negative. Figure 3.2 - Conventional current In other conductive materials, the electric current flow direction is due to the flow of charged particles in both directions at the same time. Electric currents in electrolytes are flows of electrically charged atoms (ions), which exist in both positive and negative varieties. For example, an electrochemical cell may be constructed with salt water (a solution of sodium chloride) on one side of a membrane and pure water on the other. The membrane lets the positive sodium ions pass, but not the negative chloride ions, so a net current results. Electric currents in plasma are flows of electrons as well as positive and negative ions. In ice and in certain solid electrolytes, flowing protons constitute the electric current. To simplify this situation, the original definition of conventional current still stands. There are also materials where the electric current is due to the flow of electrons and yet it is conceptually easier to think of the current as due to the flow of positive "holes" (the spots that should have an electron to make the conductor neutral). This is the case in a p-type semiconductor.

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Random Drift All materials are composed of atoms, each of which is capable of being ionised. If some form of energy, such as heat, is applied to a material, some electrons acquire sufficient energy to move to a higher energy level. As a result, some electrons are freed from their parent atom's which then becomes ions. Other forms of energy, particularly light or an electric field, will cause ionisation to occur.

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The number of free electrons resulting from ionisation is dependent upon the quantity of energy applied to a material, as well as the atomic structure of the material. At room temperature some materials, classified as conductors, have an abundance of free electrons. Under a similar condition, materials classified as insulators have relatively few free electrons. In a study of electric current, conductors are of major concern. Conductors are made up of atoms that contain loosely bound electrons in their outer orbits. Due to the effects of increased energy, these outermost electrons frequently break away from their atoms and freely drift throughout the material. The free electrons, also called mobile electrons, take a path that is not predictable and drift about the material in a haphazard manner. Consequently such a movement is termed random drift. It is important to emphasize that the random drift of electrons occurs in all materials. The degree of random drift is greater in a conductor than in an insulator.

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Directed Drift Associated with every charged body there is an electrostatic field. Bodies that are charged alike repel one another and bodies with unlike charges attract each other. An electron will be affected by an electrostatic field in exactly the same manner as any negatively charged body. It is repelled by a negative charge and attracted by a positive charge. If a conductor has a difference in potential impressed across it, as shown in figure 3.3, a direction is imparted to the random drift. This causes the mobile electrons to be repelled away from the negative terminal and attracted toward the positive terminal. This constitutes a general migration of electrons from one end of the conductor to the other. The directed migration of mobile electrons due to the potential difference is called directed drift.

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Figure 3.3 - Directed drift

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The directed movement of the electrons occurs at a relatively low velocity (rate of motion in a particular direction). The effect of this directed movement, however, is felt almost instantaneously, as explained by the use of figure 3.3. As a difference in potential is impressed across the conductor, the positive terminal of the battery attracts electrons from point A. Point A now has a deficiency of electrons. As a result, electrons are attracted from point B to point A. Point B has now developed an electron deficiency, therefore, it will attract electrons. This same effect occurs throughout the conductor and repeats itself from points D to C. At the same instant the positive battery terminal attracted electrons from point A, the negative terminal repelled electrons toward point D. These electrons are attracted to point D as it gives up electrons to point C. This process is continuous for as long as a difference of potential exists across the conductor. Though an individual electron moves quite slowly through the conductor, the effect of a directed drift occurs almost instantaneously. As an electron moves into the conductor at point D, an electron is leaving at point A. This action takes place at approximately the speed a light (186,000 Miles Per Second).

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FI Li Figure 3.4 - Effect of directed drift.

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Magnitude of Current Flow Electric current has been defined as the directed movement of electrons. Directed drift, therefore, is current and the terms can be used interchangeably. The expression directed drift is particularly helpful in differentiating between the random and directed motion of electrons. However, current flow is the terminology most commonly used in indicating a directed movement of electrons. L! L

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The magnitude of current flow is directly related to the amount of energy that passes through a conductor as a result of the drift action. An increase in the number of energy carriers (the mobile electrons) or an increase in the energy of the existing mobile electrons would provide an increase in current flow. When an electric potential is impressed across a conductor, there is an increase in the velocity of the mobile electrons causing an increase in the energy of the carriers. There is also the generation of an increased number of electrons providing added carriers of energy. The additional number of free electrons is relatively small, hence the magnitude of current flow is primarily dependent on the velocity of the existing mobile electrons. The magnitude of current flow is affected by the difference of potential in the following manner. Initially, mobile electrons are given additional energy because of the repelling and attracting electrostatic field. If the potential difference is increased, the electric field will be stronger, the amount of energy imparted to a mobile electron will be greater, and the current will be increased. If the potential difference is decreased, the strength of the field is reduced, the

L energy supplied to the electron is diminished, and the current is decreased.

Measurement of Current The magnitude of current is measured in amperes. A current of one ampere is said to flow L when one coulomb of char We passes a point in one second. Remember, one coulomb is equal to the charge of 6.28 x 101 electrons.

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Frequently, the ampere is much too large a unit for measuring current. Therefore, the milliampere (mA), one-thousandth of an ampere, or the microampere (µA), one-millionth of an ampere, is used. The device used to measure current is called an ammeter and will be discussed in detail in a later module. A current of 1 Amp is flowing when a quantity of 1 Coulomb of charge flows for 1 second. The current I in amperes can be calculated with the following equation:

n Where: n Q is the electric charge in coulombs (ampere seconds) t

is the time in seconds

It follows that:

=.-t

and

t rj

J

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Electrical Resistance

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It is known that the directed movement of electrons constitutes a current flow. It is also known that the electrons do not move freely through a conductor's crystalline structure. Some materials offer little opposition to current flow, while others greatly oppose current flow. This opposition to current flow is known as resistance (R), and the unit of measure is the ohm. The standard of

measure for one ohm is the resistance provided at zero degrees Celsius by a column of mercury having a cross-sectional area of one square millimetre and a length of 106.3

centimetres.

A conductor has one ohm of resistance when an applied potential of one volt produces a current of one ampere. The symbol used to represent the ohm is the Greek letter omega (ca). Resistance, although an electrical property, is determined by the physical structure of a material. The resistance of a material is governed by many of the same factors that control current flow. Therefore, in a subsequent discussion, the factors that affect current flow will be used to assist in the explanation of the factors affecting resistance.

Conductance

Electricity is a study that is frequently explained in terms of opposites. The term that is the opposite of resistance is conductance. Conductance is the ability of a material to pass electrons. The factors that affect the magnitude of resistance are exactly the same for conductance, but they affect conductance in the opposite manner. Therefore, conductance is directly proportional to area, and inversely proportional to the length of the material. The temperature of the material is definitely a factor, but assuming a constant temperature, the conductance of a material can be calculated. The unit of conductance is the mho (G), which is ohm spelled backwards. Recently the term mho has been redesignated siemens (S). Whereas the symbol used to represent resistance (R) is the Greek letter omega (92), the symbol used to represent conductance (G) is (S). The relationship that exists between resistance (R) and conductance (G) or (S) is a reciprocal one. A reciprocal of a number is 'one' divided by that number. In terms of resistance and conductance:

R=-, G= G R

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Electrical Laws Faraday's Law Faraday's law of induction states that the induced electromotive force in a closed loop of wire is directly proportional to the time rate of change of magnetic flux through the loop.

J

Ohm's Law An electrical circuit, the current passing through a conductor between two points is directly proportional to the potential difference (i.e. voltage drop or voltage) across the two points, and inversely proportional to the resistance between them. Kirchhoff's Laws Current Law -At any point in an electrical circuit where charge density is not changing in time, the sum of currents flowing towards that point is equal to the sum of currents flowing away from that point. Voltage Law - The directed sum of the electrical potential differences around any closed circuit must be zero. Lens's Law The induced current in a loop is in the direction that creates a magnetic field that opposes the change in magnetic flux through the area enclosed by the loop. That is, the induced current tends to keep the original magnetic flux through the field from changing.

L. J

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Electrical Fundamentals 3.4 Generation of Electricity

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Knowledge Levels - Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2

basic knowledge levels.

The knowledge level indicators are defined as follows:

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3

n

• •

A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's

instructions.

• The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents

L

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Module 3.4 Generation of Electricity How Voltage is Produced Voltage Produced by Friction Voltage Produced by Pressure Voltage Produced by Heat Voltage Produced by Light

Voltage Produced by Chemical Action 1-7

Voltage Produced by Magnetism

5 5 6 6 7 7

9 10

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EASA 66 Reference

Level

Generation of Electricity

3.4

1

Production of electricity by the following methods: light, heat, friction, pressure, chemical action, magnetism and motion

71

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L Module 3.4 Generation of Electricity

rk L

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How Voltage is Produced It has been demonstrated that a charge can be produced by rubbing a rubber rod with fur. Because of the friction involved, the rod acquires electrons from the fur, making it negative; the fur becomes positive due to the loss of electrons. These quantities of charge constitute a difference of potential between the rod and the fur. The electrons which make up this difference

of potential are capable of doing work if a discharge is allowed to occur.

To be a practical source of voltage, the potential difference must not be allowed to dissipate, but must be maintained continuously. As one electron leaves the concentration of negative charge, another must be immediately provided to take its place or the charge will eventually diminish to the point where no further work can be accomplished. A voltage source, therefore, is a device which is capable of supplying and maintaining voltage while some type of electrical apparatus is connected to its terminals. The internal action of the source is such that electrons are continuously removed from one terminal, keeping it positive, and simultaneously supplied to the second terminal which maintains a negative charge. Presently, there are six known methods for producing a voltage or electromotive force (EMF). Some of these methods are more widely used than others, and some are used mostly for specific applications. Following is a list of the six known methods of producing a voltage.

L L

• Friction - Voltage produced by rubbing certain materials together. • Pressure (piezoelectricity) - Voltage produced by squeezing crystals of certain substances • Heat (thermoelectricity) - Voltage produced by heating the joint (junction) where two unlike metals are joined. • Light (photoelectricity) - Voltage produced by light striking photosensitive (light sensitive) substances. • Chemical Action - Voltage produced by chemical reaction in a battery cell. • Magnetism - Voltage produced in a conductor when the conductor moves through a magnetic field, or a magnetic field moves through the conductor in such a manner as to cut the magnetic lines of force of the field.

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Voltage Produced by Friction The first method discovered for creating a voltage was that of generation by friction. The development of charges by rubbing a rod with fur is a prime example of the way in which a voltage is generated by friction. Because of the nature of the materials with which this voltage is generated, it cannot be conveniently used or maintained. For this reason, very little practical use has been found for voltages generated by this method.

In the search for methods to produce a voltage of a larger amplitude and of a more practical nature, machines were developed in which charges were transferred from one terminal to another by means of rotating glass discs or moving belts. The most notable of these machines is the Van de Graaff generator. It is used today to produce potentials in the order of millions of volts for nuclear research. As these machines have little value outside the field of research, their theory of operation will not be described here.

Voltage Produced by Pressure One specialized method of generating an EMF utilizes the characteristics of certain ionic crystals such as quartz, Rochelle salts, and tourmaline. These crystals have the remarkable ability to generate a voltage whenever stresses are applied to their surfaces. Thus, if a crystal of quartz is squeezed, charges of opposite polarity will appear on two opposite surfaces of the crystal. If the force is reversed and the crystal is stretched, charges will again appear, but will be of the opposite polarity from those produced by squeezing. If a crystal of this type is given a vibratory motion, it will produce a voltage of reversing polarity between two of its sides. Quartz or similar crystals can thus be used to convert mechanical energy into electrical energy. This phenomenon, called the piezoelectric effect, is shown in figure 4.1. Some of the common devices that make use of piezoelectric crystals are microphones, phonograph cartridges, and oscillators used in radio transmitters, radio receivers, and sonar equipment. This method of generating an EMF is not suitable for applications having large voltage or power requirements, but is widely used in sound and communications systems where small signal voltages can be effectively used. QUARTZ CRYSTAL COMPRESSED

MOLECULES OF NON-CRYSTALLIZED MATTER

(A)

P:4 �

LECTRON

FLOW

(C) 4

QUARTZ CRYSTAL DECOMPRESSED

XN

ELECTRON

MOLECULES OF CRYSTALLIZED MATTER ( )

FLOW

(D)

Figure 4.1 - (A) Non-crystallized structure; (B) crystallized structure; (C) compression of a crystal; (D) decompression of a crystal.

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Crystals of this type also possess another interesting property, the "converse piezoelectric effect." That is, they have the ability to convert electrical energy into mechanical energy. A voltage impressed across the proper surfaces of the crystal will cause it to expand or contract its surfaces in response to the voltage applied.

Voltage Produced by Heat When a length of metal, such as copper, is heated at one end, electrons tend to move away from the hot end toward the cooler end. This is true of most metals. However, in some metals, such as iron, the opposite takes place and electrons tend to move toward the hot end. These characteristics are illustrated in figure 4.2. The negative charges (electrons) are moving through the copper away from the heat and through the iron toward the heat. They cross from the iron to the copper through the current meter to the iron at the cold junction. This device is generally referred to as a thermocouple

COLD

JUIIClloN

1

Li Figure 4.2 - Voltage produced by heat.

r

Thermocouples have somewhat greater power capacities than crystals, but their capacity is still very small if compared to some other sources. The thermoelectric voltage in a thermocouple depends mainly on the difference in temperature between the hot and cold junctions. Consequently, they are widely used to measure temperature, and as heat-sensing devices in

automatic temperature control equipment. Thermocouples generally can be subjected to much greater temperatures than ordinary thermometers, such as the mercury or alcohol types.

U

Voltage Produced by Light When light strikes the surface of a substance, it may dislodge electrons from their orbits around the surface atoms of the substance. This occurs because light has energy, the same as any moving force. Some substances, mostly metallic ones, are far more sensitive to light than others. That is, more electrons will be dislodged and emitted from the surface of a highly sensitive metal, with a given amount of light, than will be emitted from a less sensitive substance. Upon losing electrons, the photosensitive (light-sensitive) metal becomes positively charged, and an electric force is created. Voltage produced in this manner is referred to as a photoelectric voltage.

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The photosensitive materials most commonly used to produce a photoelectric voltage are various compounds of silver oxide or copper oxide. A complete device which operates on the photoelectric principle is referred to as a "photoelectric cell." There are many different sizes and types of photoelectric cells in use, and each serves the special purpose for which it is designed. Nearly all, however, have some of the basic features of the photoelectric cells shown in figure 4.3. PHOTOSENSITIVE

OXIDE SURFACE

SILVER

(: .LIGHTSOURCE

2

i

SEMITRANSPARENT {,.) LAYER PASSES LIGHT AND COLLECTS"? 6 PHOTOELECTRONS

E LE CTROH FLOW ry-

PHOTOSENSITIVE COPPER OXIDE PURE COPPER BASE LAVER4t

(B) Figure 4.3 - Voltage produced by light.

Li

The cell (figure 4.3 view A) has a curved light-sensitive surface focused on the central anode. When light from the direction shown strikes the sensitive surface, it emits electrons toward the anode. The more intense the light, the greater the number of electrons emitted. When a wire is connected between the filament and the back, or dark side of the cell, the accumulated electrons will flow to the dark side. These electrons will eventually pass through the metal of the reflector and replace the electrons leaving the light-sensitive surface. Thus, light energy is converted to a flow of electrons, and a usable current is developed. The cell (figure 4.3 view B) is constructed in layers. A base plate of pure copper is coated with light-sensitive copper oxide. An extremely thin semitransparent layer of metal is placed over the copper oxide. This additional layer serves two purposes: • It permits the penetration of light to the copper oxide. • It collects the electrons emitted by the copper oxide. An externally connected wire completes the electron path, the same as in the reflector-type cell. The photocell's voltage is used as needed by connecting the external wires to some other device, which amplifies (enlarges) it to a usable level. I 4-8 TTS Integrated Training System © Copyright 2010

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The power capacity of a photocell is very small. However, it reacts to light-intensity variations in an extremely short time. This characteristic makes the photocell very useful in detecting or accurately controlling a great number of operations. For instance, the photoelectric cell, or some form of the photoelectric principle, is used in television cameras, automatic manufacturing process controls, door openers, burglar alarms, and so forth. Ft i

i.. I_

I U

Voltage Produced by Chemical Action Voltage may be produced chemically when certain substances are exposed to chemical action. If two dissimilar substances (usually metals or metallic materials) are immersed in a solution that produces a greater chemical action on one substance than on the other, a difference of potential will exist between the two. If a conductor is then connected between them, electrons will flow through the conductor to equalize the charge. This arrangement is called a primary cell. The two metallic pieces are called electrodes and the solution is called the electrolyte. The

voltaic cell illustrated in figure 4.4 is a simple example of a primary cell. The difference of

potential results from the fact that material from one or both of the electrodes goes into solution in the electrolyte, and in the process, ions form in the vicinity of the electrodes. Due to the electric field associated with the charged ions, the electrodes acquire charges. ZINC ELECTRODE . .

COPPER ELECTRODE

r!

I Figure 4.4 - Voltaic cell.

U

The amount of difference in potential between the electrodes depends principally on the metals used. The type of electrolyte and the size of the cell have little or no effect on the potential difference produced. There are two types of primary cells, the wet cell and the dry cell. In a wet cell the electrolyte is a liquid. A cell with a liquid electrolyte must remain in an upright position and is not readily transportable. An automotive battery is an example of this type of cell. The dry cell, much more commonly used than the wet cell, is not actually dry, but contains an electrolyte mixed with other materials to form a paste. Torches and portable radios are commonly powered by dry cells.

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Batteries are formed when several cells are connected together to increase electrical output.

Voltage Produced by Magnetism Magnets or magnetic devices are used for thousands of different jobs. One of the most useful and widely employed applications of magnets is in the production of vast quantities of electric power from mechanical sources. The mechanical power may be provided by a number of different sources, such as gasoline or diesel engines, and water or steam turbines. However, the final conversion of these source energies to electricity is done by generators employing the principle of electromagnetic induction. These generators, of many types and sizes, are discussed in other modules in this series. The important subject to be discussed here is the fundamental operating principle of all such electromagnetic-induction generators. To begin with, there are three fundamental conditions which must exist before a voltage can be produced by magnetism. • There must be a conductor in which the voltage will be produced. • There must be a magnetic field in the conductor's vicinity. • There must be relative motion between the field and conductor. The conductor must be moved so as to cut across the magnetic lines of force, or the field must be moved so that the lines of force are cut by the conductor. In accordance with these conditions, when a conductor or conductors move across a magnetic field so as to cut the lines of force, electrons within the conductor are propelled in one direction or another. Thus, an electric force, or voltage, is created. In figure 4.5, note the presence of the three conditions needed for creating an induced voltage.

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(A)

(F)

(C)

direction of motion reversed

Figure 4.5 - Voltage produced by magnetism. • A magnetic field exists between the poles of the C-shaped magnet. • There is a conductor (copper wire). • There is a relative motion. The wire is moved back and forth across the magnetic field. • In figure 4.5 view A, the conductor is moving toward the front of the page and the electrons move from left to right. The movement of the electrons occurs because of the magnetically induced EMF acting on the electrons in the copper. The right-hand end becomes negative, and the left-hand end positive. The conductor is stopped at view B, motion is eliminated (one of the three required conditions), and there is no longer an induced EMF. Consequently, there is no longer any difference in potential between the two ends of the wire. The conductor at view C is moving away from the front of the page. An induced EMF is again created. However, note carefully that the reversal of motion has caused a reversal of direction in the induced EMF. If a path for electron flow is provided between the ends of the conductor, electrons will leave the

negative end and flow to the positive end. This condition is shown in part view D. Electron flow

will continue as long as the EMF exists. In studying figure 4.5,it should be noted that the induced EMF could also have been created by holding the conductor stationary and moving the magnetic field back and forth. The more complex aspects of power generation by use of mechanical motion and magnetism are discussed later in Chapter 14 - DC Motor/Generator Theory.

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3.5 DC Sources of Electricity

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Knowledge Levels --- Category A, B1, B2 and C Aircraft Maintenance Licence

basic knowledge levels. The knowledge level indicators are defined as follows:

fl

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. examples. The applicant should be able to use typical terms.

•

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • examples. subject. subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • •

A detailed knowledge of the theoretical and practical aspects of the subject.

manner.

Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. and specific examples. • • describing the subject. instructions. • corrective action where appropriate.

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Table of Contents

Module 3.5 DC Sources of Electricity Introduction The Cell Primary and Secondary Cells Electrochemical Action Primary Cell Chemistry Secondary Cell Chemistry

5 5 5 7 7 8 9

Polarization of the Cell

10

Local Action Types of Cells

11 11

15

Other Types of Cells Secondary Wet Cells Cell Capacity Cells in Series and Parallel Battery Construction Battery Internal Resistance Battery Maintenance Capacity and Rating of Batteries

Battery Charging

Thermocouples Photocells

18 19 20 22 30 31 32

33 35 44

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DC Sources of Electricity 3.5 Construction and basic chemical action of: primary cells, secondary cells, lead acid cells, nickel cadmium cells, other alkaline cells Cells connected in series and parallel Internal resistance and its affect on a battery Construction, materials and operation of thermocouples Operation of photo-cells

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Module 3.5 DC Sources of Electricity Introduction The purpose of this chapter is to introduce and explain the basic theory and characteristics of batteries. The batteries which are discussed and illustrated have been selected as representative of many models and types which are used in aircraft today. No attempt has been made to cover every type of battery in use, however, after completing this chapter you will have a good working knowledge of the batteries which are in general use. First, you will learn about the building block of all batteries, the cell. The explanation will explore the physical makeup of the cell and the methods used to combine cells to provide useful voltage, current, and power. The chemistry of the cell and how chemical action is used to convert chemical energy to electrical energy are also discussed.

i r

In addition, the care, maintenance, and operation of batteries, as well as some of the safety precautions that should be followed while working with and around batteries are discussed. Batteries are widely used as sources of direct-current electrical energy in automobiles, boats, aircraft, ships, portable electric/electronic equipment, and lighting equipment. In some instances, they are used as the only source of power; while in others, they are used as a secondary or standby power source. A battery consists of a number of cells assembled in a common container and connected together to function as a source of electrical power.

The Cell A cell is a device that transforms chemical energy into electrical energy. The simplest cell, known as either a galvanic or voltaic cell, is shown in Figure 5.1. It consists of a piece of carbon (C) and a piece of zinc (Zn) suspended in a jar that contains a solution of water (H20) and sulphuric acid (H2SO4) called the electrolyte.

U Figure 5.1 - Simple voltaic or galvanic cell.

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The cell is the fundamental unit of the battery. A simple cell consists of two electrodes placed in a container that holds the electrolyte.

n j

In some cells the container acts as one of the electrodes and, in this case, is acted upon by the electrolyte. This will be covered in more detail later. Electrodes The electrodes are the conductors by which the current leaves or returns to the electrolyte. In the simple cell, they are carbon and zinc strips that are placed in the electrolyte; while in the dry cell (Figure 5.2), they are the carbon rod in the centre and zinc container in which the cell is assembled.

Garton

(graphite) c lectrocle

surr.'unded

by carbon

. nDn-o)ndlctinc,

tube

black and

manganese dloxlde

is the

oath rode.

Ion transfer is

accomallshed in a paste of

ammonium chloriide-

n

arfid zinc chl rids

Zinc metal sleeve- is the, ante. Figure 5.2 - Dry cell, cross-sectional view. In a discharging battery or galvanic cell (drawing) the cathode is the positive terminal, where conventional current flows out. This outward current is carried internally by positive ions moving from the electrolyte to the positive cathode (chemical energy is responsible for this "uphill" motion). It is continued externally by electrons moving inwards, negative charge moving one way amounting to positive current flowing the other way. The anode is the negative terminal, where conventional current flows in, and electrons out.

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Electrolyte The electrolyte is the solution that acts upon the electrodes. The electrolyte, which provides a path for electron flow, may be a salt, an acid, or an alkaline solution. In the simple galvanic cell, the electrolyte is in a liquid form. In the dry cell, the electrolyte is a paste. Container The container which may be constructed of one of many different materials provides a means of holding (containing) the electrolyte. The container is also used to mount the electrodes. In the voltaic cell the container must be constructed of a material that will not be acted upon by the electrolyte.

Primary and Secondary Cells Primary Cell A primary cell is one in which the chemical action eats away one of the electrodes, usually the negative electrode. When this happens, the electrode must be replaced or the cell must be discarded. In the galvanic-type cell, the zinc electrode and the liquid electrolyte are usually replaced when this happens. In the case of the dry cell, it is usually cheaper to buy a new cell. Secondary Cell A secondary cell is one in which the electrodes and the electrolyte are altered by the chemical action that takes place when the cell delivers current. These cells may be restored to their original condition by forcing an electric current through them in the direction opposite to that of discharge. The automobile storage battery is a common example of the secondary cell.

Electrochemical Action If a load (a device that consumes electrical power) is connected externally to the electrodes of a cell, electrons will flow under the influence of a difference in potential across the electrodes from the anode (negative electrode), through the external conductor to the cathode (positive electrode). A cell is a device in which chemical energy is converted to electrical energy. This process is called electrochemical action. The voltage across the electrodes depends upon the materials from which the electrodes are made and the composition of the electrolyte. The current that a cell delivers depends upon the resistance of the entire circuit, including that of the cell itself. The internal resistance of the cell depends upon the size of the electrodes, the distance between them in the electrolyte, and the resistance of the electrolyte. The larger the electrodes and the closer together they are in the electrolyte (without touching), the lower the internal resistance of the cell and the more current the cell is capable of supplying to the load.

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Primary Cell Chemistry When a current flows through a primary cell having carbon and zinc electrodes and a diluted solution of sulphuric acid and water (combined to form the electrolyte), the following chemical reaction takes place. The electron flow through the load is the movement of electrons from the negative electrode of the cell (zinc) and to the positive electrode (carbon). This causes fewer electrons in the zinc and an excess of electrons in the carbon. The hydrogen ions (H2) from the sulphuric acid are attracted to the carbon electrode. Since the hydrogen ions are positively charged, they are attracted to the negative charge on the carbon electrode. This negative charge is caused by the excess of electrons. The zinc electrode has a positive charge because it has lost electrons to the carbon electrode. This positive charge attracts the negative ions (S04) from the sulphuric acid. The negative ions combine with the zinc to form zinc sulphate. This action causes the zinc electrode to be eaten away. Zinc sulphate is a greyish-white substance that is sometimes seen on the battery post of an automobile battery.

fl

The process of the zinc being eaten away and the sulphuric acid changing to hydrogen and zinc sulphate is the cause of the cell discharging. When the zinc is used up, the voltage of the cell is reduced to zero. In Figure 5.2 you will notice that the zinc electrode (the case) is labelled negative and the carbon electrode is labelled positive. This represents the current flow outside the cell from positive to negative.

nI

1

The zinc combines with the sulphuric acid to form zinc sulphate and hydrogen. The zinc sulphate dissolves in the electrolyte (sulphuric acid and water) and the hydrogen appears as gas bubbles around the carbon electrode. As current continues to flow, the zinc gradually dissolves and the solution changes to zinc sulphate and water. The carbon electrode does not enter into the chemical changes taking place, but simply provides a return path for the current.

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Secondary Cell Chemistry As stated before, the differences between primary and secondary cells are, the secondary cell can be recharged and the electrodes are made of different materials. The secondary cell shown in Figure 5.3 uses sponge lead as the anode and lead peroxide as the cathode. This is the leadacid type cell and will be used to explain the general chemistry of the secondary cell. Later in the chapter when other types of secondary cells are discussed, you will see that the materials which make up the parts of a cell are different, but that the chemical action is essentially the

same.

SOLUTION

SULFURIC ACID H2SO4

AND WATER H2O

da

SPONGE LEAD

LEAD PEROXIDE Pb0 2

Pb

L

(B) DISCHARGING

(A) CHARGED

DECREASING

DECREASING

SPONGE LEAD LEAD PEROXIDE

INCREASING INCREASING LEAD SULFATE LEAD SULFATE Pb + PbSO4 PbO-+ PbSO4

IL

Li

L (D) CHARGING INCREASING SPONGE LEAD DECREASING

INCREASING LEAD PEROXIDE

PbSO4 + Pb

PbSO4 + PbG2

LEAD SULFATE

L

DECREASING LEAD SULFATE

SPONGE LEAD

(C) DISCHARGED MINIMUM MINIMUM SPONGE LEAD LEAD PEROXIDE MAXIMUM MAXIMUM LEAD SULFATE LEAD SULFATE PbSO1 + Pb PbSG1+ Pb0 2

F�j LEAD PEROXIDE

LEAD SULFATE

Figure 5.3 - Secondary cell.

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Figure 5.3 view A shows a lead-acid secondary cell that is fully charged. The anode is pure sponge lead, the cathode is pure lead peroxide, and the electrolyte is a mixture of sulphuric acid and water. Figure 5.3 view B shows the secondary cell discharging. A load is connected between the cathode and anode; electrons flow negative to positive as shown. This electron flow creates the same process as was explained for the primary cell with the following exceptions: • In the primary cell the zinc anode was eaten away by the sulphuric acid. In the secondary cell the sponge-like construction of the anode retains the lead sulphate formed by the chemical action of the sulphuric acid and the lead. • In the primary cell the carbon cathode was not chemically acted upon by the sulphuric acid. In the secondary cell the lead peroxide cathode is chemically changed to lead sulphate by the sulphuric acid. When the cell is fully discharged it will be as shown in Figure 5.3 view C. The cathode and anode retain some lead peroxide and sponge lead but the amounts of lead sulphate in each is maximum. The electrolyte has a minimum amount of sulphuric acid. With this condition no further chemical action can take place within the cell. As you know, the secondary cell can be recharged. Recharging is the process of reversing the chemical action that occurs as the cell discharges. To recharge the cell, a voltage source, such as a generator, is connected as shown in Figure 5.3 view D. The negative terminal of the voltage source is connected to the cathode of the cell and the positive terminal of the voltage source is connected to the anode of the cell. With this arrangement the lead sulphate is chemically changed back to sponge lead in the cathode, lead peroxide in the anode, and sulphuric acid in the electrolyte. After all the lead sulphate is chemically changed, the cell is fully charged as shown in Figure 5.3 view A. Once the cell has been charged, the discharge-charge cycle may be repeated. Notice in the above paragraph that the Anode and Cathode appear to have changed polarity. This is because a cell being recharged is an electrolytic cell (rather than a voltaic or galvanic cell, as it was when discharging). In an electrolytic cell, the anode is positive, and the cathode is negative.

Polarization of the Cell The chemical action that occurs in the cell while the current is flowing causes hydrogen bubbles to form on the surface of the anode. This action is called polarization. Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance. When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases. A cell that is heavily polarized has no useful output. There are several methods to prevent polarization or to depolarise the cell. One method uses a vent on the cell to permit the hydrogen to escape into the air. A disadvantage of this method is that hydrogen is not available to reform into the electrolyte 5-10 TTS Integrated Training System

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during recharging. This problem is solved by adding water to the electrolyte, such as in an automobile battery. A second method is to use material that is rich in oxygen, such as manganese dioxide, which supplies free oxygen to combine with the hydrogen and form water. A third method is to use a material that will absorb the hydrogen, such as calcium. The calcium releases hydrogen during the charging process. All three methods remove enough hydrogen so that the cell is practically free from polarization.

Local Action When the external circuit is removed, the current ceases to flow, and, theoretically, all chemical action within the cell stops. However, commercial zinc contains many impurities, such as iron, carbon, lead, and arsenic. These impurities form many small electrical cells within the zinc electrode in which current flows between the zinc and its impurities. Thus, the chemical action continues even though the cell itself is not connected to a load.

Local action may be prevented by using pure zinc (which is not practical), by coating the zinc with mercury, or by adding a small percentage of mercury to the zinc during the manufacturing process. The treatment of the zinc with mercury is called amalgamating (mixing) the zinc. Since mercury is many times heavier than an equal volume of water, small particles of impurities weighing less than mercury will float to the surface of the mercury. The removal of these impurities from the zinc prevents local action. The mercury is not readily acted upon by the acid. When the cell is delivering current to a load, the mercury continues to act on the impurities in the zinc. This causes the impurities to leave the surface of the zinc electrode and float to the surface of the mercury. This process greatly increases the storage life of the cell.

Types of Cells The development of new and different types of cells in the past decade has been so rapid that it is virtually impossible to have a complete knowledge of all the various types. A few recent developments are the silver-zinc, nickel-zinc, nickel-cadmium, silver-cadmium, organic and inorganic lithium, and mercury cells. Primary Dry Cell The dry cell is the most popular type of primary cell. It is ideal for simple applications where an inexpensive and non-critical source of electricity is all that is needed. The dry cell is not actually dry. The electrolyte is not in a liquid state, but is a moist paste. If it should become totally dry, it would no longer be able to transform chemical energy to electrical energy. The construction of a common type of dry cell is shown in Figure 5.4. These dry cells are also referred to as Leclanche' cells. The internal parts of the cell are located in a cylindrical zinc container. This zinc container serves as the negative electrode (anode) of the cell. The container is lined with a non-conducting material, such as blotting paper, to separate the zinc from the paste. A carbon electrode is located in the centre, and it serves as the positive terminal (cathode) of the cell. The paste is a mixture of several substances such as ammonium chloride, powdered coke, ground carbon, manganese dioxide, zinc chloride, graphite, and water.

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Positive terminal n

Protective casing Electrolyte paste

Pitch seal

(ammonium chloride

Air space

n

and zinc chloride)

Carbon and manganese

Zinc

dioxide mixture

Separator

Carbon rod

n

Negative terminal Figure 5.4 - Cutaway view of the general-purpose dry cell. This electrolyte paste also serves to hold the cathode rigid in the centre of the cell. When the paste is packed in the cell, a small space is left at the top for expansion of the electrolytic paste caused by the depolarisation action. The cell is then sealed with a cardboard or plastic seal. Since the zinc container is the anode, it must be protected with some insulating material to be electrically isolated. Therefore, it is common practice for the manufacturer to enclose the cells in cardboard and metal containers. The dry cell (Figure 5.4) is basically the same as the simple voltaic cell (wet cell) described earlier, as far as its internal chemical action is concerned. The action of the water and the ammonium chloride in the paste, together with the zinc and carbon electrodes, produces the voltage of the cell. Manganese dioxide is added to reduce polarization when current flows and zinc chloride reduces local action when the cell is not being used. A cell that is not being used (sitting on the shelf) will gradually deteriorate because of slow internal chemical changes (local action). This deterioration is usually very slow if cells are properly stored. If unused cells are stored in a cool place, their shelf life will be greatly preserved. Therefore, to minimize deterioration, they should be stored in refrigerated spaces.

fl

The cell is sealed at the top to keep air from entering and drying the electrolyte. Care should be taken to prevent breaking this seal. ri

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The Leclanche Cell Georges Leclanche invented and patented in 1866 his battery, the Leclanche cell. It contained a

conducting solution (electrolyte) of ammonium

L•

chloride, a cathode (positive terminal) of carbon, a depolarizer of manganese dioxide, and an anode (negative terminal) of zinc. The Leclanche battery was essentially a self-contained version of an earth battery, and fairly copied its design. The Leclanche battery (or wet cell as it was referred to) was the forerunner of the modern dry cell zinccarbon battery.

Figure 5.5 - The Leclanche Cell

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77 Ile

TERMI/VA4

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ELEC-

I

; �I

TRODE

U r-� L

The Daniell Cell The Daniell cell, also called the gravity cell or crowfoot cell was invented in 1836 by John Frederic Daniell, who was a British chemist and meteorologist. The Daniell cell was a great improvement over and is somewhat safer than the voltaic cell used in the early days of battery development. The Daniel cell's theoretical voltage is 1.1 volts. The Daniel proper consists of a central zinc anode dipping into a porous earthenware pot containing zinc sulphate solution. The porous pot is, in turn, immersed in a solution of copper sulphate contained in a copper can, which acts as the cell's cathode. The use of a porous barrier prevents the copper ions in the copper sulphate solution from reaching the zinc anode and undergoing reduction. This would render the cell ineffective by bringing the battery to equilibrium without driving a current.

Figure 5.6 - The Daniell Cell

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Mercuric-Oxide Zinc Cell The mercuric-oxide zinc cell (mercury cell) is a primary cell that was developed during World War II. Two important assets of the mercury cell are its ability to produce current for a long period of time and a long shelf life when compared to the dry cell shown in Figure 5.4.The mercury cell also has a very stable output voltage and is a power source that can be made in a small physical size. With the birth of the space program and the development of small transceivers and miniaturized equipment, a power source of small size was needed. Such equipment requires a small cell which is capable of delivering maximum electrical energy at a constant discharge voltage. The mercury cell, which is one of the smallest cells, meets these requirements. Present mercury cells are manufactured in three basic types as shown in Figure 5.7. The wound-anode type, shown in Figure 5.7 view A, has an anode composed of a corrugated zinc strip with a paper absorbent. The zinc is mixed with mercury, and the paper is soaked in the electrolyte which causes it to swell and press against the zinc and make positive contact. This process ensures that the electrolyte makes contact with the cathode.

Zinc Anod v

j At,tvloho+ Cup IRiuf44Qr

Figure 5.7 - Mercury cell If the anode and cathode of a cell are connected together without a load, a short circuit condition exists. Short circuits (shorts) can be very dangerous. They cause excessive heat,

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pressure, and current flow which may cause serious damage to the cell or be a safety hazard to personnel. Warning: Do not short the mercury cell. Shorted mercury cells have exploded with considerable force. U

Other Types of Cells There are many different types of prima ry cells. Because of such factors as cost, size, ease of replacement, and voltage or current needs, many types of primary cells have been developed. The following is a brief description of some of the primary cells in use today.

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The Manganese Dioxide-Alkaline-Zinc Cell is similar to the zinc-carbon cell except for the electrolyte used. This type of cell offers better voltage stability and longer life than the zinccarbon type. It also has a longer shelf life and can operate over a wide temperature range. The manganese dioxide-alkaline-zinc cell has a voltage of 1.5 volts and is available in a wide range of sizes. This cell is commonly referred to as the alkaline cell. The Magnesium-Manganese Dioxide Cell uses magnesium as the anode material. This allows a higher output capacity over an extended period of time compared to the zinc-carbon cell. This cell produces a voltage of approximately 2 volts. The disadvantage of this type of cell is the production of hydrogen during its operation. The Lithium-Organic Cell and the Lithium-Inorganic Cell are recent developments of a new line of high-energy cells. The main advantages of these types of cells are very high power, operation over a wide temperature range, they are lighter than most cells, and have a remarkably long shelf life of up to 20 years. Warning: Lithium cells contain toxic materials under pressure. Do not puncture, recharge, short-circuit, expose to excessively high temperatures, or incinerate. Use these batteries/cells only in approved equipment. Do not throw in bin.

n

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Disposable Cells These are not designed to be rechargeable - i.e. primary cells. "Disposable" may also imply that special disposal procedures must take place for proper disposal according to regulation, depending on battery type. • Zinc-carbon: mid cost, used in light drain applications. • Zinc-chloride: similar to zinc-carbon but slightly longer life. • Alkaline: alkaline/manganese "long life" batteries widely used in both light-drain and heavy-drain applications. • Silver-oxide: commonly used in hearing aids, watches, and calculators. • Lithium Iron Disulphide: commonly used in digital cameras. Sometimes used in watches and computer clocks. Very long life (up to ten years in wristwatches) and capable of delivering high currents but expensive. Will operate in sub-zero temperatures. • Lithium-Thionyl Chloride: used in industrial applications, including computers, electric meters and other devices which contain volatile memory circuits and act as a "carryover" voltage to maintain the memory in the event of a main power failure. Other applications include providing power for wireless gas and water meters. The cells are rated at 3.6 Volts and come in 1/2AA, AA, 2/3A, A, C, D & DD sizes. They are relatively expensive, but have a long shelf life, losing less than 10% of their capacity in ten years. • Mercury: formerly used in digital watches, radio communications, and portable electronic instruments. Manufactured only for specialist applications due to toxicity. • Zinc-air: commonly used in hearing aids. Lid

(negative terminal)

Plastic

sealing ring Anode (zinc powder)

j

Figure 5.8 - Zinc-air cell • •

Nickel Oxyhydroxide: Ideal for applications that use bursts of high current, such as digital cameras. They will last two times longer than alkaline batteries in digital cameras. Paper: In August 2007, a research team at RPI (led by Drs. Robert Linhardt, Pulickel M. Ajayan, and Omkaram Nalamasu) developed a paper battery with aligned carbon nanotubes, designed to function as both a lithium-ion battery and a super-capacitor,

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using ionic liquid, essentially a liquid salt, as electrolyte. The sheets can be rolled, twisted, folded, or cut into numerous shapes with no loss of integrity or efficiency, or stacked, like printer paper (or a voltaic pile), to boost total output. As well, they can be made in a variety of sizes, from postage stamp to broadsheet. Their light weight and low cost make them attractive for portable electronics, aircraft, and automobiles, while their ability to use electrolytes in blood make them potentially useful for medical devices such as pacemakers. In addition, they are biodegradable, unlike most other disposable cells.

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Rechargeable Cells Also known as secondary batteries or accumulators. The National Electrical Manufacturers Association has estimated that U.S. demand for rechargeables is growing twice as fast as demand for non-rechargeables. There are a few main types: •

Nickel-cadmium (NiCd): Best used for motorized equipment and other high-discharge, short-term devices. NiCd batteries can withstand even more drain than NiMH; however, the mAh rating is not high enough to keep a device running for very long, and the memory effect is far more severe. • Nickel-metal hydride (NiMH): Best used for high-tech devices. NiMH batteries can last up to four times longer than alkaline batteries because NiMH can withstand high current for a long while. • Rechargeable alkaline: Uses similar chemistry as non-rechargeable alkaline batteries and are best suited for similar applications. Additionally, they hold their charge for years, unlike NiCd and NiMH batteries. • Lithium Ion (Li-Ion): Continuing in the tradition of modern battery chemistries, the lithium ion battery has an increased energy density and can provide a respectable amount of current. High discharge rates don't significantly reduce its capacity, nor does it lose very much capacity after each cycle, still retaining 80% of its energy capacity after 500 recharge cycles. This is a volatile technology, early versions were prone to exploding in the labs. It is the volatile nature of lithium that gives this battery its punch, though. These benefit come with a price, of course (perhaps to pay for equipment damaged in the research?). • Fuel Cells: The fuel cell isn't so much a battery as it is a catalytic chemical engine that creates electricity from hydrogen and oxygen. The fuel is typically a variation of hydrogen, such as the hydrocarbon fuels methanol,

natural gas, or even gasoline.

The output of the fuel cell is electricity and water.

Figure 5.9 -- The Fuel Cell

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Secondary Wet Cells Secondary cells are sometimes known as wet cells. There are four basic type of wet cells, the lead-acid, nickel-cadmium, silver-zinc, and silver-cadmium. • Lead Acid Cell The lead-acid cell is the most widely used secondary cell. The previous explanation of the secondary cell describes exactly the manner in which the lead-acid cell provides electrical power. The discharging and charging action presented in electrochemical action describes the lead-acid cell.

You should recall that the lead-acid cell has an anode of lead peroxide, a cathode of sponge lead, and the electrolyte is sulphuric acid and water. • Nickel-Cadmium Cell The nickel-cadmium cell (NiCad or NiCd) is far superior to the lead-acid cell. In comparison to lead-acid cells, these cells generally require less maintenance throughout their service life in regard to the adding of electrolyte or water. The major difference between the nickel-cadmium cell and the lead-acid cell is the material used in the cathode, anode, and electrolyte. In the nickel-cadmium cell the cathode is cadmium hydroxide, the anode is nickel hydroxide, and the electrolyte is potassium hydroxide and water. The nickel-cadmium and lead-acid cells have capacities that are comparable at normal discharge rates, but at high discharge rates the nickel-cadmium cell can deliver a larger amount of power. In addition the nickel-cadmium cell can: • Be charged in a shorter time • Stay idle longer in any state of charge and keep a full charge when stored for a longer period of time • Be charged and discharged any number of times without any appreciable damage. • Due to their superior capabilities, nickel-cadmium cells are being used extensively in many aircraft applications that require a cell with a high discharge rate.

• Silver-Zinc Cells The silver-zinc cell is used extensively to power emergency equipment. This type of cell is relatively expensive and can be charged and discharged fewer times than other types of cells. When compared to the lead-acid or nickel-cadmium cells, these disadvantages are overweighed by the light weight, small size, and good electrical capacity of the silverzinc cell. The silver-zinc cell uses the same electrolyte as the nickel-cadmium cell (potassium hydroxide and water), but the anode and cathode differ from the nickel-cadmium cell. The anode is composed of silver oxide and the cathode is made of zinc.

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L

• Silver-Cadmium Cell The silver-cadmium cell is a fairly recent development for use in storage batteries. The silver-cadmium cell combines some of the better features of the nickel-cadmium and silver-zinc cells. It has more than twice the shelf life of the silver-zinc cell and can be recharged many more times. The disadvantages of the silver-cadmium cell are high cost

and low voltage production. The electrolyte of the silver-cadmium cell is potassium hydroxide and water as in the nickel-cadmium and silver-zinc cells. The anode is silver oxide as in the silver-zinc cell and the cathode is cadmium hydroxide as in the nicad cell. You may notice that different combinations of materials are used to form the electrolyte, cathode, and anode of different cells. These combinations provide the cells with different qualities for many varied applications. f-i L

Cell Capacity The capacity of a cell relates to the amount of current that the cell is capable of supplying. The capacity will depend upon the area of the plates: the larger the area, the greater the capacity.

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The voltage produced is independent of plate size and is purely related to the materials of the cell. In Figure 5.10 the two example use identical materials but are of different sizes. The voltages produced by each cell, therefore, are identical but the capacities are different. LARGE CURRENT

u

SMALL CURREIfT CAPACITY

CAPACITY LONG LINE REPRESEXTS, THE "POSITIVE" TERMINAL

L

I

SKULL AREA PLATES

U

LARGE AREA PLATES

CELL SCH4ATIC SYMBOL

Figure 5.10 - Cell Plate Area - Current Capacity Relationship

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Cells in Series and Parallel Cells in Series If cells are connected in series, as shown in Figure 5.11, the total voltage will increase. iT BATTERY TERMINAL VOLTAGE

Figure 5.11 - Cells in Series The terminal voltages of the individual cells are added together to obtain the battery terminal voltage. The overall capacity, however, does not increase. Cells in Parallel If cells or batteries are connected in parallel, as shown in Figure 5.12, the total capacity will increase. INCREASED CAPACITY

, J

11

I

Figure 5.12 - Cells in Parallel

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Cells in Series-Parallel Figure 5.13 depicts a battery network supplying power to a load requiring both a voltage and a current greater than one cell can provide. To provide the required 4.5 volts, groups of three 1.5volt cells are connected in series. To provide the required 1/2 ampere of current, four series groups are connected in parallel, each supplying 1/8 ampere of current.

LOAD rU

Figure 5.13 - Schematic of series-parallel connected cells. U

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The connections shown have been used to illustrate the various methods of combining cells to form a battery. Series, parallel, and series-parallel circuits will be covered in detail in the next chapter, "Direct Current." Some batteries are made from primary cells. When a primary-cell battery is completely discharged, the entire battery must be replaced. Because there is nothing else that can be done

to primary cell batteries, the rest of the discussion on batteries will be concerned with batteries made of secondary cells.

L

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Battery Construction The Lead-Acid Cell The basic lead-acid cell consists of two sets of plates, one of which is negative and the other positive. They are interleaved and prevented from coming into contact with each other by porous separators. The separators have high insulation qualities but permit the unobstructed circulation of the electrolyte at the plate surfaces. The basic lead-acid cell components are shown in Figure 5.14.

11

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Figure 5.14 - Lead-Acid Cell Components VENT PLUG

FILLER OPENING IN CELL COVER

TERMINAL POST

lE

TERMINAL

CONNECTOR PLATE LINK

ST RAP

CONNECTOR

CONTAINER

rd EGAT IVE

POSITIVE

PLATE

PLATE

SEPARATOR CASE

RIB

SEDIMENT SPACE

Figure 5.15 - Lead-acid battery construction p

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U

L

CELL ELEMENT PARTLY ASSEMBLED Figure 5.16 - Lead-acid battery plate arrangement. The positive plates are made up of grids of lead and antimony filled with lead peroxide. The negative plates are made up of similar grids, but filled with spongy lead. The electrolyte is a solution of sulphuric acid and water in contact with both sets of plates. The type of cell construction permits the electrolyte to circulate freely and also provides a path for sediment to settle at the bottom of the cell. When an external circuit is connected to a fully charged cell, electrons flow from the negative lead plates, via the circuit, to the positive lead peroxide plates. As the electrons leave the negative plates, positive ions form. These attract negative sulphate ions from the sulphuric acid of the electrolyte. This causes lead sulphate to form on the negative plates. The electrons arriving at the positive plates, from the external circuit, drive negative oxygen ions from the lead peroxide into the electrolyte. These combine with hydrogen, which has lost sulphate ions, to form water. The positive lead ions that are left on the positive plates also attract and combine with sulphate ions from the electrolyte to form lead sulphate on the positive plates. Once lead sulphate collects on both the positive and negative plates and the electrolyte becomes diluted by the water, which has formed in it, the cell is considered discharged.

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1

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A discharged cell is recharged using a direct current of the correct voltage. When the positive plates of the cell are connected to the positive of the charging source and the negative plates to the negative of the source, electrons are drawn from the positive plates and forced onto the negative plates. Electrons arriving at the negative plates drive negative sulphate ions out of the lead sulphate back into the electrolyte. The sulphate ions join with hydrogen to form sulphuric acid. When electrons flow from the positive plates they leave positively charged lead ions. These attract oxygen from the water in the electrolyte to form lead peroxide on the plates. When the cell is fully charged the positive plates again become lead peroxide and the negative plates lead. The electrolyte becomes a high concentration of sulphuric acid. The specific gravity of the electrolyte of a fully charged cell is approximately 1.260. This falls to about 1.150 when the cell is completely discharged. These values will depend upon the manufacturer's instructions.

7

The specific, gravity, therefore, is a good indication of the state of charge of the cell and is measured using a hydrometer. Using the rubber bulb, enough electrolyte is drawn up into the hydrometer, to float the float. The specific gravity is then indicated by the calibration mark on the float at the surface of the electrolyte. This is shown in Figure 5-17.

FLOAT

1.100----1.1501.100 1.250 1.300 1.350

s

1.400

01

►

Figure 5.17 - The Hydrometer During the charging of the cell hydrogen gas is released from the electrolyte and bubbles to the surface. As the cell nears full charge more hydrogen is released and the bubbling increases. A vent is, therefore, incorporated in the cell cap.

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The voltage of a fully charged cell is approximately 2.2 volts (2 volts nominal) and in the discharged state 1.8 volts.

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The electrolyte level should be just above the top of the plates and the level will generally drop over a period of use due to evaporation and gassing. The level can be adjusted by topping up with distilled water after removal of the vent cap.

Li

Generally lead-acid batteries are made up of cells in a common case so that cells cannot be removed individually as shown in Figure 5.18.

L'

Figure 5.18 - A Typical Lead-Acid Battery

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

The Nickel-Cadmium Cell Aircraft engines, particularly turbines, require extremely high current for starting. High rate discharges of lead-acid batteries causes their output voltage to fall, due to the increased internal resistance caused by the build-up of sulphate deposits. This drawback led to the development of the alkaline cell for aircraft use. The nickel-cadmium, or ni-cad, battery has a very distinct advantage in that its internal resistance is very low. Its output voltage, therefore, remains almost constant until it is nearly totally discharged. The low resistance also allows high charging rates without damage. The ni-cad cell has positive plates made from powdered nickel which is fused, or sintered, to a porous nickel mesh. The mesh is then impregnated with nickel hydroxide.

The negative plates are of the same construction but are impregnated with cadmium hydroxide. Separators of nylon and cellophane, in the form of a continuous strip wound between the plates, keeps the plates from touching each other. Cellophane is used because it has low electrical resistivity and also acts as a gas barrier preventing oxygen, given off at the positive plates during overcharge, from passing to the negative plates. If the oxygen were allowed to reach the negative plates it would combine with active cadmium, reduce cell voltage and produce heat as a result of chemical reaction. The cell construction is shown in Figure 5.19, where the complete plate group is mounted in a sealed plastic container.

t l t

17,

CELLOPHANE

l f

NYLON

NYLON

l i,.

Figure 5.19 - Nickel-Cadmium Cell Construction

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governed by the statement on page 2 of this Chapter.

Integrated Training System

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Designed in association with the club66pro.co.uk question practice aid

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VENT AND FILLER CAP

TERMINAL U

SEPARATOR

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Figure 5.20 - Nickel-cadmium cell. The electrolyte is an alkaline solution of potassium hydroxide and distilled or de-ionized water with a specific gravity of 1.24 to 1.30. The specific gravity of the electrolyte does not change during charge or discharge so it cannot be used to indicate the state of charge.

1-7

L

LU

L

The electrolyte does not play an active part in the chemical reaction and is used only to provide a path for current flow. During charging of the cell an exchange of ions takes place. Oxygen is removed from the negative plates and added to the positive plates, the electrolyte acting as an ionized conductor.

The positive plates are, therefore, brought to a higher state of oxidation.

When the cell is fully charged all the oxygen is driven out of the negative plates, leaving only metallic cadmium, and the positive plates are highly oxidized nickel hydroxide. The electrolyte is forced out of both sets of plates during charging so that the electrolyte level in the cell rises. The electrolyte level is, therefore, only checked and any water added when the cell is fully charged.

r1

L.

Towards the end of the charging process and during overcharging, gassing occurs as a result of

h

electrolysis. This only reduces the water content of the electrolyte.

U

During discharge the chemical action is reversed. The positive plates gradually lose oxygen to become less oxidized and the negative plates regain lost oxygen and change to cadmium hydroxide.

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The plates absorb electrolyte so that the level in the cell falls but it should always cover the top of the plates. The charge and discharge levels are shown in Figure 5.21.

Charged level Discharged level

Li

Figure 5.21 - Nickel-Cadmium cell electrolyte levels The discharge and charging cycle of a ni-cad cell produces high temperatures which, if not correctly monitored, can break down the cellophane gas barrier. This creates a short circuit allowing current flow to increase. More heat is produced, causing further break down. The condition is aggravated by the internal resistance of the cell falling as the temperature rises. These factors all contribute to a process known as "thermal runaway", which ultimately results in the destruction of the cell. The ni-cad electrolyte would be contaminated and its specific gravity reduced if it were to be exposed to the carbon dioxide in the air. The atmosphere must, therefore, be kept out of a nicad cell. Three basic types of ni-cad cell are, therefore, produced: a)

The sealed type where the cell is completely sealed, as used in small capacity batteries.

b)

The semi-sealed type where the cell is almost fully sealed but has a safety pressure valve.

c)

The semi-open type which has a non-return valve, allowing the cell to gas yet preventing the electrolyte from being contaminated by the air. This type is used in the main aircraft battery.

The individual ni-cad cell produces an open circuit voltage of between 1.55 and 1.80 volts, depending on the manufacturer. n

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1

Although the nickel-cadmium battery has become the preferred type in today's aircraft, there are also the nickel-iron and silver-zinc types of alkaline cell. Silver-zinc rechargeable batteries have been used in the space programme, where size and weight factors greatly outweigh initial cost. The capacity of each cell is added together to obtain the total capacity. In effect the area of the L% plates has been increased. The voltage, on the other hand, does not increase.

L

L

Figure 5.22 -- Examples of NiCad Batteries

Li

LI Li Module 3.5 DC Sources of Electricity

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Battery Internal Resistance Each cell in a battery has a certain internal resistance. The terminal voltage of the battery when it is off load is not affected by this internal resistance. In Figure 5.23 the battery has been drawn with its cells in series with the total internal resistance of the battery.

© C p i 2 0

o

p

2

BATTERY TERMINAL VOLTAGE

t

C

p

Figure 5.23 - Battery Showing Cells and Internal Resistance If an external circuit is connected across the battery terminals of Figure 5.23, electrons will flow from the negative plate of the cells, through the external circuit and through the internal resistance to the positive plate of the cells. A voltage drop, or potential difference, will appear across the internal resistance due to the current flow. The voltage available to the external circuit at the battery terminals will now be the original off load terminal voltage minus the volts drop across the internal resistance. The terminal voltage will, therefore, decrease with an increase in circuit current or an increase in internal resistance.

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Battery Maintenance 1

Li

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U U1 U

The following information concerns the maintenance of secondary-cell batteries and is of a general nature. You must check the appropriate technical manuals for the specific type of battery prior to performing maintenance on any battery. Specific Gravity For a battery to work properly, its electrolyte (water plus active ingredient) must contain a certain amount of active ingredient. Since the active ingredient is dissolved in the water, the amount of active ingredient cannot be measured directly. An indirect way to determine whether or not the electrolyte contains the proper amount of active ingredient is to measure the electrolyte's specific gravity. Specific gravity is the ratio of the weight of a certain amount of a given substance compared to the weight of the same amount of pure water. The specific gravity of pure water is 1.0. Any substance that floats has a specific gravity less than 1.0. Any substance that sinks has a specific gravity greater than 1.0. The active ingredient in electrolyte (sulphuric acid, potassium hydroxide, etc.) is heavier than water. Therefore, the electrolyte has a specific gravity greater than 1.0. The acceptable range of specific gravity for a given battery is provided by the battery's manufacturer. To measure a battery's specific gravity, use an instrument called a hydrometer.

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U r U r-t

The Hydrometer A hydrometer, shown in Figure 5.24, is a glass syringe with a float inside it. The float is a hollow glass tube sealed at both ends and weighted at the bottom end, with a scale calibrated in specific gravity marked on its side. To test an electrolyte, draw it into the hydrometer using the suction bulb. Draw enough electrolyte into the hydrometer to make the float rise. Do not draw in so much electrolyte that the float rises into the suction bulb. The float will rise to a point determined by the specific gravity of the electrolyte. If the electrolyte contains a large amount of active ingredient, its specific gravity will be relatively high. The float will rise higher than it would if the electrolyte contained only a small amount of active ingredient. To read the hydrometer, hold it in a vertical position and read the scale at the point that surface of the electrolyte touches the float. Refer to the manufacturer's technical manual to determine whether or not the battery's specific gravity is within specifications.

Figure 5.24 - Hydrometer in use

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Note: Hydrometers should be flushed with fresh water after each use to prevent inaccurate readings. Storage battery hydrometers must not be used for any other purpose. Other Maintenance The routine maintenance of a battery is very simple. Terminals should be checked periodically for cleanliness and good electrical connection. The battery case should be inspected for cleanliness and evidence of damage. The level of electrolyte should be checked and if the electrolyte is low, distilled water should be added to bring the electrolyte to the proper level. Maintenance procedures for batteries are normally determined by higher authority and each command will have detailed procedures for battery care and maintenance. lE

Safety Precautions with Batteries All types of batteries should be handled with care: • never short the terminals of a battery • carrying straps should be used when transporting batteries. • protective clothing, such as rubber apron, rubber gloves, and a face shield should be worn when working with batteries. • no smoking, electric sparks, or open flames should be permitted near charging batteries. • care should be taken to prevent spilling of the electrolyte. In the event electrolyte is splashed or spilled on a surface, such as the floor or table, it should be diluted with large quantities of water and cleaned up immediately. If the electrolyte is spilled or splashed on the skin or eyes, IMMEDIATELY flush the skin or eyes with large quantities of fresh water for a minimum of 15 minutes. If the electrolyte is in the eyes, be sure the upper and lower eyelids are pulled out sufficiently to allow the fresh water to flush under the eyelids. The medical department should be notified as soon as possible and informed of the type of electrolyte and the location of the accident.

Capacity and Rating of Batteries The capacity of a battery is measured in ampere-hours. The ampere-hour capacity is equal to the product of the current in amperes and the time in hours during which the battery will supply this current. The ampere-hour capacity varies inversely with the discharge current. For example, a 400 ampere-hour battery will deliver 400 amperes for 1 hour or 100 amperes for 4 hours. Storage batteries are rated according to their rate of discharge and ampere-hour capacity. Most batteries are rated according to a 20-hour rate of discharge. That is, if a fully charged battery is completely discharged during a 20-hour period, it is discharged at the 20-hour rate. Thus, if a battery can deliver 20 amperes continuously for 20 hours, the battery has a rating of 20 amperes x 20 hours, or 400 ampere-hours. Therefore, the 20-hour rating is equal to the average current that a battery is capable of supplying without interruption for an interval of 20 hours. (Note: Aircraft batteries are rated according to a 1-hour rate of discharge). All standard batteries deliver 100 percent of their available capacity if discharged in 20 hours or more, but they will deliver less than their available capacity if discharged at a faster rate. The faster they discharge, the less ampere-hour capacity they have. 5-32 Use and/or disclosure is governed by the statement

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The low-voltage limit, as specified by the manufacturer, is the limit beyond which very little useful energy can be obtained from a battery. This low-voltage limit is normally a test used in battery shops to determine the condition of a battery.

Battery Charging It should be remembered that adding the active ingredient to the electrolyte of a discharged battery does not recharge the battery. Adding the active ingredient only increases the specific gravity of the electrolyte and does not convert the plates back to active material, and so does not bring the battery back to a charged condition. A charging current must be passed through the battery to recharge it. Batteries are usually charged in battery shops. Each shop will have specific charging procedures for the types of batteries to be charged. The following discussion will introduce you to the types of

battery charges.

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The following types of charges may be given to a storage battery, depending upon the condition of the battery: • • • • •

Initial charge Normal charge Equalizing charge Floating charge Fast charge

• Initial Charge When a new battery is shipped dry, the plates are in an uncharged condition. After the electrolyte has been added, it is necessary to charge the battery. This is accomplished by giving the battery a long low-rate initial charge. The charge is given in accordance with the manufacturer's instructions, which are shipped with each battery. • Normal Charge A normal charge is a routine charge that is given in accordance with the nameplate data during the ordinary cycle of operation to restore the battery to its charged condition.

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• Equalizing Charge An equalizing charge is a special extended normal charge that is given periodically to batteries as part of a maintenance routine. It ensures that all the sulphate is driven from the plates and that all the cells are restored to a maximum specific gravity. The equalizing charge is continued until the specific gravity of all cells, corrected for temperature, shows no change for a 4-hour period. • Floating Charge In a floating charge, the charging rate is determined by the battery voltage rather than by

a definite current value. The floating charge is used to keep a battery at full charge while

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Integrated Training System Designed in association with the club66pro,co,uk question practice aid

the battery is idle or in light duty. It is sometimes referred to as a trickle charge and is accomplished with low current.

n

• Fast Charge A fast charge is used when a battery must be recharged in the shortest possible time. The charge starts at a much higher rate than is normally used for charging. It should be used only in an emergency, as this type charge may be harmful to the battery. • Charging Rate Normally, the charging rate of aircraft storage batteries is given on the battery nameplate. If the available charging equipment does not have the desired charging rates, the nearest available rates should be used. However, the rate should never be so high that violent gassing (explained later in this text) occurs. • Charging Time The charge must be continued until the battery is fully charged. Frequent readings of specific gravity should be taken during the charge and compared with the reading taken before the battery was placed on charge.

Gassing When a battery is being charged, a portion of the energy breaks down the water in the electrolyte. Hydrogen is released at the negative plates and oxygen at the positive plates. These gases bubble up through the electrolyte and collect in the air space at the top of the cell. If violent gassing occurs when the battery is first placed on charge, the charging rate is too high. If the rate is not too high, steady gassing develops as the charging proceeds, indicating that the battery is nearing a fully charged condition. Warning: A mixture of hydrogen and air can be dangerously explosive. No smoking, electric sparks, or open flames should be permitted near charging batteries.

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Ther moc oupl es In 1821, the Germ anEston ian physi cist Thom as Johan n Seebe ck disco vered that when any c o n d u c t o r ( s u c h a

s a metal) is subjected to a thermal gradient, it will generate a voltage. This is now known as the thermoelectric effect or Seebeck effect. Any attempt to measure this voltage necessarily involves connecting another conductor to the "hot" end. This additional conductor will then also experience the temperature gradient, and develop a voltage of its own which will oppose the original. Fortunately, the magnitude of the effect depends on the metal in use. Using a dissimilar metal to complete the circuit creates a circuit in which the two legs generate different voltages, leaving a small difference in voltage available for measurement. That difference increases with temperature, and can typically be between one and seventy microvolts per degree Celsius (µV/t) for the modern range of available metal combinations. Certain combinations have become popular as industry standards, driven by cost, availability, convenience, melting point, chemical properties, stability, and output. This coupling of two metals gives the thermocouple its name. It is important to note that thermocouples measure the temperature difference between two points, not absolute temperature. In traditional applications, one of the junctions-the cold junction-was maintained at a known (reference) temperature, while the other end was attached to a probe. Having available a known temperature cold junction, while useful for laboratory calibrations, is simply not convenient for most directly connected indicating and control instruments. They incorporate into their circuits an artificial cold junction using some other thermally sensitive device, such as a thermistor or diode, to measure the temperature of the input connections at the instrument, with special care being taken to minimize any temperature gradient between terminals. Hence, the voltage from a known cold junction can be simulated, and the appropriate correction applied. This is known as cold junction compensation. Additionally, a device can perform cold junction compensation by computation. It can translate device voltages to temperatures by either of two methods. It can use values from look-up tables or approximate using polynomial interpolation. A thermocouple can produce current, which means it can be used to drive some processes directly, without the need for extra circuitry and power sources. For example, the power from a thermocouple can activate a valve when a temperature difference arises. The electric power generated by a thermocouple is a conversion of the heat energy that one must continuously supply to the hot side of the thermocouple to maintain the electric potential. The flow of heat is necessary because the current flowing through the thermocouple tends to cause the hot side to cool down and the cold side to heat up (the Peltier effect). Operation If two dissimilar metals are joined together a contact potential, which is independent of any external electrical supply, will appear at the junction. In a thermocouple two dissimilar metals are joined at both ends to form a hot junction and a cold junction. Module 3.5 DC Sources of Electricity

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In the simplest arrangement the thermocouple would be connected directly to a meter, the meter terminals being the cold junction. In an aircraft, however, the hot junction is in the engine and the meter indicator on the flight deck. If the thermocouple cold junction were to be connected to the meter by copper wires, as shown in Figure 5.25, the potential at the cold junction would be as if points "A" and "B" were joined together (provided that "A" and "B" were at the same temperature). This would still allow the meter to read the difference between V1 and V2.

ni

a

n COPPER

Figure 5.25 - Alternative thermocouple connections

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If however, the hot and cold junctions were relatively close together, the temperature difference between them would not be so great as if they were far apart. The thermocouple EMF would, therefore, be reduced and, in Figure 5.25, there would also be a problem of fluctuations in the readings. If the cold junction was in the meter itself there would be a greater temperature difference and hence a greater EMF and also less fluctuations.

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To achieve this, the connecting leads from the thermocouple to the meter must be of the same material as the thermocouple or at least have the same thermoelectric characteristics. They are called extension leads if they are of the same material and compensating leads if they are of the same characteristics. The small EMF generated by the thermocouple is not only dependent upon the temperature but also upon the metals employed. Figure 5.26 shows a graph of voltage against temperature for several common thermocouples.

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5-36 Use and/or disclosure is governed by the statement

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EIVIF (pv) I

nickel-ch

iumlcopper-nickel

iron/constantan

nickel-chromiLrrdnickel-aluminiumi

30 F-"'

20 37% pier num 13% rho ium /platinum

V

copper/constantan

10

9Q96� plafinum

1 C96i rhodium Iplatlnum

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250 L

300

750

1000

r

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1250

Figure 5.26 - Thermocouple Material Graph Nickel/chromium and nickel/aluminium are normally chosen for aircraft thermocouples due to their near linear characteristics and their long operating life at temperature of up to 11001. The nickel/chromium is the positive connection and the nickel/aluminium the negative connection. The thermocouple and its connections are housed in a protective metal sheath or probe which allows the hot junction to be exposed to the engine gases. Thermocouples can be connected in series with each other to form a thermopile, where all the hot junctions are exposed to the higher temperature and all the cold junctions to a lower temperature. Thus, the voltages of the individual thermocouple add up, which allows for a larger

L

voltage and increased power.

r--1

Materials Thermocouple materials are available in several different metallurgical formulations per type, such as: (listed in decreasing levels of accuracy and cost) Special limits of error, Standard, and Extension grades. Extension grade wire is less costly than dedicated thermocouple junction

L-

1

wire and is usually specified for accuracy over a more restricted temperature range. Extension

L

grade wire is used when the point of measurement is farther from the measuring instrument than would be financially viable for standard or special limits materials, and has a very similar thermal coefficient of EMF for a narrow range (usually encompassing ambient). In this case, a standard or special limits wire junction is tied to the extension grade wire outside of the area of

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temperature measurement for transit to the instrument. Since most modern temperature measuring instruments that utilize thermocouples are electronically buffered to prevent any significant current draw from the thermocouple, the length of the thermocouple or extension wire is irrelevant. Changes in metallurgy along the length of the thermocouple (such as termination strips or changes in thermocouple type wire) will introduce another thermocouple junction which affects measurement accuracy. Also, industry standards are that the thermocouple colour code is used for the insulation of the positive lead, and red is the negative lead. Types A variety of thermocouples are available, suitable for different measuring applications. They are usually selected based on the temperature range and sensitivity needed. Thermocouples with

low sensitivities (B, R, and S types) have correspondingly lower resolutions. Other selection

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criteria include the inertness of the thermocouple material, and whether or not it is magnetic. The thermocouple types are listed below with the positive electrode first, followed by the negative electrode. Type K (chromel-alumel) is the most commonly used general purpose thermocouple. It is inexpensive and, owing to its popularity, available in a wide variety of probes. They are available in the -200 CC to +1350 CC range. The typ e K was specified at a time when metallurgy was less advanced than it is today and, consequently, characteristics vary considerably between examples. Another potential problem arises in some situations since one of the constituent metals, nickel, is magnetic. The characteristic of the thermocouple undergoes a step change when a magnetic material reaches its Curie point. This occurs for this thermocouple at 354CC. Sensitivity is approximately 41 µV/ C.

n J

Type E (chromel-constantan) has a high output (68 µV/t) which makes it well s uited to cryogenic use. Additionally, it is non-magnetic. Type J (iron-constantan) is less popular than type K due to its limited range (-40 to +750 C). The main application is with old equipment that cannot accept modern thermocouples. J types cannot be used above 760 CC as an abrupt magnetic transformation causes permanent decalibration. The magnetic properties also prevent use in some applications. Type J thermocouples have a sensitivity of about 50 µV/CC. Type N (nicrosil-nisil) thermocouples are suitable for use at high temperatures, exceeding 1200 CC, due to their stability and ability to resist high temperature oxidation. Sensitivity is about 39 µV/CC at 900CC, slightly lower than type K. Desi gned to be an improved type K, it is becoming more popular. Types B, R, and S thermocouples use platinum or a platinum-rhodium alloy for each conductor. These are among the most stable thermocouples, but have lower sensitivity, approximately 10 µV/CC, than other types. The high cost of these thermocouple types makes them unsuitable for general use. Generally, type B, R, and S thermocouples are used only for high temperature measurements.

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LI

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

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Type B thermocouples use a platinum-rhodium alloy for each conductor. One conductor contains 30% rhodium while the other conductor contains 6% rhodium. These thermocouples are suited for use at up to 1800 t. Type B thermoc ouples produce the same output at 0"C and 42 C, limiting their use below about 50 t. Type R thermocouples use a platinum-rhodium alloy containing 13% rhodium for one conductor and pure platinum for the other conductor. Type R thermocouples are used up to

1600 `C.

Type S thermocouples use a platinum-rhodium alloy containing 10% rhodium for one conductor and pure platinum for the other conductor. Like type R, type S thermocouples are used up to 1600 C. In particular, type S is used a s the standard of calibration for the melting point of gold (1064.43 C). Type T (copper-constantan) thermocouples are suited for measurements in the -200 to 350 `C range. Often used as a differential measurem ent since only copper wire touches the probes. As both conductors are non-magnetic, type T thermocouples are a popular choice for applications such as electrical generators which contain strong magnetic fields. Type T thermocouples have a sensitivity of about 43 ltV/t.

L L

Type C (tungsten 5% rhenium -- tungsten 26% rhenium) thermocouples are suited for measurements in the 0 C to 23209C range. This the rmocouple is well-suited for vacuum furnaces at extremely high temperatures and must never be used in the presence of oxygen at temperatures above 260 C. Type M thermocouples use a nickel alloy for each wire. The positive wire contains 18% molybdenum while the negative wire contains 0.8% cobalt. These thermocouples are used in the vacuum furnaces for the same reasons as with type C. Upper temperature is limited to 1400 t. Though it is a less common type of thermocouple, look-up tables to correlate temperature to EMF (milli-volt output) are available.

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Module 3.5 DC Sources of Electricity

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club66pro.co.uk question practice aid

Thermocouple Comparison and Identification The table below describes properties of several different thermocouple types. Within the tolerance columns, T represents the temperature of the hot junction, in degrees Celsius. For example, a thermocouple with a tolerance of ±0.0025xT would have a tolerance of ±2.5 CC at 1000 1C.

Type

Temperature range 9C (continuous)

K

0 to +1100

J

0 to +700

E.J

IL

BS Colour codeANSI Colour code Brown

Yellow

Blue

Red

Yellow Blue

White Red LJ

N

0 to +1100

R

0 to +1600

S

0 to 1600

B

+200 to +1700

T

-185 to +300

E

0 to +800

Orange

Orange

White

Red

White

Not defined.

Blue White Blue

No standard use copper wire

Not defined.

Not defined.

White Blue

Blue Red

Brown

Blue

Blue

Rde

Table 5.1 -- Thermocouple comparison and wire identification ._ J

In

H 5-40 Use and/or disclosu re is governed by the statement

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[ 1 Applications L A practical thermocouple is shown in Figure 5.27.

ARMOURED CASING

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CONNECTING

LEADS CERAMIC INSULATION

HOT JUNCTION

Figure 5.27 - A practical thermocouple Two basic types of probe are employed for measuring exhaust gas temperatures in turbine engines. These are shown in Figure 5.28. COUPLE

COUPLE

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SHEATH

STAGNATION TYPE

SHEATH

RAPID RESPONSE TYPE

Figure 5.28 - Turbine engine probes U F-1

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Figure 5.29 - Examples of thermocouple hot junction assemblies The stagnation probe has a large entry port and a small exit port so that the gas is brought almost to rest, preventing errors caused by the kinetic energy of the gas flow. This type is designed for high velocity gas flow.

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The rapid response probe is designed for slow exhaust gas velocity. The gas flows from the inlet port, over the junction, to the diametrically opposite outlet port. Exhaust gas thermocouples are mounted radially around the engine tail pipe. There are usually a minimum of four. The RB 211 engine, however, has seventeen connected in a parallel arrangement which has the advantage that the failure of one or more thermocouples does not cause complete failure of the output signal. A typical thermocouple installation is shown in Figure 5.30.

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EXHAUST THERMOCOUPLE

AND HARNESS

E AIR INTAKE THERMOCOUPLE

JUNCTION BOX TO

INSTRUMENTATION AND CONTROL SYSTEM

Figure 5.30 - Thermocouple installation

U Module 3.5 DC Sources of Electricity

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Photocells

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Photocells undergo a change in their electrical parameters when exposed to light energy and are known as photoelectric devices. They are affected by light in three different ways as follows. Photo-emission:- Where the application of light causes the emission of electrons from a prepared surface as discussed in Chapter 4, the construction of which is shown in Figure 5.31. AIRTIGHT EVACUATED GLASS ENVELOPE

EXTERNAL CONNECTIONS

Figure 5.31 - The Photocell With the positive potential of a supply connected to the anode of the cell and the negative to the cathode, the current in the circuit will depend upon the amount of light falling on the device: no light, no current; high intensity light, high current. When the cell is used in an aircraft smoke detector, a projector lamp shines abeam of light past the detector cell. If no light reaches the cell, no current flows in the cell's external circuit and no warning is given. When smoke appears in the detection chamber the projector lamp beam is refracted onto the detector cell by the smoke particles. The cell conducts activating the smoke warning circuit. This is shown in Figure 5.32.

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Figure 5.32 - Smoke Detector Operation Solid state devices have now largely replaced this type of cell. Photo-voltaic:- Where the application of light causes the production of a voltage. The photo-voltaic (or solar cell), can be used to produce electrical energy for a variety of purposes. If a large number of cells are connected together to form a solar panel the power generated is limited only by the number of cells employed.

The silicon solar cell consists of a wafer of silicon which has been doped to make it a semiconductor. A thin layer of boron is then diffused into it.

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The wafer is reinforced with metal and provided with electrical contacts to enable it to be connected to other cells.

Figure 5.33 - A photovoltaic cell panel

Photons of light penetrating an atom of the cell forces electrons in the atom into the conduction band. This produces a voltage across the cell which can be used to drive a current around an externally connected circuit. There are many uses of the solar cell, from the operation of light meters in cameras to powering calculators and satellites in space.

Module 3.5 DC Sources of Electricity

5-45

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Photo-conduction:- Where a device undergoes a change of resistance with a variation in light intensity. The photo-conductive cell or light dependent resistor is a solid state device as shown in Figure 5.34.

N TYPE PHOTOCONDUCTIVE MATERIAL

PROTECTIVE GLASS CAP

P TYPE SUBSTANCE

CONNECTING PINS

Figure 5.34 - The Photo-Conductive Cell The effective area of the light collecting photo-conductive material is increased by etching it onto the substrate in a serpentine manner. When there is an increase in light intensity the additional photon bombardment releases more electrons from the atomic bond which increases the current through the device. The resistance has, therefore, decreased. The reverse occurs with a reduction in light intensity.

Figure 5.35 - A photoconductor r7

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5-46 TTS Integrated Training System disclosure is 1 Use and/or governed by the statement

Module 3.5 DC Sources of Electricity

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Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Ltd.

Knowledge Levels - Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category 81 or the category B2

basic knowledge levels.

The knowledge level indicators are defined as follows:

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

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LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject9 The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • A detailed knowledge of the theoretical and practical aspects of the subject. • A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive

manner.

Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents

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Module 3.6 DC Circuits The Basic Electric Circuit Ohm's Law Series DC Circuits Kirchhoff's Voltage Law Kirchhoff's Current Law Circuit Terms and Characteristics Internal Resistance of the Supply Parallel DC Circuits Series-Parallel DC Circuits Practice Circuit Problem

5 5 7 16 27 34 41 46 48 66 71

Redrawing Circuits for Clarity

75

Effects of Open and Short Circuits Voltage Dividers

80 83

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DC Circuits Ohms Law, Kirchhoff's Voltage and Current Laws Calculations using the above laws to find resistance, voltage and current Significance of the internal resistance of a supply

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

R1

AA/VNV 204

ET_?

50 Q R 3

Figure 6.14 - Solving for applied voltage in a series circuit.

Given: R 1= 20 ohms R2 = 30 ohms R3 = 50 ohms I = 2 amps

Solution: ET=E1+E2+E3 E1= R1 x I1 (I1= The current through resistor R1) E2=R2xI2 E3=R3xI3

Substituting:

ET=(RixI1)+(R2x12)+

(83x13)

ET=(20 ohms x 2 amps)+(30 ohms x 2amps)+ (50 ohms x2amps) ET = 40 volts + 60 volts + 100 volts ET = 200 volts

Module 3.6 DC Circuits

6-21

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Note: When you use Ohm's law, the quantities for the equation must be taken from the same part of the circuit. In the above example the voltage across R2 was computed using the current through R 2 and the resistance of R2. The value of the voltage dropped by a resistor is determined by the applied voltage and is in proportion to the circuit resistances. The voltage drops that occur in a series circuit are in direct proportion to the resistances. This is the result of having the same current flow through each resistor - the larger the ohmic value of the resistor, the larger the voltage drop across it. Summary of Series DC Circuit Characteristics The important factors governing the operation of a series circuit are listed below. These factors have been set up as a group of rules so that they may be easily studied. These rules must be completely understood before the study of more advanced circuit theory is undertaken. Rules for Series DC Circuits • The same current flows through each part of a series circuit. • The total resistance of a series circuit is equal to the sum of the individual resistances. • The total voltage across a series circuit is equal to the sum of the individual voltage drops. • The voltage drop across a resistor in a series circuit is proportional to the ohmic value of the resistor. Series Circuit Analysis To establish a procedure for solving series circuits, the following sample problems will be solved. Example: Three resistors of 5 ohms, 10 ohms, and 15 ohms are connected in series with a power source of 90 volts as shown in Figure 6.15. Find the total resistance, circuit current, voltage drop of each resistor. R 5

R2 100

ET

90 V

R3 15Q

Figure 6.15 - Solving for various values in a series circuit.

6-22

Module 3.6 DC Circuits

Use and/or disclosure is

governed statement

by

the

Use and/or disclosure is

TTS Integrated Training System nr rnnvrinhf grain

TTS Integrated Training System © Copyright 2010

governed by the statement on page 2 01 this Chapter.

Integrated Training System Designed in association with the ciub66pro.co,uk question practice aid

In solving the circuit the total resistance will be found first. Next, the circuit current will be

calculated. Once the current is known, the voltage drops and power dissipations can be calculated. Given:

R1 - 5 ohms R2 = 10 ohms R3 = 15 ohms E = 90 volts

Solution: RT = R1 + R2 + R3

RT=5 ohms + 10 ohms + 15 ohms RT = 30 ohms In

U U

rl

L

ET R

T

90 volts 30 ohms I = 3 amps El = IR1

E1=3amperes x 5 ohms E1= 15 volts E2 = IR2 Ez = 3 amperes x 10 ohms E2 = 30 volts E3 = IR3 E3 = 3 amperes x 15 ohms E3 = 45 volts

Module 3.6 DC Circuits

6-23

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Example: Four resistors, Ri = 10 ohms, R2 = 10 ohms, R3 = 50 ohms, and R4 = 30 ohms, are connected in series with a power source as shown in Figure 6.16. The current through the circuit is 1/2 ampere. • What is the battery voltage? • What is the voltage across each resistor?

n E T

?

Figure 6.16 - Computing series circuit values. Given:

R1 = 10 ohms R2= 10 ohms R3 = 50 ohms R4 = 30 ohms 1 = 0.5 amps

n 6-24 Use and/or disclosure is governed by the statement

TTS Integrated[c1 Training System rnnvrinht 001 n

TTS Integrated Training System © Copyright 2010

Module 3.6 DC Circuits

Use and/or disclosure is governed by the statement on page 2 of this Chapter.

Integrated Training System Designed in association with the club66pro.co.uk question practice aid ,i

L L

Solution (a):

ET = IRT +R,4 RT=R1+R2+R3 RT = 10 ohms + 10 ohms + 50 ohms + 30 ohms

RT = 100 ohms

L-

ET = 0.5 ampsx 100 ohms ET= 50 volts Solution (b):

E1= IR1 E1= 0.5 amperes x 10 ohms

E1= 5 volts

L F

1

E2 = IR2 E2 = 0,5 amperes x 10 ohms E2=5Volts

E3 = IR3 E3 = 0.5 amperes x 50 ohms E3 = 25 volts EA = IR4 E4 = 0.5 amperes x 30 ohms Eq = 15 volts An important fact to keep in mind when applying Ohm's law to a series circuit is to consider whether the values used are component values or total values. When the information available enables the use of Ohm's law to find total resistance, total voltage, and total current, total values must be inserted into the formula. To find total resistance:

L

L

To find total voltage:

ET=IT=RT

Module 3.6 DC Circuits

6-25

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

To find total current:

IT - E T RT Note: IT is equal to I in a series circuit. However, the distinction between IT and I in the formula should be noted. The reason for this is that future circuits may have several currents, and it will be necessary to differentiate between IT and other currents. To compute any quantity (E, I, R, or P) associated with a single given resistor, the values used in the formula must be obtained from that particular resistor. For example, to find the value of an unknown resistance, the voltage across and the current through that particular resistor must be used. To find the value of a resistor:

R_ ER IR To find the voltage drop across a resistor: n

ER 'R xR To find current through a resistor:

IR

ER

R

n

p

6-26 TTS Integrated Training System Use and/or disclosure is governed by the statement

Module 3.6 DC Circuits

Use and/or disclosure t governed by the statement

TTS Integrated Training System

© Copyright 2010

on page 2 o1 this Chapter.

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Kirchhoff's Voltage Law In 1847, G.R.Kirchhoff extended the use of Ohm's law by developing a simple concept concerning the voltages contained in a series circuit loop. Kirchhoff's voltage law states: "The algebraic sum of the voltage drops in any closed path in a circuit and the electromotive forces in that path is equal to zero. " To state Kirchhoff's law another way, the voltage drops and voltage sources in a circuit are equal at any given moment in time. If the voltage sources are assumed to have one sign (positive or negative) at that instant and the voltage drops are assumed to have the opposite sign, the result of adding the voltage sources and voltage drops will be zero.

U

Note: The terms electromotive force and EMF are used when explaining Kirchhoff's law because Kirchhoff's law is used in alternating current circuits. In applying Kirchhoff's law to direct current circuits, the terms electromotive force and EMF apply to voltage sources such as batteries or power supplies. Through the use of Kirchhoff's law, circuit problems can be solved which would be difficult, and often impossible, with knowledge of Ohm's law alone. When Kirchhoff's law is properly applied, an equation can be set up for a closed loop and the unknown circuit values can be calculated.

F1

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Module 3.6 DC Circuits

6-27

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Polarity of Voltage To apply Kirchhoff's voltage law, the meaning of voltage polarity must be understood. In the circuit shown in Figure 6.17, the conventional current is shown flowing in a clockwise direction. Notice that the end of resistor R2, into which the current flows, is marked positive (+). The end of R2 at which the current leaves is marked negative (-). These polarity markings are used to show that the end of R1 into which the current flows is at a higher positive potential than the end of the resistor at which the current leaves. Point D is more positive than point C.

4'

Figure 6.17 - Voltage polarities. Point B, which is at the same potential as point C, is labelled positive. This is to indicate that point B is more positive than point A. To say a point is positive (or negative) without stating what the polarity is based upon has no meaning. In working with Kirchhoff's law, positive and negative polarities are assigned in the direction of current flow. Application of Kirchhoff's Voltage Law Kirchhoff's voltage law can be written as an equation, as shown below: Ea

+ Eb + Ec +... En=0

where Ea, Eb, etc., are the voltage drops or EMFs around any closed circuit loop. To set up the equation for an actual circuit, the following procedure is used.

• Assume a direction of current through the circuit. (The correct direction is desirable but not necessary.) • Using the assumed direction of current, assign polarities to all resistors through which the current flows. • Place the correct polarities on any sources included in the circuit. n 6-28 Use and/or disclosure is governed by the statement

Module 3.6 DC Circuits

Use

by tdisclosure is

TTS Integrated Training System rn

on1 n

TTS Integrated Training System

© Copyright 2010

governed by the statement on page 2 of this Chapter.

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

• Starting at any point in the circuit, trace around the circuit, writing down the amount and

polarity of the voltage across each component in succession. The polarity used is the sign after the assumed current has passed through the component. Stop when the point at which the trace was started is reached.

iu

• Place these voltages, with their polarities, into the equation and solve for the desired quantity. Example: Three resistors are connected across a 50-volt source. What is the voltage across the third resistor if the voltage drops across the first two resistors are 25 volts and 15 volts? Solution: First, a diagram, such as the one shown in Figure 6.18, is drawn. Next, a direction of

current is assumed (as shown). Using this current, the polarity markings are placed at each end of each resistor and also on the terminals of the source. Starting at point A, trace around the circuit in the direction of current flow, recording the voltage and polarity of each component. Starting at point A and using the components from the circuit: ( +Ex) + (+E2) + (+E1) +(-EA) = 0 Substituting values from the circuit: U

25 volts + 15 volts +Ex - 50 volts = 0 Ex- 10volts =0 Ex = 10 volts The unknown voltage (Ex) is found to be 10 volts.

Li

L+ T

50V

EX

+

Figure 6.18 - Determining unknown voltage in a series circuit. Using the same idea as above, you can solve a problem in which the current is the unknown quantity.

Module 3.6 DC Circuits

6-29

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Example: A circuit having a source voltage of 60 volts contains three resistors of 5 ohms, 10 ohms, and 15 ohms. Find the circuit current. Solution: Draw and label the circuit (Figure 6.19). Establish a direction of current flow and assign polarities. Next, starting at any point - point A will be used in this example - write out the loop equation. R1

w%/VV\ 5c El

+

ion E2

60V

16 E3

+

A

R3

Figure 6.19 - Correct direction of assumed current. Basic Equation, starting at A E3-EA+E1+E2=0

Since E = IR, by substitution: (IxR3)-EA +(IxRi)+(lxR2)=0 Substituting values: (I x 15 ohms) - 60 volts + (I x 5 ohms) + (I x 10 ohms) = 0 Combining like terms: (I x 30 ohms) - 60 volts = 0 (I x 30 ohms) = 60 volts 1= 2 amps Since the current obtained in the above calculations is a positive 2 amps, the assumed direction of current was correct. To show what happens if the incorrect direction of current is assumed, the problem will be solved as before, but with the opposite direction of current. The circuit is redrawn showing the new direction of current and new polarities in Figure 6.20. Starting at point A the loop equation is:

6-30 Use and/or disclosure is

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Module 3.6 DC Circuits

use and/or dis ta e e is

TTS Integrated Training System

n t'.,r,.,rinh+ 9M (

Ti

TTS Integrated Training System © Copyright 2010

governed

by the st atement

on page 2 of this Chapter.

Integrated Training System

f -,

L

Designed in association with the club66pro.co.uk question practice aid

Basic Equation, starting at A E3+EA+E1+E2=0 Since E = IR, by substitution: (IxR3)+EA +(IxR1)+(IxR2)=0

U

Substituting values: (I x 15 ohms) + 60 volts + (I x 5 ohms) + (I x 10 ohms) = 0

E

Li

Combining like terms: (I x 30 ohms) + 60 volts = 0 (I x 30 ohms) = -60 volts I = -2 amps

l...l'I Et

Ion

-�-+ EA 60Y

E2

15 M

-

E3

+

A

R3 Figure 6.20 - incorrect direction of assumed current.

L

Notice that the amount of current is the same as before. The polarity, however, is negative. The negative polarity simply indicates the wrong direction of current was assumed. Should it be necessary to use this current in further calculations on the circuit using Kirchhoff's law, the negative polarity should be retained in the calculations.

Li

I

U Use and/or disclosure is

Module 3.6 DC Circuits

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Series Aiding and Opposing Sources In many practical applications a circuit may contain more than one source of EMF. Sources of EMF that cause current to flow in the same direction are considered to be series aiding and the voltages are added. Sources of EMF that would tend to force current in opposite directions are said to be series opposing, and the effective source voltage is the difference between the opposing voltages. When two opposing sources are inserted into a circuit current flow would be

in a direction determined by the larger source. Examples of series aiding and opposing sources are shown in Figure 6.21. � �II + E2

-aE1

R1 R2

SERIES AIDING

R1

SERIES OPPOSING Figure 6.21 - Aiding and opposing sources. A simple solution may be obtained for a multiple-source circuit through the use of Kirchhoff's voltage law. In applying this method, the same procedure is used for the multiple-source circuit as was used above for the single-source circuit. This is demonstrated by the following example.

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Example: Using Kirchhoff's voltage equation, find the amount of current in the circuit shown in fig 3-22.

L

V

vv�V^ Rj

+

60 Q

L

R2 --

1813V

20Q L

Figure 6.22 - Solving for circuit current using Kirchhoff's voltage equation. L

Solution: As before, a direction of current flow is assumed and polarity signs are placed on the drawing. The loop equation will be started at point A. ER2+E3+E1+ER1+E2 =0

(I x 20 ohms) + 40 volts + (-180 volts) + (I x 60 ohms) + 20 volts = 0 20 volts + 40 volts - 180 volts + (I x 20 ohms) + (I x 60 ohms) = 0 -120 volts + (I x 80 ohms) = 0 I x 80 ohms = 120 volts I = 120/80 = 1.5 amps

L!

r.._.,

Use and/or disclosure Is

Module 3.6 DC Circuits

6-33 TTS Integrated Training System

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Kirchhoff's Current Law Ohm's law states that the current in a circuit is inversely proportional to the circuit resistance. This fact is true in both series and parallel circuits. There is a single path for current in a series circuit. The amount of current is determined by the total resistance of the circuit and the applied voltage. In a parallel circuit the source current divides among the available paths. The behaviour of current in parallel circuits will be shown by a series of illustrations using

example circuits with different values of resistance for a given value of applied voltage. Part (A) of Figure 6.23 shows a basic series circuit. Here, the total current must pass through

the single resistor. The amount of current can be determined.

R1 N

"I/

IT= 5A

100

(A)

12 =5A IT = 10A R-1 1002

R2 100

11 = 5A IT = 14A (8)

Figure 6.23 - Analysis of current in parallel circuit.

50 volts

I

T = 10 ohms IT =5amps

n

6-34 TTS Integrated Training System governed by the statement

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rrn rn,- rt„hr on1n

© Copyright 2010

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L'

Part (B) of Figure 6.23 shows the same resistor (R1) with a second resistor (R2) of equal value connected in parallel across the voltage source. When Ohm's law is applied, the current flow through each resistor is found to be the same as the current through the single resistor in part (A). Given:

U

L

ES = 50 volts

Ri = 10 ohms R2 = 10 ohms Solution:

Li

`i L

I=E R E5 =E R1-ER2

Li

-,

r

I R1 =E R1 R1

iL r

I R1 =

L

50 volts 10 ohms

I R1 = 5 amps IR2

IR2

ER2 = R2

50 volts 10 ohms

IR2 =5 amps

ri

r1

It is apparent that if there is 5 amperes of current through each of the two resistors, there must be a total current of 10 amperes drawn from the source. The total current of 10 amperes, as illustrated in Figure 6.23 (B) leaves the positive terminal of the battery and flows to point a. Since point a is a connecting point for the two resistors, it is called a junction. At junction a, the total current divides into two currents of 5 amperes each.

These two currents flow through their respective resistors and rejoin at junction b. The total current then flows from junction b back to the positive terminal of the source. The source supplies a total current of 10 amperes and each of the two equal resistors carries one-half the total current.

Module 3.6 DC Circuits

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

From the previous explanation, the characteristics of current in a parallel circuit can be expressed in terms of the following general equation: IT=11 +12+.

n

.In

1.

Compare part (A) of Figure 6.24 with part (B) of the circuit in Figure 6.24. Notice that doubling the value of the second branch resistor (R2) has no effect on the current in the first branch (IR1), but does reduce the second branch current (IR2) to one-half its original value. The total circuit current drops to a value equal to the sum of the branch currents. These facts are verified by the following equations. Given: E S = 50 volts R1 =10 ohms

R2 = 20 ohms

Solution:

R

E

Es =ER1 =ER2 ER1

L1

R1 50 volts

J

10 ohms IRI =5 amps IR2

; .a

_ ER2

IR2 =

R2

IT

50 volts 20 ohms

IR2 =2.5 amps IT =I R1+IR2

I T =5amps + 2.5 amps IT =7.5 amps

6-36 Use and/or disclosure is

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Module 3.6 DC Circuits

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ITS Integrated Training System © Copyright 2010

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-i

Integrated Training System Designed in association with the ciub66pro.co.uk question practice aid

IT=7.5A E3

R1 11 11 =1A

T 50v LJ

12 2.3A R2

20C?

ion IT = 7.5A

(I

L~

A}

[I

L

(B)

L

Figure 6.24 - Current behaviour in parallel circuits. r

L

The amount of current flow in the branch circuits and the total current in the circuit shown in Figure 6.24 (B) are determined by the following computations. Given: E S = 50 volts R 1 =10 ohms Rz =10 ohms R 3 =10 ohms

L

t La

Module 3.6 DC Circuits

6-37

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Solution: E

R

ES = ER1 = ER2 = ER3

IR1 = -

IR1 =

R1 50 volts 10 ohms

I R1 = 5 amps IR2

- ER2

Rz

IR2 =50 volts

10 ohms

IR2 = 5 amps

I R3 =ER3

R3

I R3

50 volts 10 ohms

IR3 = 5 amps IT =I R1 +IR2 +IR3

I T = 5 amps + 5 amps + 5 amps

IT =15 amps Notice that the sum of the ohmic values in each circuit shown in Figure 6.24 is equal (30 ohms), and that the applied voltage is the same (50 volts). However, the total current in 6.24 (B) (15 amps) is twice the amount in 6.24 (A) (7.5 amps). It is apparent, therefore, that the manner in which resistors are connected in a circuit, as well as their actual ohmic values, affect the total current.

n

LJ

6-38 Use and/or disclosure is

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Module 3.6 DC Circuits

Use and/or disclosure is

TTS Integrated Training System (cl (nnvrinhf 9n-10

TTS Integrated Training System © Copyright 2010

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tJ

F-J

Integrated Training System

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Designed in association with the club66pro.co.uk question practice aid

The division of current in a parallel network follows a definite pattern. This pattern is described by kirchhoff`s current law which states: "The algebraic sum of the currents entering and leaving any junction of conductors is equal to zero. " This law can be stated mathematically as: Ia+lb+.

.In

+O

where: la, lb, etc., are the currents entering and leaving the junction. Currents entering the junction are considered to be positive and currents leaving the junction are considered to be negative. When solving a problem using Kirchhoff's current law, the currents must be placed into the equation with the proper polarity signs attached. Example. Solve for the value of 13 in Figure 6.25.

Given: I 1 = 10 amps

12 = 3 amps 14

=5 amps

Ia+Ib+,,.In =0

Solution: Il=3A

11 =10A

J

1►

13=?

14=SA

Figure 6.25 - Circuit for example problem. L.

The currents are placed into the equation with the proper signs. I1+I2+ 13+ I4=0

10 amps

+(-3 amps) +I3+(-5 amps) I3+2amps =0

0

13= -2 amps

13 has a value of 2 amperes, and the negative sign shows it to be a current leaving the junction. r, u

Module 3.6 DC Circuits

6-39

Use and/or disclosure is

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TTS Integrated Training System

n Cnnvrinht 9n 1n

Lr

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Example: Using Figure 6.26, solve for the magnitude and direction of 13. 12 = 3A fl

11 =6A

J

13=?

14 = 5A

Figure 6.26 - Circuit for example problem. Given: I1 =6 amps 12 = 3 amps I4=5amps Solution:

U Ia+Ib+,,, I„=0

7

I1+I2+I3+14 = 0

6 amps + (-3 amps) +I3+(-5 amps)=0 I3+(-2 amps) = 0 13= -2 amps

13 is 2 amperes and its positive sign shows it to be a current entering the junction.

6-40 6 © Copyright 2010

Module 3.6 DC Circuits

t governed he on page 2 r Chapter. o

rI

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Circuit Terms and Characteristics Before you learn about the types of circuits other than the series circuit, you should become familiar with some of the terms and characteristics used in electrical circuits. These terms and characteristics will be used throughout your study of electricity and electronics.

Reference Point A reference point is an arbitrarily chosen point to which all other points in the circuit are compared. In series circuits, any point can be chosen as a reference and the electrical potential at all other points can be determined in reference to that point. In Figure 6.27 point A shall be considered the reference point. Each series resistor in the illustrated circuit is of equal value. The applied voltage is equally distributed across each resistor. The potential at point D is 75 volts more positive than at point A. Points C and B are 50 volts and 25 volts more positive than

point A respectively. r-

--- D +75V

I+ 75V

AOV

Figure 6.27 - Reference points in a series circuit. When point B is used as the reference, as in Figure 6.28, point D would be positive 50 volts in respect.to the new reference point. The former reference point, A, is 25 volts negative in respect to point B.

U

Module 3.6 DC Circuits

6-41

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

n

a fly 25V

7''- A -25V

Figure 6.28 - Determining potentials with respect to a reference point. As in the previous circuit illustration, the reference point of a circuit is always considered to be at zero potential. Since the earth (ground) is said to be at a zero potential, the term ground is used to denote a common electrical point of zero potential. In Figure 6.29, point A is the zero reference, or ground, and the symbol for ground is shown connected to point A. Point C is 75 volts positive in respect to ground.

G +75V

E2 t 50V

l+

75V

B +25V

= 25V t'--- A OV

Figure 6.29 - Use of ground symbols. In most electrical equipment, the metal chassis is the common ground for the many electrical circuits. When each electrical circuit is completed, common points of a circuit at zero potential are connected directly to the metal chassis, thereby eliminating a large amount of connecting

6-42 Use and/or disclosure is

Module 3.6 DC Circuits TTS Integrated Training System

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0 Convrinht 2010

LJ

TTS Integrated Training System © Copyright 2010

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wire. The current passes through the metal chassis (a conductor) to reach other points of the i circuit. This is particularly useful on aircraft where the airframe can be used as the return circuit for all the aircraft's electrical systems. An example of a chassis grounded circuit is illustrated in Figure 6.30. (i

Rq

E

R2

CONDUCTING CHASSIS

Figure 6.30 - Ground used as a conductor. Most voltage measurements used to check proper circuit operation in electrical equipment are taken in respect to ground. One meter lead is attached to a grounded point and the other meter lead is moved to various test points. Open Circuit A circuit is said to be open when a break exists in a complete conducting pathway. Although an open occurs when a switch is used to de-energize a circuit, an open may also develop accidentally. To restore a circuit to proper operation, the open must be located, its cause determined, and repairs made. Sometimes an open can be located visually by a close inspection of the circuit components. Defective components, such as burned out resistors, can usually be discovered by this method. Others, such as a break in wire covered by insulation or the melted element of an enclosed fuse, are not visible to the eye. Under such conditions, the understanding of the effect an open has on circuit conditions enables a technician to make use of test equipment to locate the open component. In Figure 6.31, the series circuit consists of two resistors and a fuse. Notice the effects on circuit r;

conditions when the fuse opens.

1

Li

Module 3.6 DC Circuits

6-43

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

5V 1n')

FUSE �'

) By

(A) NORMAL CIRCUIT (NORMAL CURRENT)

-L LA

'I 5th

I

V-1)

0V

0V BLOWN FUSE

15V

(B) OPEPI CIRCUIT (DUE TO EXCESSIVE CURRENT) Figure 6.31 - Normal and open circuit conditions. (A) Normal current; (B) Excessive current. Current ceases to flow; therefore, there is no longer a voltage drop across the resistors. Each end of the open conducting path becomes an extension of the battery terminals and the voltage felt across the open is equal to the applied voltage (EA).

7 LI

An open circuit has infinite resistance. Infinity represents a quantity so large it cannot be measured. The symbol for infinity is o. In an open circuit, RT = -Short Circuit A short circuit is an accidental path of low resistance which passes an abnormally high amount of current. A short circuit exists whenever the resistance of a circuit or the resistance of a part of a circuit drops in value to almost zero ohms. A short often occurs as a result of improper wiring or broken insulation. In Figure 6.32, a short is caused by improper wiring. Note the effect on current flow. Since the resistor has in effect been replaced with a piece of wire, practically all the current flows through the short and very little current flows through the resistor. Current flows through the short (a path of almost zero resistance) and the remainder of the circuit by passing through the 10-ohm resistor and the battery. The amount of current flow increases greatly because its resistive path has decreased from 10,010 ohms to 10 ohms. Due to the excessive current flow, the 10-ohm 6-44 TTS Integrated Training System Use and/or disclosure is governed by the statement

-

a .d .W� r...,...-

Module 3.6 DC Circuits

Use and/or disclosure is governed by the statement

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.j

© Copyright 2010

on page 2 of this Chapter.

IL I

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r-,

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resistor becomes heated. As it attempts to dissipate this heat, the resistor will probably be destroyed. Figure 6.33 shows a pictorial wiring diagram, rather than a schematic diagram, to indicate how broken insulation might cause a short circuit.

R1 VVAJV

1

10,000 fD

NORMAL CURRENT

A

R1 R1 = 1090005

T

Li

1052

SHORT

r 3

EXCESSIVE CURRENT B

Figure 6.32 - Normal and short circuit conditions.

SHORT DUE TO WORN INSULATION

Figure 6.33 - Short due to broken insulation. Fii

Use and/or disclosure is

Module 3.6 DC Circuits

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Internal Resistance of the Supply A meter connected across the terminals of a good 1.5-volt battery reads about 1.5 volts. When the same battery is inserted into a complete circuit, the meter reading decreases to something less than 1.5 volts. This difference in terminal voltage is caused by the internal resistance of the battery (the opposition to current offered by the electrolyte in the battery). All sources of electromotive force have some form of internal resistance which causes a drop in terminal voltage as current flows through the source. This principle is illustrated in Figure 6.34, where the internal resistance of a battery is shown as Ri. In the schematic, the internal resistance is indicated by an additional resistor in series with the battery. The battery, with its internal resistance, is enclosed within the dotted lines of the schematic diagram. With the switch open, the voltage across the battery terminals reads 15 volts. When the switch is closed, current flow causes voltage drops around the circuit. The circuit current of 2 amperes causes a voltage drop of 2 volts across Ri. The 1-ohm internal battery resistance thereby drops the battery terminal voltage to 13 volts. Internal resistance cannot be measured directly with a meter. An attempt to do this would damage the meter.

n Figure 6.34 - Effect of internal resistance.

L.J

The effect of the source resistance on the power output of a DC source may be shown by an analysis of the circuit in Figure 6.35. When the variable load resistor (RL) is set at the zero-ohm position (equivalent to a short circuit), current (I) is calculated using the following formula:

E

100 volts

Ri

5 ohms

I=-=

= 20 amperes

n

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This is the maximum current that may be drawn from the source. The terminal voltage across the short circuit is zero volts and all the voltage is across the resistance within the source.

E ' 100V

RL

1 P

I

+oEFF.

L

EE

sf n

4

I [

16.7

9

64.3 10 66..7 ;gyp $D 30 85.7

-TER&IINAL VOLTAGE f3 L = RC5j, 8' a NC E 13P L OAD p

1 16.7 2 28.6 31 37.5

71 58.3 1 61.6

= OPEN • CIRCUIT VOLTAGE O SOURCE

RI - INIERrffi RESIST ricE OF SOURCE

7

20

D

0 279.9

14.3 11 4-09 12.5 465.8 4 44.4 11.'1 492.8 s 50 I -ja-1 &0Q 6 54.5 61 49134

-l

L

DD

L =OrJEk U$ED IN LOAD

4 tJ6 9

$0 90.9

1 =CUR RENT Fi!ZC4r1 &O URGE

16.7

1 44.4

&3 I 483.9 7.7 7.1

471.3 454.5

4

320

28.6 V.5

50 64.5

51&3

61.6

444.9

64.3 GG.7

2,9 248.5 2,2 1 1-954 1.9 1 1721

&8.9 90.9

4.7

80 85,7

E4FE, = PleRf,ENTAGE 0F EFFICIENCY (B]

C1R IJ7TA4JDSYM&O'L DESIGNATION

CI [ART

U

SI ..

AP LE

024E3J310

L

20 30 RL iOI1P1 )

40

50

(C)

GRJtPH

Figure 6.35 - Effect of source resistance on power output. If the load resistance (RL) were increased (the internal resistance remaining the same), the current drawn from the source would decrease. Consequently, the voltage drop across the internal resistance would decrease. At the same time, the terminal voltage applied across the

load would increase and approach a maximum as the current approaches zero amps.

Module 3.6 DC Circuits

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Parallel DC Circuits The discussion of electrical circuits presented up to this point has been concerned with series circuits in which there is only one path for current. There is another basic type of circuit known as the parallel circuit with which you must become familiar. Where the series circuit has only one path for current, the parallel circuit has more than one path for current. Ohm's law and Kirchhoff's law apply to all electrical circuits, but the characteristics of a parallel DC circuit are different than those of a series DC circuit. Parallel Circuit Characteristics A parallel circuit is defined as one having more than one current path connected to a common voltage source. Parallel circuits, therefore, must contain two or more resistances which are not connected in series. An example of a basic parallel circuit is shown in Figure 6.36.

L. J

L.

7 --

S

7

S

r

PATH 2 Figure 6.36 - Example of a basic parallel circuit. Start at the voltage source (Es) and trace anticlockwise around the circuit. Two complete and separate paths can be identified in which current can flow. One path is traced from the source, through resistance R1, and back to the source. The other path is from the source, through resistance R2, and back to the source.

t. j

7

n

Use and/or disclosure is governed by the statement

TTS Integrated Training System 0 Rnrnrrinht 911111

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d by t governed by I he statement on page 2 of this chapter.

II

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Voltage in a Parallel Circuit You have seen that the source voltage in a series circuit divides proportionately across each resistor in the circuit. In a parallel circuit, the same voltage is present in each branch. (A branch is a section of a circuit that has a complete path for current.) In Figure 6.36 this voltage is equal to the applied voltage (ES). This can be expressed in equation form as: Es = ER1 = ER2 Voltage measurements taken across the resistors of a parallel circuit, as illustrated by Figure 6.37 verify this equation. Each meter indicates the same amount of voltage. Notice that the voltage across each resistor is the same as the applied voltage.

Figure 6.37 - Voltage comparison in a parallel circuit. Example: Assume that the current through a resistor of a parallel circuit is known to be 4.5 milliamperes (4.5 mA) and the value of the resistor is 30,000 ohms (30 kU). Determine the source voltage. The circuit is shown in Figure 6.38. Given:

R2 = 30, 000 ohms (3OkQ) IR2 = 4.5 milliamps (4.5mA or.0045 amps) Solution: E = IR ER2 =.0045 amp x 30,000 ohms ER2 = 135 volts

Module 3.6 DC Circuits

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

R2 30k fl

L

R3

P

Figure 6.38 - Example problem parallel circuit. Since the source voltage is equal to the voltage of a branch: ES = ER2

ES = 135 volts To simplify the math operation, the values can be expressed in powers of ten as follows: 30,000 ohms = 30 x 103 ohms 4,SmA=4.5x10-3 amps ER2 =(4.5x10-3) amps x(30x103) ohms ER2 = (4.5 x 30 x 10-3 X103 ) volts

(10-3 x103 =10-3+3 =101 =1) ER2 =(4.5x30x1) volts ER2 =135 volts ES =ER2

E S =135 volts

El

n 6-50 TTS Integrated Training System Use andlor disclosure is governed by the statement

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Use and/or disclosure is governed by the statement

TTS Integrated Training System [c)

cnnvrinht 9010

© Copyright 2010

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{

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Resistance in a Parallel Circuit

In the example diagram, Figure 6.39, there are two resistors connected in parallel across a 5-

1

volt battery. Each has a resistance value of 10 ohms. A complete circuit consisting of two parallel paths is formed and current flows as shown. 9A

r

L 0_5A

100

&V

105

O.5A

r TI L

Figure 6.39 - Two equal resistors connected in parallel.

L

Computing the individual currents shows that there is one-half of an ampere of current through

each resistance. The total current flowing from the battery to the junction of the resistors, and returning from the resistors to the battery, is equal to 1 ampere.

The total resistance of the circuit can be calculated by using the values of total voltage (ET) and total current (IT). NOTE: From this point on the abbreviations and symbology for electrical quantities will be used in example problems.

LI

Given: ET=5V

IL

L I ri

L

LI

IT= 1A Solutio n:

ET RT

IT

RT

5V 1A

RT =5 0

R= E

I

This computation shows the total resistance to be 5 ohms; one-half the value of either of the two resistors. Module 3.6 DC Circuits

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Since the total resistance of a parallel circuit is smaller than any of the individual resistors, total resistance of a parallel circuit is not the sum of the individual resistor values as was the case in a series circuit. The total resistance of resistors in parallel is also referred to as equivalent resistance (Req). The terms total resistance and equivalent resistance are used interchangeably. There are several methods used to determine the equivalent resistance of parallel circuits. The best method for a given circuit depends on the number and value of the resistors. For the circuit described above, where all resistors have the same value, the following simple equation is used: Reg

n

R N

i_

Reg = equivalent parallel resistance R = ohmic valu a of one resistor N = number of resistors This equation is valid for any number of parallel resistors of equal value. Example. Four 40-ohm resistors are connected in parallel. What is their equivalent resistance?

Given:

7

R1+ R2 + R3+ R4 Rt=40 3 Solution:

R eq

L_

R N 405

Reg

4

Reg =10 Q Figure 6.40 shows two resistors of unequal value in parallel. Since the total current the equivalent resistance can be calculated.

6-52 Use and/or disclosure is governed by the statement

Module 3.6 DC Circuits

is shown, H

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P- ";M'+ on TTS Integrated 'Training System nIn

TTS Integrated Training System © Copyright 2010

governed by the statement on page 2 of this Chapter.

11 1

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ES

R1 3S?

ISA Figure 6.40 - Example circuit with unequal parallel resistors.

Given: E$=30V IT= 15A Solution: Reg = E$ IT

Reg =

30 V 15A

Reg =252 v

The equivalent resistance of the circuit shown in Figure 6.40 is smaller than either of the two resistors (R 1, R2). An important point to remember is that the equivalent resistance of a parallel circuit is always less than the resistance of any branch. Equivalent resistance can be found if you know the individual resistance values and the source

voltage. By calculating each branch current, adding the branch currents to calculate total current, and dividing the source voltage by the total current, the total can be found. This method, while effective, is somewhat lengthy. A quicker method of finding equivalent resistance

is to use the general formula for resistors in parallel:

l

1 = 1+ 1+ 1 +... 1 Rn Reg R1 R2 R3

L If you apply the general formula to the circuit shown in Figure 6.40 you will get the same value

for equivalent resistance (2Q) as was obtained in the previous calculation that used source

L-j

6-53

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voltage and total current. Given: R1

3S

l

R2=60 Solution:

7 i~

Convert the fractions to a common denominator.

n

II

n

Since both sides are reciprocals (divided into one), disregard the reciprocal function.

n J

Reg =2Q The formula you were given for equal resistors in parallel Reg =R ) is a simplification of the general formula for resistors in parallel 1_ 1+1 Reg

R1

R2

+ 1 +.., I R3 Rn

L.

There are other simplifications of the general formula for resistors in parallel which can be used to calculate the total or equivalent resistance in a parallel circuit. Reciprocal Method - This method is based upon taking the reciprocal of each side of the equation. This presents the general formula for resistors in parallel as:

f

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fII L

ri

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L

R eq =

1 1

1

1

R1

R2

R„

This formula is used to solve for the equivalent resistance of a number of unequal parallel resistors. You must find the lowest common denominator in solving these problems. Example: Three resistors are connected in parallel as shown in Figure 6.41. The resistor values are: R1 = 20 ohms, R2 = 30 ohms, R3 = 40 ohms. What is the equivalent resistance? (Use the reciprocal method.)

E

R3 40 52

Figure 6.41 - Example parallel circuit with unequal branch resistors. Given: f

R 1 =20Q R2 = 305 R3 = 40Q

L

H 6-55

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Solution: Req

R eg =

1

1

1

R1

R2

R3

1

+

200 Reg =

6

1 1

30Q

+

1200 R

+

1 4 120Q

1

4052

+

3

120

=

1 eq

13

120

Product Over the Sum Method - A convenient method for finding the equivalent, or total, resistance of two parallel resistors is by using the following formula. R eq =

R1X R2 Rt + R2

This equation, called the product over the sum formula, is used so frequently it should be committed to memory. Example. What is the equivalent resistance of a 20-ohm and a 30-ohm resistor connected in parallel, as in Figure 6.42?

Figure 6.42 - Parallel circuit with two unequal resistors.

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Given: �

R 1 =205 R2 =305 Solution: Reg

R1

+ R2

20Q x 30Q

Reg

20 + 305 _ 600 Reg 50 Reg =120

L

t L

Module 3.6 DC Circuits

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Equivalent Parallel Circuits In the study of electricity, it is often necessary to reduce a complex circuit into a simpler form.

Any complex circuit consisting of resistances can be redrawn (reduced) to a basic equivalent circuit containing the voltage source and a single resistor representing total resistance. This process is called reduction to an equivalent circuit. Figure 6.43 shows a parallel circuit with three resistors of equal value and the redrawn equivalent circuit. The parallel circuit shown in part A shows the original circuit. To create the equivalent circuit, you must first calculate the equivalent resistance.

R2 45D

R3 450

(A)

E

b ov

(B) Figure 6.43 - Parallel circuit with equivalent circuit.

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Solution:

7

U

L

Once the equivalent resistance is known, a new circuit is drawn consisting of a single resistor (to represent the equivalent resistance) and the voltage source, as shown in part B. Rules for Parallel DC Circuits • The same voltage exists across each branch of a parallel circuit and is equal to the source voltage. • The current through a branch of a parallel network is inversely proportional to the amount of resistance of the branch. • The total current of a parallel circuit is equal to the sum of the individual branch currents of the circuit. • The total resistance of a parallel circuit is found by the general formula:

1

.'

1

1

1

ReqR1

R2

Rn

or one of the formulas derived from this general formula. Solving Parallel Circuit Problems Problems involving the determination of resistance, voltage, current, and power in a parallel circuit are solved as simply as in a series circuit. The procedure is the same - (1) draw the circuit diagram, (2) state the values given and the values to be found, (3) select the equations to be used in solving for the unknown quantities based upon the known quantities, and (4) substitute the known values in the equation you have selected and solve for the unknown value.

L1

Example: A parallel circuit consists of five resistors. The value of each resistor is known and the current through R1 is known. You are asked to calculate the value for total resistance, total power, total current, source voltage, the power used by each resistor, and the current through resistors R2, R3,R4, and R5.

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Given: R1 = 200

R2=300 R3 = 180

Rq = 180 R5 = 180 IR>. = 9A Find: RT, ES , IT, PT, IR2, IR3, IR4, IR5.

This may appear to be a large amount of mathematical manipulation. However, if you use the step-by-step approach, the circuit will fall apart quite easily. The first step in solving this problem is for you to draw the circuit and indicate the known values as shown in Figure 6.44.

a

Es

C6 L_S

Figure 6.44 - Parallel circuit problem. There are several ways to approach this problem. With the values you have been given, you could first solve for RT, the power used by R1, or the voltage across R1, which you know is equal to the source voltage and the voltage across each of the other resistors. Solving for RT or the power used by R1 will not help in solving for the other unknown values. Once the voltage across RI is known, this value will help you calculate other unknowns. Therefore the logical unknown to solve for is the source voltage (the voltage across R1).

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Given: R1 = 200 IR1=9A

L

ER1= ES

Solution: ES=R1 xIR1

ES=9A x200 ES = 180V Now that source voltage is known, you can solve for current in each

branch.

Given: ES = 180V R2 = 300 R3 = 18n Rq = 18

R5=180 Solution: ES El

IR2

= R2

IRZ = IR2

I R3

IR3

180V 305

=6A ES R

3

180 V 18 0

IR3 =10 A Since R3 = R4 = Rs and the voltage across each branch is the same: IR4 =10A

L

IR5 =10A

i

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Solving for total resistance. Given: R1 = 20Q R2 = 30Q

R3 = 18Q R4= 18Q

R5 = 18Q Solution: RT =Reg 1

1

Reg

R1

1

1

RT

205

+1

+ 1+ 1+ 1

R3 R4 R 2 1 + 1+ +

30Q

185

R5

1+ 1 18Q

1.B:Q

1 _ 9+6+10+10+105 RT 180 (LCD) RT =455

180 180 RTT45Q RT=4S

n

An alternate method for solving for RT can be used. By observation, you can see that R3, R 4, and R5 are of equal ohmic value. Therefore an equivalent resistor can be substituted for these three resistors in solving for total resistance. Given: R3 =R4= R5 = 185

Solution: R

Regl = R

N 185 Regl = 3 Reg, = 6S

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The circuit can now be redrawn using a resistor labelled Regi in place of R3, R4, and R5 as shown in Figure 6.45.

R1 200 190V

{_i

Figure 6.45 - First equivalent parallel circuit. An equivalent resistor can be calculated and substituted for R I and R2 by use of the product over the sum formula.

t L Given: R1 = 205 R2 = 305

Solution: Reg R1 +R2 205 x 30Q 2052 +3052 600 R 2 50 -9 Re92 =125 Re92

The circuit is now redrawn again using a resistor labelled R5g2 in place of Ri and R2 as shown in Figure 6.46.

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1

Es

Req 2

Req 1

12Q

6c

18tiV

Figure 6.46 - Second equivalent parallel circuit. You are now left with two resistors in parallel. The product over the sum method can now be used to solve for total resistance. Given:

a

Regl = 60

Reg, =125 RT = Reg Solution:

n

_ R1 x R 2

Reg R1

+ R2

RT =Regl X Reg2 Reg? + Reg2

RT

6cx12Q 60+129

RT = 72 1 RT =4Q This agrees with the solution found by using the general formula for solving for resistors in parallel. The circuit can now be redrawn as shown in Figure 6.47 and total current can be calculated.

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;

Li

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Req

Es

4c

L L

Figure 6.47 - Parallel circuit redrawn to final equivalent circuit.

f,

Given: ES =160V PIT = 4Q

Solution:

This solution can be checked by using the values already calculated for the branch currents. T� L Given: {..�

I R1 =9A I R2 =6A

I R3 =10A L IR4 =10A I R5 =10A Solution: LIT =IR1 +IR2+,..IRn IT =9A+6A+10A+10A+10A LIT =45A

II

L

Module 3.6 DC Circuits

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Series-Parallel DC Circuits In the preceding discussions, series and parallel DC circuits have been considered separately. The technician will encounter circuits consisting of both series and parallel elements. A circuit of this type is referred to as a combination circuit. Solving for the quantities and elements in a combination circuit is simply a matter of applying the laws and rules discussed up to this point.

Solving Combination-Circuit Problems The basic technique used for solving DC combination-circuit problems is the use of equivalent circuits. To simplify a complex circuit to a simple circuit containing only one load, equivalent circuits are substituted (on paper) for the complex circuit they represent. To demonstrate the method used to solve combination circuit problems, the network shown in Figure 6.48 (A) will be

lf

j

used to calculate various circuit quantities, such as resistance, current and voltage.

60V

Es

T 60V

R2 200 Es

60V (A)

I

Figure 6.48 - Example combination circuit. Examination of the circuit shows that the only quantity that can be computed with the given information is the equivalent resistance of R2 and R3. Given:

R2=20Q R3 = 300

t i.

1..

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Solution: R2 x R3

Regl=R2+ R3 Reg,

(Product over thesum)

_ _ 205 x 305 205 +305

Reg 1 = 600 Q0 Regi=122 Now that the equivalent resistance for R2 and R3 has been calculated, the circuit can be redrawn as a series circuit as shown in Figure 6.48 (B). The equivalent resistance of this circuit (total resistance) can now be calculated. r

-,

L i

L

Given: RI=80 Reg,=120

(Resistors in series)

Solution: Reg =R1+Regl

Reg = 80 +125 Reg = 205 r,

Li

0r

RT =205

The original circuit can be redrawn with a single resistor that represents the equivalent resistance of the entire circuit as shown in Figure 6.48 (C).

U

To find total current in the circuit:

Given: ii

ES =60V

RT = 205 r7

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Solution: IT = E S

RT

IT

265

(Ohm's Law)

IT =3A

To find the voltage dropped across R1, R2, and R3, refer to Figure 6.48 (B). Reg1 represents the parallel network of R2 and R3. Since the voltage across each branch of a parallel circuit is equal, the voltage across Real (Eeg1) will be equal to the voltage across R2 (ER2 ) and also equal to the voltage across R3 (ER3). Given: IT =3A R1 =BQ Regi=12S

(Current through each p art of a series circuit is equal to total current)

Solution: ER1=I1xR1 ER1=3Ax8Q ER1 = 24V ER2 =

eg

ER3 = E l

Eeg1= I T x Regl

Eegl = 3A x 125 Eegl = 36V

ER2 =36V ER3 =36V

To find power used by RI: Given: ER1= 24V IT = 3A

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Solution:

ci r-t

PR1=ER1XIT

PR1 = 24V x 3A PR1=72W

UI

To find the current through R2 and R3, refer to the original circuit, Figure 6.48 (A). You know ER2 and ER3 from previous calculation. Given: ER2 = 36V

ER3 = 36V R2 = 20C) R2=30 C) Solution: I R2 =

2 R2

(Ohm's Law)

36V U

I R2 =

U

IR2

IR3 r, U

U

U`

20Q

=1.8A _ _ ER 3 R3

36V

I R3 I R3

J 305 =1.2A

Now that you have solved for the unknown quantities in this circuit, you can apply what you have learned to any series, parallel, or combination circuit. It is important to remember to first look at the circuit and from observation make your determination of the type of circuit, what is known, and what you are looking for. A minute spent in this manner may save you many

unnecessary calculations.

U

Having computed all the currents and voltages of Figure 6.48 a complete description of the operation of the circuit can be made. The total current of 3 amps leaves the negative terminal of the battery and flows through the 8-ohm resistor (R1). In so doing, a voltage drop of 24 volts occurs across resistor Ri. At point A, this 3-ampere current divides into two currents. Of the total current, 1.8 amps flows through the 20-ohm resistor. The remaining current of 1.2 amps flows

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from point A, down through the 30-ohm resistor to point B. This current produces a voltage drop of 36 volts across the 30-ohm resistor. (Notice that the voltage drops across the 20- and 30ohm resistors are the same.) The two branch currents of 1.8 and 1.2 amps combine at junction B and the total current of 3 amps flows back to the source. The action of the circuit has been completely described with the exception of power consumed, which could be described using the values previously computed. It should be pointed out that the combination circuit is not difficult to solve. The key to its solution lies in knowing the order in which the steps of the solution must be accomplished.

p

p

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Practice Circuit Problem Figure 6.49 is a typical combination circuit. To make sure you understand the techniques of solving for the unknown quantities, solve for ERt. R.

R2

300 lal

900Q

R C-1 1

AN 4000

RS U

ES

RS R

400 0

--- 300V

11

9kC2

•--

�

4000

Eg

d

4

300V

(B) Re4 2 U

E

r

- 300V

200!2

R 11 1 kf24

_ ES

T

300V

-4

Figure 6.49 - Combination practice circuit. 1.J

It is not necessary to solve for all the values in the circuit to compute the voltage drop across resistor R1 (E R1). First look at the circuit and determine that the values given do not provide enough information to solve for ER1 directly. If the current through R1 (IR1) is known, then ER1 can be computed by applying the formula: ER1= R1 X IRi The following steps will be used to solve the problem.

I `.

The total resistance (RT) is calculated by the use of equivalent resistance. Given: R1 = 3000 R2 _= 1000 Solution:

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Reg1 = R1 + R2

Reg1 = 3000 + 1000 Reg1 = 4000 Redraw the circuit as shown in Figure 6.49 (B). Solution: R

eq2

R eq2 =

R

(Equal resistors

N 4000

in parallel)

2

Reg2 = 2000 Solution:

Re92 = 2000 Redraw the circuit as shown in Figure 6.49 (C). Given: Reg2 = 2000

R4= 1k0 Solution: Req = Re02 + R4 ReQ = 2000 +1kO

Reg = 1, 2k0 The total current (IT) is now computed. Given: ES = 300V Req = LAC)

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Solution: IT = $eq

_ 300V I T ^ 1.2k52 I T = 25OmA Solve for the voltage dropped across Reg2. This represents the voltage dropped across the network R1, R2, and R3 in the original circuit. Given:

R,g = 200Q IT = 250mA

Li

Solution: EReo2 = Reg2 X IT

EReg2 = 2000 x 250mA

EReg2 = 50V Solve for the current through Reg1. (Reg1 represents the network R1 and R2 in the original circuit.) Since the voltage across each branch of a parallel circuit is equal to the voltage across the equivalent resistor representing the circuit: Given: EReg2 = EReg1 EReg1 = 50V

ReQ1 = 4000

Solution: IRe g 1 =

IR eg 1 =

ERegi Reg1

50V 4000

I Reg 1=125mA Solve for the voltage dropped across R1 (the quantity you were asked to find). Since Reg1 represents the series network of R1 and R2 and total current flows through each resistor in a series circuit, IR1 must equal IReg1.

Given:

i U

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IR1 = 125mA R1 = 3000

Solution: ER1= IR1 X R1 ER1= 125mA x 3000 ER1 = 37.SV

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Redrawing Circuits for Clarity You will notice that the schematic diagrams you have been working with have shown parallel circuits drawn as neat square figures, with each branch easily identified. In actual practice the wired circuits and more complex schematics are rarely laid out in this simple form. For this reason, it is important for you to recognize that circuits can be drawn in a

variety of ways, and to learn some of the techniques for redrawing them into their simplified

form. When a circuit is redrawn for clarity or to its simplest form, the following steps are used. • Trace the current paths in the circuit. • Label the junctions in the circuit. • Recognize points which are at the same potential. • Visualize a rearrangement, "stretching" or "shrinking," of connecting wires. • Redraw the circuit into simpler form (through stages if necessary). To redraw any circuit, start at the source, and trace the path of current flow through the circuit.

At points where the current divides, called junctions, parallel branches begin. These junctions

are key points of reference in any circuit and should be labelled as you find them. The wires in circuit schematics are assumed to have no resistance and there is no voltage drop along any wire. This means that any unbroken wire is at the same voltage all along its length, until it is interrupted by a resistor, battery, or some other circuit component. In redrawing a circuit, a wire can be "stretched" or "shrunk" as much as you like without changing any electrical characteristic of the circuit. Figure 6.50 (A) is a schematic of a circuit that is not drawn in the box-like fashion used in previous illustrations. To redraw this circuit, start at the voltage source and trace the path for current to the junction marked (a). At this junction the current divides into three paths. If you were to stretch the wire to show the three current paths, the circuit would appear as shown in Figure 6.50 (B).

L1

II

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(a)

A (a)

(a)

(a)

(B) Figure 6.50 - Redrawing a simple parallel circuit. While these circuits may appear to be different, the two drawings actually represent the same circuit. The drawing in Figure 6.50 (B) is the familiar box-like structure and may be easier to work with. Figure 6.51(A) is a schematic of a circuit shown in a box-like structure, but may be misleading. This circuit in reality is a series-parallel circuit that may be redrawn as shown in Figure 6.51 (B). The drawing in part (B) of the figure is a simpler representation of the original circuit and could be reduced to just two resistors in parallel.

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(a)

(A) (a)

ES

R3

(B) Figure 6.51 - Redrawing a simple series-parallel circuit.

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Redrawing a Complex Circuit

Figure 6.52 (A) shows a complex circuit that may be redrawn for clarification in the following

steps.

(A)

(B)

(C) Figure 6.52 - Redrawing a complex circuit. NOTE: As you redraw the circuit, draw it in simple box-like form. Each time you reach a junction, a new branch is created by stretching or shrinking the wires. Start at the positive terminal of the voltage source. Current flows through R1 to a junction and divides into three paths; label this junction (a). Follow one of the paths of current through R2 and R3 to a junction where the current divides into two more paths. This junction is labelled (b). The current through one branch of this junction goes through R5 and back to the source. (The most direct path.) Now that you have completed a path for current to the source, return to the last junction, (b). Follow current through the other branch from this junction. Current flows from junction (b) through R4 to the source. All the paths from junction (b) have been traced. Only one path from junction (a) has been completed. You must now return to junction (a) to complete the other two paths. From junction (a) the current flows through R7 back to the source. (There are no additional branches on this path.) Return to junction (a) to trace the third path from this junction.

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Current flows through R 6 and R8 and comes to a junction. Label this junction (c). From junction (c)

one path for current is through R 9 to the source. The other path for current from junction (c)

is through R10 to the source. All the junctions in this circuit have now been labelled. The circuit and the junction can be redrawn as shown in Figure 6.52 (C). It is much easier to recognize the series and parallel paths in the redrawn circuit. T ``-'

What is the total resistance of the circuit shown in Figure 6.53? (Hint: Redraw the circuit to simplify and then use equivalent resistances to compute for RT.) R2 192

R3 1OQ

Es

R4 30

Figure 6.53 - Simplification circuit problem. What is the total resistance of the circuit shown in Figure 6.54?

I

j

�

I

I

I

I

Es I

I L'

I

I

-

+-50 V

1

R-)

6o5-

R2

40C

r-1

U F1

Figure 6.54 - Source resistance in a parallel circuit. What effect does the internal resistance have on the rest of the circuit shown in Figure 6.54?

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Effects of Open and Short Circuits Earlier in this chapter the terms open and short circuits were discussed. The following discussion deals with the effects on a circuit when an open or a short occurs. The major difference between an open in a parallel circuit and an open in a series circuit is that in the parallel circuit the open would not necessarily disable the circuit. If the open condition occurs in a series portion of the circuit, there will be no current because there is no complete path for current flow. If, on the other hand, the open occurs in a parallel path, some current will still flow in the circuit. The parallel branch where the open occurs will be effectively disabled, total resistance of the circuit will increase, and total current will decrease. To clarify these points, Figure 6.55 illustrates a series parallel circuit. First the effect of an open

_J

in the series portion of this circuit will be examined. Figure 6.55 (A) shows the normal circuit,

RT = 40 ohms and IT = 3 amps. In Figure 6.55 (B) an open is shown in the series portion of the circuit, there is no complete path for current and the resistance of the circuit is considered to be infinite.

--

R1

\,-

20 C

_L ES

Rz

T

100 a

129V

(B)

(A) _A/v\R1 20)

--

tj

12 UV

(C) Figure 6.55 - Series-parallel circuit with opens.

In Figure 6.55 (C) an open is shown in the parallel branch of R3. There is no path for current

through R3. In the circuit, current flows through R 1 and R2 only. Since there is only one path for current flow, R1 and R2 are effectively in series.

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Under these conditions RT = 1200 and IT = 1 amp. As you can see, when an open occurs in a

parallel branch, total circuit resistance increases and total circuit current decreases.

A short circuit in a parallel network has an effect similar to a short in a series circuit. In general, the short will cause an increase in current and the possibility of component damage regardless of the type of circuit involved. To illustrate this point, Figure 6.56 shows a series-parallel network in which shorts are developed. In Figure 6.56 (A) the normal circuit is shown. RT = 40 ohms and IT=3amps.

(A)

I

r1 Wv

20 3

ES 120V

R2

100 in ell

00

(C) Figure 6.56 - Series-parallel circuit with shorts. In Figure 6.56 (B), R1 has shorted. R1 now has zero ohms of resistance. The total of the resistance of the circuit is now equal to the resistance of the parallel network of R2 and R3, or 20 ohms. Circuit current has increased to 6 amps. All of this current goes through the parallel network (R2, R3) and this increase in current would most likely damage the components. In Figure 6.56 (C), R3 has shorted. With R3 shorted there is a short circuit in parallel with R2. The short circuit routes the current around R2, effectively removing R2 from the circuit. Total circuit resistance is now equal to the resistance of R1, or 20 ohms. r1

As you know, R2 and R3 form a parallel network. Resistance of the network can be calculated as follows:

Module 3.6 DC Circuits

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Given:

r)

R2 = 1000 R3=0Q Solution:

R2 XR3 Reg =

R eq = R eg

R2 + R3 1000 X 00

10052 + OS

= 00

`_

j

The total circuit current with R3 shorted is 6 amps. All of this current flows through R1 and would most likely damage R1. Notice that even though only one portion of the parallel network was shorted, the entire paralleled network was disabled. Opens and shorts alike, if occurring in a circuit, result in an overall change in the equivalent resistance. This can cause undesirable effects in other parts of the circuit due to the corresponding change in the total current flow. A short usually causes components to fail in a circuit which is not properly fused or otherwise protected. The failure may take the form of a burned-out resistor, damaged source, or a fire in the circuit components and wiring.

i

Fuses and other circuit protection devices are installed in equipment circuits to prevent damage caused by increases in current. These circuit protection devices are designed to open if current increases to a predetermined value. Circuit protection devices are connected in series with the circuit or portion of the circuit that the device is protecting. When the circuit protection device opens, current flow ceases in the circuit.

n

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Voltage Dividers Most electrical and electronics equipment use voltages of various levels throughout their circuitry. One circuit may require a 90-volt supply, another a 150-volt supply, and still another a 180-volt supply. These voltage requirements could be supplied by three individual power sources. This L

method is expensive and requires a considerable amount of room. The most common method of supplying these voltages is to use a single voltage source and a voltage divider. Before

voltage dividers are explained, a review of what was discussed earlier concerning voltage references may be of help. As you know, some circuits are designed to supply both positive and negative voltages. Perhaps now you wonder if a negative voltage has any less potential than a positive voltage. The answer is that 100 volts is 100 volts. Whether it is negative or positive does not affect the feeling you get when you are shocked.

U

Voltage polarities are considered as being positive or negative in respect to a reference point, usually ground. Figure 6.57 will help to illustrate this point.

U

L L ES

Es

100V

100V

r

L

(A)

(B) Figure 6.57 - Voltage polarities.

L

Figure 6.57 (A) shows a series circuit with a voltage source of 100 volts and four 50-ohm

t L

resistor R1. The current in this circuit determined by Ohm's law is 0.5 amp. Each resistor develops (drops) 25 volts. The five tap-off points indicated in the schematic are points at which

resistors connected in series. The ground, or reference point, is connected to one end of

Module 3.6 DC Circuits

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001f)

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the voltage can be measured. As indicated on the schematic, the voltage measured at each of the points from point E to point A starts at 25 volts and becomes more negative in 25 volt steps to a value of positive zero volts. In Figure 6.57 (B), the ground, or reference point has been moved to point B. The current in the circuit is still 0.5 amp and each resistor still develops 25 volts. The total voltage developed in the circuit remains at 100 volts, but because the reference point has been changed, the voltage at point A is negative 25 volts. Point E, which was at positive 100 volts in Figure 6.57 (A), now has a voltage of positive 75 volts. As you can see the voltage at any point in the circuit is dependent on three factors; the current through the resistor, the ohmic value of the resistor, and the reference point in the circuit. A typical voltage divider consists of two or more resistors connected in series across a source voltage (Es). The source voltage must be as high or higher than any voltage developed by the voltage divider. As the source voltage is dropped in successive steps through the series resistors, any desired portion of the source voltage may be "tapped off" to supply individual voltage requirements. The values of the series resistors used in the voltage divider are determined by the voltage and current requirements of the loads. Figure 6.58 is used to illustrate the development of a simple voltage divider. The requirement for this voltage divider is to provide a voltage of 25 volts and a current of 910 milliamps to the load from a source voltage of 100 volts. Figure 6.58 (A) provides a circuit in which 25 volts is available at point B. If the load was connected between point B and ground, you might think that the load would be supplied with 25 volts. This is not true since the load connected between point B and ground forms a parallel network of the load and resistor R1. (Remember that the value of resistance of a parallel network is always less than the value of the smallest resistor in the network.)

n L. �

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f this Chapter.

1-?

11

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-0

G + 100V

R U

750

Es

1Q01f

0B

+25V

0

AOV

R1 25D

(A)

+ 100V R 75 0.

_ Es 100V

--`Q + 26V LOAD +25V

910 ma OV

(B) Figure 6.58 - Simple voltage divider.

r,

Since the resistance of the network would now be less than 25 ohms, the voltage at point B would be less than 25 volts. This would not satisfy the requirement of the load. To determine the size of resistor used in the voltage divider, a rule-of-thumb is used. The

current in the divider resistor should equal approximately 10 percent of the load current. This current, which does not flow through any of the load devices, is called bleeder current. Given this information, the voltage divider can be designed using the following steps. Determine the load requirement and the available voltage source. ES = 100V Eload = 25V Ilaad = 91OmA

U r -, U

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Select bleeder current by applying the 10% rule-of-thumb. IR1= 10% X'load

IR1=.1x910mA IR1= 91mA Calculate bleeder resistance. ERI I R1

R1

R1=

25V 91mA

R1= 274.730 The value of R1 may be rounded off to 275 ohms: R1 = 2750

Calculate the total current (load plus bleeder). IT = Iload+ IR1

IT= 910mA + 91mA

IT= 1A (rounded off) Calculate the resistance of the other divider resistor(s).

ER2 = Es - ER1 ER2 =100V-25V ER2 =75V R2 =

ER2

IT 75V 1A

R2 =7552

The voltage divider circuit can now be drawn as shown in

Figure 6.58 (B). i

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Multiple-Load Voltage Dividers A multiple-load voltage divider is shown in Figure 6.59. An important point that was not emphasized before is that when using the 10% rule-of-thumb to calculate the bleeder current,

you must take 10% of the total load current.

LJ rl

+175V 3OmA LOAD 3

n Es 285V

O

+150V 1 OmA LOAD 2

-'I

f Li

1 +9OV 10mA LOAD 1

Figure 6.59 - Multiple-load voltage divider. Given the information shown in Figure 6.59, you can calculate the values for the resistors needed in the voltage-divider circuits. The same steps will be followed as in the previous voltage divider problem.

i

U

L Module 3.6 DC Circuits

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Given: Load 1: E = 90V I = 10mA

..a

Load 2:E = 150V

1

I = 10mA

Load 3: E = 175V I = 3OmA E$ = 285V

,

J

The bleeder current should be 10% of the total load current. Solution: IR1 =

n

10% x I (load total)

IR1 = 10% x (10mA + 10mA + 3OmA)

IRl = SmA

J

Since the voltage across R1 (ER1) is equal to the voltage requirement for load 1, Ohm's law can be used to calculate the value for R1. Solution: R= ER11

IR1

90V Rj=SmA R1 `

18kc

The current through R2 (IR2) is equal to the current through Rtplus the current through load 1. Solution:

in

1

n

IR2 = IRl + Iloadl

IR2 = SmA + 10mA Ift2 = 15mA The voltage across R2 (ER2) is equal to the difference between the voltage requirements of load 1 and load 2.

7, Use and/or disclosure is governed by the statement

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Eload2 - Eloadl ER2 =150V - 90V ER2 = 60V ER2 =

L r L

L

Ohm's law can now be used to solve for the value of R2.

Solution: R2 = ER2 IR2

R

2

60V 1SmA

R2 = 4kQ L

The current through R3 (IR3) is equal to the current through

R2 plus the current through load 2.

IR3 = IR2 + I1oad2

IR3 = 15mA + 10mA IR3 = 25mA The voltage across R3 (ER3) equals the difference between and load 2.

the voltage requirement of load 3

ER3 = Eload3 - Eload2

ER3 = 175V - 150V

ER3= 25V Li Ohm's law can now be used to solve for the value of R 3. Solution: R3 = R3

ER3 IR3

25V 2SmA

R3 =1kQ The current through R4 (lR4) is equal to the current through 1R4 is equal to total circuit current (1 T).

R3 plus the current through load 3.

IR4 = IR3 + Iload3

IR4 = 25mA + 3OmA

L

IR4=S5mA

r-�

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The voltage across R4 requirement of load 3.

(ER4)

equals the difference between the source voltage and the voltage

ER4 = ES - Eload3

ER4 = 285V - 175V ER4= 110V Ohm's law can now be used to solve for the value of R 4. Solution: R4 = ER4 1R4

110V 55mA R 4 = 2kQ R4

With the calculations just explained, the values of the resistors used in the voltage divider are as follows: Rl = 18kQ RZ = 4k0 R3 = 1kc R4 = 2k0

1

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Voltage Divider with Positive and Negative Voltage Requirements

In many cases the load for a voltage divider requires both positive and negative voltages. r;

L

Positive and negative voltages can be supplied from a single source voltage by connecting the ground (reference point) between two of the divider resistors. The exact point in the circuit at which the reference point is placed depends upon the voltages required by the loads. For example, a voltage divider can be designed to provide the voltage and current to three loads from a given source voltage. Given:

u

Load 1: E _ -25V I =3OOmA Load 2; E _ +SOV

I = SOmA Load 3: E _ +250V I = 1OOmA ES= 310V

Li

The circuit is drawn as shown in Figure 6.60. Notice the placement of the ground reference point. The values for resistors R1, R3, and R4 are computed exactly as was done in the last example. IR1 is the bleeder current and can be calculated as follows:

L rl,

L

L

Module 3.6 DC Circuits

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+0 A

R 4

R

n

0 E

+250V 900mA LOAD 3

1

310V

+60V

0

5OmA LOAD 2 A

4 -25V

-0

n

300mA LOAD I i

Figure 6.60 - Voltage divider providing both positive and negative voltages. Calculate the value of R1. Solution: n 1

R 1= ER1 IRI _ 25V

1 45mA RI = 5565 Calculate the current through R2 using Kirchhoff's current law. At point A:

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IR1 + I1oad1 + IR2 + Iload2 + Iload3 - 0

L' r-

45mA + 300mA + IR2 - SOMA - 1OOmA = 0 345mA + IR2 - 150mA = 0 195mA + IR2 = 0 IR2= -195mA (or 195mA leaving point A) Since ER2 = E load 2, you can calculate the value of R2. Solution: R2 = ER2 IR2

50V 195mA R 2 = 2565 R2

4I

Calculate the current through R3. IR3 = IR2 + Iload2

u

IR3 = 195mA + 50mA IR3 = 245mA

The voltage across R3 (ER3) equals the difference between the voltage requirements of loads 3 and 2. Solution: L

F-1,

ER3 = Eload3 - Eload2

ER3 = 250V - 50V ER3 = 200V

Li

Li

Calculate the value of R3. Solution:

R3=

ER3

IR3

_ 200V R3 245mA R3 = 8165 U

Calculate the current through R4.

ri

U

Module 3.6 DC Circuits

6-93

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IRA = IR3 + Iload3 IR4

7

= 245mA + 100mA

IN = 345mA The voltage across ER4 equals the source voltage (Es) minus the voltage requirement of load 3 and the voltage requirement of load 1. Remember Kirchhoff's voltage law which states that the sum of the voltage drops and EMFs around any closed loop is equal to zero. Solution: ER4 = Eg - Eload3 - Eloadl ER4 = 310V - 250V - 25v

ER4 = 35V

Calculate the value of R4.

n

Solution: n

R4 = ER4 IR4

r-. J

R4 =

R4

35V 345mA L. J

=101.40

With the calculations just explained, the values of the resistors used in the voltage divider are as follows: �

R1 =55160 R2 =2560 R3 =8160

R4 =1010 n

From the information just calculated, any other circuit quantity, such as power, total current, or resistance of the load, could be calculated.

U, J

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Li

Practical Application of Voltage Dividers In actual practice the computed value of the bleeder resistor does not always come out to an even value. Since the rule-of-thumb for bleeder current is only an estimated value, the bleeder resistor can be of a value close to the computed value. (If the computed value of the resistance were 510 ohms, a 500-ohm resistor could be used.) Once the actual value of the bleeder resistor is selected, the bleeder current must be recomputed. The voltage developed by the bleeder resistor must be equal to the voltage requirement of the load in parallel with the bleeder resistor. The value of the remaining resistors in the voltage divider is computed from the current through the remaining resistors and the voltage across them. These values must be used to provide the required voltage and current to the loads. If the computed values for the divider resistors are not even values; series, parallel, or seriesparallel networks can be used to provide the required resistance.

iL

Example: A voltage divider is required to supply two loads from a 190.5 volts source. Load 1 requires +45 volts and 210 milliamps; load 2 requires +165 volts and 100 milliamps. Calculate the bleeder current using the rule-of-thumb.

iL

Given:

Iload 1= 210rA LJ

L

iI

Iload2 = 100iA Solution: IR1

= 10% x (210mA + 100mA)

IR1 = 31mA Calculate the ohmic value of the bleeder resistor.

L.

Given: ER1=45V (Eloadl )

IRI = 31mA Solution: R1 = ER1 IR1

R1

45V 31mA

R 1 =1451.60

Li Module 3.6 DC Circuits

6-95

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

n

Li

Since it would be difficult to find a resistor of 1451.6 ohms, a practical choice for R1 is 1500 ohms. Calculate the actual bleeder current using the selected value for R1. Given: =45V IR1= 1.SkQ ER1

Solution: I R1

R 1

I R1 =

45V 1.5k

I R1 = 30mA Using this value for IR1, calculate the resistance needed for the next divider resistor. The current

(1R2) is equal to the bleeder current plus the current used by load 1. Given: IR1 = 3OmA I1oad 1= 210mA

Solution: IR2 = IR1+ IloadI

IR2 = 30mA + 210mA IR4 = 240mA The voltage across R2 (ER2) is equal to the difference between the voltage requirements of loads 2 and1, or 120 volts. Calculate the value of R2. Given:

n L. �

ER2 = 120V IR2 = 21OmA Solution:

7

ti

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6"96

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on page 2 of this Chapter,

n

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LR2 =ER2 IR2

LI

120V 240mA

R2

R2 = 500Q

The value of the final divider resistor is calculated with IR3 (IR2 + I load 2) equal to 340 mA and E R3 (Es - E load 2) equal to 25.5V. Given: ER3= 25,SV C� i?

IR3 = 34OmA

U

Solution:

L

L

R3 -ER3 IR3

R3

_ 25.5V 340mA

R3 =750

L n

A 75-ohm resistor may not be easily obtainable, so a network of resistors equal to 75 ohms can be used in place of R3. Any combination of resistor values adding up to 75 ohms could be placed in series to develop the required network. For example, if you had two 37.5-ohm resistors, you could connect them in series to get a network of 75 ohms. One 50-ohm and one 25-ohm resistor or seven 10-ohm and one 5-ohm resistor could also be used. A parallel network could be constructed from two 150-ohm resistors or three 225-ohm resistors. Either of these parallel networks would also be a network of 75 ohms.

L

The network used in this example will be a series-parallel network using three 50-ohm resistors. With the information given, you should be able to draw this voltage divider network.

Once the values for the various divider resistors have been selected, you can compute the power used by each resistor using the methods previously explained. When the power used by each resistor is known, the wattage rating required of each resistor determines the physical size and type needed for the circuit. This circuit is shown in Figure 6.61.

U

I

Module 3.6 DC Circuits

6-97

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so

`~5

R3 so a

R4

ID

.50 Q

L. i

Es

190.5V R2 Soo U

-t-165V 1OUmA

LOAD 2 + 45V Z10MA LOAD 1

Figure 6.61 - Practical example of a voltage divider.

11

n

L

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TTS Integrated Training System Module 3 Licence Category B1/B2

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Electrical Fundamentals 3.7 Resistance/Resistor

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Module 3.7 Resistance/Resistor

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Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Ltd.

Knowledge Levels - Category A, B1, B2 and C Aircraft Maintenance Licence

7

H

Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2

basic knowledge levels.

The knowledge level indicators are defined as follows:

7

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

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LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • •

A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents

}

U

S iL

Module 3.7 Resistance/Resistor

5

(a) Resistivity Electrical Resistance Standard Colour Code Systems Resistors in Series and Parallel Operation and use of Potentiometers and Rheostats Operation of the Wheatstone Bridge

5 5 7 10 15 25 30

(b) Conductance Electrical Resistors Resistor Wattage Rating Construction of Potentiometers

33 33 34 35 36

S [2

Module 3.7 Resistance/Resistor

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Module 3.7 Enabling Objectives EASA 66 ReferenceLevel

Objective

J Resistance/Resistor

3.7 (a)

2

(b)

1

Resistance and affecting factors

it

Specific resistance Resistor colour code, values and tolerances, preferred values, wattage ratings Resistors in series and parallel Calculation of total resistance using series, parallel and series parallel combinations Operation and use of potentiometers and rheostats Operation of Wheatstone Bridge Positive and negative temperature coefficient conductance Fixed resistors, stability, tolerance and limitations, methods of construction Variable resistors, thermistors, voltage dependent resistors Construction of potentiometers and rheostats Construction of Wheatstone Bridge

lf

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Module 3.7 Resistance/Resistor

LHa) Resistivity Electrical resistivity (also known as specific electrical resistance) is a measure of how

strongly a material opposes the flow of electric current. A low resistivity indicates a material that readily allows the movement of electrical charge. The SI unit of electrical resistivity is the ohm

metre.

It differs from resistance, in that it depends only on the material, and is a property of the material, and is independent of the dimensions of the conductor. The electrical resistivity p (rho) of a material is given by

RA e Where: L; {

U

p is the static resistivity (measured in ohm metres, Q-m); R is the electrical resistance of a uniform specimen of the material (measured in ohms, _Q e is the length of the piece of material (measured in metres, m); A is the cross-sectional area of the Figure 7.1 - Dimensions of a conductor specimen (measured in square metres,

);

m2).

The unit of resistivity is thus the ohm-metre; values may be obtained from tables where they are usually quoted at 0CC. The resistivities of some of the more common materials in electrical use are shown in table 7.1.

L

Resistivity is temperature dependant, with most materials increasing in resistivity as temperature increases. This is called a positive temperature coefficient. Some materials, including all semiconductors, have a negative temperature coefficient. Carbon is a semiconductor material.

r- 7

Module 3.7 Resistance/Resistor

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RESISTIVITY Material

x

T

I metre m

Silver

1.51

Copper

1.59

Gold

2.04

Aluminium

2.45

Platinum

9.81

Iron

8.90

Hard Steel

46

Mercury

94

Manganin

41

Constantan

49

Nickrome

110

Carbon

_

7000

RESISIVITY RELATIVE TO COPPER

TEMPERATURE COEFFICIENT X10-4 PER °C

0.95

41

1.00

43

1.28

40

1.54

45

6.17

39.2

5.60

65

28.9

16

59.2

9

26.1

0.1

30.8

0.4

69

1.5

4425

Negative

USE

Good conductors

n

J

l

Used as conductors because of their other properties Stable resistors (low temp. coefficient) Very low cost

Table 7.1: Resistivities of some common materials at 0°C The formula quoted for resistivity is usually transposed as follows:

R=pL A

This then provides the resistance of a conductor, given its resistivity, length and cross sectional area. These being the factors which affect resistance. More discussion on these factors next.

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Electrical Resistance It is known that the directed movement of electrons constitutes a current flow. It is also known that the electrons do not move freely through a conductor's crystalline structure. Some materials offer little opposition to current flow, while others greatly oppose current flow. This opposition to current flow is known as resistance (R), and the unit of measure is the ohm. The standard of measure for one ohm is the resistance provided at zero degrees Celsius by a column of

mercury having a cross-sectional area of one square millimetre and a length of 106.3

centimetres. A conductor has one ohm of resistance when an applied potential of one volt produces a current of one ampere. The symbol used to represent the ohm is the Greek letter omega (Q).

Resistance, although an electrical property, is determined by the physical structure of a material. The resistance of a material is governed by many of the same factors that control current flow. Therefore, in a subsequent discussion, the factors that affect current flow will be used to assist in the explanation of the factors affecting resistance.

Li

Factors that Affect Resistance The magnitude of resistance is determined in part by the "number of free electrons" available within the material. Since a decrease in the number of free electrons will decrease the current flow, it can be said that the opposition to current flow (resistance) is greater in a material with fewer free electrons. Thus, the resistance of a material is determined by the number of free electrons available in a material. A knowledge of the conditions that limit current flow and, therefore, affect resistance can now be used to consider how the type of material, physical dimensions, and temperature will affect the resistance of a conductor. Type of Material (Resistivity) - Depending upon their atomic structure, different materials will have different quantities of free electrons. Therefore, the various conductors used in electrical applications have different values of resistance.

U

This was discussed in the previous section under "Resistivity". Consider a simple metallic substance. Most metals are crystalline in structure and consist of atoms that are tightly bound in the lattice network. The atoms of such elements are so close together that the electrons in the outer shell of the atom are associated with one atom as much as with its neighbour. (See figure 7.2 view A). As a result, the force of attachment of an outer electron with an individual atom is practically zero. Depending on the metal, at least one electron, sometimes two, and in a few cases, three electrons per atom exist in this state. In such a case, a relatively small amount of additional electron energy would free the outer electrons from the attraction of the nucleus. At normal room temperature materials of this type have many free electrons and are good conductors. Good conductors will have a low resistance.

Module 3.7 Resistance/Resistor

7-7

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n L

(B) Figure 7.2 - Atomic spacing in conductors. If the atoms of a material are farther apart, as illustrated in figure 7.2 view B, the electrons in the outer shells will not be equally attached to several atoms as they orbit the nucleus. They will be attracted by the nucleus of the parent atom only. Therefore, a greater amount of energy is required to free any of these electrons. Materials of this type are poor conductors and therefore have a high resistance.

ii.

Silver, gold, and aluminium are good conductors. Therefore, materials composed of their atoms would have a low resistance. The element copper is the conductor most widely used throughout electrical applications. Silver has a lower resistance than copper but its cost limits usage to circuits where a high conductivity is demanded. Aluminium, which is considerably lighter than copper, is used as a conductor when weight is a major factor. Effect of Cross-Sectional Area - Cross-sectional area greatly affects the magnitude of resistance. If the cross-sectional area of a conductor is increased, a greater quantity of electrons is available for movement through the conductor. Therefore, a larger current will flow for a given amount of applied voltage. An increase in current indicates that when the crosssectional area of a conductor is increased, the resistance must have decreased. If the crosssectional area of a conductor is decreased, the number of available electrons decreases and, for a given applied voltage, the current through the conductor decreases. A decrease in current flow indicates that when the cross-sectional area of a conductor is decreased, the resistance must have increased. Thus, the resistance of a conductor is inversely proportional to its cross-sectional area. H

7

-8

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U

Effect of Conductor Length - The length of a conductor is also a factor which determines the resistance of a conductor. If the length of a conductor is increased, the amount of energy givenup increases. As free electrons move from atom to atom some energy is given off as heat. The longer a conductor is, the more energy is lost to heat. The additional energy loss subtracts from the energy being transferred through the conductor, resulting in a decrease in current flow for a given applied voltage. A decrease in current flow indicates an increase in resistance, since voltage was held constant. Therefore, if the length of a conductor is increased, the resistance increases. The resistance of a conductor is directly proportional to its length. Effect of Temperature - Temperature changes affect the resistance of materials in different ways. In some materials an increase in temperature causes an increase in resistance, whereas in others, an increase in temperature causes a decrease in resistance. The amount of change of resistance per unit change in temperature is known as the temperature coefficient. If for an increase in temperature the resistance of a material increases, it is said to have a positive temperature coefficient. A material whose resistance decreases with an increase in temperature has a negative temperature coefficient. Most conductors used in electronic applications have a positive temperature coefficient. However, carbon, a frequently used material, is a substance having a negative temperature coefficient. Several materials, such as the alloys constantan and manganin, are considered to have a zero temperature coefficient because their resistance remains relatively constant for changes in temperature. The resistance Rt at a temperature of t (°C) can be calculated from the approximation Rt=R°(1 +at) Where R. is the resistance at 0`C. a is the temperature coefficient per degree, taking O9C as the standard. For example: The field winding of a generator has a resistance of 40 S2 at Ot. What is its resistance at 50t? Resistance-Temperature coefficient of copper is 0.0043 pert at 0t (see table 7.1). Rt = Ro (1 + at) = 40(1 + 0.0043 x 50) =40x1.215= 48.65

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L

Standard Colour Code Systems 4-Band System In the standard colour code system, four bands are painted on the resistor, as shown in figure 7.3.

Tolerance 1iiiItil)lier 2nd digit '1st digit Figure 7.3 - A common 4-band resistor The colour of the first band indicates the value of the first significant digit. The colour of the second band indicates the value of the second significant digit. The third colour band represents a decimal multiplier by which the first two digits must be multiplied to obtain the resistance value of the resistor. The colours for the bands and their corresponding values are shown in Table 7.2. YEL

1st & 2nd

V4NT

3 2

bands

3rd band

ORN

x1

x10

x100

n

4

6

6

7

x1 Kx10Kx100KOM x101v1

8

9

n/an/a

Table 7.2 - Standard Colour Code for Resistors

n L. J

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m I__.._._L, nnin

Use andlor disclosure is

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Simplifying the Colour Code - Resistors are the most common components used in electronics. The technician must identify, select, check, remove, and replace resistors. Resistors and resistor circuits are usually the easiest branches of electronics to understand. The resistor colour code sometimes presents problems to a technician. It really should not, because once the resistor colour code is learned, you should remember it for the rest of your

life.

Black, brown, red, orange, yellow, green, blue, violet, gray, white - this is the order of colours you should know automatically. There is a memory aid that will help you remember the code in Li

its proper order. Each word starts with the first letter of the colours. If you match it up with the colour code, you will not forget the code.

L

Bad Boys Run Over Yellow Gardenias Behind Victory Garden Walls,

BlackBad BrownBoys Red

Run

OrangeOver YellowYellow GreenGardenias Blue

Behind

Violet

Victory

GrayGarden WhiteWalls

L L

u

Table 7.3 - Resistor colour order - aid to memory There are many other memory aid sentences that you might want to ask about from experienced technicians. We could not possibly print them here, for fear of offending someone. There is still a good chance that you will make a mistake on a resistor's colour band. Most technicians do at one time or another. If you make a mistake on the first two significant colours, it usually is not too serious. If you make a mistake on the third band, you are in trouble, because the value is going to be at least 10 times too high or too low.

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The fourth band, which is the tolerance band, usually does not present too much of a problem. If there is no fourth band, the resistor has a 20-percent tolerance; a silver fourth band indicates a 10percent tolerance; and a gold fourth band indicates a 5-percent tolerance.

ORf

=

l aOLDSLYt- -

Tolerance Band

1%

2%

0.5%

0.25%

0.1%

0.05%

5%

10%

Table 7.4 - 5th Band Colour Codes (Tolerance Band) Some older 4-band resistors that conform to military

specifications have a fifth band. The fifth band indicates the reliability level per 1,000 hours of operation as follows: Fifth band colourLevel Brown

1.0%

Red

0.1%

Orange

0.01%

Yellow

0.001%

Figure 7.4 - A 4-band resistor with a 5th band for reliability

Table 7.5 - 5th colour band - Reliability For a resistor whose the fifth band is colour coded brown, the resistor's chance of failure will not exceed 1 percent for every 1,000 hours of operation. In equipment such as the aircraft's complex computers, the reliability level is ve ry significant. For example, in a piece of equipment containing 10,000 orange fifth-band resistors, no more than one resistor will fail during 1,000 hours of operation. This is very good reliability. However, the reliability of modern manufactured resistors is now so high, that the chance of failure is well under the 0.001% of the yellow band designated resistor. Hence the 5th band is no longer used to denote reliability. The five band resistor is now used for the high tolerance, high resolution resistors, as will be explained next.

H

7-12 7-11

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5-Band System Read the colours from left to right just like for 4-band resistors. The first band is the first significant digit (1st number), the second band is the second significant digit (2nd number), the third band is the third significant digit (3rd number), the fourth band is the multiplier band (number of zeros to add to the two digit number, again this band can also be Gold or Silver to move the decimal point to the left), and the fifth band is the tolerance band. Tolerance values f orveanresstors can only be 0.05%, 0.1 %, 0.25%, 0.5% or 1 % (grey, violet, blue, green, fi brown). For most of us, we will only see 1 % tolerance resistors as the highest precision components in electronic devices. If you work on test instruments or specialized equipment, you may see some of the higher precision components.

L-- Tolerance

rvlIJItiplier

iL

3rd digit

2nd digit

'Ist digit Ff

U

U

Figure 7.5 - A modern 5-band resistor 6-Band - Temperature coefficient Occasionally, one can encounter resistors with six colour bands, the last one of which is anomalous for a tolerance class specification (orange, yellow or white). In such cases, the last band defines the worst-case temperature-dependence coefficient of the component. The codes for temperature coefficients are listed in Table 7.5. Temperature-tolerance colour-coding is used very rarely and may differ slightly among manufacturers.

L Table 7.5 -

Temperature Coefficients

IL Figure 7.6 -- A 6-band resistor

Module 3.7 Resistance/Resistor

TemlfCo 1a ert3nce rylitiltiplier 3rd digit 2nd digit '1st digit

7-13

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Combined 4-Band and 5-Band Colour Chart The same colour chart can be used to determine the value of both 4-band and 5-band resistors. On the 4-band resistor, the `3rd-band' column of the chart is ignored. 4-band color code

10K Ohms± 5%

First Digit

Second Digit

Third Digit

SLV 0.01 GLR 0.1

BLK-0 BRN-1

BLK-0 BRN-1

8 LK-0

BLK-1

RED2. YEL-4

RED-2

SLV'± 10% GLD ± S%

Coefficient

BRN-1

BRN-10

BRNt,1%

RED-2

91=0'-100. ,

#t D 2.%

mft%

YEL-4

YEL-4

GRN-6

GRN-6

GRN-6

BLU-6

BLU-6

B LU-6

V1G-7

V14-7

G1Y•8

V10-7 GRY-8

GRY-8

WHT-9

WHT-9

WHT-9

Temperature

11 ED Z WO,

YEL-10K GRN-100K

SLU-1 M V10-10M

.J

GRN ± 0.6,1/o BLU-± 0,25% V10 ± 0.1%

Figure 7.7 - Combined 4-Band, 5-Band and 6-Band Chart

n

7-14 Use and/or disclosure is governed by the statement

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Resistors in Series and Parallel Series Resistance Referring to Figure 7.8, the current in a series circuit must flow through each lamp to complete the electrical path in the circuit. Each additional lamp offers added resistance. In a series circuit, the total circuit resistance (RT) is equal to the sum of the individual resistances. As an equation:

RT = R1 + R2 + R3+ . , . Rn

NOTE: The subscript n denotes any number of additional resistances that might be in the equation.

L U

L

L

BASIC CIRCUIT

SERIES CIRCUIT

Figure 7.8 - Comparison of basic and series circuits.

Module 3.7 Resistance/Resistor

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Example: In Figure 7.9 a series circuit consisting of three resistors: one of 10 ohms, one of 15 ohms, and one of 30 ohms, is shown. A voltage source provides 110 volts. What is the total resistance?

1110 V

R3 305) Figure 7.9 - Solving for total resistance in a series circuit.

Given:

Rt = 10 ohms R2 = 15 ohms R3 = 30 ohms

n Soulution:RT=R1+R2+R3 RT = 10 ohms + 15 ohms + 30 ohms RT = 55 ohms In some circuit applications, the total resistance is known and the value of one of the circuit resistors has to be determined. The equation RT = R1 + R2 + R3 can be transposed to solve for the value of the unknown resistance. Example: In Figure 7.10 the total resistance of a circuit containing three resistors is 40 ohms. Two of the circuit resistors are 10 ohms each. Calculate the value of the third resistor (R3).

7-16 TTS Integrated Training System r-•-

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n

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105

RT

40 Q

i

R3

Li

Figure 7.10 - Calculating the value of one resistance in a series circuit. Given: R1= 40 ohms R2 = 10 ohms R3=10 ohms

r-

UT Li

Solution: RT=R1+R2+R3 (Subtract R1 + R2 from both sides of the equation) RT-R1-R2=R3

R3=RT-R1-R2 R3 = 40 ohms - 10 ohms - 10 ohms R3 = 40 ohms - 20 ohms R3 = 20 ohms L

i

L_1

`. 1

Module 3.7 Resistance/Resistor

7-17

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Parallel Resistance In the example diagram, Figure 7.11, there are two resistors connected in parallel across a 5volt battery. Each has a resistance value of 10 ohms. A complete circuit consisting of two parallel paths is formed and current flows as shown. 1A

7 5V

1oc

Figure 7.11 - Two equal resistors connected in parallel. Computing the individual currents shows that there is one-half of an ampere of current through each resistance. The total current flowing from the battery to the junction of the resistors, and returning from the resistors to the battery, is equal to 1 ampere. The total resistance of the circuit can be calculated by using the values of total voltage (ET) and total current (IT). NOTE: From this point on the abbreviations and symbology for electrical quantities will be used in example problems. Given: ET=5V IT= 1 A

I

Solution: R=E

I

RT =ET IT _5V RT 1A RT =592 This computation shows the total resistance to be 5 ohms; one-half the value of either of the two resistors.

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Since the total resistance of a parallel circuit is smaller than any of the individual resistors, total resistance of a parallel circuit is not the sum of the individual resistor values as was the case in a series circuit. The total resistance of resistors in parallel is also referred to as equivalent resistance (Req). The terms total resistance and equivalent resistance are used

interchangeably. There are several methods used to determine the equivalent resistance of parallel circuits. The best method for a given circuit depends on the number and value of the resistors. For the circuit described above, where all resistors have the same value, the following simple equation is used: R eq - N

Reg = equivalent parallel resistance R = ohmic value of one resistor N = number of resistors This equation is valid for any number of parallel resistors of equal value. Example. Four 40-ohm resistors are connected in parallel. What is their equivalent resistance? L

Given:

Irl

+ R2 + R3 + R4 R1=40Q R1

Solution: R Reg N R eg

4

Reg =10 Q

1! Figure 7.12 shows two resistors of unequal value in parallel. Since the total current is shown, the equivalent resistance can be calculated,

Module 3.7 Resistance/Resistor

7-19

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ES

1OA

R1

30V

5A

R2

6-2

3Q

15A 1

0

Figure 7.12 - Example circuit with unequal parallel resistors. Given: ES= 30'V IT= 15A Solution:

R eq Req =

ES IT 30V 15A

Req =20

The equivalent resistance of the circuit shown in Figure 7.12 is smaller than either of the two resistors (R 1, R2). An important point to remember is that the equivalent resistance of a parallel circuit is always less than the resistance of any branch. Equivalent resistance can be found if you know the individual resistance values and the source voltage. By calculating each branch current, adding the branch currents to calculate total current, and dividing the source voltage by the total current, the total can be found. This method, while effective, is somewhat lengthy. A quicker method of finding equivalent resistance is to use the general formula for resistors in parallel:

7-20 Use and/or disclosure is governed by the statement

1

1

R eg

R1

+1

R2

+ 1

R3

+

1

Rn

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Use and/or disclosure is

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If you apply the general formula to the circuit shown in Figure 6.40 you will get the same value for equivalent resistance (2f2) as was obtained in the previous calculation that used source voltage and total current. L Given: UI'

R1=30

R2 = 65 Solution: 1

1

1

Reg

R1

R2

1 Reg

1 3Q

1 6Q

[1

U Convert the fractions to a common denominator.

Since both sides are reciprocals (divided into one), disregard the reciprocal function. Reg =20 The formula you were given for equal resistors in parallel R (Reg =-) is a simplification of the general formula for resistors in parallel 1 Reg

=

1 R1

+1+1+... R2

R3

1 R.

There are other simplifications of the general formula for resistors in parallel which can be used to calculate the total or equivalent resistance in a parallel circuit.

L

7-21 Module 3.7 Resistance/Resistor

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Reciprocal Method - This method is based upon taking the reciprocal of each side of the equation. This presents the general formula for resistors in parallel as: R eq -

1 1

R1

1

R2

t_-�

1 R n

This formula is used to solve for the equivalent resistance of a number of unequal parallel resistors. You must find the lowest common denominator in solving these problems. Example: Three resistors are connected in parallel as shown in Figure 7.13. The resistor values

l

are: R1 = 20 ohms, R2 = 30 ohms, R3 = 40 ohms. What is the equivalent resistance? (Use the reciprocal method.)

R, 200

ES

]C

3052

452 1

Figure 7.13 - Example parallel circuit with unequal branch resistors. Given: R1 = 200 R2 = 305 R3 = 400

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!"...

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!�

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Solution:

R eg =

1

1 R1

Reg =

f..i

Reg =

L U

R eq

1 R3 1

1 20Q

1 300

1 400

1 6

120Q

r7

1 R2

+

4

1200

+

3

1200

1 13 120 120

Product Over the Sum Method - A convenient method for finding the equivalent, or total, resistance of two parallel resistors is by using the following formula.

R�9

1 R1

Li

+

R2

R2

This equation, called the product over the sum formula, is used so frequently it should be committed to memory. Example: What is the equivalent resistance of a 20-ohm and a 30-ohm resistor connected in

I

parallel, as in Figure 7.14?

L

u u

Figure 7.14 - Parallel circuit with two unequal resistors.

Module 3.7 Resistance/Resistor

7-23

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Given: R1 = 202 R2 = 305

Solution:

n

R1x R2 Reg yR1 +R2 Reg

205 x 305 205 +305

,1

600

Reg

o

Reg =125 n

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Operation and use of Potentiometers and Rheostats A potentiometer is a variable tapped resistor that can be used as a voltage divider.

Li Potentiometer

Variable Resistor

Rheostat

Figure 7.15 - Schematic symbol for a potentiometer. The arrow represents the moving terminal, called the wiper. U

A form of potentiometer is used as an instrument to measure the potential (or voltage) in a

circuit by tapping off a fraction of a known voltage from a resistive slide wire and comparing it

with the unknown voltage by means of a galvanometer. The sliding tap of the potentiometer is adjusted and the galvanometer briefly connected to both the sliding tap and the unknown potential; the deflection of the galvanometer is observed and the sliding tap adjusted until the galvanometer no longer deflects. At that point the galvanometer is drawing no current from the unknown source, and the magnitude of voltage can be calculated from the position of the sliding contact. This null balance method is a fundamental technique of electrical metrology.

U

As an electrical component, potentiometer (or'pot' for short) describes a three-terminal resistor with a sliding contact that forms an adjustable voltage divider. If all three terminals are used, it can act as a variable voltage divider. If only two terminals are used (one side and the wiper), it acts as a variable resistor or rheostat. Potentiometers are commonly used as controls for electrical devices such as a volume control of a radio. Potentiometers operated by a mechanism can be used as position transducers, for example, in a joystick. Potentiometer as Measuring Instrument Before the introduction of the calibratable (sprung) moving coil meter, potentiometers were used in measuring voltage, hence the '-meter' part of their name. Today this method is confined to standards work, and is not normally used in other areas of electronics. The original potentiometer is a type of bridge circuit for measuring voltages by comparison between a small fraction of the voltage which could be precisely measured, then balancing the two circuits to get null current flow which could be precisely measured. The word itself derives from the phrase "voltage potential," and "potential" was used to refer to "strength." The original potentiometers are divided into four main classes listed below. Constant Current Potentiometer This is used for measuring voltages below 1.5 volts. In this circuit, the unknown voltage is connected across a section of resistance wire the ends of which are connected to a standard electrochemical cell that provides a constant current through the wire, The unknown EMF, in

Module 3.7 Resistance/Resistor

7-25

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series with a galvanometer, is then connected across a variable-length section of the resistance wire using a sliding contact(s). The sliding contact is moved until no current flows into or out of the standard cell, as indicated by a galvanometer in series with the unknown EMF. The voltage across the selected section of wire is then equal to the unknown voltage. All that remains is to calculate the unknown voltage from the current and the fraction of the length of the resistance wire that was connected to the unknown EMF. The galvanometer does not need to be calibrated, as its only function is to read zero. When the galvanometer reads zero, no current is drawn from the unknown electromotive force and so the reading is independent of the source's internal resistance. Because the resistance wire can be made very uniform in cross-section and resistivity, and the

position of the wiper can be measured easily, this method can be analyzed to accurately

determine the uncertainties in the measurement. When measuring potentials larger than that

produced by a standard cell, an external voltage divider is used to scale the measured voltage down to approximately 1 volt for measurement by the potentiometer; the uncertainties due to the voltage divider construction and the load placed on the source by the voltage divider then become part of the uncertainty of the overall measurement. Constant Resistance Potentiometer The constant resistance potentiometer is a variation of the basic idea in which a variable current is fed through a fixed resistor. These are used primarily for measurements in the millivolt and microvolt range.

t iJ

Microvolt Potentiometer This is a form of the constant resistance potentiometer described above but designed to minimize the effects of contact resistance and thermal EMF. This equipment is satisfactorily used down to readings of 10 nV or so. Thermocouple Potentiometer Another development of the standard types was the `thermocouple potentiometer' especially modified for performing temperature measurements with thermocouples. Potentiometers for use with thermocouples also measure the temperature at which the thermocouple wires are connected, so that cold-junction compensation may be applied to correct the apparent measured EMF to the standard cold-junction temperature of 0 degrees C. Potentiometer as an Electronic Component A potentiometer is a potential divider, a three terminal resistor where the position of the sliding connection is user adjustable via a knob or slider. Potentiometers are sometimes provided with one or more switches mounted on the same shaft. For instance, when attached to a volume control, the knob can also function as an on/off switch at the lowest volume. Ordinarily potentiometers are rarely used to directly control anything of significant power (more than a watt). Instead they are used to adjust the level of analogue signals (e.g. volume controls on audio equipment), and as control inputs for electronic circuits. For example, a light dimmer uses a potentiometer to control the switching of a triac and so indirectly control the brightness of lamps. � ._i

7-26 r--,:

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Rheostats

A rheostat is a two-terminal variable resistor. Often these are designed to handle much higher voltage and current. Typically these are constructed as a resistive wire wrapped to form a toroid coil with the wiper moving over the upper surface of the toroid, sliding from one turn of the wire to the next. Sometimes a rheostat is made from resistance wire wound on a heat resisting cylinder with the slider made from a number of metal fingers that grip lightly onto a small portion of the turns of resistance wire. The 'fingers' can be moved along the coil of resistance wire by a sliding knob thus changing the 'tapping' point. They are usually used as variable resistors rather than variable potential dividers.

fl

Ir

L

Figure 7.16 - A high power toroidal wire-wound rheostat. Any three-terminal potentiometer can be used as a two-terminal variable resistor, by not connecting to the 3rd terminal. It is common practice to connect the wiper terminal to the unused end of the resistance track to reduce the amount of resistance variation caused by dirt on the track. Applications of Potentiometers Potentiometers are widely used as user controls, and may control a very wide variety of equipment functions. The widespread use of pots in consumer electronics has declined in the 1990s, with digital controls now more common. However they remain in use in many applications. 2 of the most common applications are as volume controls and as position sensors.

r-? L

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7-27

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Audio Control Sliding potentiometers ("faders") One of the most common uses for

modern low-power potentiometers

is as audio control devices. Both sliding pots (also known as faders) and rotary potentiometers (commonly called knobs) are regularly used to adjust loudness, frequency attenuation and other characteristics of audio signals.

The 'log pot' is used as the volume control in audio amplifiers, where it is also called an "audio taper pot", because the amplitude response of the human ear is also logarithmic. It ensures that, Figure 7.17 - Sliding potentiometers. on a volume control marked 0 to 10, for example, a setting of 5 sounds half as loud as a setting of 10. There is also an anti-log pot or reverse audio taper which is simply the reverse of a log pot. It is almost always used in a ganged configuration with a log pot, for instance, in an audio balance control. Potentiometers used in combination with filter networks act as tone controls. Transducers Potentiometers are also very widely used as a part of position transducers because of the simplicity of construction and because they can give a large output signal.

7-28 Use and/or disclosure is governed by the statement

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Theory of Operation

A potentiometer with a resistive load, showing equivalent fixed resistors for clarity, is shown in figure 7.18.

Figure 7.18 - The potentiometer and its equivalent circuit as a voltage divider The potentiometer can be used as a potential divider (or voltage divider) to obtain a manually adjustable output voltage at the slider (wiper) from a fixed input voltage applied across the two ends of the pot. This is the most common use of pots. One of the advantages of the potential divider compared to a variable resistor in series with the source is that, while variable resistors have a maximum resistance where some current will always flow, dividers are able to vary the output voltage from maximum (Vs) to ground (zero volts) as the wiper moves from one end of the pot to the other. There is, however, always a small amount of contact resistance.

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In addition, the load resistance is often not known and therefore simply placing a variable resistor in series with the load could have a negligible effect or an excessive effect, depending on the load.

Module 3.7 Resistance/Resistor

7-29

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Operation of the Wheatstone Bridge A Wheatstone bridge is a measuring instrument invented by Samuel Hunter Christie in 1833 and improved and popularized by Sir Charles Wheatstone in 1843. It is used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which includes the unknown component. Its operation is similar to the original potentiometer except

that in potentiometer circuits the meter used is a sensitive galvanometer.

The Basic Bridge Circuit The fundamental concept of the Wheatstone Bridge is two voltage dividers, both fed by the same input, as shown to the right. The circuit output is taken from both voltage divider outputs, as shown here.

R1 IN

4

R3

OUT R2

Y R4

Figure 7.19 -- The basic Wheatstone Bridge circuit In its classic form, a galvanometer (a very sensitive DC current meter) is connected between the output terminals, and is used to monitor the current flowing from one voltage divider to the other. If the two voltage dividers have exactly the same ratio (R1/R2 = R3/R4), then the bridge is said to be balanced and no current flows in either direction through the galvanometer. If one of the resistors changes, even a little bit in value, the bridge will become unbalanced and current will flow through the galvanometer. Thus, the galvanometer becomes a very sensitive indicator

J

of the balance condition.

7-30

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Using the Wheatstone Bridge In its basic application, a DC voltage (E) is applied to the Wheatstone Bridge, and a galvanometer (G) is used to monitor the balance condition. The values of Ri and R3 are precisely known, but do not have to be identical. R2 is a calibrated variable resistance, whose current value may be read from a dial or scale.

Figure 7.20 - The practical Wheatstone Bridge An unknown resistor, Rx, is connected as the fourth side of the circuit, and power is applied. R2 is adjusted until the galvanometer, G, reads zero current. At this point, Rx = R2xR3/R1. This circuit is most sensitive when all four resistors have similar resistance values. However, the circuit works quite well in any event. If R2 can be varied over a 10:1 resistance range and R1 is of a similar value, we can switch decade values of R3 into and out of the circuit according to the range of value we expect from RX. Using this method, we can accurately measure any value of Rx by moving one multiple-position switch and adjusting one precision potentiometer. Applications of the Wheatstone Bridge It is not possible to cover all of the practical variations and applications of the Wheatstone Bridge, let alone all types of bridges, in a single Web page. Sir Charles Wheatstone invented many uses himself, and others have been developed, along with many variations, since that time. One very common application in industry today is to monitor sensor devices such as strain gauges. Such devices change their internal resistance according to the specific level of strain (or pressure, temperature, etc.), and serve as the unknown resistor RX. However, instead of trying to constantly adjust R2 to balance the circuit, the galvanometer is replaced by a circuit that can be calibrated to record the degree of imbalance in the bridge as the value of strain or other condition being applied to the sensor. A second application is used by electrical power distributors to accurately locate breaks in a power line. The method is fast and accurate, and does not require a large number of field technicians.

Module 3.7 Resistance/Resistor

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7-32 TTS Integrated Training System Use and/or disclosure is governed by the statement

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Conductance Electricity is a study that is frequently explained in terms of opposites. The term that is the opposite of resistance is conductance. Conductance is the ability of a material to pass electrons. The factors that affect the magnitude of resistance are exactly the same for conductance, but they affect conductance in the opposite manner. Therefore, conductance is directly proportional to area, and inversely proportional to the length of the material. The temperature of the material is definitely a factor, but assuming a constant temperature, the conductance of a material can be calculated. The unit of conductance is the mho (G), which is ohm spelled backwards. Recently the term

mho has been redesignated Siemens (S). Whereas the symbol used to represent resistance (R) is the Greek letter omega (S2), the symbol used to represent conductance (G) is (S). The relationship that exists between resistance (R) and conductance (G) or (S) is a reciprocal one. A reciprocal of a number is one divided by that number. In terms of resistance and conductance:

1 -1 '

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1 =

Positive and Negative Coefficients of Conductance Since conductance is merely the reciprocal of resistance, it is temperature dependant. However, the reciprocal nature of its relationship with resistance means that where a material has a positive temperature coefficient of resistance, it will have a negative temperature coefficient of conductance, and vice versa.

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Module 3.7 Resistance/Resistor

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Electrical Resistors

Resistance is a property of every electrical component. At times, its effects will be undesirable. However, resistance is used in many varied ways. Resistors are components manufactured to possess specific values of resistance. They are manufactured in many types and sizes. When drawn using its schematic representation, a resistor is shown as a series of jagged lines, as illustrated in figure 7.21. TYPICAL RESISTOR

TYPE

SYMBOL

A FIXED CARBON FIXED WIREWOUND (TAPPED) C

ADJUSTABLE W REWOUND

D POTENTIOMETERQ--AVr

E RHEOSTAT

Figure 7.21 - Types of resistors. Composition of Resistors One of the most common types of resistors is the moulded composition, usually referred to as the carbon resistor. These resistors are manufactured in a variety of sizes and shapes. The chemical composition of the resistor determines its ohmic value and is accurately controlled by the manufacturer in the development process. They are made in ohmic values that range from one ohm to millions of ohms. The physical size of the resistor is related to its wattage rating, which is the ability of resistor to dissipate heat caused by the resistance. Carbon resistors, as you might suspect, have as their principal ingredient the element carbon. In the manufacturer of carbon resistors, fillers or binders are added to the carbon to obtain various resistor values. Examples of these fillers are clay, Bakelite, rubber, and talc. These fillers are doping agents and cause the overall conduction characteristics to change.

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Carbon resistors are the most common resistors found because they are easy to manufacturer,

inexpensive, and have a tolerance that is adequate for most electrical and electronic U {. i

applications. Their prime disadvantage is that they have a tendency to change value as they age. One other disadvantage of carbon resistors is their limited power handling capacity. The disadvantage of carbon resistors can be overcome by the use of WIREWOUND resistors (fig. 1-29 (B) and (C)). Wire-wound resistors have ve accurate values and possess a higher current handling capability than carbon resistors. The material that is frequently used to manufacture wire-wound resistors is German silver which is composed of copper, nickel, and

zinc. The qualities and quantities of these elements present in the wire determine the resistivity

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of the wire. (The resistivity of the wire is the measure or ability of the wire to resist current. Usually the percent of nickel in the wire determines the resistivity.) One disadvantage of the wire-wound resistor is that it takes a large amount of wire to manufacture a resistor of high ohmic value, thereby increasing the cost. A variation of the wire-wound resistor provides an exposed surface to the resistance wire on one side. An adjustable tap is attached to this side. Such resistors, sometimes with two or more adjustable taps, are used as voltage dividers in power supplies and other applications where a specific voltage is desired to be "tapped" off. Fixed and Variable Resistors There are two kinds of resistors, Fixed and Variable. The fixed resistor will have one value and will never change (other than through temperature, age, etc.). The resistors shown in A and B of figure 7.21 are classed as fixed resistors. The tapped resistor illustrated in B has several fixed taps and makes more than one resistance value available. The sliding contact resistor shown in C has an adjustable collar that can be moved to tap off any resistance within the ohmic value range of the resistor. There are two types of variable resistors, one called a potentiometer and the other a rheostat (see views D and E of fig. 7.21) An example of the potentiometer is the volume control on your radio, and an example of the rheostat is the dimmer control for the dash lights in an automobile. There is a slight difference between them. Rheostats usually have two connections, one fixed and the other moveable. Any variable resistor can properly be called a rheostat. The potentiometer always has three connections, two fixed and one moveable. Generally, the rheostat has a limited range of values and a high current-handling capability. The potentiometer has a wide range of values, but it usually has a limited current-handling capability. Potentiometers are always connected as voltage dividers.

Resistor Wattage Rating When a current is passed through a resistor, heat is developed within the resistor. The resistor U must be capable of dissipating this heat into the surrounding air; otherwise, the temperature of the resistor rises causing a change in resistance, or possibly causing the resistor to burn out. The ability of the resistor to dissipate heat depends upon the design of the resistor itself. This r ability to dissipate heat depends on the amount of surface area which is exposed to the air. A resistor designed to dissipate a large amount of heat must therefore have a large physical size. The heat dissipating capability of a resistor is measured in Watts. Some of the more common wattage ratings of carbon resistors are: one-eighth watt, one-fourth watt, one-half watt, one watt, and two Watts. In some of the newer state-of-the-art circuits of today, much smaller wattage resistors are used. Generally, the type that you will be able to physically work with are of the values given. The higher the wattage rating of the resistor the larger is the physical size.

Module 3.7 Resistance/Resistor

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Resistors that dissipate very large amounts of power (watts) are usually wire-wound resistors. Wire-wound resistors with wattage ratings up to 50 watts are not uncommon.

Construction of Potentiometers Figure 7.22 - Construction of a wire-wound circular potentiometer. The resistive element (1) of the shown device is trapezoidal, giving a non-linear relationship between resistance and turn angle. The wiper (3) rotates with the axis (4), providing the changeable resistance between the wiper contact (6) and the fixed contacts

(5) and (9). The vertical position of the axis is fixed in

the body (2) with the ring (7) (below) and the bolt (8)

(above).

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A potentiometer is constructed using a flat semi-circular graphite resistive element, with a sliding contact (wiper). The wiper is connected through another sliding contact to the third terminal. On panel pots, the wiper is usually the centre terminal. For single turn pots, this wiper typically travels just under one revolution around the contact. 'Multi-turn' potentiometers also exist, where the resistor element may be helical and the wiper may move 10, 20, or more complete revolutions. Besides graphite, other materials may be used to make the resistive element. These may be resistance wire, or carbon particles in plastic, or a ceramic/metal mixture called cermet.

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Figure 7.23 - A typical single turn potentiometer

One form of rotary potentiometer is called a string pof. It is a multi-turn potentiometer with an attached reel of wire turning against a spring. It is convenient for measuring movement and therefore acts as a position transducer. In a linear slider pot, a sliding control is provided instead of a dial control. The word linear also describes the geometry of the resistive element which is a rectangular strip, not semi-circular as in a rotary potentiometer. Because of the large opening for the wiper and knob, this type of pot has a greater potential for getting contaminated. Potentiometers can be obtained with either linear or logarithmic laws (or "tapers"). 7-36 TTS Integrated Training System Use and/or disclosure is governed by the statement

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1 1 1 111111

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Figure 7.24 - PCB mount trimmer potentiometers, or "trimpots", intended for infrequent adjustment.

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Linear Taper Potentiometer A linear taper potentiometer has a resistive element of constant cross-section, resulting in a device where the resistance between the contact (wiper) and one end terminal is proportional to the distance between them. Linear taper describes the electrical characteristic of the device, not the geometry of the resistive element. Linear taper potentiometers are used when an approximately proportional relation is desired between shaft rotation and the division ratio of the potentiometer; for example, controls used for adjusting the centreing of (an analogue) cathoderay oscilloscope. Logarithmic Potentiometer A logarithmic taper potentiometer has a resistive element that either'tapers' in from one end to the other, or is made from a material whose resistivity varies from one end to the other. This results in a device where output voltage is a logarithmic (or inverse logarithmic depending on

type) function of the mechanical angle of the pot. Most (cheaper) "log" pots are actually not logarithmic, but use two regions of different, but constant, resistivity to approximate a logarithmic law. A log pot can also be simulated with a linear pot and an external resistor. True log pots are significantly more expensive. Logarithmic taper potentiometers are often used in connection with audio amplifiers.

Module 3.7 Resistance/Resistor

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7-38 TTS Integrated Training System Use and/or disclosure is governed by the statement

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Module 3 Licence Category B1/B2 Electrical Fundamentals 3.8 Power

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Module 3.8 Power

8-1 on page 2 of this Chapter.

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Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Ltd.

Knowledge Levels - Category A, 131, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows:

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the

subject.

• The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • •

A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents r~.

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Module 3.8 Power Introduction Power Rating Power Conversion and Efficiency Power in a Series Circuit Power Transfer and Efficiency

5 5 11 12 14 16

Power in a Parallel Circuit Power in the Voltage Divider

17 18

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Module 3.8 Power

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Module 3.8 Enabling Objectives Objective

EASA 66 Reference

Level

Power Power, work and energy (kinetic and potential) Dissipation of power by a resistor Power formula Calculations involving power, work and energy

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Module 3.8 Power iL Fli Li

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Introduction Power, whether electrical or mechanical, pertains to the rate at which work is being done. Work is done whenever a force causes motion. When a mechanical force is used to lift or move a weight, work is done. However, force exerted without causing motion, such as the force of a compressed spring acting between two fixed objects, does not constitute work. Previously, it was shown that voltage is an electrical force, and that voltage forces current to flow in a closed circuit. However, when voltage exists but current does not flow because the circuit is open, no work is done. This is similar to the spring under tension that produced no motion. When voltage causes electrons to move, work is done. The instantaneous rate at which this work is done is called the electric power rate, and is measured in Watts. A total amount of work may be done in different lengths of time. For example, a given number of electrons may be moved from one point to another in 1 second or in 1 hour, depending on the rate at which they are moved. In both cases, total work done is the same. However, when the work is done in a short time, the wattage, or instantaneous power rate, is greater than when the same amount of work is done over a longer period of time. As stated, the basic unit of power is the watt. Power in watts is equal to the voltage across a circuit multiplied by current through the circuit. This represents the rate at any given instant at which work is being done. The symbol P indicates electrical power. Thus, the basic power formula is P = E x 1, where E is voltage and I is current in the circuit. The amount of power changes when either voltage or current, or both voltage and current, are caused to change.

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In practice, the only factors that can be changed are voltage and resistance. In explaining the different forms that formulas may take, current is sometimes presented as a quantity that is

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changed. Remember, if current is changed, it is because either voltage or resistance has been

Figure 8.9 shows a basic circuit using a source of power that can be varied from 0 to 8 volts and a graph that indicates the relationship between voltage and power. The resistance of this circuit is 2 ohms; this value does not change. Voltage (E) is increased (by f increasing the voltage source), in steps of 1 volt, from 0 volts to 8 volts. By applying Ohm's law, the current (I) is determined for each step of voltage. For instance, when E is 1 volt, the current L..; is:

Module 3.8 Power

8-5

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I

=E

R

1 volt 2 ohms I = 0,5 ampere VARIABLE POWER SUPPLY 0 - 8 VOLTS

R (FIXED) 2 OHMS

P EI P E2

P

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1

2

3

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Figure 8.1 - Graph of power related to changing voltage. i

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Power (P), in watts, is determined by applying the basic power formula: P=ExI P=Ivolt x0.5ampere P=O.Swatt

When E is increased to 2 volts: F)

IuE R 2 volts

2 ohms I = 1 amp ere and P=ExI

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P = 2 volts x 1 ampere P = 2 watts When E is increased to 3 volts:

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I =E R 3 volts 2 ohms I = 1.5 amperes

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and P=ExI P = 3 volts x 1.5 ampere

P=4.5watts

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You should notice that when the voltage was increased to 2 volts, the power increased from 0.5 watts to 2 watts or 4 times. When the voltage increased to 3 volts, the power increased to 4.5 watts or 9 times. This shows that if the resistance in a circuit is held constant, the power varies directly with the square of the voltage.

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Another way of proving that power varies as the square of the voltage when resistance is held constant is:

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E R

Since:

By substitution in: P=ExI You get;

P=Ex

ER lf

ExE R

Or: Therefore:

P = E2 R

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Another important relationship may be seen by studying Figure 8.10. Thus far, power has been calculated with voltage and current (P = E x I), and with voltage and resistance

P

E2 R

Referring to Figure 8.10, note that power also varies as the square of current just as it does with voltage. Thus, another formula for power, with current and resistance as its factors, is p = 12R. This can be proved by: Since:

E=IxR

By substituition in: P = E x I You get:

P=IxRxI

Or:

P=IxIxR

Therefore:

P = 12 x R

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(VARIABLE POWER SUPPLY) 0 - 8 VOLTS

R (FIXED)

2 OHMS

P = EI

P = I2R WATTS

13

2

5

1

1.5

2

2.5

3

3.5

4

I (AMPS)

Figure 8.2 - Graph of power related to changing current. Up to this point, four of the most important electrical quantities have been discussed. These are voltage (E), current (I), resistance (R), and power (P). You must understand the relationships which exist among these quantities because they are used throughout your study of electricity. In the preceding paragraphs, P was expressed in terms of alternate pairs of the other three basic quantities E, I, and R. In practice, you should be able to express any one of these quantities in terms of any two of the others. Figure 8.3 is a summary of 12 basic formulas you should know. The four quantities E, I, R, and P are at the centre of the figure. Adjacent to each quantity are three segments. Note that in each segment, the basic quantity is expressed in terms of two other basic quantities, and no two segments are alike.

Module 3.8 Power

8-9

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n

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Figure 8.3 - Summary of basic formulas. For example, the formula wheel in Figure 8.3 could be used to find the formula to solve the following problem:

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A circuit has a voltage source that delivers 6 volts and the circuit uses 3 watts of power. What is the resistance of the load?

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Since R is the quantity you have been asked to find, look in the section of the wheel that has R in the centre. The segment

E2 F contains the quantities you have been given. The formula you would use is

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RE2 P The problem can now be solved.

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Module 3.8 Power

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Given:

E = 6 volts P = 3 watts

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Soultion: R = E P (6 volts)2 3 watts ,,

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Power Rating Electrical components are often given a power rating. The power rating, in watts, indicates the rate at which the device converts electrical energy into another form of energy, such as light, heat, or motion. An example of such a rating is noted when comparing a 150-watt lamp to a 100-watt lamp. The higher wattage rating of the 150-watt lamp indicates it is capable of converting more electrical energy into light energy than the lamp of the lower rating. Other common examples of devices with power ratings are soldering irons and small electric motors. In some electrical devices the wattage rating indicates the maximum power the device is designed to use rather than the normal operating power. A 150-watt lamp, for example, uses 150 watts when operated at the specified voltage printed on the bulb. In contrast, a device such as a resistor is not normally given a voltage or a current rating. A resistor is given a power rating in watts and can be operated at any combination of voltage and current as long as the power rating is not exceeded. In most circuits, the actual power used by a resistor is considerably less than the power rating of the resistor because a 50% safety factor is used. For example, if a resistor normally used 2 watts of power, a resistor with a power rating of 3 watts would be used.

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Resistors of the same resistance value are available in different wattage values. Carbon resistors, for example, are commonly made in wattage ratings of 1/8, 1/4, 1/2, 1, and 2 watts. The larger the physical size of a carbon resistor the higher the wattage rating. This is true because a larger surface area of material radiates a greater amount of heat more easily. When resistors with wattage ratings greater than 5 watts are needed, wirewound resistors are used. Wirewound resistors are made in values between 5 and 200 watts. Special types of wirewound resistors are used for power in excess of 200 watts. As with other electrical quantities, prefixes may be attached to the word watt when expressing very large or very small amounts of power. Some of the more common of these are the kilowatt (1,000 watts), the megawatt (1,000,000 watts), and the milliwatt (1/1,000 of a watt).

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Power Conversion and Efficiency The term power consumption is common in the electrical field. It is applied to the use of power in the same sense that gasoline consumption is applied to the use of fuel in an automobile. Another common term is `power conversion'. Power is used by electrical devices and is converted from one form of energy to another. An electrical motor converts electrical energy to mechanical energy. An electric light bulb converts electrical energy into light energy and an electric range converts electrical energy into heat energy. Power used by electrical devices is measured in energy. This practical unit of electrical energy is equal to 1 watt of power used continuously for 1 hour. The term kilowatt hour (kWh) is used more extensively on a daily basis and is equal to 1,000 watt-hours. The efficiency of an electrical device is the ratio of power converted to useful energy divided by the power consumed by the device. This number will always be less than one (1.00) because of the losses in any electrical device. If a device has an efficiency rating of 0.95, it effectively transforms 95 watts into useful energy for every 100 watts of input power. The other 5 watts are lost to heat, or other losses which cannot be used. Calculating the amount of power converted by an electrical device is a simple matter. You need to know the length of time the device is operated and the input power or horsepower rating. Horsepower, a unit of work, is often found as a rating on electrical motors. One horsepower is equal to 746 watts. Example: A 3/4-hp motor operates 8 hours a day. How much power is converted by the motor per month? How many kWh does this represent? Given:

t = 8 hrs x 30 days P=3/4hp Solution: Convert horsepower to watts P = hp x 746 watts P=3/4x746 watts P = 559 watts Convert watts to watt-hours P = work x time P = 559 watts x 8 x 30 P = 134,000 watt-hours per month (NOTE: These figures are rounded to the nearest 1000.) 8-12 TTS Integrated Training System © Copyright 2010

Module 3.8 Power

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To convert to kWh

P=Power in watt-hours 1000 P=

134, 000 in watt-hours

1000

P = 134 kWh If the motor actually uses 137 kWh per month, what is the efficiency of the motor? Given: Power converted = 134 kWh per month

L El

Power used = 137 kWh per month

Solution:

EFF =

Power converted Powerused

I

u

F1 U

U

n iU

fl

EFF =134 kWh per month 137 kWh per month EFF =,978 (Rounded to three figures)

Module 3.8 Power

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Power in a Series Circuit Each of the resistors in a series circuit consumes power which is dissipated in the form of heat. Since this power must come from the source, the total power must be equal to the power consumed by the circuit resistances. In a series circuit the total power is equal to the SUM of the power dissipated by the individual resistors. Total power (PT) is equal to: PT=Py+P2+P3...Pn Example: A series circuit consists of three resistors having values of 5 ohms, 10 ohms, and 15 ohms. Find the total power when 120 volts is applied to the circuit. (See Figure 8.4)

r-n

L1

- ET 120V

R3

155 Figure 8.4 - Solving for total power in a series circuit. Given:

R1 = 5 ohms R2 = 10 ohms

R3 = 15 ohms E = 120 volts Solution: The total resistance is found first.

RT=R1+R2+R3 RT = 5 ohms + 10 ohms + 15 ohms RT = 30 ohms By using the total resistance and the applied voltage, the circuit current is calculated.

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120 volts

L

30 ohms

I = 4 amps

lL

By means of the power formulas, the power can be calculated for each resistor: For R1: P1 =I2xR1

P1 =(4amps)2x5ohms P1 = 80 watts For R2: P2 =I2xR2 P2 = ( 4 amps)2 x 10 ohms P2 = 160 watts For R3: P3 =I2xR3 P3 =(4amps)2x15ohms P3 = 240 watts

For total power: PT =P1 +P2+P3

PT = 80 watts + 160 watts + 240 watts

PT = 480 watts To check the answer, the total power delivered by the source can be calculated: L

Psource = I source x E source P source =

4 amps x 120 volts Psotnrce = 480 watts r-7 f

The total power is equal to the sum of the power used by the individual resistors. Rule for Series DC Circuits The total power in a series circuit is equal to the sum of the individual powers used by each circuit component.

L

8-15 Module 3.8 Power

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Power Transfer and Efficiency Maximum power is transferred from the source to the load when the resistance of the load is equal to the internal resistance of the source. When the load resistance is 5 ohms, matching the source resistance, the maximum power of 500 watts is developed in the load. The efficiency of power transfer (ratio of output power to input power) from the source to the load increases as the load resistance is increased. The efficiency approaches 100 percent as the load resistance approaches a relatively large value compared with that of the source, since less power is lost in the source. The efficiency of power transfer is only 50 percent at the maximum power transfer point (when the load resistance equals the internal resistance of the source). The efficiency of power transfer approaches zero efficiency when the load resistance is relatively small compared with the internal resistance of the source. The problem of a desire for both high efficiency and maximum power transfer is resolved by a compromise between maximum power transfer and high efficiency. Where the amounts of power involved are large and the efficiency is important, the load resistance is made large relative to the source resistance so that the losses are kept small. In this case, the efficiency is high. Where the problem of matching a source to a load is important, as in communications circuits, a strong signal may be more important than a high percentage of efficiency. In such cases, the efficiency of power transfer should be only about 50 percent; however, the power transfer would be the maximum which the source is capable of supplying. L.

You should now understand the basic concepts of series circuits. The principles which have been presented are of lasting importance. Once equipped with a firm understanding of series circuits, you hold the key to an understanding of the parallel circuits to be presented next.

n

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Power in a Parallel Circuit Power computations in a parallel circuit are essentially the same as those used for the series circuit. Since power dissipation in resistors consists of a heat loss, power dissipations are additive regardless of how the resistors are connected in the circuit. The total power is equal to the sum of the power dissipated by the individual resistors. Like the series circuit, the total

power consumed by the parallel circuit is:

PT=PI+P2+,,, Pn Example: Find the total power consumed by the circuit in Figure 8.5.

50y

R1

R2

10c!

25Q

L

Figure 8.5 - Example parallel circuit. Given: R1=10c IRS= 5A R2 = 25 Q

Ii

L

IR2= 2A R3=50Q IR3= 1A Solution:

i Module 3.8 Power

8-17

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

P=I2R PR1 = (IR1 ) 2 x R1 PR1=(5A)2x 100 PR; Z30rv0'tii

PR2=(IR2)2xR2

PR2=(2A)2x25Q

PR2 = 10OW

PR3=(IR3)2xR3 PR3=(1A)2x 500 PR3 = SOW PT = PRI + PR2 +PR3

PT = 25OW +

SOW

PT = 40OW

Solution:

PT=ESM.M.F.

Current (I)

Ampere (A)

>magnetic flux (t)Weber (Wb)

Resistance (R) Ohm (Q) Conductance (C)Siemens (S) E.M.F. = I x R

Quantity

>Reluctance (R) >Permeance (A) >

Table 10.2 - Comparison of electrical and magnetic terms

Unit Ampere-turn (At) Ampere-turn/Weber (At/Wb) Webers/Ampere-turn

(Wb/At)

M.M.F. =(D x R

The magnetic circuit differs from the electric circuit in the following important respects:a)

The current in the electric circuit is confined to a defined path by insulating material on the circuit conductors; the flux in the magnetic circuit cannot be restrained in this manner, since there is no known "insulator" for magnetic flux (not even a vacuum) - the flux can only be "lured" into the desired path by making the latter of low reluctance.

b)

The resistance of an electric circuit is almost constant, the reluctance of a magnetic circuit, on the other hand, varies over a wide range by reason of changes in permeability which decreases rapidly as saturation point is approached.

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Module 3.10 Magnetism

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Hysteresis Saturation point in a material being magnetized is reached when an increase in magnetic field strength produces only a small increase in flux density. At this stage all the magnetic domains (groups of atoms with the electron orbits aligned, which can be thought of as little magnets) in the material are aligned and the increase in flux density is only that which would occur in free

space. The effect is clearly shown by the graphs of B/H curves for a number of ferromagnetic materials. 20

r

t•6 TI

Nrt

I.

1.4

0

`1

L

1.2

rt 08

1 1000

2000

300{)

4000

5000

1

6000

H ornpre/r elre

Figure 10.29 - B-H Curves These curves have been drawn on the assumption that the iron had no trace of magnetism at

the commencement. If, however, the iron is already magnetized to some extent, the new L

magnetism may aid or oppose that which exists. If it opposes the existing magnetism it is found that the change in flux density lags behind the magnetic field strength. This effect is called "Hysteresis" and it is usually studied by considering a complete cycle of magnetism, which entails magnetizing in one direction of polarity, then in the opposite direction and finally in the initial direction again. A typical graph for a sample of iron is shown, the arrows indicating the direction of magnetism from the commencement. It should be noted that in this case the iron has been magnetized to saturation in both directions.

L Module 3.10 Magnetism

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Figure 10.30 - Hysteresis Loop The graph, known as a "Hysteresis Loop", makes clear the magnetic properties of the material concerned. The intercept on the B axis (r) is the "Residual" flux density when H has been reduced to zero and is called the "Remanence" of the material. The intercept on the H axis (c) is the "Coercive Force" required to reduce the residual flux to zero and is called the coercivity of the material. The three properties of a magnetic material, permeability, remanence and coercivity, indicate its usefulness for a particular application. For example, a suitable material for permanent magnets would have high coercivity and high remanence; a suitable material for electromagnets would have high permeability but low remanence and low coercivity. Typical Hysteresis Loops are:B

H hard steel Figure 10.31 - Hysteresis Loops for Soft Iron and Hard Steel The area enclose by the Hysteresis Loop is a measure of the energy wasted (converted to heat) in magnetizing and demagnetizing a material. The wasted energy is known as "Hysteresis Loss".

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Summary of Magnetism Terms and Symbols

Li Li r

Name

Description

Magnetic Flux

A measure of quantity of magnetism, taking into account the strength and the extent of a magnetic

0

Weber (Wb)

B

Teslas (T)

field. The amount of magnetic flux through a unit area Magnetic Flux Density taken perpendicular to the direction of the magnetic flux. Also called magnetic induction. Calculated by magnetic flux divided by cross sectional area A vector quantity indicating the ability of a magnetic Magnetic magnetic permeability of the space where the field exists. It is measured in amperes per meter. Also called magnetic intensity.

force)

Magneto Motive Force Any physical cause that produces magnetic flux

L

SI Unit

Reluctance

AmpereHturns/Metre (At/m)

MMFAmpere-turns ( At) R

Ampere( (At/Wb)

The constant value No is known as the magnetic constant or the permeability of vacuum, and has the exact or definedvalue Po = 4rrx10-7 H/m.

p

Henries/metre (H/m)

The degree to Wiich a material admits a flow of magnetism. Theinverse of reluctance.

A

Webers per Ampere-turn' (Wb/A)

The force whichin iron or steel produces a slowness difficulty in irrparting magnetism to it, and also interposes an olstacle to the return of a bar to its natural state whn active magnetism has ceased. A form of Magnets field strength.

H

Amperes/Metre (gym)

The magnetic flex density remaining in a material, especially a feromagnetic material, after removal of the magnetizincfield. Good permanent magnets have a high depee of remanence. Also called retentivity, or rsidual magnetism.

BTeslas (T)

A measure of the opposition to magnetic flux, analogous to electric resistance. The ability of a substance to allow magnetism to

Permeance

y rL

Remanence

Hysteresis

The magnetizabn of a material such as iron depends not ory on the magnetic field it is exposed to but on prevics exposures to magnetic fields.

C

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TTS Integrated Training System Module 3 Licence Category B1/B2

IL

Electrical Fundamentals 3.11 Inductance/Inductor

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Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Ltd.

Knowledge Levels - Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows:

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

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LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • •

A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents I

L

L

Module 3.11 Inductance/Inductor Characteristics of Inductance Electromotive Force (EMF) Self-Inductance Factors Affecting Coil Inductance Unit of Inductance Energy in an Inductor Growth and Decay of Current in an LR Series Circuit UR Time Constant Power Loss in an Inductor Mutual Inductance Adding Inductors Applications of Inductors

Inductor construction

5 5 5 7 10 14 14 15 17 18 20 22 24

25

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L

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Use andror disclosure is

Module 3.11 Inductance/Inductor

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Module 3.11 Enabling Objectives n

Objective EASA 66 ReferenceLevel Inductance/Inductor Faraday's Law Action of inducing a voltage in a conductor moving in a magnetic field Induction principles Effects of the following on the magnitude of an induced voltage: magnetic field strength, rate of change of flux, number of conductor turns Mutual induction The effect the rate of change of primary current and mutual inductance has on inducted voltage Factors affecting mutual inductance: number of turns in coil, physical size of coil, permeability of coil, position of coils with respect to each other Lenz's Law and polarity determining rules Back EMF, self induction Saturation point Principle uses of inductors

n i.

3.11

2

l

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L

Module 3.11 Inductance/Inductor Characteristics of Inductance

11

Inductance is the characteristic of an electrical circuit that opposes the starting, stopping, or a change in value of current. The above statement is of such importance to the study of inductance that it bears repeating. Inductance is the characteristic of an electrical conductor that opposes change in current. The symbol for inductance is L and the basic unit of inductance is the HENRY (H). One Henry is equal to the inductance required to induce one volt in an inductor by a change of current of one ampere per second.

r-

You do not have to look far to find a physical analogy of inductance. Anyone who has ever had

to push a heavy load (wheelbarrow, car, etc.) is aware that it takes more work to start the load moving than it does to keep it moving. Once the load is moving, it is easier to keep the load moving than to stop it again. This is because the load possesses the property of inertia. Inertia is the characteristic of mass which opposes a change in velocity. Inductance has the same effect on current in an electrical circuit as inertia has on the movement of a mechanical object. It requires more energy to start or stop current than it does to keep it flowing. L r-,

F-I

Electromotive Force (EMF) You have learned that an electromotive force is developed whenever there is relative motion between a magnetic field and a conductor.

{1

L r-1

3.

L

Electromotive force is a difference of potential or voltage which exists between two points in an electrical circuit. In generators and inductors the EMF is developed by the action between the magnetic field and the electrons in a conductor. This is shown in Figure 11.1, 11.2, 11.3 and 11.4.

i

L

{l

Figure 11.1 - Generation of an EMF in an electrical conductor

Module 3.11 Inductance/Inductor

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When a magnetic field moves through a stationary metallic conductor, electrons are dislodged from their orbits. The electrons move in a direction determined by the movement of the magnetic lines of flux. This is shown below:

IDIRECTION OF I

�---MAGNETIC FLUX LINE

MOVEMENT of FLUX

Figure 11.2 - Generation of an EMF in an electrical conductor The electrons move from one area of the conductor into another area. The area that the electrons moved from has fewer negative charges (electrons) and becomes positively charged. The area the electrons move into becomes negatively charged. This is shown below:

1

I

4

1

(+±\e\e, g11lee ++18\01I e 10 e I

+

1

l

1

1 I

r

Figure 11.3 - Generation of an EMF in an electrical conductor. The difference between the charges in the conductor is equal to a difference of potential (or voltage). This voltage caused by the moving magnetic field is called electromotive force (EMF). In simple terms, the action of a moving magnetic field on a conductor can be compared to the action of a broom. Consider the moving magnetic field to be a moving broom. As the magnetic broom moves along (through) the conductor, it gathers up and pushes electrons before it, as shown below:

I

-r

4

eee

+ +

+

Gpe 0

Figure 11.4 - Generation of an EMF in an electrical conductor

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;

The area from which electrons are moved becomes positively charged, while the area into

L which electrons are moved becomes negatively charged. The potential difference between these two areas is the electromotive force or EMF. U

Li Li L

Self-Inductance Even a perfectly straight length of conductor has some inductance. As you know, current in a conductor produces a magnetic field surrounding the conductor. When the current changes, the magnetic field changes. This causes relative motion between the magnetic field and the conductor, and an electromotive force (EMF) is induced in the conductor. This EMF is called a self-induced EMF because it is induced in the conductor carrying the current. The EMF produced by this moving magnetic field is also referred to as back electromotive force (backEMF). The polarity of the counter electromotive force is in the opposite direction to the applied voltage of the conductor. The overall effect will be to oppose a change in current magnitude. This effect is summarized by Lenz's law which states that: The induced EMF in any circuit is always in a direction to oppose the effect that produced it. If the shape of the conductor is changed to form a series of loops, then the electromagnetic field around each portion of the conductor cuts across some other portion of the same conductor. This is shown in its simplest form in Figure 11.5.

fl

U L U

Figure 11.6 - A simple inductor, with a variable current

U

A length of conductor is looped so that two portions of the conductor lie next to each other. When the conventional current is flowing in the conductor it produces a magnetic field around all portions of the conductor. For simplicity, the magnetic field (expanding lines of flux) is shown in a single plane that is perpendicular to the loops. With increasing current, the flux field expands outward from the centre of the inductor, cutting across the loops of the conductor. This results in an induced EMF in the loops. Note that the induced EMF is in the opposite direction to (in opposition to) the battery current and voltage, as stated in Lenz's law.

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motion or force F I magnetic field 5 J

induced current !

Figure 11.6 - Self-inductance IA

The direction of this induced voltage may be determined by applying the right-hand rule for generators (see Figure 11.5). This rule is applied to one of the loops of the conductor. This rule states that if you point the thuMb of your right hand in the direction of relative Motion of the conductor and your First finger in the direction of the magnetic Field, your seCond finger, extended as shown, will now indicate the direction of the induced conventional Current which will generate the induced voltage (back-EMF) as shown.

n

In Figure 11,5, if the current were to be reduced, the flux field would be collapsing. Applying the right-hand rule in this case shows that the reversal of flux movement has caused a reversal in the direction of the induced voltage. The induced voltage is now in the same direction as the battery voltage. The most important thing for you to note is that the self-induced voltage opposes both changes in current. That is, when the current is increasing, this voltage tries to delay the build-up of current by opposing the battery voltage. When the current is decreasing, it tries to keep the current flowing in the same direction by aiding the battery voltage.

7

--,

Thus, from the above explanation, you can see that when a current is building up it produces an expanding magnetic field. This field induces an EMF in the direction opposite to the actual flow of current. This induced EMF opposes the growth of the current and the growth of the magnetic field. If the increasing current had not set up a magnetic field, there would have been no opposition to its growth. The whole reaction, or opposition, is caused by the creation or collapse of the magnetic field, the lines of which as they expand or contract, cut across the conductor and develop the back-EMF.

n

Since all circuits have conductors in them, you can assume that all circuits have inductance. However, inductance has its greatest effect only when there is a change in current. Inductance does not oppose current, only a change in current. Where current is constantly changing as in an AC circuit, inductance has more effect. To increase the property of inductance, the conductor can be formed into a greater number of loops or coils. A coil is also called an inductor. Figure 11.7 shows a conductor formed into a coil. Current through one loop produces a magnetic field that encircles the loop in the direction as shown in Figure 11.7 (A). As current increases, the magnetic field expands and cuts all the

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loops as shown in Figure 11.7 (B). The current in each loop affects all other loops. The field

cutting the other loop has the effect of increasing the opposition to a current change.

(B) Figure 11.7 - Inductance.

U

Inductors are classified according to core type. The core is the centre of the inductor just as the core of an apple is the centre of an apple. The inductor is made by forming a coil of wire around a core. The core material is normally one of two basic types: soft-iron or air. An iron-core inductor and its schematic symbol (which is represented with lines across the top of it to indicate the presence of an iron core) are shown, in Figure 11.8 (A). The air-core inductor may be nothing more than a coil of wire, but it is usually a coil formed around a hollow form of some nonmagnetic material such as cardboard. This material serves no purpose other than to hold the shape of the coil. An air-core inductor and its schematic symbol are shown in Figure 11.8 (B).

L

INDUCTOR, IRON CORE

INDUCTOR, AIR CARE

(A) L

(B)

Figure 11.8 - Inductor types and schematic symbols.

Module 3.11 Inductance/Inductor

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Factors Affecting Coil Inductance There are several physical factors which affect the inductance of a coil. They include the number of turns in the coil, the diameter of the coil, the coil length, the type of material used in the core, and the number of layers of winding in the coils. Inductance depends entirely upon the physical construction of the circuit, and can only be measured with special laboratory instruments. Of the factors mentioned, consider first how the number of turns affects the inductance of a coil. Figure 11.9 shows two coils. Coil (A) has two turns and coil (B) has four turns. In coil (A), the flux field set up by one loop cuts one other loop. In coil (B), the flux field set up by one loop cuts three other loops. Doubling the number of turns in the coil will produce a field twice as strong, if the same current is used. A field twice as strong, cutting twice the number of turns, will induce four times the voltage. Therefore, it can be

said that the inductance varies as the square of the number of turns.

(A)

(B)

Figure 11.9 - Inductance factor (turns). The second factor is the coil diameter. In Figure 11.10 you can see that the coil in view B has twice the diameter of coil view A. Physically, it requires more wire to construct a coil of large diameter than one of small diameter with an equal number of turns. Therefore, more lines of force exist to induce a counter EMF in the coil with the larger diameter. Actually, the inductance of a coil increases directly as the cross-sectional area of the core increases. Recall the formula for the area of a circle: A = Trr2. Doubling the radius of a coil increases the inductance by a factor of four. 20

(A)

(B)

Figure 11.10 - Inductance factor (diameter).

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The third factor that affects the inductance of a coil is the length of the coil. Figure 11.11 shows two examples of coil spacings. Coil (A) has three turns, rather widely spaced, making a relatively long coil. A coil of this type has few flux linkages, due to the greater distance between each turn. Therefore, coil (A) has a relatively low inductance. Coil (B) has closely spaced turns, making a relatively short coil. This close spacing increases the flux linkage, increasing the inductance of the coil. Doubling the length of a coil while keeping the same number of turns halves the value of inductance.

--- 2 L

U

U (A)

(B)

Figure 11.11 - Inductance factor (coil length). Closely wound The fourth physical factor is the type of core material used with the coil. Figure 11.12 shows two U F U

coils: Coil (A) with an air core, and coil (B) with a soft-iron core. The magnetic core of coil (B) is a better path for magnetic lines of force than is the nonmagnetic core of coil (A). The soft-iron magnetic core's high permeability has less reluctance to the magnetic flux, resulting in more magnetic lines of force. This increase in the magnetic lines of force increases the number of lines of force cutting each loop of the coil, thus increasing the inductance of the coil. It should now be apparent that the inductance of a coil increases directly as the permeability of the core material increases.

Module 3.11 Inductance/Inductor

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(A) AIR CORE

(B) SOFT-IRON CORE

Figure 11.12 - Inductance factor (core material). SOFT-IRON CORE Another way of increasing the inductance is to wind the coil in layers. Figure 11.13 shows three cores with different amounts of layering. The coil in Figure 11.13 (A) is a poor inductor compared to the others in the figure because its turns are widely spaced and there is no layering. The flux movement, indicated by the dashed arrows, does not link effectively because there is only one layer of turns. A more inductive coil is shown in Figure 11.13 (B). The turns are closely spaced and the wire has been wound in two layers. The two layers link each other with a greater number of flux loops during all flux movements. Note that nearly all the turns, such as X, are next to four other turns (shaded). This causes the flux linkage to be increased. n

I

(A)

(B)

(C) Figure 11.13 - Coils of various inductances.

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A coil can be made still more inductive by winding it in three layers, as shown in Figure 11.13

(C). The increased number of layers (cross-sectional area) improves flux linkage even more.

Note that some turns, such as Y, lie directly next to six other turns (shaded). In actual practice, layering can continue on through many more layers. The important fact to remember, however, is that the inductance of the coil increases with each layer added.

L

Where,

�f

= Inductance of coil in Henrys N = Number of turns in wire coil (straight wire = 1) L

�1= Permeability of core material (absolute, not relative) jLt = Relative permeability, dimensionless

1 for air)

fLo = 1.26 x 10 _g T-mfAt permeability of free space A = Area of coil in square meters = 7tr'

I

Average length of coil in meters Figure 11.14 - The formula for the inductance of a coil

As you have seen, several factors can affect the inductance of a coil, and all of these factors are variable. They are all physical factors. Many differently constructed coils can have the same inductance. The important information to remember, however, is that inductance is dependent upon the degree of linkage between the wire conductor(s) and the electromagnetic field. In a straight length of conductor, there is very little flux linkage between one part of the conductor and another. Therefore, its inductance is extremely small. It was shown that conductors become much more inductive when they are wound into coils. This is true because there is maximum flux linkage between the conductor turns, which lie side by side in the coil. Li It should be noted that he above formula is just an approximation, and is valid only within a normal working range of current. Beyond a certain current level, the core of the inductor becomes saturated and the permeability of the core medium changes. Thus the inductance of the inductor cannot be predicted with as much ease as the formula may suggest.

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Unit of Inductance As stated before, the basic unit of inductance (L) is the Henry (H), named after Joseph Henry, the co-discoverer with Faraday of the principle of electromagnetic induction. An inductor has an inductance of 1 Henry if an EMF of 1 volt is induced in the inductor when the current through the inductor is changing at the rate of 1 ampere per second. The relationship between the induced voltage, the inductance, and the rate of change of current with respect to time is stated mathematically as:

i

f

Eind=L lesi At where Eind is the induced EMF in volts; L is the inductance in henrys; and Al is the change in current in amperes occurring in At seconds. The symbol A (Greek letter delta), means "a change in ... .". The Henry is a large unit of inductance and is used with relatively large inductors. With small inductors, the millihenry is used. (A millihenry is equal to 1 x 10-3 Henry, and one Henry is equal to 1,000 millihenrys.) For still smaller inductors the unit of inductance is the microhenry (pH). (A pH = 1 x 10-6H, and one Henry is equal to 1,000,000 microhenrys.)

Energy in an Inductor The energy (measured in joules, in SI) stored by an inductor is equal to the amount of work required to establish the current through the inductor, and therefore the magnetic field. This is given by:

&9 tored ==

1

2 LP

where L is inductance and I is the current flowing through the inductor. n

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r7

L Growth and Decay of Current in an LR Series Circuit n

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When a battery is connected across a "pure" inductance, the current builds up to its final value at a rate determined by the battery voltage and the internal resistance of the battery. The current build-up is gradual because of the counter EMF generated by the self-inductance of the coil. When the current starts to flow, the magnetic lines of force move outward from the coil. These lines cut the turns of wire on the inductor and build up a counter EMF that opposes the EMF of the battery. This opposition causes a delay in the time it takes the current to build up to a steady value. When the battery is disconnected, the lines of force collapse. Again these lines cut the turns of the inductor and build up an EMF that tends to prolong the flow of current. A voltage divider containing resistance and inductance may be connected in a circuit by means of a special switch, as shown in Figure 11.15 (A). Such a series arrangement is called an LR series circuit. S. CLO$ES THE

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Figure 11.15 - Growth and decay of current in an LR series circuit.

1-fl L Modu[e 3.11 Inductance/inductor

11-15

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When switch S1 is closed (as shown), a voltage E. appears across the voltage divider. At this instant the current will attempt to increase to its maximum value. However, this instantaneous current change causes coil L to produce a back-EMF, which is opposite in polarity and almost equal to the EMF of the source. This back EMF opposes the rapid current change. Figure 11.15 (B) shows that at the instant switch S1 is closed, there is no measurable growth current (Ig), a

minimum voltage drop is across resistor R, and maximum voltage exists across inductor L.

As current starts to flow, a voltage (eR) appears across R, and the voltage across the inductor is reduced by the same amount. The fact that the voltage across the inductor (L) is reduced means that the growth current (ig) is increased and consequently eR is increased. Figure 11.15 (B) shows that the voltage across the inductor (eL) finally becomes zero when the growth current (i0) stops increasing, while the voltage across the resistor (eR) builds up to a value equal to the source voltage (Es). Electrical inductance is like mechanical inertia, and the growth of current in an inductive circuit can be likened to the acceleration of a boat on the surface of the water. The boat does not move at the instant a constant force is applied to it. At this instant all the applied force is used to overcome the inertia of the boat. Once the inertia is overcome the boat will start to move. After a while, the speed of the boat reaches its maximum value and the applied force is used up in overcoming the friction of the water against the hull. When the battery switch (Si) in the LR circuit of Figure 11.15 (A) is closed, the rate of the current increase is maximum in the inductive circuit. At this instant all the battery voltage is used in overcoming the EMF of self-induction which is a maximum because the rate of change of current is maximum. Thus the battery voltage is equal to the drop across the inductor and the voltage across the resistor is zero. As time goes on more of the battery voltage appears across the resistor and less across the inductor. The rate of change of current is less and the induced EMF is less. As the steady-state condition of the current is approached, the drop across the inductor approaches zero and all of the battery voltage is "dropped" across the resistance of the circuit. Thus the voltages across the inductor and the resistor change in magnitude during the period of growth of current the same way the force applied to the boat divides itself between the effects of inertia and friction. In both examples, the force is developed first across the inertia/inductive effect and finally across the friction/resistive effect.

Figure 11.15 (C) shows that when switch S2 is closed (source voltage Es removed from the circuit), the flux that has been established around the inductor (L) collapses through the windings. This induces a voltage el- in the inductor that has a polarity opposite to Es and is essentially equal to Es in. magnitude. The induced voltage causes decay current (id) to flow in resistor R in the same direction in which current was flowing originally (when S 1 was closed). A voltage (eR) that is initially equal to source voltage (Es) is developed across R. The voltage across the resistor (eR) rapidly falls to zero as the voltage across the inductor (et_) falls to zero due to the collapsing flux. Just as the example of the boat was used to explain the growth of current in a circuit, it can also be used to explain the decay of current in a circuit. When the force applied to the boat is removed, the boat still continues to move through the water for a while, eventually coming to a stop. This is because energy was being stored in the inertia of the moving boat. After a period of 11-16 TTS Integrated Training System Use and/or disclosure is governed by the statement

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time the friction of the water overcomes the inertia of the boat, and the boat stops moving. Just

as inertia of the boat stored energy, the magnetic field of an inductor stores energy. Because of i iC U

this, even when the power source is removed, the stored energy of the magnetic field of the inductor tends to keep current flowing in the circuit until the magnetic field collapse.

UR Time Constant The UR time constant is a valuable tool for use in determining the time required for current in

j

L

U

an inductor to reach a specific value. As shown in Figure 11.16, one L/R time constant is the time required for the current in an inductor to increase to 63 percent (actually 63.2%) of the maximum current. Each time constant is equal to the time required for the current to increase by 63.2 percent of the difference in value between the current flowing in the inductor and the maximum current. Maximum current flows in the inductor after five L/R time constants are completed. The following example should clear up any confusion about time constants. Assume that maximum current in an LR circuit is 10 amperes. As you know, when the circuit is energized, it takes time for the current to go from zero to 10 amperes. When the first time constant is completed, the current in the circuit is equal to 63.2% of 10 amperes. Thus the amplitude of current at the end of 1 time constant is 6.32 amperes. �---� --- GROWTH 10

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Figure 11.16 - L/R time constant.

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In a similar way to the voltage increase and decay in a capacitor, the current increase in an inductor is logarithmic, and the current decay is an exponential decay. During the second time constant, current again increases by 63.2% (0.632) of the difference in

value between the current flowing in the inductor and the maximum current. This difference is ,-,

T

f

10 amperes minus 6.32 amperes and equals 3.68 amperes; 63.2% of 3.68 amperes is 2.32 amperes. This increase in current during the second time constant is added to that of the first time constant. Thus, upon completion of the second time constant, the amount of current in the LR circuit is 6.32 amperes + 2.32 amperes = 8.64 amperes. During the third constant, current again increases:

LI

Module 3.11 Inductance/Inductor

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10 amperes - 8.64 amperes =136 ampere s 1.36 amperes x .632 = 0.860 ampere 8,64 amperes +0,860ampere = 9.50amperes During the fourth time constant, current again increases: 10 amperes - 9,50 amperes = 0.5 ampere

n

0.5 ampere x .632 = 0.316 ampere 9.50 amperes + 0.316ampere = 9.82amperes

l

During the fifth time constant, current increases as before: 10amperes - 9.82amperes = 0.18ampere 0,18 ampere x ,632 = 0,114 ampere 9.82 amperes + ,114 ampere = 9.93 amperes Thus, the current at the end of the fifth time constant is almost equal to 10.0 amperes, the maximum current. For all practical purposes the slight difference in value can be ignored. When an LR circuit is de-energized, the circuit current decreases (decays) to zero in five time constants at the same rate that it previously increased. If the growth and decay of current in an LR circuit are plotted on a graph, the curve appears as shown in Figure 11.16. Notice that current increases and decays at the same rate in five time constants. The value of the time constant in seconds is equal to the inductance in henrys divided by the circuit resistance in ohms. The formula used to calculate one L/R time constant is:

Time Constant (TQ in seconds = L (in henrys )

R(inohms) n

Power Loss in an Inductor Since an inductor (coil) consists of a number of turns of wire, and since all wire has some resistance, every inductor has a certain amount of resistance. Normally this resistance is small. It is usually neglected in solving various types of ac circuit problems because the reactance of the inductor (the opposition to alternating current, which will be discussed later) is so much greater than the resistance that the resistance has a negligible effect on the current. However, since some inductors are designed to carry relatively large amounts of current, considerable power can be dissipated in the inductor even though the amount of resistance in the inductor is small. This power is wasted power and is called copper loss. The copper loss of 11-18 TTS Integrated Training System Use and/or disclosure is governed by the statement

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an inductor can be calculated by multiplying the square of the current in the inductor by the resistance of the winding (12R).

L

In addition to copper loss, an iron-core coil (inductor) has two iron losses. These are called hysteresis loss and eddy-current loss. Hysteresis loss is due to power that is consumed in reversing the magnetic field of the inductor core each time the direction of current in the inductor

changes.

Eddy-current loss is due to heating of the core by circulating currents that are induced in the iron core by the magnetic field around the turns of the coil. These currents are called eddy currents and circulate within the iron core only. All these losses dissipate power in the form of heat. Since this power cannot be returned to the electrical circuit, it is lost power.

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Module 3.11 Inductance/Inductor

11-19

;. j

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Mutual Inductance Whenever two coils are located so that the flux from one coil links with the turns of the other coil, a change of flux in one coil causes an EMF to be induced in the other coil. This allows the energy from one coil to be transferred or coupled to the other coil. The two coils are said to be coupled or linked by the property of mutual inductance (M). The amount of mutual inductance depends on the relative positions of the two coils. This is shown in Figure 11.17. If the coils are separated a considerable distance, the amount of flux common to both coils is small and the mutual inductance is low. Conversely, if the coils are close together so that nearly all the flux of one coil links the turns of the other, the mutual inductance is high. The mutual inductance can be increased greatly by mounting the coils on a common iron core.

Li

n

L,

(A) INDUCTORS CLOSE -- LARGE M

(B) INDUCTORS FAR APART - SMALL M

(C) INDUCTOR AXES PERPENDICULAR- NO M Figure 11.17 - The effect of position of coils on mutual inductance (M).

71

Two coils are placed close together as shown in Figure 11.18. Coil 1 is connected to a battery through switch S, and coil 2 is connected to an ammeter (A). When switch S is closed as in Figure 11.18 (A), the current that flows in coil 1 sets up a magnetic field that links with coil 2, causing an induced voltage in coil 2 and a momentary deflection of the ammeter. When the current in coil 1 reaches a steady value, the ammeter returns to zero. If switch S is now opened as in Figure 11.18 (B), the ammeter (A) deflects momentarily in the opposite direction, indicating a momentary flow of current in the opposite direction in coil 2, This current in coil 2 is produced by the collapsing magnetic field of coil 1.

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n

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Figure 11.18 - Mutual inductance. Factors Affecting Mutual Inductance The mutual inductance of two adjacent coils is dependent upon the physical dimensions of the two coils, the number of turns in each coil, the distance between the two coils, the relative positions of the axes of the two coils, and the permeability of the cores. The coefficient of coupling between two coils is equal to the ratio of the flux cutting one coil to the flux originated in the other coil. If the two coils are so positioned with respect to each other so that all of the flux of one coil cuts all of the turns of the other, the coils are said to have a unity coefficient of coupling. It is never exactly equal to unity (1), but it approaches this value in certain types of coupling devices. If all of the flux produced by one coil cuts only half the turns of the other coil, the coefficient of coupling is 0.5. The coefficient of coupling is designated by the letter K. The mutual inductance between two coils, L1 and L2, is expressed in terms of the inductance of each coil and the coefficient of coupling K. As a formula:

M=K L1L2 where:

LU

M= Mutual inductance in henrys

K= Coefficient of coupling L1, L2 = Inductance of coil in henrys

n U

Module 3.11 Inductance/Inductor

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Example problem: One 1 0-H coil and one 20-H coil are connected in series and are physically close enough to each other so that their coefficient of coupling is 0.5. What is the mutual inductance between the coils? Use the formula: M=K L1L2 M = 0.5 (10H)(20H) M = 0.5 200 H M = 0.5 x 14.14H M = 7.07H

Adding Inductors Series Inductors without Magnetic Coupling When inductors are well shielded or are located far enough apart from one another, the effect of mutual inductance is negligible. If there is no mutual inductance (magnetic coupling) and the inductors are connected in series, the total inductance is equal to the sum of the individual inductances. As a formula:

r-,

LT =L1 +L2+L3+,.L-n

n

where LT is the total inductance; L 1, L2, L3 are the inductances of L1, L2, L3i and Ln means that any number (n) of inductors may be used. The inductances of inductors in series are added together like the resistances of resistors in series. Series Inductors With Magnetic Coupling When two inductors in series are so arranged that the field of one links the other, the combined inductance is determined as follows: LT = L1+L2 ±2M

where:

L T = The total inductan ce L 1.L 2 = The inductances of L1.L2

M = The mutu al induc tanc e between the two inductors The plus sign is used with M when the magnetic fields of the two inductors are aiding each other, as shown in Figure 11.1 9.The minus sign is used with M when the magnetic field of the two inductors oppose each other, as shown in Figure 11.20. The factor 2M accounts for the influence of L1 on L2 and L2 on L1.

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7,

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Figure 11.19 - Series inductors with aiding fields. ,

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Figure 11.20 - Series inductors with opposing fields. Example problem: A 10-H coil is connected in series with a 5-H coil so the fields aid each other. Their mutual inductance is 7 H. What is the combined inductance of the coils? Usetheformula:

U

LT =L1+L2+2M

LT=10H+5H+2(7H) LT=29H Module 3.11 Inductance/Inductor

11-23

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

J

Parallel Inductors without Coupling The total inductance (LT) of inductors in parallel is calculated in the same manner that the total resistance of resistors in parallel is calculated, provided the coefficient of coupling between the coils is zero. Expressed mathematically:

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1 1 =1 +1+1... + LT L1 L2 L3 LM Note that the formula for parallel inductors with coupling is not provided. This is because it is not possible for magnetic coupling between parallel conductors to take place, as it would require lines of flux from one conductor to cross over the lines of flux from the other conductor, and this is contrary to the rules that predict the behaviour of lines of flux. Whilst there may be some

disruption due to the distortion of the lines of flux when influenced by neighbouring inductors, there are too many variables to take into account, and the effect on inductance is so minimal, that a formula is neither possible nor necessary. n L.J

Applications of Inductors Inductors are used extensively in analogue circuits and signal processing. Inductors in conjunction with capacitors and other components form tuned circuits which can emphasize or filter out specific signal frequencies. This can range from the use of large inductors as chokes in power supplies, which in conjunction with filter capacitors remove residual hum or other fluctuations from the direct current output, to such small inductances as generated by a ferrite bead or torus around a cable to prevent radio frequency interference from being transmitted down the wire. Smaller inductor/capacitor combinations provide tuned circuits used in radio reception and broadcasting, for instance.

L.- J

Figure 11.21 - A choke with two 47mH windings, such as might be found in a power supply. Two (or more) inductors which have coupled magnetic flux form a transformer, which is a fundamental component of every electric utility power grid. The efficiency of a transformer decreases as the frequency increases but size can be decreased as well; for this reason, aircraft used 400 hertz alternating current rather than the usual 50 or 60 hertz, allowing a great savings in weight from the use of smaller transformers.

n

An inductor is used as the energy storage device in some switchmode power supplies. The inductor is energized for a specific fraction of the regulator's switching frequency, and deenergized for the remainder of the cycle. This energy transfer ratio determines the input-voltage to output-voltage ratio. This XL is used in complement with an active semiconductor device to maintain very accurate voltage control. Inductors are also employed in electrical transmission systems, where they are used to intentionally depress system voltages or limit fault current. In this field, they are more commonly referred to as reactors.

71

n

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As inductors tend to be larger and heavier than other components, their use has been reduced in modern equipment; solid state switching power supplies eliminate large transformers, for instance, and circuits are designed to use only small inductors, if any; larger values are simulated by use of gyrator circuits.

Inductor construction An inductor is usually constructed as a coil of conducting material, typically copper wire,

wrapped around a core either of air or of ferromagnetic material. Core materials with a higher permeability than air confine the magnetic

field closely to the inductor, thereby increasing the inductance. 4 in

Li

Inductors come in many shapes. Most are constructed as enamel coated wire wrapped around a ferrite bobbin with wire exposed on the outside, while some enclose the wire completely in ferrite and are

called shielded . Some Inductors nave an adjustable core, which b

enables changing of the inductance. Inductors used to block very lg requencles are some lrrtes ma a wl a wire passing roug a t th h ferrite cylinder or bead. Figure 11.22 - Inductors. Major scale in centimetres

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Small inductors can be etched directly onto a printed circuit board by laying out the trace in a spiral pattern. Small value inductors can also be built on integrated circuits using the same processes that are used to make transistors. In these cases, aluminium interconnect is typically used as the conducting material. However, practical constraints make it far more common to use a circuit called a "gyrator" which uses a capacitor and active components to behave similarly to an inductor.

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Figure 11.23 - Various forms of iron cored inductors

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u Module 3.11 Inductance/Inductor

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Intentionally Blank

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fi 11

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Module 3 Licence Category B1/B2 Electrical Fundamentals 3.12 DC Motor/Generator Theory

L U

U-

Li Module 3.12 DC Motor/Generator Theory

12-1

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Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Ltd.

Knowledge Levels - Category A, B1, B2 and C Aircraft

Maintenance Licence

Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or

3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows:

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

n n

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical

examples.

• The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • A detailed knowledge of the theoretical and practical aspects of the subject. • A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive

manner.

Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects.

•

The applicant should be able to give a detailed description of the subject using theoretical fundamentals

and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's

instructions.

• The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents

iL

Fl

L Li L'

Module 3.12 DC Motor/Generator Theory

5

DC Generators Introduction Principles of Operation Commutation Motor Reaction in a Generator Armature Losses Construction Features of D.C. Generators Types of D.C. Generators Three-Wire Generators Armature Reaction Generator Ratings D.C. Generator Maintenance Field Excitation

5 5 5 12 13 14 16 22 26 27 29 30 32

DC Motors Introduction Principles of Operation Basic D.C. Motor D.C. Motor Construction Armature Reaction Types of D.C. Motors Back-EMF Types of Duty Reversing Motor Direction Motor Speed Energy Losses in D.C. Motors Inspection and Maintenance of D.C. Motors

33 33 33 36 39 40 42 45 46 46 48 49 51

Starter-Generator Systems

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Module 3.12 Enabling Objectives Objective

EASA 66 Reference

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DC Motor/Generator Theory Basic motor and generator theory Construction and purpose of components in DC generator Operation of, and factors affecting output and direction of current flow in DC generators Operation of, and factors affecting output power, torque, speed and direction of rotation of DC motors Series wound, shunt wound and compound motors

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Starter Generator construction

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Module 3.12 DC Motor/Generator Theory

L DC Generators Introduction Energy for the operation of most electrical equipment in an aeroplane depends upon the elec-

trical energy supplied by a generator. A generator is any machine which converts mechanical energy into electrical energy by electromagnetic induction. A generator designed to produce alternating-current energy is called an AC generator, or alternator, a generator which produces direct-current energy is called a DC generator. Both types operate by inducing an AC voltage in coils by varying the amount and direction of the magnetic flux cutting through the coil For aeroplanes equipped with direct-current electrical systems, the DC generator is the regular source of electrical energy. One or more DC generators, driven by the engine, supply electrical energy for the operation of all units in the electrical system, as well as energy for charging the

battery. The number of generators used is determined by the power requirement of a particular

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aeroplane. In most cases, only one generator is driven by each engine, but in some large aeroplanes, two generators are driven by a single engine. Aircraft equipped with alternatingcurrent systems use electrical energy supplied by AC generators, also called alternators.

Principles of Operation A generator is a machine that converts mechanical energy into electrical energy by using the principle of magnetic induction. This principle is explained as follows: Whenever a conductor is moved within a magnetic field in such a way that the conductor cuts across magnetic lines of flux, voltage is generated in the conductor.

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The amount of voltage generated depends on (1) the strength of the magnetic field, (2) the angle at which the conductor cuts the magnetic field, (3) the speed at which the conductor is moved, and (4) the length of the conductor within the magnetic field. The polarity of the voltage depends on the direction of the magnetic lines of flux and the direction of movement of the conductor. To determine the direction of current in a given situation, the right-hand rule for generators is used. This rule is explained in the following

manner.

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Extend the thumb, first finger, and second finger of your right hand at right angles to one another, as shown in Figure 12.1. Point your thuMb in the direction the conductor Movement. Point your First finger in the direction of magnetic Flux (from north to south). Your seCond finger will then point in the direction of conventional Current flow in an external circuit to which the voltage is applied.

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Figure 12.1 - Right-hand rule for generators. When lines of magnetic force are cut by a conductor passing through them, voltage is induced in the conductor. The strength of the induced voltage is dependent upon the speed of the conductor and the strength of the magnetic field. If the ends of the conductor are connected to form a complete circuit, a current is induced in the conductor. The conductor and the magnetic field make up an elementary generator. This simple generator is illustrated in figure 12.2, together with the components of an external generator circuit which collect and use the energy produced by the simple generator. The loop of wire (A and B of figure 12.2) is arranged to rotate in a magnetic field. When the plane of the loop of wire is parallel to the magnetic lines of force, the voltage induced in the loop causes a current to flow in the direction indicated by the arrows in figure 12.2. The voltage induced at this position is maximum, since the wires are cutting the lines of force at right angles and are thus cutting more fines of force per second than in any other position relative to the magnetic field.

Figure 12.2 - Inducing maximum voltage in an elementary generator

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As the loop approaches the vertical position shown in figure 12.3, the induced voltage decreases because both sides of the loop (A and B) are approximately parallel to the lines of force and the rate of cutting is reduced.

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Figure 12.3 - Inducing minimum voltage in an elementary generator

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Figure 12.4 - Inducing maximum voltage in an elementary generator

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Figure 12.5 - Inducing a minimum voltage in the opposite direction When the loop is vertical, no lines of force are cut since the wires are momentarily travelling loop continues the number of lines of force cut increases until the loop has rotated an additional F-

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90°to a horizontal plane. As shown in figure 12.5, the number of lines of force cut and the induced voltage once again are maximum. The direction of cutting, however, is in the opposite direction to that occurring in figure 12.3 and 12.5, so the direction (polarity) of the induced voltage is reversed. As rotation of the loop continues the number of lines of force having been cut again decreases, and the induced voltage becomes zero at the position shown in figure 12.5, since the wires A and B are again parallel to the magnetic line' of force. If the voltage induced throughout the entire 360°of rotation is plotted, the curve shown in figure 12.6 results. This voltage is called an alternating voltage because of its reversal from positive to negative values - first in one direction and then in the other.

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Maximum Figure 12.6 -- Output of an elementary generator To use the voltage generated in the loop for producing a current flow in an external circuit, some means must be provided to connect the loop of wire in series with the external circuit. Such an electrical connection can be effected by opening the loop of wire and connecting its two ends to two metal rings, called slip rings, against which two metal or carbon brushes ride. The brushes are connected to the external circuit. By replacing the slip rings of the basic AC generator with two half-cylinders, called a commutator, a basic DC generator (figure 12.7), is obtained, In this illustration the black side of the coil is connected to the black segment and the white side of the coil to the white segment. The segments are insulated from each other. The two stationary brushes are placed on opposite sides of the commutator and are so mounted that each brush contacts each segment of the commutator as the latter revolves simultaneously with the loop. The rotating parts of a DC generator (coil and commutator) are called an armature. The generation of an EMF by the loop rotating in the magnetic field is the same for both AC and DC generators, but the action of the commutator produces a DC voltage. This generation of a DC voltage is described as follows for the various positions of the loop rotating in a magnetic field, with reference to figure 12.8.

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Figure 12.7 - Basic DC generator L

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The loop in position A of figure 12.8 is rotating clockwise, but no lines of force are cut by the coil sides and no EMF is generated. The black brush is shown coming into contact with the black segment of the commutator, and the white brush is just coming into contact with the white segment. In position B of figure 12.8, the flux is being cut at a maximum rate and the induced EMF i5 maximum. At this time, the black brush is contacting the black segment and the white brush is contacting the white segment. The deflection of the meter is toward the right, indicating the polarity of the output voltage. At position C of figure 12.8, the loop has completed 180°of rotation. Again, no flux lines are being cut and the output voltage is zero. The important condition to observe at position C is the action of the segments and brushes. The black brush at the 180°angle is contacting both black and white segments on one side of the commutator, and the white brush is contacting both segments on the other side of the commutator. After the loop rotates slightly past the 180° points the black brush is contacting only the white segment and the white brush is contacting only the black segment. Because of ibis switching of commutator elements, the black brush is always in contact with the coil side moving downward, and the white brush is always in contact with the coil side moving upward. Though the current actually reverses its direction in the loop in exactly the same way as in the AC generator, commutator action causes the current to flow always iin the same direction through the external circuit or meter.

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A graph of one cycle of operation is shown in figure 12.8. The generation of the EMF for positions A, B and C is the same as for the basic AC generator, but at position D, commutator action reverses the current in the external circuit, and the second half-cycle has the same waveform as the first half-cycle.

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Figure 12.8 - Operation of a basic DC generator

The progress of commutation is sometimes called rectification, since rectification is the converting of an AC voltage to a DC voltage. At the instant that each brush is contacting two segments on the commutator (positions A, C, and E in figure 12.8), a direct short circuit is produced. If an EMF were generated in the loop at this time, a high current would flow in the circuit, causing an arc and thus damaging the commutator. For this reason, the brushes must be placed in the exact position where the short will occur when the generated EMF is zero. This position is called the neutral plane. The voltage generated by the basic DC generator in figure 12.8 varies from zero to its maximum value twice for each revolution of the loop. This variation of voltage is called `ripple," and may be reduced by using more loops, or coils, as shown in A of figure 12.9. As the number of loops is increased, the variation between maximum and minimum values of voltage is reduced (B of figure 12.9), and the output voltage of the generator approaches a steady DC value. In A of figure 12.9 the number of commutator segments is increased in direct proportion to the number of loops; that is, there are two segments for one loop, four segments for two loops, and eight segments for four loops.

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Figure 12.9 - Increasing the number of coils reduces the ripple in the voltage The voltage induced in a single-turn loop is small. Increasing the number of loops does not increase the maximum value of generated voltage, but increasing the number of turns in each loop will increase this value. Within narrow limits, the output voltage of a DC generator is determined by the product of the number of turns per loop, the total flux per pair of poles in the machine, and the speed of rotation of the armature. An AC generator, or alternator, and a DC generator are identical as far as the method of generating voltage in the rotating loop is concerned. However, if the current is taken from the loop by slip rings, it is an alternating current, and the generator is called an AC generator, or alternator. If the current is collected by a commutator, it is direct current, and the generator is called a DC generator. L

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Commutation Commutation is the process by which a DC voltage output is taken from an armature that has an ac voltage induced in it. You should remember from our discussion of the elementary DC generator that the commutator mechanically reverses the armature loop connections to the external circuit. This occurs at the same instant that the voltage polarity in the armature loop reverses. A DC voltage is applied to the load because the output connections are reversed as each commutator segment passes under a brush. The segments are insulated from each other. In Figure 12.10, commutation occurs simultaneously in the two coils that are briefly shortcircuited by the brushes. Coil B is short-circuited by the negative brush. Coil Y, the opposite coil,

is short-circuited by the positive brush. The brushes are positioned on the commutator so that each coil is short-circuited as it moves through its own electrical neutral plane. As you have seen previously, there is no voltage generated in the coil at that time. Therefore, no sparking can occur between the commutator and the brush. Sparking between the brushes and the commutator is an indication of improper commutation. Improper brush placement is the main cause of improper commutation. n

LOAD Figure 12.10 - Commutation of a DC generator.

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Motor Reaction in a Generator r;

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When a generator delivers current to a load, the armature current creates a magnetic force that opposes the rotation of the armature. This is called motor reaction. A single armature conductor is represented in Figure 12.11, view A. When the conductor is stationary, no voltage is generated and no current flows. Therefore, no force acts on the conductor. When the conductor is moved downward (Figure 12.11, view B) and the circuit is completed through an external load, current flows through the conductor in the direction indicated. This sets up lines of flux around the conductor in a clockwise direction.

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The interaction between the conductor field and the main field of the generator weakens the field above the conductor and strengthens the field below the conductor. The main field consists of lines that now act like stretched rubber bands. Thus, an upward reaction force is produced that acts in opposition to the downward driving force applied to the armature conductor. If the current in the conductor increases, the reaction force increases. Therefore, more force must be applied to the conductor to keep it moving. With no armature current, there is no magnetic (motor) reaction. Therefore, the force required to

turn the armature is low. As the armature current increases, the reaction of each armature conductor against rotation increases. The actual force in a generator is multiplied by the number of conductors in the armature. The driving force required to maintain the generator armature speed must be increased to overcome the motor reaction. The force applied to turn the armature must overcome the motor reaction force in all DC generators. The device that

provides the turning force applied to the armature is called the prime mover. The prime mover

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may be an electric motor, a gasoline engine, a steam turbine, or any other mechanical device that provides turning force.

Armature Losses In DC generators, as in most electrical devices, certain forces act to decrease the efficiency. These forces, as they affect the armature, are considered as losses and may be defined as follows: • 12R, or copper loss in the winding • Eddy current loss in the core • Hysteresis loss (a sort of magnetic friction) Copper Losses The power lost in the form of heat in the armature winding of a generator is known as copper loss. Heat is generated any time current flows in a conductor. Copper loss is an 12R loss, which increases as current increases. The amount of heat generated is also proportional to the resistance of the conductor. The resistance of the conductor varies directly with its length and inversely with its cross-sectional area. Copper loss is minimized in armature windings by using large diameter wire.

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Eddy Current Losses The core of a generator armature is made from soft iron, which is a conducting material with desirable magnetic characteristics. Any conductor will have currents induced in it when it is rotated in a magnetic field. These currents that are induced in the generator armature core are called eddy currents. The power dissipated in the form of heat, as a result of the eddy currents, is considered a loss. Eddy currents, just like any other electrical currents, are affected by the resistance of the material in which the currents flow. The resistance of any material is inversely proportional to its cross-sectional area. Figure 12.12, view A, shows the eddy currents induced in an armature core that is a solid piece of soft iron. Figure 12.12, view B, shows a soft iron core of the same size, but made up of several small pieces insulated from each other. This process is called lamination. The currents in each piece of the laminated core are considerably less than in the solid core because the resistance of the pieces is much higher. (Resistance is inversely proportional to cross-sectional area.) The currents in the individual pieces of the laminated core are so small that the sum of the individual currents is much less than the total of eddy currents in the solid iron core.

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N

S SOLID CORE

A

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Figure 12.12 - Eddy currents in DC generator armature cores. As you can see, eddy current losses are kept low when the core material is made up of many thin sheets of metal. Laminations in a small generator armature may be as thin as 1/64 inch.

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The laminations are insulated from each other by a thin coat of lacquer or, in some instances, simply by the oxidation of the surfaces. Oxidation is caused by contact with the air while the laminations are being annealed. The insulation value need not be high because the voltages induced are very small. Most generators use armatures with laminated cores to reduce eddy current losses. Hysteresis Losses Hysteresis loss is a heat loss caused by the magnetic properties of the armature. When an armature core is in a magnetic field, the magnetic particles of the core tend to line up with the magnetic field. When the armature core is rotating, its magnetic field keeps changing direction. The continuous movement of the magnetic particles, as they try to align themselves with the magnetic field, produces molecular friction. This, in turn, produces heat. This heat is transmitted to the armature windings. The heat causes armature resistances to increase.

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To compensate for hysteresis losses, heat-treated silicon steel laminations are used in most DC generator armatures. After the steel has been formed to the proper shape, the laminations are heated and allowed to cool. This annealing process reduces the hysteresis loss to a low value.

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Construction Features of D.C. Generators Generators used on aircraft may differ somewhat in design since they are made by various manufacturers. All, however, are of the same general construction and operate similarly. The major parts, or assemblies, of a DC generator are a field frame (or yoke), a rotating armature, and a brush assembly. The parts of a typical aircraft generator are shown in figure 12.13.

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end frame

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Figure 12.13 -- A typical 24-volt aircraft generator Field Frame The field frame is also called the yoke, which is the foundation or frame for the generator. The frame has two functions: It completes the magnetic circuit between the poles and acts as a mechanical support for the other parts of the generator. In A of figure 12.14, the frame for a two pole generator is shown in cross sectional view. A four-pole generator frame is shown in B of figure 12.14.

1

In small generators, the frame is made of one piece of iron, but in larger generators, it is usually made up of two parts bolted together. The frame has high magnetic properties and, together with the pole pieces, forms the major part of the magnetic circuit. The field poles, shown in figure 12.14, are bolted to the inside of the frame and form a core on which the field coil windings are mounted. The poles are usually laminated to reduce eddy current losses and serve the same purpose as the iron core of an electromagnet; that is, they concentrate the lines of force produced by the field coils. The entire frame, including field poles, is made from highquality magnetic iron or sheet steel.

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a Figure 12.14 - A two-pole and four-pole frame assembly A practical DC generator uses electromagnets instead of permanent magnets. To produce a magnetic field of the necessary strength with permanent magnets would greatly increase the physical size of the generator.

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The field coils are made up of many turns of insulated wire and are usually wound on a form which fits over the iron core of the pole to which it is securely fastened (figure 12.15). The exciting current, which is used to produce the magnetic field and which flows through the field i + coils, is obtained from an external source or from the generated AC of the machine, No electrical connection exists between the windings of the field coils and the pole pieces. r-L

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Figure 12.15 - A field coil removed from a field pole Most field coils are connected in such a manner that the poles show alternate polarity. Since there is always one north pole for each south pole, there must always be an even number of poles in any generator. Note that the pole pieces in figure 12.14 project from the frame. Because air offers a great amount of reluctance to the magnetic field, this design reduces the length of the air gap between the poles and the rotating armature and increases the efficiency of the generator. When the pole pieces are made to project as shown in figure 12.14, they are called salient poles. Armature The armature assembly consists of armature coils wound on an iron core, a commutator, and associated mechanical parts. Mounted on a. shaft, it rotates through the magnetic field produced by the field coils. The core of the armature acts as an iron conductor in the magnetic field and, for this reason, is laminated to prevent the circulation of eddy currents. There are two general kinds of armatures: the ring (or Gramme ring) and the drum. Figure 12.16 shows a ring type armature made up of an iron core, an eight section winding, and an eightsegment commutator. This kind of armature is rarely used; most generators use the drum-type armature.

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Figure 12.16 -An eight-section, ring-type armature

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A drum type armature (figure 12.17) has coils placed in slots in the core, but there is no electrical connection to the coils and core. The use of slots increases the mechanical safety of the armature. Usually, the coils are held in place in the slots by means of wooden or fiber wedges. The connections of the individual coils, called coil ends, are brought out to individual

segments on the commutator.

Commutator

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Figure 12.17 - A drum-type armature F1

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Drum-type armatures are wound with either of two types of windings - the lap winding or the wave winding. The lap winding is illustrated in Figure 12.18, view A

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This type of winding is used in DC generators designed for high-current applications. The windings are connected to provide several parallel paths for current in the armature. For this reason, lap-wound armatures used in DC generators require several pairs of poles and brushes.

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ARMATURE

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LAP WINDING

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WAVE WINDING Figure 12.18 - Types of windings used on drum-type armatures. Figure 12.18, view B, shows a wave winding on a drum-type armature. This type of winding is used in DC generators employed in high-voltage applications. Notice that the two ends of each coil are connected to commutator segments separated by the distance between poles. This configuration allows the series addition of the voltages in all the windings between brushes. This type of winding only requires one pair of brushes. In practice, a practical generator may have several pairs to improve commutation. Figure 12.19 shows a cross-sectional view of a typical commutator. The commutator is located at the end of an armature and consists of wedge-shaped segments of hard-drawn copper, insulated from each other by thin sheets of mica. The segments are held in place by steel Vrings or clamping flanges fitted with bolts. Rings of mica insulate the segments from the flanges. The raised portion of each segment is called a riser, and the leads from the armature coils are soldered to the risers. When the segments have no risers, the leads are soldered to short slits in the ends of the segments. The brushes ride on the surface of the commutator, forming the electrical contact between the armature coils and the external circuit. A flexible, braided-copper conductor, commonly called a pigtail, connects each brush to the external circuit.

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Slots

Back V-ring with mica inner and

outer rings for insulation

Figure 12.19 -- Commutator with portion removed to show construction

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The brushes, usually made of high-grade carbon and held in place by brush holders insulated from the frame, are free to slide up and clown in their holders in order to follow any irregularities in the surface of the commutator. The brushes are usually adjustable so that the pressure of the brushes on the commutator can be varied and the position of the brushes with respect to the segments can be adjusted. The constant making and breaking of connections to the coils in which a voltage is being induced necessitates the use of material for brushes which has a definite contact resistance.

Also, this material must be such that the friction between the commutator and the brush is low, to prevent excessive wear. For these reasons, the material commonly used for brushes is high grade carbon. The carbon must be soft enough to prevent undue wear of the commutator and yet hard enough to provide reasonable brush life. Since the contact resistance of carbon is fairly high, the brush must be quite large to provide a large area of contact. The commutator surface is highly polished to reduce friction as much as possible. Oil or grease must never be used on a commutator, and extreme care must be used when cleaning it to avoid marring or scratching the

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Types of D.C. Generators There are three types of AC generators: series wound, shunt-wound, and shunt-series or compound wound. The difference in type depends on the relationship of the field winding to the external circuit. Series-Wound D.C. Generators The field winding of a series generator is connected in series with the external circuit, called the load (figure 12.20), The field coils are composed of a few turns of large ire; the magnetic field strength depends more on the current flow rather than the number of turns in the coil. Series generators have very poor voltage regulation under changing load, since the greater the current through the field coils to the external circuit the greater the induced EMF and the greater the terminal or output voltage. Therefore, when the load is increased, the voltage increases;

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likewise, when the load is decreased the voltage decreases.

The output voltage of a series-wound generator may be controlled by a rheostat in parallel with the field windings as shown in A of figure 12.20. Since the series-wound generator has such poor regulation, it is never employed as an aeroplane generator.

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B Figure 12.20 - Diagram and schematic of a series-wound generator Generators in aeroplanes have field windings which are connected either in shunt or in compound.

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Shunt-Wound D.C. Generators A generator having a field winding connected in parallel with the external circuit is called a shunt generator, as shown in A and B of figure 12.21. The field coils of a shunt generator contains many turns of small wire; the magnetic strength is derived from the large number of turns rather than the current strength through the coils. If a constant voltage is desired, the shunt-wound generator is not suitable for rapidly fluctuating loads. An increase in load causes a decrease in the terminal or output voltage, and any decrease in load causes an increase in terminal voltage; once the armature and the load are connected in series, all current flowing in the external circuit passes through the armature winding. Because of the resistance in the armature winding, there is a voltage drop (IR drop = current x resistance). As the load increases, the armature current increases and the IR drop in the armature increases. The voltage delivered to the terminals is the difference between the induced voltage and the voltage drop; therefore, there is a decrease in terminal voltage. This decrease in voltage causes a decrease in field strength, because the current in the field coils decreases in proportion to the decrease in terminal voltage; with a weaker field, the voltage is further decreased. When the load decreases, the output voltage increases accordingly, and a larger current flows in the windings. This action is cumulative, so the output voltage continues to rise to a point called field saturation, after which there is no further increase in output voltage. The terminal voltage of a shunt generator can be controlled by means of a rheostat inserted in series with the field windings as shown in A of figure 12.21. As the resistance is increased, the field current is reduced; consequently, the generated voltage is reduced also. For a given setting of the field rheostat, the terminal voltage at the armature brushes will be approximately equal to the generated voltage minus the IR drop produced by the load current in the armature; thus, the voltage at the terminals of the generator will drop as the load is applied. Certain voltage-sensitive devices are available which automatically adjust the field rheostat to compensate for variations in load. When these devices are used, the terminal voltage remains essentially constant.

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Figure 12.21 -- A shunt-wound generator

U Module 3.12 DC Motor/Generator Theory

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Compound-Wound DC Generators A compound-wound generator combines a series winding and a shunt winding in such a way that the characteristics of each are used to advantage. The series field coils are made of a relatively small number of turns of large copper conductor, either circular or rectangular in cross section and are connected in series with the armature circuit. These coils are mounted on the same poles on which the shunt field coils are mounted and therefore, contribute a magneto-motive-force which influences the main field flux of the generator. A diagrammatic and a schematic illustration of a compound-wound generator is shown in A and B of figure 12.22. If the ampere-turns of the series field act in the same direction as those of the shunt field, the combined magneto-motive-force is equal to the sum of the series and shunt field components.

Load is added to a compound generator in the same manner in which load is added to a shunt

generator, by increasing the number of parallel paths across the generator terminals. Thus, the decrease in total load resistance with added load is accompanied by an increase in armature circuit and series-field circuit current

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Field coil

Compound wound

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14

A

Figure 12.22 - A compound-wound generator

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The effect of die additive series field is that of increased field flux with increased load. The extent of the increased field flux depends on the degree of saturation of the field as determined by the shunt field current. Thus, the terminal voltage of the generator may increase or decrease with load, depending on the influence of the series load coils. This influence is referred to as the degree of compounding. A flat-compound generator is one in which the no-load and full-load voltages have the same value; whereas an under-compound generator has a full-load voltage less than the no-load value, and an over-compound generator has a lull-load voltage which is higher than the no-load value. Changes in terminal voltage with increasing load depends upon the degree of compounding.

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If the series field aids the shunt field, the generator is said to be cumulative-compounded (B of

figure 12.22) if the series field opposes the shunt field, the machine is said to be differentially compounded, and is called a differential generator.

Compound generators are usually designed to be over-compounded. This feature permits varied degrees of compounding by connecting a variable shunt across the series field. Such a shunt is sometimes called a diverter. Compound generators are used where voltage regulation is of prime importance.

L f

Differential generators have somewhat the same characteristics as series generators in that they are essentially constant-current generators. However, they generate rated voltage at no load, the voltage dropping materially as the load current increases.

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Figure 12.23 - Generator characteristics T"i

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Constant-current generators are ideally suited as power sources for electric arc welders and are used almost universally in electric arc welding. If the shunt field of a compound generator is connected across both the armature and the series field, it is known as a long-shunt connection, but if the shunt field is connected across the armature alone, it is called a short-shunt connection. These connections produce essentially the same generator characteristics. A summary of the characteristics of the various types of generators discussed is shown graphically in figure 12.23. These characteristics have been further simplified, for comparison, in figure 12.24. Actual curves are seldom, if ever, as perfect as shown.

U

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A. SHUNT -WOUND

DC GENERATOR

B. SERIES -WOUND w

DC GENERATOR

0a

LOAD CUMIr ti I

C. COMPCUND-WOUND DC GENERATOR

Figure 12.24 - Voltage output characteristics of the series-, shunt-, and compound-wound DC generators.

Three-Wire Generators Some DC generators, called three-wire generators, are designed to deliver 210 volts, or 120 volts from either side of a neutral wire (figure 12.25). This is accomplished by connecting a reactance coil to opposite sides of the commutator, with the neutral connected to the midpoint of the reactance coil. Such a reactance coil acts as a low-loss voltage divider. If resistors were used, the IR loss would be prohibitive unless the two loads were perfectly matched. The coil is built into some generators as part of the armature, with the midpoint connected to a single slip ring which the neutral contacts by means of a brush. In other generators, the two connections to the commutator are connected, in turn, to two slip rings, and the reactor is located outside the generator. In either case, the load unbalance on either side of the neutral must not be more than 25 percent of the rated current output of the generator. The three-wire generator permits simultaneous operation of 120-volt lighting circuits and 240-volt motors from the same generator.

'7 ni Figure 12.25 - Three wire generator

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Armature Reaction Current flowing through the armature sets up electromagnetic fields in the windings. These new fields tend to distort or bend the magnetic flux between the poles of the generator from a straight line path. Since armature current increases with load, the distortion becomes greater with an increase in load. This distortion of the magnetic field is called armature reaction and is illustrated in figure 12.26. it

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C. The interactive effect of both fields Figure 12.26 -Armature reaction

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Armature windings of a generator are spaced in such a way that, during rotation of the armature, there are certain positions when the brushes contact two adjacent segments, thereby shorting the armature windings to these segments. Usually, when the magnetic field is not distorted, there is no voltage being induced in the shorted windings, and, therefore, no harmful

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results occur from the shorting of the windings. However, when the field is distorted, a voltage is induced in these shorted windings and sparking takes place between the brushes and the commutator segments. Consequently, the commutator becomes pitted. The wear on the brushes becomes excessive, and the output of the generator is reduced. To correct this condition, the brushes are set so that the plane of the coils which are shorted by the brushes is perpendicular to the distorted magnetic field, which is accomplished by moving the brushes forward in the direction of rotation. This operation is called shifting the brushes to the neutral plane, or plane of commutation. The neutral plane is the position where the plane of the two opposite coils is perpendicular to the magnetic field in the generator. On a few generators, the

brushes can be shifted manually ahead of the normal neutral plane to the neutral plane caused by field distortion. On nonadjustable brush generators, the manufacturer sets the brushes for minimum sparking. Interpoles may be used to counteract some of the effects of field distortion. Since shifting the brushes is inconvenient and unsatisfactory, especially when the speed and load of the generator are changing constantly. An interpole is a pole placed between the main poles of a generator. For example, a four-pole generator has four interpoles, which are north and south poles, alternately, as are the main poles. A four-pole generator with interpoles is shown in figure 12.27. An interpole has the same polarity as the next main pole in the direction of rotation. The magnetic flux produced by an interpole causes the current in the armature to change direction as an armature winding passes under it. This cancels the electromagnetic fields about the armature windings. The magnetic strength of the interpoles varies with the load on the generator and since field distortion varies with the load the magnetic field of the interpoles counteracts the effects of the field set up around the armature windings and minimizes field distortion.

Figure 12.27 - Generator with interpoles

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Thus, the interpole tends to keep the neutral plant in the same position for all loads on the

generator; therefore, field distortion is reduced by the interpoles and the efficiency of output, and service life of the brushes are improved.

Generator Ratings A generator is rated in power output. Since a generator is designed to operate at a specified voltage, the rating usually is given as the number of amperes the generator can safely supply at its rated voltage. Generator rating and performance data are stamped on the name plate attached to the

generator. When replacing a generator, it is important to choose one of the proper rating. The rotation of generators is termed either clockwise or anticlockwise as viewed from the driven end. Usually the direction of rotation is stamped on the data plate. If no direction is stamped on the plate, the rotation may be marked by an arrow on the cover plate of the brush housing. It is important that a generator with the correct direction of rotation be used; otherwise the voltage will be reversed. U

The speed of an aircraft engine varies from idle RPM to takeoff RPM. However, during the major portion of a flight, it is at a constant cruising speed. The generator drive is usually geared to revolve the generator between 1 % and 11/2 times the engine crankshaft speed. Most aircraft generators have a speed at which they begin to produce their normal voltage. Termed the n, `coming-in" speed, it is usually about 1,500 RPM. U i

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D.C. Generator Maintenance Inspection The following information about the inspection and maintenance of DC generator systems is general in nature because of the large number of differing aircraft generator systems. These procedures are for familiarization only. Always follow the applicable manufacturer's instructions for a given generator system. In general, the inspection of the generator installed in the aircraft should include the following items: 1.

Security of generator mounting.

2.

Condition of electrical connections.

3.

Dirt and oil in the generator. If oil is present, check engine oil seal. Blow out dirt with compressed air. Condition of generator brushes. Generator operation. Voltage regulator operation.

4. 5. 6.

A detailed discussion of items 4. 5 and 6 is presented in the following paragraphs.

Condition of Generator Brushes Sparking of brushes quickly reduces the effect of brush area in contact with the commutator bars. The degree of such sparking should be determined. Excessive wear warrants a detailed inspection. The following information pertains to brush seating, brush pressure, high-mica condition, and brush wear. Manufacturers usually recommend the following procedures to seat bru4hes which do not make good contact with slip rings or commutators. The brush should be lifted sufficiently to permit the insertion of a strip of No. 000, or finer, sandpaper under the brush, rough side out (figure 12.28). Pull sandpaper in the direction of armature rotation, being careful to keep the ends of the sandpaper as close to the slip ring or commutator surface as possible in order to a sold rounding the edges of the brush. When pulling the sandpaper back to the starting point, the brush should be raised so it does not ride on the sandpaper. The brush should be sanded only in the direction of rotation. After the generator has run for a short period, brushes should be inspected to make sure that pieces of sand have not become embedded in the brush and are collecting copper. Under no circumstances should emery cloth or similar abrasives be used for seating brushes (or smoothing commutators), since they contain conductive materials which will cause arcing between brushes and commutator bars.

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No. 000 sandpaper (Sand side next to bnish)

Figure 12.28 - Seating brushes with sandpaper Excessive pressure will cause rapid wear of brushes. Too little pressure, however, will allow "bouncing" of the brushes, resulting in burned and pitted surfaces. A carbon-graphite or light metalized brush should exert a pressure of 11/2 to 21/2 PSI on the commutator. The pressure recommended by the manufacturer should be checked with the use of a spring scale graduated in ounces. Brush spring tension is usually adjusted between 32 to 36 ounces; however, the tension may differ slightly for each specific generator. When a spring scale is used, the measurement of the pressure which a brush exerts on the commutator is read directly on the scale. The scale is applied at the point of contact between U the spring arm and the top of the brush, with the brush installed in the guide. The scale is drawn up until the arm just lifts off the brush surface. At this instant, the force on the scale should be rT' read. 1r7 ;

Flexible low-resistance pigtails are provided on most heavy current carrying brushes, and their connections should be securely made and checked at frequent intervals The pigtails should never be permitted to alter or restrict the free motion of the brush. 4_i

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The purpose of the pigtail is to conduct the current, rather than subjecting the brush spring to currents which would alter its spring action by overheating. The pigtails also eliminate any

possible sparking to the brush guides caused by the movement of the brushes within the holder,

thus minimizing side wear of the brush.

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Carbon dust resulting from brush sanding should be thoroughly cleaned from all parts of the generators after a sanding operation. Such carbon dust has been the cause of several serious fires as well as costly damage to the generator.

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Operation over extended periods of time often results in the mica insulation between commutator bars protruding almost to the surface of the bars. This condition is called "high mica" and interferes with the contact of the brushes to the commutator. Whenever this condition exists, or if the armature has been turned on a lathe, carefully undercut the mica insulation to a depth equal to the width of the mica, or approximately 0.020 inch. Each brush should be a specified length to work properly. If a brush is too short the contact it makes with the commutator will be faulty, which can also reduce the spring force holding the brush in place. Most manufacturers specify the amount of wear permissible from a new brush length. When a brush has worn to the minimum length permissible, it must be replaced. Some special generator brushes should not be replaced because of a slight grooving of the face of the brush. These grooves are normal and will appear in a DC and AC generator brushes which are installed in some models of aircraft generators. These brushes have two cores made of a harder material with a higher expansion rate than the material used in the main body of the brush. (Usually, the main body of the brush face rides on the commutator. However, at certain temperatures, the cores extend and wear through any film on the commutator.

Field Excitation

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When a DC voltage is applied to the field windings of a DC generator, current flows through the windings and sets up a steady magnetic field. This is called field excitation (or fieldflash). This excitation voltage can be produced by the generator itself or it can be supplied by an outside source, such as a battery. A generator that supplies its own field excitation is called a selfexcited generator. Self-excitation is possible only if the field pole pieces have retained a slight amount of permanent magnetism, called residual magnetism. When the generator starts rotating, the weak residual magnetism causes a small voltage to be generated in the armature. This small voltage applied to the field coils causes a small field current. Although small, this field current strengthens the magnetic field and allows the armature to generate a higher voltage. The higher voltage increases the field strength, and so on. This process continues until the output voltage reaches the rated output of the generator. Generator field flashing is required when generator voltage does not build up and the generating system (including the voltage regulator) does not have fieldflash capability. This condition is usually caused by insufficient residual magnetism in the generator fields. In some cases, a generator that has been out-of-service for an extended period may lose its residual magnetism and require flashing. Residual magnetism can be restored by flashing the field thereby causing a current surge in the generator. To restore the generator's fieldflash capability, it is necessary to manually flash the field. This is done by applying jump leads connected to a suitably sized battery, to the field terminals for a few moments, to restore the residual magnetism. Ideally the procedure should be done whilst turning the generator on start-up, but this is not always possible for safety reasons.

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Motors

Introduction Most devices in aircraft from the starter to the automatic pilot depend upon mechanical energy furnished by direct-current motors. A direct-current motor is a rotating machine which transforms direct current energy into mechanical energy. It consists of two principal parts - a field assembly and an armature assembly. The armature is the rotating part in which current-carrying wires are acted upon by the magnetic field.

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Principles of Operation Stated very simply, a DC motor rotates as a result of two magnetic fields interacting with each other. The armature of a DC motor acts like an electromagnet when current flows through its coils. Since the armature is located within the magnetic field of the field poles, these two Whenever a current-carrying wire is placed in the field of a magnet, a force acts on the wire. The force is not one of attraction or repulsion; however, it is at right angles to the wire and also at right angles to the magnetic field set up by the magnet. The action of the force upon a current-carrying wire placed in a magnetic field is shown in figure 967. A wire is located between two permanent magnets. The lines of force in the magnetic field are from the north pole to the south pole. When no current flows, as in diagram A, no force is exerted on the wire, but when current flows through the wire, a magnetic field is set up about it, as shown in figure 12.29. The direction of the field depends on the direction of current flow. Current in one direction creates a clockwise field about the wire, and current in the other direction, an anticlockwise field.

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Wire with current

l tant field and direction

located in a magnetic field

and accompanying field

of force on wire

A

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C

Figure 12.29 - Force on a current-carrying wire

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Since the current-carrying wire produces a magnetic field, a reaction occurs between the field about the wire and the magnetic field between the magnets. When the current flows in a direction to create a anticlockwise magnetic field about the wire, this field and the field between the magnets add or reinforce at the bottom of the wire because the lines of force are in the same direction. At the top of the wire, they subtract or neutralize, since the lines of force in the two fields are opposite in direction. Thus, the resulting field at the bottom is strong and the one at the top is weak. Consequently, the wire is pushed upward as shown in diagram C of figure 12.29. The wire is always pushed away from the side where the field is strongest.

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If current flow through the wire were reversed in direction, the two fields would add at the top and subtract at the bottom. Since a wire is always pushed away from the strong field, the wire would be pushed downs Force Between Parallel Conductors Two wires carrying current in the vicinity of one another exert a force on each other because of their magnetic fields. An end view of two conductors is shown in figure 12.30.

A

Figure 12.30 - Fields surrounding parallel conductors In A, electron flow in both conductors is toward the reader, and the magnetic fields are anticlockwise around the conductors between the wires, the fields cancel because the directions of the two fields oppose each other. The wires are forced in the direction of the weaker field, toward each other. This force is one of attraction. In B, the electron flow in the two wires is in opposite directions. The magnetic fields are, therefore, clockwise in one and anticlockwise in the other, as shown. The fields reinforce each other between the wires, and the wires are forced in the direction of the weaker field, away from each other. This force is one of repulsion. To summarize: Conductors carrying current in the same direction tend to be drawn together; conductors carrying current in opposite directions tend to be repelled from each other. Developing Torque If a coil in which current is flowing is placed in a magnetic field, a force is produced which will cause the coil to rotate. In the coil shown in figure 12.31, conventional current flows inward on side A and outward on side B.

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Torque

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Figure 12.31 - Developing a torque The magnetic field about B is anticlockwise and that about A, clockwise. As previously explained, a force will develop which pushes side B downward. At the same time, the field of the L magnets and the field about A, in which the current is inward, will add at the bottom and subtract at the top. Therefore, A will move upward. The coil will thus rotate until its plane is perpendicular to the magnetic lines between the north and south poles of the magnet, as U indicated in figure 12.31 by the white coil at right angles to the black coil. The tendency of a force to produce rotation is called torque. When the steering wheel of a car is U turned, torque is applied. The engine of an airplane gives torque to the propeller.

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UTorque is developed also by the reacting magnetic fields about the current-carrying coil justdescribed. This is the torque which turns the coil. The left-hand motor rule can be used to determine the direction a current carrying wire will U move in a magnetic field. As illustrated in figure 12.32, if the First finger of the left band is pointed in the direction of the magnetic Field and the seCond finger in the direction of Conventional Current flow, the thuMb will indicate the direction the current-carrying wire will Move.

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Direction

of Current Figure 12-32 - Left-hand rule for motors. The amount of torque developed in a coil depends upon several factors: the strength of the magnetic field, the number of turns in the coil, and the position of the coil in the field. Magnets are made of special steel which produces a strong field. Since there is a torque acting on each turn, the greater the number of turns on the coil, the greater the torque. In a coil carrying a steady current located in a uniform magnetic field, the torque will vary at successive positions of rotation. When the plane of the coil is parallel to the lines of force, the torque is zero. When its plane cuts the lines of force at right angles, the torque is 100 percent. At intermediate positions, the torque ranges between zero and 100 percent. n

Basic D.C. Motor A coil of wire through which the current flows will rotate when placed in a magnetic field. This is the technical basis governing the construction of a DC motor. Figure 12.32 shows a coil mounted in a magnetic field in which it can rotate. However, if the connecting wires from the battery were permanently fastened to the terminals of the coil and there was a flow of current, the coil would rotate only until it lined itself up with the magnetic field. Then, it would stop, because the torque at that point would he zero. A motor, of course, must continue rotating.

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Torque

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Figure 12.32 - Basic DC motor operation

It is necessary therefore, to design a device that will reverse the current in the coil just at the time the coil becomes parallel to the lines of force. This will create torque again and cause the coil to rotate. If the current reversing device set up to reverse the current each time the coil is about to stop, the coil can be made to continue rotating as long as desired. One method of doing this is to connect the circuit so that, as the coil rotates, each contact slides off the terminal to which it connects and slides onto the terminal of opposite polarity. In other words, the coil contacts switch terminals continuously as the coil rotates, preserving the torque and keeping the coil rotating. In figure 12.32, the coil terminal segments are labelled A and B.

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As the coil rotates, the segments slide onto and past the fixed terminals or brushes. With this arrangement, the direction of current in the side of the coil next to the north seeking pole flows toward the reader, and the force acting on that side of the coil turns it downward. The part of the motor which changes the current from one wire to another is called the commutator.

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When the coil is positioned as shown in A of figure 12.32, current will flow from the positive terminal of the battery to the positive brush, to segment B of the commutator, through the loop i Li to segment A of the commutator, to the negative brush, and then, back to the negative terminal

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of the battery. By using the left-hand motor rule, it is seen that the coil will rotate anticlockwise.

L The torque at this position of the coil is maximum since the greatest number of lines of force are being cut by the coil.

Li When the coil has rotated 900 to the position shown in B of figure 12.32. segments A and B of

the commutator no longer make contact with the battery circuit and no current can flow through

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the coil. At this position, the torque has reached a minimum value, since a minimum number of lines of force are being cut. However, the momentum of the coil carries it beyond this position until the segments again make contact with the brushes, and current again enters the coil; this time, though, it enters through segment A and leaves through segment B. However, since the positions of segments A and B have also been reversed, the effect of the current is as before, the torque acts in the same direction, and the coil continues its anticlockwise rotation. On

passing through the position shown in C of figure 12.32, the torque again reaches maximum.

Continued rotation carries the coil again to a position of minimum torque, as in D of figure 12.32. At this position the brushes no longer carry current, but once more the momentum rotates the coil to the point where current enters through segment B and have, through A. Further rotation brings the coil to the starting point and, thus, one revolution is completed.

The switching of the coil terminals from the positive to the negative brushes occurs twice per

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revolution of the coil.

The torque in a motor containing only a single coil is neither continuous nor very effective, for there are two positions where there is actually no torque at all. To overcome this, a practical DC motor contains a large number of coils wound on the armature. These coils are so spaced that, for any position of the armature, there will be coils near the poles of the magnet. This makes the torque both continuous and strong. The commutator, likewise, contains a large number of segments instead of only two.

i1

1

The armature in a practical motor is not placed between the poles of a permanent magnet but between those of an electromagnet, since a much stronger magnetic field can be furnished. The core is usually made of a mild or annealed steel, which can be magnetized strongly by induction. The current magnetizing the electromagnet is from the same source that supplies the current to the armature, n

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D.C. Motor Construction The major parts in a practical motor are the armature assembly, the field assembly, the brush assembly, and the end frame. (See figure 12.33.)

Ent! i'ht e

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Figure 12.33 - Cutaway view of a practical DC motor Armature Assembly The armature assembly contains a laminated, soft iron core, coils, and a commutator, all mounted on a rotating steel shaft. Laminations made of stacks of soft iron, insulated from each other, form the armature core. Solid iron is not used, since a solid iron core revolving in the magnetic field would heat and use energy needlessly. The armature windings are insulated copper wire, which are inserted in slots insulated with fibre paper (fish paper) to protect the windings. The ends of the windings are connected to the commutator segments. Wedges or steel bands hold the windings in place to prevent them from flying out of the slots when the armature is rotating at high speeds. The commutator consists of a large number of copper segments insulated from each other and the armature shaft by pieces of mica. Insulated wedge

rings hold the segments in place.

L

Field Assembly The field assembly consists of the field frame, the pole pieces and the field coils. The field frame is located along the inner wall of the motor housing. It contains laminated soft steel pole pieces on which the field coil is wound. A coil, consisting of several turns of insulated wire, fits over each pole piece and, together with the pole, constitutes a field pole. Some motors have as few as two poles, others as many as eight.

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The brush assembly consists of the brushes and their holders. The brushes are usually small blocks of graphitic carbon, since this material has a long service life and also causes minimum wear to the commutator. The holders permit some play in the brushes so they can follow any irregularities in the surface of the commutator and make good contact. Springs hold the brushes firmly against the commutator. A commutator and two pairs of brushes are shown in figure 12.34.

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ll Figure 12.34 - Commutator and brushes End Frame The end frame is the part of the motor opposite the commutator usually the end frame is designed so that it can be connected to the unit to be driven. The bearing for the drive end is also located in the end frame. Sometimes the end frame is made a part of the unit driven by the motor. When this is done, the bearing on the drive end may be located in any one of a number of places.

Armature Reaction You will remember that the subject of armature reaction was covered in the section on DC generators. The reasons for armature reaction and the methods of compensating for its effects are basically the same for DC motors as for DC generators. Figure 12.36 reiterates for you the distorting effect that the armature field has on the flux between the pole pieces. Notice, however, that the effect has shifted the neutral plane backward, against the direction of rotation. This is different from a DC generator, where the neutral plane shifted forward in the direction of rotation.

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U

U

U

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Figure 12.35 - Armature reaction.

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As before, the brushes must be shifted to the new neutral plane. As shown in figure 12.35, the shift is anticlockwise. Again, the proper location is reached when there is no sparking from the brushes. Compensating windings and interpoles, two more "old" subjects, cancel armature reaction in DC motors. Shifting brushes reduces sparking, but it also makes the field less effective. Cancelling armature reaction eliminates the need to shift brushes in the first place. Compensating windings and interpoles are as important in motors as they are in generators.

L. Lj L

Compensating windings are relatively expensive; therefore, most large DC motors depend on interpoles to correct armature reaction. Compensating windings are the same in motors as they are in generators. Interpoles, however, are slightly different. The difference is that in a generator the interpole has the same polarity as the main pole ahead of it in the direction of rotation. In a motor the interpole has the same polarity as the main pole following it. The interpole coil in a motor is connected to carry the armature current the same as in a generator. As the load varies, the interpole flux varies, and commutation is automatically corrected as the load changes. It is not necessary to shift the brushes when there is an increase or decrease in load. The brushes are located on the no-load neutral plane. They remain in that position for all conditions of load. The DC motor is reversed by reversing the direction of the current in the armature. When the armature current is reversed, the current through the interpole is also reversed. Therefore, the interpole still has the proper polarity to provide automatic commutation.

7 L Modu[e 3.12 DC Motor/Generator Theory

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Types of D.C. Motors There are three basic types of DC motors: (1) series motors, (2) shunt motors, and (3) compound motors. They differ largely in the method in which their field and armature coils are connected. Series DC Motor In the series motor, the field windings, consisting of a relatively few turns of heavy wire, are connected in series with the armature winding. Both a diagrammatic and a schematic illustration of a series motor is shown in figure 12.36. The same current flowing through the field winding also flows through the armature winding. Any increase in currents therefore, strengthens the magnetism of both the field and the armature.

n J 11

Armature

Wire

0

Comifutatar

B Schematic

A

Diagrammatic

Figure 12.36 - Series motor Because of the low resistance in the field winding, the series motor is able to draw a large current in starting. This starting current, in passing through both the field and armature windings, produces a high starting torque, which is the series motor's principal advantage. The speed of a series motor is dependent upon the load. Any change in load is accompanied by a substantial change in speed. A series motor will run at high speed when it has a light load and at low speed with a heavy load. If the load is removed entirely, the motor may operate at such a high speed that the armature will fly apart. If high starting torque is needed under heavy load conditions, series motors have many applications. Series motors are often used in aircraft as engine starters and for raising and lowering landing gear, cowl flaps, and wing flaps.

12-42

Module 3.12 DC Motor/Generator Theory Use and/or disclosure is

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Shunt D.C. Motor

In the shunt motor the field winding is connected in parallel or in shunt with the armature

Ll

winding. (See figure 12.37). The resistance in the field winding is high. Since the field winding is connected directly across the power supply, the current through the field is constant. The field current does not vary with motor speed as in the series motor and, therefore, the torque of the shunt motor will vary only with the current through the armature. The torque developed at starting is less than that developed by a series motor of equal size.

t

E

Diagrammatic

Figure 12.37 - Shunt motor The speed of the shunt motor varies very little with changes in load. When all load is removed, it assumes a speed slightly higher than the loaded speed. This motor is particularly suitable for use when constant speed is desired and when high starting torque is not needed.

I

L r-,

L r f

L

Compound D.C. Motor A compound motor has two field windings, as shown in figure 12.38. One is a shunt field connected in parallel with the armature; the other is a series field that is connected in series with the armature. The shunt field gives this type of motor the constant speed advantage of a regular shunt motor. The series field gives it the advantage of being able to develop a large torque when the motor is started under a heavy load. It should not be a surprise that the compound motor has both shuntand series-motor characteristics.

F-71 U

l

-l, )I LJ

Module 3.12 DC Motor/Generator Theory

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INPUT STAGE

SERIES FIELD

INPUT VOLTAGE

O--Iti SHUNT FIELD

LONG

n ~J

SERIES FIELD

SHUNT FIELD SHORT

ARMATURE I

(A) LONG SHUNT

I

ARM ATURE

(B) SHORT SHUNT

Figure 12.38 - Compound-wound DC motor. When the shunt field is connected in parallel with the series field and armature, it is called a "long shunt" as shown in figure 12.38, (view A). Otherwise, it is called a "short shunt", as shown in figure 12.38, (view B). The shunt winding is composed of many turns of fine wire and is connected in parallel with the armature winding. The series winding consists of a few turns of large wire and is connected in series with the armature winding. The starting torque is higher than in the shunt motor but lower than in the series motor. Variation of speed with load is less than in a series-wound rid motor hut greater than in a shunt motor. The compound motor is used whenever the combined characteristics of the series and shunt motors are desired.

17

Like the compound generator the compound motor has both series and shunt field windings. The series winding may either aid the shunt wind (cumulative compound) or oppose the shunt winding (differential compound). The starting and load characteristics of the cumulative-compound motor are somewhere between those of the series and those of the shunt motor. Because of the series field, the cumulative-compound motor has a higher starting torque than a shunt motor. Cumulative-compound motors are used in driving machines which are subject to sudden changes in load. They are also used where a high starting torque is desired, but a series motor cannot be used easily. l.S

In the differential compound motor, an increase in load creates an increase'in current and a decrease in total flux in this type of motor. These two tend to offset each other and the result is a practically constant speed. However, since an increase in load tends to decrease the field strength, the speed characteristic becomes unstable. Rarely is this type of motor used in aircraft systems. A graph of the variation in speed with changes of load of the various types of DC motors is shown in figure 12.39

12-44

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J

© Copyright 2010

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U 11

Differential cotupcundiisg Shunt connection

CummulatLv r)

L

compounding

2 Series

O

connection

4 0 4n

Output power or load

10,E

Figure 12.39 - Load characteristics of DC motors r;

Back-EMF The armature resistance of a small, 28-volt DC motor is extremely low, about 0.1 ohm. When the armature is connected across the 28-volt source, current through the armature will apparently be

E

28 0.1 ® 280 alrnpolm.s.,

This high value of current flow is not only impracticable but also unreasonable, especially when the current drain, during normal operation of a motor, is found to be about 4 amperes. This is because the current through a motor armature during operation is determined by more factors than ohmic resistance. r

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u r

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When the armat ure in a motor rotates in a magne tic field a voltag e is induce

d in its windings. This voltage is called the back or Back-EMF (electromotive force) and is opposite in direction to the voltage applied to the motor from the external source. Back-EMF opposes the current which causes the armature to rotate. The current flowing through the armature, therefore, decreases as the back-EMF increases. The faster the armature rotates, the greater the Back-EMF. For this reason, a motor connected to a battery may draw a fairly high current on starting, but as the armature speed increases the current flowing through the armature decreases. At rated speed,

the Back-EMF may be only a few volts less than the battery voltage. Then, if the load on the

motor is increased, the motor will slow down, less Back-EMF will be generated, and the current drawn from the external source will increase. In a shunt motor, the back-EMF affects only the current in the armature, since the field is connected in parallel across the power source. As the motor slows down and the back-EMF decreases, more current flows through the armature, but the magnetism in the field is unchanged. When the series motor slows down, the Back-EMF

12-45

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decreases and more current flows through the field and the armature, thereby strengthening their magnetic fields. Because of these characteristics, it is more difficult to stall a series motor than a shunt motor.

Types of Duty Electric motors are called upon to operate under various conditions. Some motors are used for intermittent operation; others operate continuously. Motors built for intermittent duty can be operated for short periods only and, then, must be allowed to cool before being operated again. If such a motor is operated for long periods under full load, the motor will be overheated. Motors built for continuous duty may be operated at rated power for long periods.

n

Reversing Motor Direction By reversing the direction of current flow in either the armature or the field windings, the direction of a motor's rotation may be reversed. This will reverse the magnetism of either the armature or the magnetic field in which the armature rotates. If the wires connecting the motor to an external source are interchanged, the direction of rotation will not be reversed, since changing these wires reverses the magnetism of both field and armature and leases the torque in the same direction as before. One method for reversing direction of rotation employs two field windings wound in opposite directions on the same pole. This type of motor is called a split field motor. Figure 12.40 shows a series motor with a split field winding. n

J

7

n

Figure 12.40 - Split field series motor The single pole, double-throw switch makes it possible to direct current through either of the two

windings. When the switch is placed in the lower position, current flows through the lower field

winding, creating a north pole at the lower field winding and at the lower pole piece, and a south pole at the upper pole piece. When the switch is placed in the upper position, current flows through the upper field winding, the magnetism of the field is reversed, and the armature rotates

n

fl 12-46 Use and/or disclosure is

governed by the statement

Module 3.12 DC Motor/Generator Theory Module 3.12 DC Motor/Generator Theory TTS Integrated Training System Use and/or disclosure is

TTS Integrated Training System © Copyright 2010

governed by the statement

on page 2 of this Chapter.

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Li

in the opposite direction. Some split field motors are built with two separate field windings would

on alternate poles. The armature in such a motor, a four-pole reversible motor, rotates in one

direction when current flows through the windings of one set of opposite pole pieces, and in the opposite direction when current flows through the other set of windings. Another method of direction reversal, called the switch method, employs a double-pole, doublethrow switch which changes the direction of current flows in either the armature or the field. In the illustration of the switch method shown in figure 12.41, current direction may be reversed through the field but not through the armature. When the switch is thrown to the "up" position, current flows through the field winding to establish a north pole at the right side of the motor and a south pole at the left side of the motor. When the switch is thrown to the "down" position, this polarity is reversed and the armature rotates in the opposite direction.

L

F

Li

D41NV I I

Dt>table pole doul,lt- throw mvitc-h

L (

Figure 12.41 - Switch method of reversing motor direction

Module 3.12 DC Motor/Generator Theory

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Motor Speed Motor speed can be controlled by varying the current in the field windings. When the amount of current flowing through the field windings is increased, the field strength increases, but the motor slows down since a greater amount of back-EMF is generated in the armature windings. When the field current is decreased, the field strength decreases, and the motor speeds up because the back-EMF is reduced. A motor in which speed can be controlled is called a variable-speed motor. It may be either a shunt or series motor. In the shunt motor, speed is controlled by a rheostat in series with the field windings (figure 12.42). The speed depends on the amount of current which flows through the rheostat to the field windings. To increase the motor speed, the resistance in the rheostat is increased, which decreases the field current. As a result, there is a decrease in the strength of the magnetic field and in the back-EMF. This momentarily increases the armature current and the torque. The motor will then automatically speed up until the back-EMF increases and causes the armature current to decrease to its former value. When this occurs, the motor will operate at a higher fixed speed than before. F rttm,

"1111,it field Pole

Figure 12.42 - Shunt motor with variable speed control To decrease the motor speed, the resistance of the rheostat is decreased. More current flows through the field windings and increases the strength of the field; then, the back-EMF increases momentarily and decreases the armature current. As a result, the torque decreases and the motor slows down until the back-EMF decreases to its former value; then the motor operates at a lower fixed speed than before.

12-48

Module 3.12 DC Motor/Generator Theory Use and/or disclosure is

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Designed in association with the club66pro.co.uk question practice aid

In the series motor (figure 9-82), the rheostat speed control is connected either in parallel or in series with the motor field, or in parallel with the armature. When the rheostat is set for maximum resistance, the motor speed is increased in the parallel armature connection by a decrease in current. When the rheostat resistance is maximum in the series connection, motor speed is reduced by a reduction in voltage across the motor. For above normal speed operation, the rheostat is in parallel with the series field. Part of the series field current is bypassed and the motor speeds up.

Fast Slow

Below normal speed

Normal speed

Above normal speed

Figure 12.43 - Controlling the speed of a series DC motor

Energy Losses in D.C. Motors

F1 L

U

Losses occur when electrical energy is converted to mechanical energy (in the motor), or mechanical energy is converted to electrical energy (in the generator). For the machine to be efficient, these losses must be kept to a minimum. Some losses are electrical, others are mechanical. Electrical losses are classified as copper losses and iron losses; mechanical losses occur in overcoming the friction of various parts of the machine. Copper losses occur when electrons are forced through the copper windings of the armature and the field. These losses are proportional to the square of the current. They are sometimes called 12R losses, since they are due to the power dissipated in the form of beat in the resistance of the field and armature windings.

Iron losses are subdivided in hysteresis and eddy current losses. Hysteresis losses are caused by the armature revolving in an alternating magnetic field. It, therefore, becomes magnetized H first in One direction and then in the other. The residual magnetism of the iron or steel of which J the armature is made causes these losses. Since the field magnets are always magnetized in one direction (DC field), they have no hysteresis losses. �i

U Eddy current losses occur because the iron core of the armature is a conductor revolving in a

magnetic field. This sets up an e.m.f. across portions of the core, causing currents to flow U Use and/or disclosure is governed by the statement

Module 3.12 DC Motor/Generator Theory

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within the core. These currents heat the core and, if they become excessive, may damage the windings. As far as the output is concerned, the power consumed by eddy currents is a loss. To reduce eddy currents to a minimum, a laminated core usually is used. A laminated core is made of thin sheets of iron electrically insulated from each other. The insulation between laminations reduces eddy current, because it is "transverse" to the direction in which the currents tend to flow. However, it has no effect on the magnetic circuit. The thinner the laminations, the more effectively this method reduces eddy current losses.

t_j

El

7

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U Inspection and Maintenance of D.C. Motors Use the following procedures to make inspection and maintenance cheeks:

LI U11

[" I

U

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allation. Check all wiring, connections, terminals, fuses, and switches for general condition and security. Keep motors clean and mounting bolts tight.

2. 1. Ch 3. ec 4. k the correct spring tension, and procedures for replacing brushes are given in the op applicable manufacturer's instructions era 5. Inspect commutator for cleanliness, pitting, scoring, roughness, corrosion or burning. tio Check for high mica (if the copper wears down below the mica the mica will insulate n the brushes from the commutator). Clean dirty commutators with a cloth moistened of with the recommended cleaning solvent. Polish rough or corroded commutators with the fine sandpaper (000 or finer) and blow out with compressed air. Never use emery uni paper since it contains metallic particles which may cause shorts. Replace the motor t if the commutator is burned, badly pitted, grooved, or worn to the extent that the mica dri insulation is flush with the commutator surface. ve 6. Inspect all exposed wiring for evidence of overheating. Replace the motor if the n insulation on leads or windings is burned, cracked, or brittle. by 7. Lubricate only if called for by the manufacturer's instructions covering the motor. the Most motors used in today's aeroplanes require no lubrication between overhauls. mo 8. Adjust and lubricate the gearbox or unit which the motor drives, in accordance with tor the applicable manufacturer's instructions covering the unit. in ac When trouble develops in a DC motor system, check first to determine the source of the trouble. cor Replace the motor only when the trouble is due to a defect in the motor itself. In most cases, the da failure of a motor to operate is caused by a defect in the external electrical circuit, or by nc mechanical failure in the mechanism driven by the motor. e wit Check the external electrical circuit for loose or dirty connections and for improper connection of h wiring. Look for open circuits, grounds, and shorts by following the applicable manufacturer's the circuit-testing procedure. If the fuse is not blown, failure of the motor to operate is usually due to an inst open circuit. A blown fuse usually indicates an accidental ground or short circuit. The ruct chattering of the relay switch which controls the motor is usually caused by a low battery. When the ions battery is low, the open-circuit voltage of the battery is sufficient to close the relay, but with the cov heavy current draw of the motor, the voltage drops below the level required to hold the relay closed. erin When the relay opens, the voltage in the battery increases enough to close the relay again. This g cycle repeats and causes chattering, which is very harmful to the relay switch, due to the heavy thecurrent causing an arc which will burn the contacts. spe cifiCheck the unit driven by the motor for failure of the unit or the mechanism. If the motor has c failed as a result of a failure in the driven unit, the fault must be corrected before installing a new inst motor. Module 3.12 DC Motor/Generator Theory

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

If it has been determined that the fault is in the motor itself (by checking for correct voltage at the motor terminals and for failure of the driven unit), inspect the commutator and brushes. A dirty commutator or defective or binding brushes may result in poor contact between brushes and commutator Clean the commutator, brushes, and brush holders with a cloth moistened with the recommended cleaning solvent. If brushes are damaged or worn to the specified minimum length, install new brushes in accordance with the applicable manufacturer's instructions covering the motor. If the motor still fails to operate, replace it with a serviceable motor,

Manual and Automatic Starters Because the DC resistance of most motor armatures is low (0.05 to 0.5 ohm), and because the back EMF does not exist until the armature begins to turn, it is necessary to use an external starting resistance in series with the armature of a DC motor to keep the initial armature current

to a safe value. As the armature begins to turn, back EMF increases; and, since the back EMF

H

7

H

opposes the applied voltage, the armature current is reduced. The external resistance in series with the armature is decreased or eliminated as the motor comes up to normal speed and full voltage is applied across the armature. Controlling the starting resistance in a DC motor is accomplished either manually, by an operator, or by any of several automatic devices. The automatic devices are usually just switches controlled by motor speed sensors. Automatic starters are not covered in detail in this module.

] l

II

Starter-Generator Systems Several types of turbine powered aircraft are equipped with starter systems which utilize a prime mover having the dual function of engine starting and of supplying power to the aircraft's electrical system. Starter-generator units are basically compound-wound machines with compensating windings and interpoles, and are permanently coupled with the appropriate engine via a drive shaft and gear train. For starting purposes, the unit functions as a fully compounded motor, the shunt winding being supplied with current via a field changeover relay. When the engine reaches self-sustaining speed and the starter circuit is isolated from the power supply, the changeover relay is also automatically de-energized and its contacts connect the shunt-field winding to a voltage regulator. The relay contacts also permit DC to flow through the shunt winding to provide initial excitation of the field. Thus, the machine functions as a conventional DC generator, its output being connected to the bus bar on reaching the regulated level.

12-52 Use and/or disclosure is governed by the statement

Module 3.12 DC Motor/Generator Theory

Use and/or disclosure is

TTS Integratedr.r Training System ...

,r,♦0 1) 1A

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n n.

TTS Integrated Training System

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r

TTS Integrated Training System

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Module 3 Licence Category B1/B2 U

Electrical Fundamentals 3.13 AC Theory

U

U Module 3.13 AC Theory

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Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Ltd.

Knowledge Levels - Category A, B1, B2 and C Aircraft Maintenance Licence

L.

Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2

11

basic knowledge levels.

The knowledge level indicators are defined as follows:

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

IL

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3

n

• A detailed knowledge of the theoretical and practical aspects of the subject. • A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents L H

Module 3.13 AC Theory What is Alternating Current (AC)?

5 5

Direct vs Alternating Current

5

AC Waveforms Other Waveforms Measurements of AC magnitude Simple AC circuit calculations AC phase Single and 3-Phase Principles Three-Phase Power Systems Phase Rotation Three-Phase Y and A Configurations

6 9 12 18 20 22 28 33 35

U

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Module 3.13 AC Theory

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Module 3.13 Enabling Objectives Objective

EASA 66 Reference

Level

AC Theory Sinusoidal waveform: phase, period, frequency, cycle Instantaneous, average, root mean square, peak, peak to

3.13

2

peak current values and calculations of these values, in

relation to voltage, current and power Triangular/Square waves Single/3 phase principles

L.

n

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13-4 Use and/or disclosure is governed by the statement

iL

Module 3.13 AC Theory TTS Integrated Training System

n Rnnurinht 9n1n

TTS Integrated Training System 0 Copyright 2010

Use and/or disclosure is governed by the statement on page 2 of this Chapter.

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Module 3.13 AC Theory What is Alternating Current (AC)?

Direct Current (DC), is electricity flowing in a constant direction, and/or possessing a voltage with constant polarity. DC is the kind of electricity made by a battery (with definite positive and negative terminals), or the kind of charge generated by rubbing certain types of materials against each other. As useful and as easy to understand as DC is, it is not the only "kind" of electricity in use. Certain sources of electricity (most notably, rotary electro-mechanical generators) naturally produce voltages alternating in polarity, reversing positive and negative over time. Either as a voltage switching polarity or as a current switching direction back and forth, this "kind" of U

electricity is known as Alternating Current (AC).

DIRECT CURRENT (DC)

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L I

ALTERNATING CURRENT (AC)

I 1 - -

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Figure 13.1 - DC and AC comparison

Direct vs Alternating Current Whereas the familiar battery symbol is used as a generic symbol for any DC voltage source, the circle with the wavy line inside is the generic symbol for any AC voltage source.

1"1

L

One might wonder why anyone would bother with such a thing as AC. It is true that in some cases AC holds no practical advantage over DC. In applications where electricity is used to dissipate energy in the form of heat, the polarity or direction of current is irrelevant, so long as there is enough voltage and current to the load to produce the desired heat (power dissipation). However, with AC it is possible to build electric generators, motors and power distribution

systems that are far more efficient than DC, and so we find AC used predominately across the

world in high power applications. To explain the details of why this is so, a bit of background knowledge about AC is necessary. If a machine is constructed to rotate a magnetic field around a set of stationary wire coils with the turning of a shaft, AC voltage will be produced across the wire coils as that shaft is rotated,

in accordance with Faraday's Law of electromagnetic induction. This is the basic operating

-1

principle of an AC generator, also known as an alternator:

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13-5 Module 3.13 AC Theory Use and/or disclosure Is governed by the statement

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AC Waveforms

LJ

When an alternator produces AC voltage, the voltage switches polarity over time, but does so in a very particular manner. When graphed over time, the "wave" traced by this voltage of alternating polarity from an alternator takes on a distinct shape, known as a sine wave.

7

(the sine wave)

Time

)-

Figure 13.2 - The sine wave, as produced by an AC generator. In the voltage plot from an electromechanical alternator, the change from one polarity to the other is a smooth one, the voltage level changing most rapidly at the zero ("crossover") point and most slowly at its peak. If we were to graph the trigonometric function of "sine" over a horizontal range of 0 to 360 degrees, we would find the exact same pattern as in Table 13.1.

Angle (°)

sin(angle)

waveAngle (°)sin(angle)

wave

0

0.0000

zero

180

0.0000

zero

15

0.2588

+

195

-0.2588

-

30

0.5000

+

210

-0.5000

-

45

0.7071

+

225

-0.7071

-

60

0.8660

+

240

-0.8660

-

75

0.9659

+

255

-0.9659

-

90

1.0000

+ peak

270

-1.0000

- peak

105

0.9659

+

285

-0.9659

-

120

0.8660

+

300

-0.8660

-

135

0.7071

+

315

-0.7071

-

150

0.5000

+

330

-0.5000

-

165

0.2588

+

345

0.2588

-

180 0.0000 zero 360 Table 13.1 - Trigonometric "sine" function

0.0000

zero

The reason why an electromechanical alternator outputs sine-wave AC is due to the physics of its operation. The voltage produced by the stationary coils by the motion of the rotating magnet 13-6 TTS Integrated Training System

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is proportional to the rate at which the magnetic flux is changing perpendicular to the coils

(Faraday's Law of Electromagnetic Induction). That rate is greatest when the magnet poles are closest to the coils, and least when the magnet poles are furthest away from the coils. Mathematically, the rate of magnetic flux change due to a rotating magnet follows that of a sine function, so the voltage produced by the coils follows that same function. If we were to follow the changing voltage produced by a coil in an alternator from any point on the sine wave graph to that point when the wave shape begins to repeat itself, we would have marked exactly one cycle of that wave. This is most easily shown by spanning the distance between identical peaks, but may be measured between any corresponding points on the graph. The degree marks on the horizontal axis of the graph represent the domain of the trigonometric sine function, and also the angular position of our simple two-pole alternator shaft

as it rotates.

f-4

0

one wave cycle

90

270

---(

60

(0)

90

1"270

one wave cycle

Alternator shaft position (degrees)

Go ta)

-H

No

Figure 13.3 - Alternator voltage as function of shaft position (time).

U

Since the horizontal axis of this graph can mark the passage of time as well as shaft position in degrees, the dimension marked for one cycle is often measured in a unit of time, most often seconds or fractions of a second. When expressed as a measurement, this is often called the period of a wave. The period of a wave in degrees is always 360, but the amount of time one period occupies depends on the rate voltage oscillates back and forth. A more popular measure for describing the alternating rate of an AC voltage or current wave than period is the rate of that back-and-forth oscillation. This is called frequency. The modern unit for frequency is the Hertz (abbreviated Hz), which represents the number of wave cycles completed during one second of time. In the United States of America, the standard power-line frequency is 60 Hz, meaning that the AC voltage oscillates at a rate of 60 complete back-and-

forth cycles every second. In Europe, where the power system frequency is 50 Hz, the AC U i -, J

voltage only completes 50 cycles every second. A radio station transmitter broadcasting at a frequency of 100 MHz generates an AC voltage oscillating at a rate of 100 million cycles every second. Prior to the canonization of the Hertz unit, frequency was simply expressed as "cycles per second." Older meters and electronic equipment often bore frequency units of "CPS" (Cycles Per Second) instead of Hz. Many people believe the change from self-explanatory units like CPS to Hertz constitutes a step backward in clarity. A similar change occurred when the unit of

f' Module 3.13 AC Theory

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"Celsius" replaced that of "Centigrade" for metric temperature measurement. The name Centigrade was based on a 100-count ("Centi-") scale ("-grade") representing the melting and boiling points of H2O respectively The name Celsius on the other hand fives no hint as to the,g unit's origin or meaning.

lf

Period and frequency are mathematical reciprocals of one another. That is to say, if a wave has a period of 10 seconds, its frequency will be 0.1

Hz, or 1/10 of a cycle per second:

Frequency in Hertz =

1

Period in seconds

An instrument called an oscilloscope, Figure 13.4, is used to display a changing voltage over

time on a graphical screen. You may be familiar with the appearance of an ECG or EKG

1. 4

(electrocardiograph) machine, used by physicians to graph the oscillations of a patient's heart over time. The ECG is a special-purpose oscilloscope expressly designed for medical use. General-purpose oscilloscopes have the ability to display voltage from virtually any voltage source, plotted as a graph with time as the independent variable. The relationship between period and frequency is very useful to know when displaying an AC voltage or current waveform on an oscilloscope screen. By measuring the period of the wave on the horizontal axis of the oscilloscope screen and reciprocating that time value (in seconds), you can determine the frequency in Hertz.

OSCILLOSCOPE n T.T

1,

1 6: c� iu l : o r 7

Frequency =

--

( J l

�yr/

a-

;.

�y

1 period

=

1

16 ms

= 62.5 Hz

Figure 13.4 - Time period of sinewave is shown on oscilloscope. Voltage and current are by no means the only physical variables subject to variation overtime. Much more common to our everyday experience is sound, which is nothing more than the alternating compression and decompression (pressure waves) of air molecules, interpreted by 13-8

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our ears as a physical sensation. Because alternating current is a wave phenomenon, it shares

many of the properties of other wave phenomena, like sound. For this reason, sound (especially structured music) provides an excellent analogy for relating AC concepts.

Other Waveforms While electromechanical alternators and many other physical phenomena naturally produce sine waves, this is not the only kind of alternating wave in existence. Other "waveforms" of AC are commonly produced within electronic circuitry. Here are but a few sample waveforms and L

their common designations in figure 13.5. Square wave

Triangle wave

U

l-*- one wave cycle

-H

k one wave cycle

--- *`1

Sawtooth wave r

sl Figure 13.5 - Some common waveshapes (waveforms).

U ..

L t

These waveforms are by no means the only kinds of waveforms in existence. They're simply a few that are common enough to have been given distinct names. Even in circuits that are supposed to manifest "pure" sine, square, triangle, or sawtooth voltage/current waveforms, the real-life result is often a distorted version of the intended waveshape. Some waveforms are so complex that they defy classification as a particular "type" (including waveforms associated with many kinds of musical instruments). Generally speaking, any waveshape bearing close resemblance to a perfect sine wave is termed sinusoidal, anything different being labeled as non-sinusoidal. Being that the waveform of an AC voltage or current is crucial to its impact in a circuit, we need to be aware of the fact that AC waves come in a variety of shapes.

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Mark-to-Space Ratio The high time of the pulse waveform is called the mark, while the low time is called the space. The mark and space do not need to be of equal duration. The mark-to-space ratio is given by:

period T

s pace mark

Y

I

HIGH

I LOW t Figure 13.6 -- Rectangular waveform's Mark and Space

mark space ratio Y HIGH time LOW time MARY. SPACE RATIO = 0.5

MARK SPACE RATIO = 3.0

mark

mark

V HIGH

HIGH

L- LOW

L LOW

t

-1.t

Figure 13.7: Examples of different Mark-to-Space ratios

IT

A mark-to-space ratio = 1.0 means that the high and low times are equal, while a marktospace ratio = 0.5 indicates that the high time is half as long as the low time; A mark-to-space ratio of 3.0 corresponds to a longer high time, in this case, three times as long as the space.

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Duty Cycle

Another way of describing the same types of waveform uses the duty cycle, where:

duty cycle = HIGH time x100% period When the duty cycle is less than 50%, the high time is shorter than the low time, and so on.

L L; F It,

L

ri

L

Module 3.13 AC Theory

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Measurements of AC magnitude So far we know that AC voltage alternates in polarity and AC current alternates in direction. We also know that AC can alternate in a variety of different ways, and by tracing the alternation over time we can plot it as a "waveform." We can measure the rate of alternation by measuring the time it takes for a wave to evolve before it repeats itself (the "period"), and express this as cycles per unit time, or "frequency." In music, frequency is the same as pitch, which is the essential property distinguishing one note from another. However, we encounter a measurement problem if we try to express how large or small an AC quantity is. With DC, where quantities of voltage and current are generally stable, we have little trouble expressing how much voltage or current we have in any part of a circuit. But how do you grant a single measurement of magnitude to something that is constantly changing?

j

One way to express the intensity, or magnitude (also called the amplitude), of an AC quantity is to measure its peak height on a waveform graph. This is known as the peak or crest value of an AC waveform.

Time Figure 13.8 - Peak voltage of a waveform Another way is to measure the total height between opposite peaks. This is known as the peakto-peak (P-P) value of an AC waveform.

Time ---0 Figure 13.9 - Peak-to-peak voltage of a waveform Unfortunately, either one of these expressions of waveform amplitude can be misleading when comparing two different types of waves. For example, a square wave peaking at 10 volts is obviously a greater amount of voltage for a greater amount of time than a triangle wave peaking at 10 volts. The effects of these two AC voltages powering a load would be quite different.

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10V i

11

Time (same load resistance) 4

( r

a pe k)

(peak) more heat energy

dissipated

less heat energy dissipated

Figure 13.10 - A square wave produces a greater heating effect than the same peak voltage triangle wave. One way of expressing the amplitude of different waveshapes in a more equivalent fashion is to mathematically average the values of all the points on a waveform's graph to a single, aggregate number. This amplitude measure is known simply as the average value of the waveform. If we average all the points on the waveform algebraically (that is, to consider their sign, either positive or negative), the average value for most waveforms is technically zero, because all the positive points cancel out all the negative points over a full cycle: Figure below

L

True average value of all points (considering their signs) is zero! Figure 13.11 - The average value of a sinewave is zero This, of course, will be true for any waveform having equal-area portions above and below the "zero" line of a plot. However, as a practical measure of a waveform's aggregate value, "average" is usually defined as the mathematical mean of all the points' absolute values over a cycle. In other words, we calculate the practical average value of the waveform by considering all points on the wave as positive quantities, as if the waveform looked like that shown in figure 13.12.

IL

13-13

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+

+

4+

+_+

+

A

a

Practical average of points, all values assumed to be positive. Figure 13.12 - Waveform seen by AC "average responding" meter Polarity-insensitive mechanical meter movements (meters designed to respond equally to the positive and negative half-cycles of an alternating voltage or current) register in proportion to the waveform's (practical) average value, because the inertia of the pointer against the tension of the spring naturally averages the force produced by the varying voltage/current values over n time. Conversely, polarity-sensitive meter movements vibrate uselessly if exposed to AC voltage or current, their needles oscillating rapidly about the zero mark, indicating the true (algebraic) average value of zero for a symmetrical; waveform. When the "average" value of a waveform is referenced in this text, it will be assumed that the "practical" definition of average is intended unless otherwise specified. Another method of deriving an aggregate value for waveform amplitude is based on the waveform's ability to do useful work when applied to a load resistance. Unfortunately, an AC measurement based on work performed by a waveform is not the same as that waveform's "average" value, because the power dissipated by a given load (work performed per unit time) is not directly proportional to the magnitude of either the voltage or current impressed upon it. Rather, power is proportional to the square of the voltage or current applied to a resistance (P = V2/R, and p = 12R). Although the mathematics of such an amplitude measurement might not be straightforward, the utility of it is. Consider a bandsaw and a jigsaw, two pieces of modern woodworking equipment. Both types of saws cut with a thin, toothed, motor-powered metal: blade to cut wood. But while the bandsaw uses a continuous motion of the blade to cut, the jigsaw uses a back-and-forth motion. The comparison of alternating current (AC) to direct current (DC) may be likened to the comparison of these two saw types.

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Bandsaw

Jigsaw

Lj

blade

m oti on,

Mood T

blade

motion

(analogous to DC)

(analogous to AC)

Figure 13.13 - Bandsaw-jigsaw analogy of DC vs AC The problem of trying to describe the changing quantities of AC voltage or current in a single, aggregate measurement is also present in this saw analogy: how might we express the speed f', of a jigsaw blade? A bandsaw blade moves with a constant speed, similar to the way DC voltage pushes or DC current moves with a constant magnitude. A jigsaw blade, on the other hand, moves back and forth, its blade speed constantly changing. What is more, the back-andforth motion of any two jigsaws may not be of the same type, depending on the mechanical design of the saws. One jigsaw might move its blade with a sine-wave motion, while another with a triangle-wave motion. To rate a jigsaw based on its peak blade speed would be quite misleading when comparing one jigsaw to another (or a jigsaw with a handsaw!). Despite the fact that these different saws move their blades in different manners, they are equal in one respect: they all out wood, and a quantitative comparison of this common function can serve as a common basis for which to rate blade speed. Picture a jigsaw and bandsaw side-by-side, equipped with identical blades (same tooth pitch, angle, etc.), equally capable of cutting the same thickness of the same type of wood at the same rate. We might say that the two saws were equivalent or equal in their cutting capacity. Might this comparison be used to assign a "bandsaw equivalent" blade speed to the jigsaw's back-and-forth blade motion; to relate the wood-cutting effectiveness of one to the other? This is the general idea used to assign a "DC equivalent" measurement to any AC voltage or current: whatever magnitude of DC voltage or current would produce the same amount of heat energy dissipation through an equal resistance.

F 13-15

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-o-5 A R

NI S --

- .-

LOV

RMS - -5A RM SN

50

power dissipated Equal p o wet dissipa t through equal resistance loads

5A

-

50W

power dissipated

Figure 13.14 - An RMS voltage produces the same heating effect as a the same DC voltage In the two circuits above, we have the same amount of load resistance (2 )) dissipating the same amount of power in the form of heat (50 watts), one powered by AC and the other by DC. Because the AC voltage source pictured above is equivalent (in terms of power delivered to a load) to a 10 volt DC battery, we would call this a "10 volt" AC source. More specifically, we would denote its voltage value as being 10 volts RMS. The qualifier "RMS" stands for root mean square, the algorithm used to obtain the DC equivalent value from points on a graph (essentially, the procedure consists of squaring all the positive and negative points on a waveform graph, averaging those squared values, then taking the square root of that average to obtain the final answer). Sometimes the alternative terms equivalent or DC equivalent are used instead of "RMS," but the quantity and principle are both the same. RMS amplitude measurement is the best way to relate AC quantities to DC quantities, or other AC quantities of differing waveform shapes, when dealing with measurements of electric power. For other considerations, peak or peak-to-peak measurements may be the best to employ. For instance, when determining the proper size of wire (ampacity) to conduct electric power from a source to a load, RMS current measurement is the best to use, because the principal concern with current is overheating of the wire, which is a function of power dissipation caused by current through the resistance of the wire. However, when rating insulators for service in highvoltage AC applications, peak voltage measurements are the most appropriate, because the principal concern here is insulator "flashover" caused by brief spikes of voltage, irrespective of time. Peak and peak-to-peak measurements are best performed with an oscilloscope, which can capture the crests of the waveform with a high degree of accuracy due to the fast action of the cathode-ray-tube in response to changes in voltage. For RMS measurements, analogue meter movements (D'Arsonval, Weston, iron vane, electrodynamometer) will work so long as they have been calibrated in RMS figures. Because the mechanical inertia and dampening effects of an electromechanical meter movement makes the deflection of the needle naturally proportional to the average value of the AC, not the true RMS value, analog meters must be specifically calibrated (or mis-calibrated, depending on how you look at it) to indicate voltage or current in RMS units. The accuracy of this calibration depends on an assumed waveshape, usually a sine wave.

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Electronic meters specifically designed for RMS measurement are best for the task. Some

instrument manufacturers have designed ingenious methods for determining the RMS value of

r_T

any waveform. One such manufacturer produces "True-RMS" meters with a tiny resistive heating element powered by a voltage proportional to that being measured. The heating effect of that resistance element is measured thermally to give a true RMS value with no mathematical calculations whatsoever, just the laws of physics in action in fulfillment of the definition of RMS. The accuracy of this type of RMS measurement is independent of waveshape. For "pure" waveforms, simple conversion coefficients exist for equating Peak, Peak-to-Peak,

U

Average (practical, not algebraic), and RMS measurements to one another. RMS = 0.707 (Peak) AVG = 0.637 (Peak) P-P = 2 (Peak)

RMS = Peak AVG = Peak P-P = 2 (Peak)

RMS = 0.577 (Peak) AVG = 0.5 (Peak)

P-P = 2 (Peak)

Figure 13.15 - Conversion factors for common waveforms. In addition to RMS, average, peak (crest), and peak-to-peak measures of an AC waveform, there are ratios expressing the proportionality between some of these fundamental measurements. The crest factor of an AC waveform, for instance, is the ratio of its peak (crest) value divided by its RMS value. The form factor of an AC waveform is the ratio of its RMS value divided by its average value. Square-shaped waveforms always have crest and form factors equal to 1, since the peak is the same as the RMS and average values. Sinusoidal waveforms have an RMS value of 0.707 (the reciprocal of the square root of 2) and a form factor of 1.11 (0.707/0.636). Triangle- and sawtooth-shaped waveforms have RMS values of

0.577 (the reciprocal of square root of 3) and form factors of 1.15 (0.577/0.5). Bear in mind that the conversion constants shown here for peak, RMS, and average amplitudes of sine waves, square waves, and triangle waves hold true only for pure forms of these waveshapes. The RMS and average values of distorted waveshapes are not related by the

same ratios.

iL

13-17 Module 3.13 AC Theory

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RMS -_999 -------------- AVG = ???

; .1

P-P = 2 (Peak) Figure 13.16 - Arbitrary waveforms have no simple conversions. This is a very important concept to understand when using an analog meter movement to measure AC voltage or current. An analog movement, calibrated to indicate sine-wave RMS amplitude, will only be accurate when measuring pure sine waves. If the waveform of the voltage or current being measured is anything but a pure sine wave, the indication given by the meter will not be the true RMS value of the waveform, because the degree of needle deflection in an analog meter movement is proportional to the average value of the waveform, not the RMS. RMS meter calibration is obtained by "skewing" the span of the meter so that it displays a small multiple of the average value, which will be equal to be the RMS value for a particular waveshape and a particular waveshape only. Since the sine-wave shape is most common in electrical measurements, it is the waveshape assumed for analog meter calibration, and the small multiple used in the calibration of the meter is 1.1107 (the form factor: 0.707/0.636: the ratio of RMS divided by average for a sinusoidal waveform). Any waveshape other than a pure sine wave will have a different ratio of RMS and average values, and thus a meter calibrated for sine-wave voltage or current will not indicate true RMS when reading a non-sinusoidal wave. Bear in mind that this limitation applies only to simple, analog AC meters not employing "True-RMS" technology.

,7 jj

Simple AC circuit calculations AC circuit measurements and calculations can get very complicated due to the complex nature of alternating current in circuits with inductance and capacitance. However, with simple circuits (figure below) involving nothing more than an AC power source and resistance, the same laws and rules of DC apply simply and directly.

Lj

400 Sa Figure 13.17 - AC circuit calculations for resistive circuits are the same as for DC.

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If

l. .i

Integrated Training System

U i�

Designed in association with the club66pro.co.uk question practice aid

R

1-

l=

Rt+R,+R,

Rtotal = I l S? E total 'total =

'total

Rtotal

Ertt ='totalRl

V

'total = 10 mA

1 kS?

ER3

ER? ='totalR3 Ea,=5V

ERt=IV i s

= 10

= Itotal R.3

ER;=4V

Series resistances still add, parallel resistances still diminish, and the Laws of Kirchhoff and Ohm still hold true. Actually, as we will discover later on, these rules and laws always hold true, it's just that we have to express the quantities of voltage, current, and opposition to current in more advanced mathematical forms. With purely resistive circuits, however, these complexities of AC are of no practical consequence, and so we can treat the numbers as though we were dealing with simple DC quantities. Because all these mathematical relationships still hold true, we can make use of the "table" method of organizing circuit values just as with DC: RI

R,

R3

E

1

5

4

I

10111

10m

tom

R

100

500

Total

400

10

Volts

tom

Amps

1k

Ohms

Table 13.3 - Ohms Law in tabular format One major caveat needs to be given here: all measurements of AC voltage and current must be expressed in the same terms (peak, peak-to-peak, average, or RMS). If the source voltage is given in peak AC volts, then all currents and voltages subsequently calculated are cast in terms

of peak units. If the source voltage is given in AC RMS volts, then all calculated currents and voltages are cast in AC RMS units as well. This holds true for any calculation based on Ohm's Laws, Kirchhoff's Laws, etc. Unless otherwise stated, all values of voltage and current in AC circuits are generally assumed to be RMS rather than peak, average, or peak-to-peak. In some areas of electronics, peak measurements are assumed, but in most applications (especially industrial electronics) the assumption is RMS.

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AC phase Things start to get complicated when we need to relate two or more AC voltages or currents that are out of step with each other. By "out of step," I mean that the two waveforms are not synchronized: that their peaks and zero points do not match up at the same points in time. The graph in figure 13.16 illustrates an example of this.

AB

AB

AB

AB Figure 13.18 - Out of phase waveforms

The two waves shown above (A versus B) are of the same amplitude and frequency, but they are out of step with each other. In technical terms, this is called a phase shift. Earlier we saw how we could plot a "sine wave" by calculating the trigonometric sine function for angles ranging from 0 to 360 degrees, a full circle. The starting point of a sine wave was zero amplitude at zero degrees, progressing to full positive amplitude at 90 degrees, zero at 180 degrees, full negative at 270 degrees, and back to the starting point of zero at 360 degrees. We can use this angle scale along the horizontal axis of our waveform plot to express just how far out of step one wave is with another.

degrees A

0

90

180

(0)

270

1.J 90

360

180

1 B

0

90

(0)

270

360

1 180

270

360 (0;

90

180

270

360

(0)

degrees Figure 13.19 - Wave A leads wave B by 450

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L;

r�

The shift between these two waveforms is about 45 degrees, the "A" wave being ahead of the "B" � L wave. A sampling of different phase shifts is given in the following graphs to better illustrate this concept.

Phase shift = 90 degrees A is ahead of B (A "leads" B)

Phase shift = 90 degrees

B is ahead of A (B "leads" A)

Phase shift = 180 degrees A and B waveforms are mirror-images of each other L Phase shift = 0 degrees A and B waveforms are in perfect step with each other Figure 13.20 - Examples of phase shifts. Because the waveforms in the above examples are at the same frequency, they will be out of step by the same angular amount at every point in time. For this reason, we can express phase shift for two or more waveforms of the same frequency as a constant quantity for the entire wave, and not just an expression of shift between any two particular points along the waves. That is, it is safe to say something like, "voltage 'A' is 45 degrees out of phase with voltage 'B'." Whichever waveform is ahead in its evolution is said to be leading and the one behind is said to

be lagging.

L

Phase shift, like voltage, is always a measurement relative between two things. There's really no such thing as a waveform with an absolute phase measurement because there's no known universal reference for phase. Typically in the analysis of AC circuits, the voltage waveform of the power supply is used as a reference for phase, that voltage stated as "xxx volts at 0 degrees." Any other AC voltage or current in that circuit will have its phase shift expressed in terms relative to that source voltage.

Module 3.13 AC Theory

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This is what makes AC circuit calculations more complicated than DC. When applying Ohm's Law and Kirchhoff's Laws, quantities of AC voltage and current must reflect phase shift as well as amplitude. Mathematical operations of addition, subtraction, multiplication, and division must operate on these quantities of phase shift as well as amplitude. Fortunately, there is a mathematical system of quantities called 'complex numbers' ideally suited for this task of representing amplitude and phase. However, this is beyond the scope of the EASA Part-66 syllabus, and will not be discussed.

Single and 3-Phase Principles Single-Phase Power Systems Depicted in Figure 13.21 is a very simple AC circuit. If the load resistor's power dissipation were substantial, we might call this a "power circuit" or "power system" instead of regarding it as just a regular circuit. The distinction between a "power circuit" and a "regular circuit" may seem arbitrary, but the practical concerns are definitely not.

load

load

#1

#2

Figure 13.21; Single phase power system schematic diagram. One such concern is the size and cost of wiring necessary to deliver power from the AC source to the load. Normally, we do not give much thought to this type of concern if we're merely analyzing a circuit for the sake of learning about the laws of electricity. However, in the real world it can be a major concern. If we give the source in the above circuit a voltage value and also give power dissipation values to the two load resistors, we can determine the wiring needs for this particular circuit.

load

load

#1

#2

3

P = 10 kW P= 1O kW Figure 13.22: As a practical matter, the wiring for the 20 kW loads at 120 Vac is rather substantial (167 A).

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120 V 1= 83.33 A

(for each load resistor)

lrotal

1load#l + Ito;

1mt"11

(83.33 A) + (83.33 A)

#

Ptotal

= (10 kW) + (10 kW)

Pw'tal = 20 kW

ltoaal = 166.67 A 83.33 amps for each load resistor in Figure 13.22 adds up to 166.66 amps total circuit current. This is no small amount of current, and would necessitate copper wire conductors of at least 1/0 gage. Such wire is well over 1/4 inch (6 mm) in diameter, weighing over 300 pounds per thousand feet. Bear in mind that copper is not cheap either! It would be in our best interest to find ways to minimize such costs if we were designing a power system with long conductor lengths. One way to do this would be to increase the voltage of the power source and use loads built to dissipate 10 kW each at this higher voltage. The loads, of course, would have to have greater resistance values to dissipate the same power as before (10 kW each) at a greater voltage than before. The advantage would be less current required, permitting the use of smaller, lighter, and cheaper wire.

load

240 V

#1

U

ti i

P= 1OkW

. load #2 P= 10 kW

Figure 13.23 - Same 10 kW loads at 240 Vac requires less substantial wiring than at 120 Vac (83 A).

U Module 3.13 AC Theory

13-23

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I- P E 10 kW 240V I = 41.67 A

7

�..J

(for each load resistor)

Itot ul = /load#l + Ilo #a

Ptotol = (10 kW) + (10 kW)

Itotal = (41.67 A) + (41.67 A)

Ptotal = 20 kW

n Itot;tl = 83.33 A

lr..,

Now our total circuit current is 83.33 amps, half of what it was before. We can now use number 4 gage wire, which weighs less than half of what 1/0 gage wire does per unit length. This is a considerable reduction in system cost with no degradation in performance. This is why power distribution system designers elect to transmit electric power using very high voltages (many thousands of volts): to capitalize on the savings realized by the use of smaller, lighter, cheaper wire. However, this solution is not without disadvantages. Another practical concern with power circuits is the danger of electric shock from high voltages. Again, this is not usually the sort of thing we concentrate on while learning about the laws of electricity, but it is a very valid concern in the real world, especially when large amounts of power are being dealt with. The gain in efficiency realized by stepping up the circuit voltage presents us with increased danger of electric shock. Power distribution companies tackle this problem by stringing their power lines along high poles or towers, and insulating the lines from the supporting structures with large, porcelain insulators. At the point of use (the electric power consumer), there is still the issue of what voltage to use for powering loads. High voltage gives greater system efficiency by means of reduced conductor current, but it might not always be practical to keep power wiring out of reach at the point of use the way it can be elevated out of reach in distribution systems. This tradeoff between efficiency and danger is one that European power system designers have decided to risk, all their households and appliances operating at a nominal voltage of 240 volts instead of 120 volts as it is in North America. That is why tourists from America visiting Europe must carry small step-down transformers for their portable appliances, to step the 240 VAC (volts AC) power down to a more suitable 120 VAC. Is there any way to realize the advantages of both increased efficiency and reduced safety hazard at the same time? One solution would be to install step-down transformers at the endpoint of power use, just as the American tourist must do while in Europe. However, this would be expensive and inconvenient for anything but very small loads (where the transformers can

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be built cheaply) or very large loads (where the expense of thick copper wires would exceed the

expense of a transformer).

An alternative solution would be to use a higher voltage supply to provide power to two lower voltage loads in series. This approach combines the efficiency of a high-voltage system with the safety of a low-voltage system.

--- 83.33 A

load

#1

t20V

10 kW

240 V

240V

load + #2

120 V

10 kW

83.33 A 1 Figure 13.24 - Series connected 120 Vac loads, driven by 240 Vac source at 83.3 A total

current.

Ui l r�

n

Notice the polarity markings (+ and -) for each voltage shown, as well as the unidirectional arrows for current. For the most part, I've avoided labeling "polarities" in the AC circuits we've been analyzing, even though the notation is valid to provide a frame of reference for phase. In later sections of this chapter, phase relationships will become very important, so I'm introducing this notation early on in the chapter for your familiarity. The current through each load is the same as it was in the simple 120 volt circuit, but the currents are not additive because the loads are in series rather than parallel. The voltage across each load is only 120 volts, not 240, so the safety factor is better. Mind you, we still have a full 240 volts across the power system wires, but each load is operating at a reduced voltage. If anyone is going to get shocked, the odds are that it will be from coming into contact with the conductors of a particular load rather than from contact across the main wires of a power

system. There's only one disadvantage to this design: the consequences of one load failing open, or being turned off (assuming each load has a series on/off switch to interrupt current) are not good. Being a series circuit, if either load were to open, current would stop in the other load as well. For this reason, we need to modify the design a bit.

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i83.33A

240 V

83.3 3 A -(♦ Figure 13.25 - Addition of neutral conductor allows loads to be individually driven.

Etotil =(120V L00)+ (120V /00) Etot"1t = 240 V L 0°

1=

1=

P E

10 kW 120 V

1= 83.33 A

Ptotal = (10 kW) + (10 kW) Pt0t:1 I = 20 kW

(for each load resistor)

Instead of a single 240 volt power supply, we use two 120 volt supplies (in phase with each other!) in series to produce 240 volts, then run a third wire to the connection point between the loads to handle the eventuality of one load opening. This is called a split-phase power system. Three smaller wires are still cheaper than the two wires needed with the simple parallel design, so we're still ahead on efficiency. The astute observer will note that the neutral wire only has to carry the diff erence of current between the two loads back to the source. In the above case, with perfectly "balanced" loads consuming equal amounts of power, the neutral wire carries zero current. Notice how the neutral wire is connected to earth ground at the power supply end. This is a common feature in power systems containing "neutral" wires, since grounding the neutral wire ensures the least possible voltage at any given time between any "hot" wire and earth ground. An essential component to a split-phase power system is the dual AC voltage source. Fortunately, designing and building one is not difficult. Since most AC systems receive their power from a step-down transformer anyway (stepping voltage down from high distribution levels to a user-level voltage like 120 or 240), that transformer can be built with a center-tapped secondary winding.

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Step-down transformer with center-tapped secondary winding 120 V 120 V

l

240 V

7

Figure 13.26 - American 120/240 Vac power is derived from a center tapped utility transformer. If the AC power comes directly from a generator (alternator), the coils can be similarly centertapped for the same effect. The extra expense to include a center-tap connection in a transformer or alternator winding is minimal. Here is where the (+) and (-) polarity markings really become important. This notation is often used to reference the phasings of multiple AC voltage sources, so it is clear whether they are aiding ("boosting") each other or opposing ("bucking") each other. If not for these polarity markings, phase relations between multiple AC sources might be very confusing. Note that the split-phase sources in the schematic (each one 120 volts @ 0°), with polarity marks (+) to (-) just like series-aiding batteries can alternatively be represented as such.

f

240 V

U

L 0°

U'

Figure 13.27 - Split phase 120/240 Vac source is equivalent to two series aiding 120 Vac

sources. To mathematically calculate voltage between "hot" wires, we must subtract voltages, because L their polarity marks show them to be opposed to each other: r

C-1 i

U FAI u

Il

Polar 120 L 0C - 120/ 1800 240L00

Rectangular 120 + jO V -(-120+j0)V 240+jOV

If we mark the two sources' common connection point (the neutral wire) with the same polarity mark (-), we must express their relative phase shifts as being 180° apart. Otherwise, we'd be

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denoting two voltage sources in direct opposition with each other, which would give 0 volts between the two "hot" conductors. Why am I taking the time to elaborate on polarity marks and phase angles? It will make more sense in the next section! Power systems in American households and light industry are most often of the split-phase variety, providing so-called 120/240 VAC power. The term "split-phase" merely refers to the split-voltage supply in such a system. In a more general sense, this kind of AC power supply is called single phase because both voltage waveforms are in phase, or in step, with each other. The term "single phase" is a counterpoint to another kind of power system called "polyphase" which we are about to investigate in detail. The advantages of polyphase power systems are more obvious if one first has a good understanding of single phase systems.

Three-Phase Power Systems Split-phase power systems achieve their high conductor efficiency and low safety risk by splitting up the total voltage into lesser parts and powering multiple loads at those lesser voltages, while drawing currents at levels typical of a full-voltage system. This technique, by the way, works just as well for DC power systems as it does for single-phase AC systems. Such systems are usually referred to as three-wire systems rather than split-phase because "phase" is a concept restricted to AC. But we know from our experience with vectors and complex numbers that AC voltages don't always add up as we think they would if they are out of phase with each other. This principle, applied to power systems, can be put to use to make power systems with even greater conductor efficiencies and lower shock hazard than with split-phase. Suppose that we had two sources of AC voltage connected in series just like the split-phase system we saw before, except that each voltage source was 120° out of phase with the other:

4 83.33A L0°

120 V °

"hat"

L 0

120 V 2120°

"neutral"

#

°

L 0

load • 120V #2 "

4

1J

+ 207.85 V L -30°

L 120°

" hot

83.33 A L 120°

Figure 13.28 - Pair of 120 Vac sources phased 120°, similar to split-phase. Since each voltage source is 120 volts, and each load resistor is connected directly in parallel with its respective source, the voltage across each load must be 120 volts as well. Given load currents of 83.33 amps, each load must still be dissipating 10 kilowatts of power. However, n 13-28

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voltage between the two "hot" wires is not 240 volts (120 L 0° - 120 L 180°) because the phase

difference between the two sources is not 180°. Instead, the voltage is: EtotaI

= (120 V L 0°) - (120 V Z 1200)

Erotal = 207.85 V L -30° Nominally, we say that the voltage between "hot" conductors is 208 volts (rounding up), and thus the power system voltage is designated as 120/208. If we calculate the current through the "neutral" conductor, we find that it is not zero, even with balanced load resistances. Kirchhoff's Current Law tells us that the currents entering and exiting the node between the two loads must be zero.

-

83.33 A L 0°

„hot„ load

#1 "neutral " '

'neutral

load #2

+

120V LO°

Node 120 V L 120°

"hot" -

83.33 A 212.0°

Figure 13.29 - Neutral wire carries a current in the case of a pair of 120° phased sources. -11oad#1

- 'load#2 - 'neutral = 0

-'neutral

'lond#l + ll0: #2

'neutral = -1lord#1 -'lo;rl±2

r;

'neutral

= - (83.33 A L 0°) - (83.33 A L 1200)

'neutral = 83.33 A L 240° or

83.33 A L -120°

So, we find that the "neutral" wire is carrying a full 83.33 amps, just like each "hot" wire. Note that we are still conveying 20 kW of total power to the two loads, with each load's "hot" wire carrying 83.33 amps as before. With the same amount of current through each "hot" wire, we must use the same gage copper conductors, so we haven't reduced system cost over the splitphase 120/240 system. However, we have realized a gain in safety, because the overall voltage between the two "hot" conductors is 32 volts lower than it was in the split-phase system (208 volts instead of 240 volts).

U

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The fact that the neutral wire is carrying 83.33 amps of current raises an interesting possibility: since it's carrying current anyway, why not use that third wire as another "hot" conductor, powering another load resistor with a third 120 volt source having a phase angle of 240°? That way, we could transmit more power (another 10 kW) without having to add any more conductors. Let's see how this might look. -4

8a. 3 AL00

load #1

120 V L°

load #3

913 3 A L '2400°

L20 V L 240° 1. 20 + Z L20°

120 V L0 kW

208 V L }o

L20 V 10 kW

7T 1

Ioad #2

\

�.

t. .J

120 V_

L0 kW

--S3.13AL L20° Figure 13.30 - With a third load phased 120° to the other two, the currents are the same as for two loads. Let's survey the advantages of a three-phase power system over a single-phase system of equivalent load voltage and power capacity. A single-phase system with three loads connected directly in parallel would have a very high total current (83.33 times 3, or 250 amps.

120V

250 A --

load

load

load

#1

#2

#3

10 kW

10 kW

10 kW

Figure 13.31 - For comparison, three 10 Kw loads on a 120 Vac system draw 250 A. J

This would necessitate 3/0 gauge copper wire (very large!), at about 510 pounds per thousand feet, and with a considerable price tag attached. If the distance from source to load was 1000 feet, we would need over a half-ton of copper wire to do the job. On the other hand, we could build a split-phase system with two 15 kW, 120 volt loads. (Figure below)

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�-- 125AL0°

"hot„ load #1

120V L0° .4

Li

120 V

L 180°

" neutral "r Ioad #2 0A

120V 15 kW

+

240 V 120 V 15 kW

L 0°

„ hot„

- 125AL 1800 Figure 13.32 - Split phase system draws half the current of 125 A at 240 Vac compared to 120

Vac system.

LJ ri

Our current is half of what it was with the simple parallel circuit, which is a great improvement. We could get away with using number 2 gage copper wire at a total mass of about 600 pounds, figuring about 200 pounds per thousand feet with three runs of 1000 feet each between source and loads. However, we also have to consider the increased safety hazard of having 240 volts present in the system, even though each load only receives 120 volts. Overall, there is greater potential for dangerous electric shock to occur. When we contrast these two examples against our three-phase system (Figure above), the advantages are quite clear. First, the conductor currents are quite a bit less (83.33 amps versus 125 or 250 amps), permitting the use of much thinner and lighter wire. We can use number 4 gage wire at about 125 pounds per thousand feet, which will total 500 pounds (four runs of 1000 feet each) for our example circuit. This represents a significant cost savings over the split-phase system, with the additional benefit that the maximum voltage in the system is lower (208 versus 240).

ti

L i

Li

One question remains to be answered: how in the world do we get three AC voltage sources whose phase angles are exactly 120° apart? Obviously we can't center-tap a transformer or alternator winding like we did in the split-phase system, since that can only give us voltage waveforms that are either in phase or 180° out of phase. Perhaps we could figure out some way to use capacitors and inductors to create phase shifts of 120°, but then those phase shifts would depend on the phase angles of our load impedances as well (substituting a capacitive or inductive load for a resistive load would change everything!). The best way to get the phase shifts we're looking for is to generate it at the source: construct the AC generator (alternator) providing the power in such a way that the rotating magnetic field passes by three sets of wire windings, each set spaced 120° apart around the circumference of the machine as in Figure 13.33.

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:.J

Three phase alternator winding winding

Single-phase alternator (a) winding

winding lb

(b,)

2a

3a

winding

f � . winding 3b

Figure 13.33 - (a) Single-phase alternator, (b) Three-phase alternator. Together, the six "pole" windings of a three-phase alternator are connected to comprise three winding pairs, each pair producing AC voltage with a phase angle 1200 shifted from either of the other two winding pairs. The interconnections between pairs of windings (as shown for the single-phase alternator: the jumper wire between windings 1 a and 1 b) have been omitted from the three-phase alternator drawing for simplicity. In our example circuit, we showed the three voltage sources connected together in a "Y" configuration (sometimes called the "star" configuration), with one lead of each source tied to a common point (the node where we attached the "neutral" conductor). The common way to depict this connection scheme is to draw the windings in the shape of a "Y" like Figure 13.34.

n

0 - 4' 120V

Figure 13.34 - Alternator "Y" configuration. The "Y" configuration is not the only option open to us, but it is probably the easiest to understand at first. More to come on this subject later in the chapter.

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Phase Rotation Let's take the three-phase alternator design laid out earlier (Figure 13.35) and watch what

happens as the magnet rotates.

L

winding winding 2a 3a

C..t

ran

r'

w i ing b wi41ding

Figure 13.35 - Three-phase alternator

I

The phase angle shift of 120° is a function of the actual rotational angle shift of the three pairs of windings (Figure 13.36). If the magnet is rotating clockwise, winding 3 will generate its peak instantaneous voltage exactly 120° (of alternator shaft rotation) after winding 2, which will hits its peak 120° after winding 1. The magnet passes by each pole pair at different positions in the rotational movement of the shaft. Where we decide to place the windings will dictate the amount of phase shift between the windings' AC voltage waveforms. If we make winding 1 our "reference" voltage source for phase angle (0°), then winding 2 will have a phase angle of -120° (120° lagging, or 240° leading) and winding 3 an angle of -240° (or 120° leading). This sequence of phase shifts has a definite order. For clockwise rotation of the shaft, the order is 12-3 (winding 1 peaks first, them winding 2, then winding 3). This order keeps repeating itself as long as we continue to rotate the alternator's shaft.. phase sequence: 1 -2-3-1

1

2

-2-3-1

-2-3

3

TIME - Figure 13.26 - Clockwise rotation phase sequence: 1-2-3 However, if we reverse the rotation of the alternator's shaft (turn it counter-clockwise), the magnet will pass by the pole pairs in the opposite sequence. Instead of 1-2-3, we'll have 3-2-1. Now, winding 2's waveform will be leading 120° ahead of 1 instead of lagging, and 3 will be another 120° ahead of 2.

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phase sequence3 - 2 - 1 - 3 - 2 - 1 - 3- 2- 1

3

2

1

TIME Figure 13.37 - Anticlockwise rotation phase sequence: 3-2-1.

The order of voltage waveform sequences in a polyphase system is called phase rotation or phase sequence. If we're using a polyphase voltage source to power resistive loads, phase rotation will make no difference at all. Whether 1-2-3 or 3-2-1, the voltage and current magnitudes will all be the same. There are some applications of three-phase power, that depend on having phase rotation being one way or the other. We've investigated how phase rotation is produced (the order in which pole pairs get passed by the alternator's rotating magnet) and how it can be changed by reversing the alternator's shaft rotation. However, reversal of the alternator's shaft rotation is not usually an option open to an end-user of electrical power supplied by a nationwide grid ("the" alternator actually being the combined total of all alternators in all power plants feeding the grid). There is a much easier way to reverse phase sequence than reversing alternator rotation: just exchange any two of the three "hot" wires going to a three-phase load. This trick makes more sense if we take another look at a running phase sequence of a threephase voltage source: 1-2-3 rotation: 1-2-3-1-2-3-1-2-3-1-2-3-1-2-3 ... 3-2-1 rotation: 3-2-1-3-2-1-3-2-1-3-2-1-3-2-1 What is commonly designated as a "1-2-3" phase rotation could just as well be called "2-3-1" or "3-1-2," going from left to right in the number string above. Likewise, the opposite rotation (3-21) could just as easily be called "2-1-3" or "1-3-2." Starting out with a phase rotation of 3-2-1, we can try all the possibilities for swapping any two of the wires at a time and see what happens to the resulting sequence in Figure below.

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Original 1-2-3

phase rotation

End result

1

U

(wires 1 and 2 swapped)

2

3

3

1

1

phase rotation = 2-1-3

(wires 2 and 3 swapped) phase rotation = 1-3-2

(wires 1 and 3 swapped) phase rotation = 3-2-1 Figure 13.38 - All possibilities of swapping any two wires. No matter which pair of "hot" wires out of the three we choose to swap, the phase rotation ends up being reversed (1-2-3 gets changed to 2-1-3, 1-3-2 or 3-2-1, all equivalent). Li

Three-Phase Y and A Configurations Initially we explored the idea of three-phase power systems by connecting three voltage sources together in what is commonly known as the "Y" (or "star") configuration. This configuration of voltage sources is characterized by a common connection point joining one side of each source.

120v

0°

120 v

L 120°

F

IL I ( Li

Figure 13.39 - Three-phase "Y" connection has three voltage sources connected to a common point.

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If we draw a circuit showing each voltage source to be a coil of wire (alternator or transformer winding) and do some slight rearranging, the "Y" configuration becomes more obvious in Figure13.40.

"line,"

"lure„ -1

"neutral"

"fine" Figure 13.40 - Three-phase, four-wire "Y" connection uses a "common" fourth wire. The three conductors leading away from the voltage sources (windings) toward a load are typically called lines, while the windings themselves are typically called phases. In a Yconnected system, there may or may not be a neutral wire attached at the junction point in the middle, although it certainly helps alleviate potential problems should one element of a threephase load fail open, as discussed earlier.

3-phase, 3-wire "Y" connection

n

1

'lineop

Ll

"fine"

n (no "neutral" wire)

117

�_J

"line" Figure 13.41 - Three-phase, three-wire "Y" connection does not use the neutral wire. When we measure voltage and current in three-phase systems, we need to be specific as to where we're measuring. Line voltage refers to the amount of voltage measured between any two line conductors in a balanced three-phase system. With the above circuit, the line voltage is roughly 208 volts. Phase voltage refers to the voltage measured across any one component (source winding or load impedance) in a balanced three-phase source or load. For the circuit shown above, the phase voltage is 120 volts. The terms line current and phase current follow the same logic: the former referring to current through any one line conductor, and the latter to current through any one component.

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Y-connected sources and loads always have line voltages greater than phase voltages, and line currents equal to phase currents. If the Y-connected source or load is balanced, the line voltage will be equal to the phase voltage times the square root of 3:

For "Y" circuits: 3

Elilte lEi ne

_.

EE,l,ase

lph Ise

However, the "Y" configuration is not the only valid one for connecting three-phase voltage

Li

source or load elements together. Another configuration is known as the "Delta," for its geometric resemblance to the Greek letter of the same name (A). Take close notice of the polarity for each winding in Figure below.

,line„ ,` line,`

Li

„liner, Figure 13.42 - Three-phase, three-wire A connection has no common. At first glance it seems as though three voltage sources like this would create a short-circuit, electrons flowing around the triangle with nothing but the internal impedance of the windings to hold them back. Due to the phase angles of these three voltage sources, however, this is not the case. One quick check of this is to use Kirchhoff`s Voltage Law to see if the three voltages around the loop add up to zero. If they do, then there will be no voltage available to push current around and around that loop, and consequently there will be no circulating current. Starting with the top winding and progressing counter-clockwise, our KVL expression looks something like this:

(120V L00)+(120V L 240°)+(120V L 120°)

Does it all equal 0 ? Yes ± Indeed, if we add these three vector quantities together, they do add up to zero. Another way to verify the fact that these three voltage sources can be connected together in a loop without

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resulting in circulating currents is to open up the loop at one junction point and calculate voltage across the break.

�.Elreak

should equal 0 V

Figure 13.43 - Voltage across open 0 should be zero. Starting with the right winding (120 V @ 120°) and progressing counter-clockwise, our KVL equation looks like this: (120 V L 120°) + (120 L 0°) + (120 V L 240°) + EL,renl: = 0

0+ErfefI=0 Erleok = 0 Sure enough, there will be zero voltage across the break, telling us that no current will circulate within the triangular loop of windings when that connection is made complete. Having established that a A-connected three-phase voltage source will not burn itself to a crisp due to circulating currents, we turn to its practical use as a source of power in three-phase circuits. Because each pair of line conductors is connected directly across a single winding in a A circuit, the line voltage will be equal to the phase voltage. Conversely, because each line conductor attaches at a node between two windings, the line current will be the vector sum of the two joining phase currents. Not surprisingly, the resulting equations for a 0 configuration are as follows:

Ford ("delta') circuits: Eliite = Ephyse

/line =_\/ 3

'pJi se

Let's see how this works in an example circuit: (Figure 13,44)

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r'

L20 V L 2400

`(

'Y Z 120"

LO kW 4-

s' L0 M

Figure 13.44 - The load on the A source is wired in a A. U

With each load resistance receiving 120 volts from its respective phase winding at the source, the current in each phase of this circuit will be 83.33 amps: 1= P

E

1=

10kW

120 V

1= 83.3 3 A (for each load resistor and source winding)

'line = V

L U

ki

{

'line

3

'phase

3

(83.33 A)

Y

1lilia = 144.34 A So each line current in this three-phase power system is equal to 144.34 amps, which is substantially more than the line currents in the Y-connected system we looked at earlier. One might wonder if we've lost all the advantages of three-phase power here, given the fact that we have such greater conductor currents, necessitating thicker, more costly wire. The answer is no. Although this circuit would require three number 1 gage copper conductors (at 1000 feet of distance between source and load this equates to a little over 750 pounds of copper for the whole system), it is still less than the 1000+ pounds of copper required for a single-phase system delivering the same power (30 kW) at the same voltage (120 volts conductor-toconductor). One distinct advantage of a A-connected system is its lack of a neutral wire. With a Yconnected system, a neutral wire was needed in case one of the phase loads were to fail open (or be turned off), in order to keep the phase voltages at the load from changing. This is not necessary (or even possible!) in a A-connected circuit. With each load phase element directly

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connected across a respective source phase winding, the phase voltage will be constant regardless of open failures in the load elements. Perhaps the greatest advantage of the A-connected source is its fault tolerance. It is possible for one of the windings in a A-connected three-phase source to fail open without affecting load voltage or current!

Figure 13.45 - Even with a source winding failure, the line voltage is still 120 V, and load phase voltage is still 120 V. The only difference is extra current in the remaining functional source windings. The only consequence of a source winding failing open for a A-connected source is increased phase current in the remaining windings. Compare this fault tolerance with a Y-connected system suffering an open source winding in Figure below.

Figure 13.46 - Open "Y" source winding halves the voltage on two loads of a A connected load. With a A-connected load, two of the resistances suffer reduced voltage while one remains at the original line voltage, 208. A Y-connected load suffers an even worse fate (Figure below) with the same winding failure in a Y-connected source.

f

n

13-40 use and/or disclosure is governed by the statement

eO..i a.i�Ok--

!1

Module 3.13 AC Theory Module 3.13 AC Theory

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L20 V L 0°

_

L20 V L 1200

L04 V `k A- L0_1 V

winding

failed open!

0

r

Figure 13.47 - Open source winding of a "Y-Y" system halves the voltage on two loads, and looses one load entirely. In this case, two load resistances suffer reduced voltage while the third loses supply voltage completely! For this reason, A-connected sources are preferred for reliability. However, if dual voltages are needed (e.g. 120/208) or preferred for lower line currents, Y-connected systems are the configuration of choice.

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Module 3.13 AC Theory

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Electrical Fundamentals 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits (l

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Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

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Copyright Notice O Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Total Training Support Ltd.

Knowledge Levels - Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows:

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LJ

LEVEL 3 • A detailed knowledge of the theoretical and practical aspects of the subject. • A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits Inductive and Capacitive Reactance Reactance, Impedance and Power Relationships in AC Circuits Ohms Law for AC

Power in AC Circuits

Power Factor Series RLC Circuits Parallel RLC Circuits Resonant Circuits

5 5 12 18

20 26 29 32 39

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Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

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Module 3.14 Enabling Objectives Objective

EASA 66 ReferenceLevel

Resistive (R), Capacitive (C) and Inductive (L) Circuits Phase relationship of voltage and current in L, C and R circuits, parallel, series and series parallel Power dissipation in L, C and R circuits Impedance, phase angle, power factor and current calculations True power, apparent power and reactive power calculations

3.14

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L i

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits Inductive and Capacitive Reactance You have already learned how inductance and capacitance individually behave in a direct current circuit. In this chapter you will be shown how inductance, capacitance, and resistance affect alternating current.

I

L r

L

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L ri L

Inductance and Alternating Current This might be a good place to recall what you learned about phase in chapter 1. When two things are in step, going through a cycle together, falling together and rising together, they are in phase. When they are out of phase, the angle of lead or lag-the number of electrical degrees by which one of the values leads or lags the other-is a measure of the amount they are out of step. The time it takes the current in an inductor to build up to maximum and to fall to zero is important for another reason. It helps illustrate a very useful characteristic of inductive circuitsthe current through the inductor always lags the voltage across the inductor. A circuit having pure resistance (if such a thing were possible) would have the alternating current through it and the voltage across it rising and failing together. This is illustrated in Figure 14.1 (A), which shows the sine waves for current and voltage in a purely resistive circuit having an AC source. The current and voltage do not have the same amplitude, but they are in phase. In the case of a circuit having inductance, the opposing force of the back-EMF would be enough to keep the current from remaining in phase with the applied voltage. You learned that in a dc circuit containing pure inductance the current took time to rise to maximum even though the full applied voltage was immediately at maximum. Figure 14.1 (B) shows the wave forms for a purely inductive AC circuit in steps of quarter-cycles.

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(A)

1 1

4 CYCLE

CYCLE

aCYCLE

7

lf

0°

90°

( $0°

270'

350°

Figure 14.1 - Voltage and current waveforms in an inductive circuit. With an AC voltage, in the first quarter-cycle (0° to 909 the applied AC voltage is continually increasing. If there was no inductance in the circuit, the current would also increase during this first quarter-cycle. You know this circuit does have inductance. Since inductance opposes any change in current flow, no current flows during the first quarter-cycle. In the next quarter-cycle (90°to 1809 the voltage decreases back to zero; c urrent begins to flow in the circuit and reaches a maximum value at the same instant the voltage reaches zero. The applied voltage now begins to build up to maximum in the other direction, to be followed by the resulting current. When the voltage again reaches its maximum at the end of the third quarter-cycle (2709 all values are exactly opposite to what they were during the first half-cycle. The applied voltage leads the resulting current by one quarter-cycle or 90 degrees. To complete the full 360°cycle of the voltage, the voltage again decreases to zero and the current builds to a maximum value. You must not get the idea that any of these values stops cold at a particular instant. Until the applied voltage is removed, both current and voltage are always changing in amplitude and direction. 14-6 Use and/or governed by the disclosure is statement

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As you know the sine wave can be compared to a circle. Just as you mark off a circle into 360 degrees, you can mark off the time of one cycle of a sine wave into 360 electrical degrees. This relationship is shown in Figure 14.2. By referring to this figure you can see why the current is said to lag the voltage, in a purely inductive circuit, by 90 degrees. Furthermore, by referring to figures 7-2 and 7-1 (A) you can see why the current and voltage are said to be in phase in a purely resistive circuit. In a circuit having both resistance and inductance then, as you would expect, the current lags the voltage by an amount somewhere between 0 and 90 degrees.

90° L

180"

2700

360°

goo

L..

r_..t L r~1

2700

Figure 14.2 - Comparison of sine wave and circle in an inductive circuit. A simple memory aid to help you remember the relationship of voltage and current in an inductive circuit is the word VIL. Since V is (sometimes) the symbol for voltage, L is the symbol for inductance, and I is the symbol for current; the word VIL demonstrates that current comes after (lags) voltage in an inductor.

L

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Inductive Reactance When the current flowing through an inductor continuously reverses itself, as in the case of an AC source, the inertia effect of the back-EMF is greater than with dc. The greater the amount of inductance (L), the greater the opposition from this inertia effect. Also, the faster the reversal of current, the greater this inertial opposition. This opposing force which an inductor presents to the FLOW of alternating current cannot be called resistance, since it is not the result of friction within a conductor. The name given to it is inductive reactance because it is the "reaction" of the inductor to the changing value of alternating current. Inductive reactance is measured in ohms and its symbol is XL. As you know, the induced voltage in a conductor is proportional to the rate at which magnetic lines of force cut the conductor. The greater the rate (the higher the frequency), the greater the backEMF. Also, the induced voltage increases with an increase in inductance; the more ampere-turns, the greater the back-EMF. Reactance, then, increases with an increase of frequency and with an increase of inductance. The formula for inductive reactance is as follows:

XL = 2?fL Iher e: XL is inductive reactance in ohms. 2 x is a constant in which the Or eekletter

,

called "pi" r epr e s ents 3.1416 and 2 x r _ 6.28 approximately, f

n

is frequency of the alternating current in Hz,

L is inductance in henrys. The following example problem illustrates the computation of XL.

Given:

f = 60 Hz L=20H

Solution,X L = 2 rfL XL =628x60 Hzx20H X L = 7,536 Q

n

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Capacitors and Alternating Current The four parts of Figure 14.3 show the variation of the alternating voltage and current in a capacitive circuit, for each quarter of one cycle. The solid line represents the voltage across the capacitor, and the dotted line represents the current. The line running through the centre is the zero, or reference point, for both the voltage and the current. The bottom line marks off the time of the cycle in terms of electrical degrees. Assume that the AC voltage has been acting on the

capacitor for some time before the time represented by the starting point of the sine wave in the figure. F _7 M. XIMU11 Fowl i

1.1rAXI!.1UK POsn r/IE_

"VOLTAGE

*♦

CURRENT ZERO

ZERO

MA::C111U11 rQCc,,TiVC Ttr.1E

VOLTAGE ♦

+

GIAXIMuM URREFff

NEGATIVE U°

97`

1.3 °

tEC'

TEa'

MAXIMUM POSITIVE

IAAXFdUI.1

POSIT IVE

\�-- vc1.TAGE

VOLTAGE /

ZERO

ZERO v

MAXIMUM

CURRENT

MAX IMUM

CURRENT k

NEGATIVE

I 1 ECATIS; E TIME " 00

2700

(C)

(A)

t U

`

TIME

27O

90

TIME °

1 BO'

(B)

270C

?&'

9O,

15]"

2700

33O"

(D)

Figure 14.3 - Phase relationship of voltage and current in a capacitive circuit.

Li

At the beginning of the first quarter-cycle (0°to 909 the voltage has just passed through zero and is increasing in the positive direction. Since the zero point is the steepest part of the sine wave, the voltage is changing at its greatest rate. The charge on a capacitor varies directly with the voltage, and therefore the charge on the capacitor is also changing at its greatest rate at the beginning of the first quarter-cycle. In other words, the greatest number of electrons is moving off one plate and onto the other plate. Thus the capacitor current is at its maximum value, as part (A) of the figure shows. As the voltage proceeds toward maximum at 90 degrees, its rate of change becomes less and less, hence the current must decrease toward zero. At 90 degrees the voltage across the capacitor is maximum, the capacitor is fully charged, and there is no further movement of electrons from plate to plate. That is why the current at 90 degrees is zero. At the end of this first quarter-cycle the alternating voltage stops increasing in the positive direction and starts to decrease. It is still a positive voltage, but to the capacitor the decrease in voltage means that the plate which has just accumulated an excess of electrons must lose some electrons. The current flow, therefore, must reverse its direction. Part (B) of the figure

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

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shows the current curve to be below the zero line (negative current direction) during the second quarter-cycle (90°to 1809. At 180 degrees the voltage has dropped to zero. This means that for a brief instant the electrons are equally distributed between the two plates; the current is maximum because the rate of change of voltage is maximum. Just after 180 degrees the voltage has reversed polarity and starts building up its maximum negative peak which is reached at the end of the third quarter-cycle (180°to 270). During this third qua rter-cycle the rate of voltage change gradually decreases as the charge builds to a maximum at 270 degrees. At this point the capacitor is fully charged and it carries the full impressed voltage. Because the capacitor is fully charged there is no further exchange of electrons; therefore, the current flow is zero at this point. The conditions are exactly the same as at the end of the first quarter-cycle (909 but the polarity is reversed. Just after 270 degrees the impressed voltage once again starts to decrease, and the capacitor

I

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must lose electrons from the negative plate. It must discharge, starting at a minimum rate of

flow and rising to a maximum. This discharging action continues through the last quarter-cycle (270°to 360) until the impressed-voltage has reac hed zero. At 360 degrees you are back at the beginning of the entire cycle, and everything starts over again. If you examine the complete voltage and current curves in part D, you will see that the current always arrives at a certain point in the cycle 90 degrees ahead of the voltage, because of the charging and discharging action. You know that this time and place relationship between the current and voltage is called the phase relationship. The voltage-current phase relationship in a capacitive circuit is exactly opposite to that in an inductive circuit. The current of a capacitor leads the voltage across the capacitor by 90 degrees. You realize that the current and voltage are both going through their individual cycles at the same time during the period the AC voltage is impressed. The current does not go through part of its cycle (charging or discharging), stop, and wait for the voltage to catch up. The amplitude and polarity of the voltage and the amplitude and direction of the current are continually changing. Their positions with respect to each other and to the zero line at any electrical instantany degree between zero and 360 degrees-can be seen by reading upwards from the timedegree line. The current swing from the positive peak at zero degrees to the negative peak at 180 degrees is NOT a measure of the number of electrons, or the charge on the plates. It is a picture of the direction and strength of the current in relation to the polarity and strength of the voltage appearing across the plates. At times it is convenient to use the word "CIV" to recall to mind the phase relationship of the current and voltage in capacitive circuits. I is the symbol for current, and in the word CIV it leads, or comes before, the symbol for voltage, V. C, of course, stands for capacitor. This memory aid is similar to the "VIL" used to remember the current and voltage relationship in an inductor. The word "CIVIL" is helpful in remembering the phase relationship in both the inductor and capacitor. Since the plates of the capacitor are changing polarity at the same rate as the AC voltage, the capacitor seems to pass an alternating current. Actually, the electrons do not pass through the dielectric, but their rushing back and forth from plate to plate causes a current flow in the circuit. It is convenient, however, to say that the alternating current flows "through" the capacitor. You know this is not true, but the expression avoids a lot of trouble when speaking of current flow in a circuit containing a capacitor. By the same short cut, you may say that the capacitor does not 14-10 use and/or disclosure is governed by the slalement

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Li

pass a direct current (if both plates are connected to a dc source, current will flow only long enough to charge the capacitor). With a capacitor type of hook-up in a circuit containing both AC and dc, only the AC will be "passed" on to another circuit. You have now learned two things to remember about a capacitor: A capacitor will appear to conduct an alternating current and a capacitor will not conduct a direct current.

Li s✓

Capacitive Reactance So far you have been dealing with the capacitor as a device which passes AC and in which the only opposition to the alternating current has been the normal circuit resistance present in any conductor. However, capacitors themselves offer a very real opposition to current flow. This opposition arises from the fact that, at a given voltage and frequency, the number of electrons which go back and forth from plate to plate is limited by the storage ability-that is, the capacitanceof the capacitor. As the capacitance is increased, a greater number of electrons change plates every cycle, and (since current is a measure of the number of electrons passing a given point in a given time) the current is increased. Increasing the frequency will also decrease the opposition offered by a capacitor. This occurs because the number of electrons which the capacitor is capable of handling at a given voltage will change plates more often. As a result, more electrons will pass a given point in a given time (greater current flow). The opposition which a capacitor offers to AC is therefore inversely proportional to frequency and to capacitance. This opposition is called capacitive reactance.

You may say that capacitive reactance decreases with increasing frequency or, for a given

frequency, the capacitive reactance decreases with increasing capacitance. The symbol for capacitive reactance is Xc.

L

Now you can understand why it is said that the Xc varies inversely with the product of the frequency and capacitance. The formula is:

XC=

1 2vfC

Where: Xc is capacitive reactance in ohms f is frequency in Hertz C is capacitance in farads ❑

is 6.28 (2 X 3.1416) The following example problem illustrates the computation of X c. Given:

f=100 Hz C= SOgF

1

Solution:

27CfC Xc

1 6,28 x 100 Hz x 50,aF

1 =,03140 XC =31.89 or 32Q Xc

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Reactance, Impedance and Power Relationships in AC Circuits Up to this point inductance and capacitance have been explained individually in AC circuits. The

rest of this chapter will concern the combination of inductance, capacitance, and resistance in AC circuits.

To explain the various properties that exist within AC circuits, the series RLC circuit will be used. Figure 14.4 is the schematic diagram of the series RLC circuit. The symbol shown in Figure 14.4 that is marked E is the general symbol used to indicate an AC voltage source.

Figure 14.4 - Series RLC circuit.

7

Reactance The effect of inductive reactance is to cause the current to lag the voltage, while that of capacitive reactance is to cause the current to lead the voltage. Therefore, since inductive reactance and capacitive reactance are exactly opposite in their effects, what will be the result when the two are combined? It is not hard to see that the net effect is a tendency to cancel each other, with the combined effect then equal to the difference between their values. This resultant is called reactance; it is represented by the symbol X; and expressed by the equation X = XLXc or X = Xc - X L. Thus, if a circuit contains 50 ohms of inductive reactance and 25 ohms of capacitive reactance in series, the net reactance, or X, is 50 ohms - 25 ohms, or 25 ohms of inductive reactance. For a practical example, suppose you have a circuit containing an inductor of 100 pH in series with a capacitor of 0.001 pF, and operating at a frequency of 4 MHz. What is the value of net reactance, or X?

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Ui

Given:

f=4MHz L =100AH C =.001 A.F

L

Solution, XL = 2 fL XL =6.20x4MHzx100p.H

XL =25125 XC=

1

2zfC

XC=

XC

1 6,28 x 4 lvMHz x .001 j F _ 1 02512

XC=39.85 X=XL-XC X=25125-39.85 X = 2472.2 Q (inductive)

Now assume you have a circuit containing a 100pH inductor in series with a 0.0002pF capacitor, and operating at a frequency of 1 MHz. What is the value of the resultant reactance in this case? Given:

f =1 M Hz L =100.H

C=.0002 AF

L

Solution; XL = 2 fL. XL =6.28 x 1MHz x 100piH XL =6285 1

XTiC 1

XC=

6.28 x 1MHz x .0002pF I X C 001256 Xc 7960 X=XC-XL X=7965-6285

X = 168 Q (capacitive)

on page 2 of this Chapter.

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You will notice that in this case the inductive reactance is smaller than the capacitive reactance and is therefore subtracted from the capacitive reactance. These two examples serve to illustrate an important point: when capacitive and inductive reactance are combined in series, the smaller is always subtracted from the larger and the resultant reactance always takes the characteristics of the larger. Impedance From your study of inductance and capacitance you know how inductive reactance and capacitive reactance act to oppose the flow of current in an AC circuit. However, there is another factor, the resistance, which also opposes the flow of the current. Since in practice AC circuits containing reactance also contain resistance, the two combine to oppose the flow of current. This combined opposition by the resistance and the reactance is called the impedance,

and is represented by the symbol Z.

Since the values of resistance and reactance are both given in ohms, it might at first seem possible to determine the value of the impedance by simply adding them together. It cannot be done so easily, however. You know that in an AC circuit which contains only resistance, the current and the voltage will be in step (that is, in phase), and will reach their maximum values at the same instant. You also know that in an AC circuit containing only reactance the current will either lead or lag the voltage by one-quarter of a cycle or 90 degrees. Therefore, the voltage in a purely reactive circuit will differ in phase by 90 degrees from that in a purely resistive circuit and for this reason reactance and resistance are rot combined by simply adding them. When reactance and resistance are combined, the value of the impedance will be greater than either. It is also true that the current will not be in step with the voltage nor will it differ in phase by exactly 90 degrees from the voltage, but it will be somewhere between the in-step and the 90degree out-of-step conditions. The larger the reactance compared with the resistance, the more nearly the phase difference will approach 90° The larger the resistance compared to the reactance, the more nearly the phase difference will approach zero degrees. If the value of resistance and reactance cannot simply be added together to find the impedance, or Z, how is it determined? Because the current through a resistor is in step with the voltage across it and the current in a reactance differs by 90 degrees from the voltage across it, the two are at right angles to each other. They can therefore be combined by means of the same method used in the construction of a right-angle triangle. Assume you want to find the impedance of a series combination of 8 ohms resistance and 5 ohms inductive reactance. Start by drawing a horizontal line, R, representing 8 ohms resistance, as the base of the triangle. Then, since the effect of the reactance is always at right angles, or 90 degrees, to that of the resistance, draw the line XL, representing 5 ohms inductive reactance, as the altitude of the triangle. This is shown in Figure 14.5. Now, complete the hypotenuse (longest side) of the triangle. Then, the hypotenuse represents the impedance of the circuit.

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r-''

b

R=8OHMS

Figure 14.5 - Vector diagram showing relationship of resistance, inductive reactance, and impedance in a series circuit. One of the properties of a right triangle is:

(hypotenuse)2 = (base)2 +(altitude) 2 or,

hypotenuse II

Applied to impedance, this becomes,

(impedance) 2 = (r esis tan ce) 2 +(r eactan ce)2

[--1

L

or,

Li

or,

impedance =

1 L

(base)2 +(altitude)2

(resistance 2 +(reactance)2

z = R2+X2 Now suppose you apply this equation to check your results in the example given above. Given:

R=9S XL=5S2

Solution:

Z= R 2 Z= (�

X X L2

)2 +(SS �z

Z

64+252

Z=

89 Q

Z =9.4Q

(See the Appendix III for a square Root Table)

LJ When you have a capacitive reactance to deal with instead of inductive reactance as in the previous example, it is customary to draw the line representing the capacitive reactance in a downward direction. This is shown in Figure 14.6. The line is drawn downward for capacitive reactance to indicate that it acts in a direction opposite to inductive reactance which is drawn

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upward. In a series circuit containing capacitive reactance the equation for finding the impedance becomes: Z

+ R =.8 OHMS

'>."

lXc= 5 OHMS

Figure 14.6 - Vector diagram showing relationship of resistance, capacitive reactance, and impedance in a series circuit. In many series circuits you will find resistance combined with both inductive reactance and capacitive reactance. Since you know that the value of the reactance, X, is equal to the difference between the values of the inductive reactance, XL, and the capacitive reactance, Xc, the equation for the impedance in a series circuit containing R, XL, and Xc then becomes:

Z= R2+(XL-Xc)2 or, Z = R2 + X2 (Note: The formulas Z = R2 + XL2, Z= R2+Xc2, and Z= R

+ +X2 canoe

usedto calculate Z onlyif the resistance and reactance are connected in series.)

In Figure 14.7 you will see the method which may be used to determine the impedance in a

l

series circuit consisting of resistance, inductance, and capacitance.

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7

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(nn,rinht 9010

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Figure 14.7 - Vector diagram showing relationship of resistance, reactance (capacitive and inductive), and impedance in a series circuit. L

Assume that 10 ohms inductive reactance and 20 ohms capacitive reactance are connected in series with 40 ohms resistance. Let the horizontal line represent the resistance R. The line drawn upward from the end of R, represents the inductive reactance, XL. Represent the capacitive reactance by a line drawn downward at right angles from the same end of R. The resultant of XL and Xc is found by subtracting XL from Xc. This resultant represents the value of X. Thus: X=XC - XL X =10 ohms

Li

F1 u

The line, Z, will then represent the resultant of R and X. The value of Z can be calculated as follows: Given:

XL =10 n XC=20 0

R=40 0 Solution:

X = X c- X L X=20Q - 105 X=100

r. ,

U

Z=

+XZ

Z = (405

2 + (10 52 )

Z=

1600 + 100 0

Z

17 0 00

Z=41.20

r,

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

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Ohms Law for AC

U-U

In general, Ohm's law cannot be applied to alternating-current circuits since it does not consider the reactance which is always present in such circuits. However, by a modification of Ohm's law which does take into consideration the effect of reactance we obtain a general law which is applicable to AC circuits. Because the impedance, Z, represents the combined opposition of all the reactances and resistances, this general law for AC is,

7 r-1 tJ

,

E 7 This general modification applies to alternating current flowing in any circuit, and any one of the values may be found from the equation if the others are known. For example, suppose a series circuit contains an inductor having 5 ohms resistance and 25 ohms inductive reactance in series with a capacitor having 15 ohms capacitive reactance. If the voltage is 50 volts, what is the current? This circuit can be drawn as shown in Figure 14.8.

Figure 14.8 - Series LC circuit.

Given:

R=5r2

XL =25Q

n

XC =150

E = 50

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Solution: X= X L- X c X=25Q - 1552

X =10:2 Z= {5Q � + (10502 -J25 ++100 2 Z= 12552 Z =11.252 ITE

Z 50V 11.2 S2 I = 4.46 A

Li. F L

L)

Now suppose the circuit contains an inductor having 5 ohms resistance and 15 ohms inductive reactance in series with a capacitor having 10 ohms capacitive reactance. If the current is 5 amperes, what is the voltage? Given:

R=SQ X L =1552 XC=1052 I = 5A

r

L

Solution:

X= X L - X C

X=15Q- 1052

X=552 Z = R2 + X2

r, u

Z = 25 + 25 52

z

5052

Z=7.075

E=IZ E=SA x 7.0752

L

E = 35.35 V

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

14-19

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Power in AC Circuits You know that in a direct current circuit the power is equal to the voltage times the current, or P = E X I. If a voltage of 100 volts applied to a circuit produces a current of 10 amperes, the power is 1000 watts. This is also true in an AC circuit when the current and voltage are in phase; that is, when the circuit is effectively resistive. But, if the AC circuit contains reactance, the current will lead or lag the voltage by a certain amount (the phase angle). When the current is out of phase with the voltage, the power indicated by the product of the applied voltage and the total current gives only what is known as the apparent power. The true power depends upon the phase angle between the current and voltage. The symbol for phase angle is 0 (Theta). When an alternating voltage is impressed across a capacitor, power is taken from the source and stored in the capacitor as the voltage increases from zero to its maximum value. Then, as

the impressed voltage decreases from its maximum value to zero, the capacitor discharges and returns the power to the source. Likewise, as the current through an inductor increases from its zero value to its maximum value the field around the inductor builds up to a maximum, and when the current decreases from maximum to zero the field collapses and returns the power to the source. You can see therefore that no power is used up in either case, since the power alternately flows to and from the source. This power that is returned to the source by the reactive components in the circuit is called reactive power. In a purely resistive circuit all of the power is consumed and none is returned to the source; in a purely reactive circuit no power is consumed and all of the power is returned to the source. It follows that in a circuit which contains both resistance and reactance there must be some power dissipated in the resistance as well as some returned to the source by the reactance. In Figure 14.9 you can see the relationship between the voltage, the current, and the power in such a circuit. The part of the power curve which is shown below the horizontal reference line is the result of multiplying a positive instantaneous value of current by a negative instantaneous value of the voltage, or vice versa. As you know, the product obtained by multiplying a positive value by a negative value will be negative. Therefore the power at that instant must be considered as negative power. In other words, during this time the reactance was returning power to the source.

Figure 14.9 - Instantaneous power when current and voltage are out of phase. iJ

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The instantaneous power in the circuit is equal to the product of the applied voltage and current

through the circuit. When the voltage and current are of the same polarity they are acting

together and taking power from the source. When the polarities are unlike they are acting in

opposition and power is being returned to the source. Briefly then, in an AC circuit which

contains reactance as well as resistance, the apparent power is reduced by the power returned to the source, so that in such a circuit the net power, or true power, is always less than the apparent power. Calculating True Power in AC Circuits As mentioned before, the true power of a circuit is the power actually used in the circuit. This power, measured in watts, is the power associated with the total resistance in the circuit. To

calculate true power, the voltage and current associated with the resistance must be used. Since the voltage drop across the resistance is equal to the resistance multiplied by the current through the resistance, true power can be calculated by the formula:

True Power = G R 12 R Mere:

True Power isrneasuredinwatts I R is resistive current in amperes R is resistance in ohms

ri

L For example, find the true power of the circuit shown in Figure 14.10.

E=500V

L

Figure 14.10 - Example circuit for determining power.

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

14-21

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

R= 60 5

Given:

XL =3052

XC=1105 E=50011

Solution:

X=XC - XL X=1105 - 30Q

X=80Q Z= R2 +X2 Z = (605 )2 + (800)2 Z = 3600 + 6400 Q Z = 10,000 0

Z= 100 Q E I=Z I =S00 W 10052 I =5A

Since the current in a series circuit is the same in all parts of the circuit:

TruePower =(I R)2R TruePower = (5 A) 2 x 60 52 True Power

14-22

,,, 9 n( 4h1

E

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

Use and/or disclosure is governed by the statement nn

1500 watts

r-K-1-

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J

Calculating Reactive Power in AC Circuits The reactive power is the power returned to the source by the reactive components of the circuit. This type of power is measured in Volt-Amperes-Reactive, abbreviated VAR. Reactive power is calculated by using the voltage and current associated with the circuit reactance.

r

i '

q 1+ +k eeac Once mu ip iey te reactive current,r Itt d g reactive power can be calculated by the formula:

Reactive Power = (Ix)2X

Where: Reactive p over is me a s ur e d in volt amperes -reactive. I is reactive current in amper es. Xis total reactance in ohms. L

Another way to calculate reactive power is to calculate the inductive power and capacitive power and subtract the smaller from the larger. Reactive Power = (IL)2XL -

F

(I c)2Xc

or (IC)2XC - (IL)2XL

There: Reactive power is measuredinvoltamperes -reactive. I cis capacitive curl entin amperes. Xc is capacitive reactance in ohms. IL is inductive curieri in amperes.

E XL is inductive reactance in ohms. Either one of these formulas will work. The formula you use depends upon the values you are K

given in a circuit.

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

14-23

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For example, find the reactive power of the circuit shown in Figure 14.10.

Given:

XL =30Q Xc=110n X=eon I =5A

Since this is a series circuit, current (I) is the same in all parts of the circuit.

Solution:

Rea ctive p ower = (I X) 2 X Re active power =(5A)2 x 80Q Reactive power = 2,000 vat

To prove the second formula also works, Reactive power =(Ic)2Xc - (I L)2 XL

Rea ctive power =(5A)2 x 11052 - (SA) 2 x30 Reactive power = Z750 var - 750 var Reactive power = 2000 var Calculating Apparent Power in AC Circuits Apparent power is the power that appears to the source because of the circuit impedance. Since the impedance is the total opposition to AC, the apparent power is that power the voltage source "sees." Apparent power is the combination of true power and reactive power. Apparent power is not found by simply adding true power and reactive power just as impedance is not found by adding resistance and reactance. To calculate apparent power, you may use either of the following formulas:

14-24 Use and/or disclosure is statement governed by the

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Apparentpower =(I2 )2Z Where;

Apparent power is measured in VA (volt-amperes)

It

1z is impedance cur r ent in amperes.

r, 4

Z is impedance in ohms.

L

or

Apparentpower = (Truepower)2 + (reactive power)2 For example, find the apparent power for the circuit shown in Figure 14.10.

Z =100 n

Given:

I =5A Recall that current in a series circuit is the same in all parts of the circuit.

I

L

Solution: ApparentPowei =(Iz)2Z Apparentpower = (5 A)2 x 100 Q Apparentpower = 2500 VA or

L Given:

True power =1500 W Reactive power = 2000 var Apparentpower = (True power )2 + (reactive power)2

L

Apparentpower = (1500W)2 + (2000var)2 Apparentpower = 625 x 10 VA

App ar ent p ow er = 2500 VA

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

14-25

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Use and/or disclosure is governed by the statement ., - 0 If lhk rK.-.

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Power Factor The power factor is a number (represented as a decimal or a percentage) that represents the portion of the apparent power dissipated in a circuit. If you are familiar with trigonometry, the easiest way to find the power factor is to find the cosine of the phase angle 0. The cosine of the phase angle is equal to the power factor. You do not need to use trigonometry to find the power factor. Since the power dissipated in a circuit is true power, then:

Apparent Power x PF =True Power,

Therefore,

PF= True Power Apparent Power

If true power and apparent power are known you can use the formula shown above. Going one step further, another formula for power factor can be developed. By substituting the equations for true power and apparent power in the formula for power factor, you get:

PF= R)'P` (Iz)' Z Since current in a series circuit is the same in all parts of the circuit, IR equals lz. Therefore, in a series circuit,

PF= R 7 For example, to compute the power factor for the series circuit shown in Figure 14.10, any of the above methods may be used.

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7

Integrated Training System Designed in association with the ciub66pro.co.uk question practice aid

Given:

True Power =1500` Apparent Power = 2500 VA Solution:

PF = True Power Appar ent Power

U L

PF=

1500 W

2500 VA PF= .6 Another method: Given:

R=60 S Z=100 n

L Solution:

PF= R

z

PF= 60 Q 1005 PF= .6

i Li

If you are familiar with trigonometry you can use it to solve for angleo and the power factor by referring to the tables in appendices V and VI. r-j 1

L

Given:

R=60 0 X=80 0

Solution:

tan 0 = X R

tan & = 80 0 605 tan 8=1.333 8 = 53.1° rI

PF=cos 0 PF = .6 NOTE: As stated earlier the power factor can be expressed as a decimal or percentage. In this example the decimal number.6 could also be expressed as 60%.

L

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

14-27

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Power Factor Correction

The apparent power in an AC circuit has been described as the power the source "sees". As far as the source is concerned the apparent power is the power that must be provided to the circuit.

You also know that the true power is the power actually used in the circuit. The difference

between apparent power and true power is wasted because, in reality, only true power is consumed. The ideal situation would be for apparent power and true power to be equal. If this were the case the power factor would be 1 (unity) or 100 percent. There are two ways in which this condition can exist. (1) If the circuit is purely resistive or (2) if the circuit "appears" purely resistive to the source. To make the circuit appear purely resistive there must be no reactance. To have no reactance in the circuit, the inductive reactance (XL) and capacitive reactance (Xc) must be equal.

Remember,

X =XL - X c

Therefore, when XL =XC, X=0 The expression "correcting the power factor" refers to reducing the reactance in a circuit. The ideal situation is to have no reactance in the circuit. This is accomplished by adding capacitive reactance to a circuit which is inductive and inductive reactance to a circuit which is capacitive. For example, the circuit shown in Figure 14.10 has a total reactance of 80 ohms capacitive and the power factor was 0.6 or 60 percent. If 80 ohms of inductive reactance were added to this circuit (by adding another inductor) the circuit would have a total reactance of zero ohms and a power factor of 1 (or 100 percent). The apparent and true power of this circuit would then be equal.

l

1

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Series RLC Circuits The principles and formulas that have been presented in this chapter are used in all AC circuits. The examples given have been series circuits. This section of the chapter will not present any new material, but will be an example of using all the principles presented so far. You should follow each example problem step by step to see how each formula used depends upon the information determined in earlier steps. f' �

L The example series RLC circuit shown in Figure 14.11 will be used to solve for XL, Xc, X, Z, IT, true power, reactive power, apparent power, and power factor. r -�

The values solved for will be rounded off to the nearest whole number. First solve for XL and Xc. f = 60 Hz

Given:

L = 27mH

L

C=380 j.tF Solution:

XL =2tfl X L = 6. 28 x 60 Hz x 27 mH

XL=10 s Xc= Xr= X

c

1 2 rfc 1 6, 28 X 60 Hz x 380 it F 1 0. 143

Xc=7s

r L L

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

14-29

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R

4c E=110V f= 60Hz

Figure 14.11 - Example series

7

L 27mH

C 38OpF RLC circuit

Now solve for X Given:

L.1

XC=79 XL =105

Solution:

X= X L- X C

X=10Q - 7S2 X = 3 0 (Inductive) Use the value of X to solve for Z. Given:

X=3Q R=452

Solution:

Z=X 2+ R2 Z = (3Q)2 + (4Q)2 Z

9+165

Z

25

Z=

50

nI

This value of Z can be used to solve for total current (IT ). Given:

Z= S Q E= 110V

Solution: IT =

5o IT =22A

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IT

Inductive (L) Circuits Use and/or disclosure is

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Since current is equal in all parts of a series circuit, the value of IT can various values of power.

be used to solve for the

IT -22A

Given:

R= 4Q

X= 3S Z= 5Q Solution: True Power =(IR)2R True Power = (22 A) 2 x 4Q

True Power =1936 W Reactive power =(I =(Ix)2X Reactive power =(22A)2 x 3n Reactive power =1452 var

Apparentpower =(Iz)2Z Apparent Power = (22A)2 x 50 Apparent Power = 2420 VA

The power factor can now be found using either apparent power a nd true power or resistance and impedance. The mathematics in this example is easier if you use impedance and resistance. Given:

R = 4Q Z

L

- 5Sa

PF= R Z PF= 4C) sn

PF =, 8cr 80%

Module 3.14 Resistive (R), Capacitive (C) and inductive (L) Circuits

14-31

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"J

fi

Parallel RLC Circuits When dealing with a parallel AC circuit, you will find that the concepts presented in this chapter for series AC circuits still apply. There is one major difference between a series circuit and a parallel circuit that must be considered. The difference is that current is the same in all parts of a series circuit, whereas voltage is the same across all branches of a parallel circuit. Because of this difference, the total impedance of a parallel circuit must be computed on the basis of the current in the circuit. You should remember that in the series RLC circuit the following three formulas were used to find reactance, impedance, and power factor:

XCorX=XC - XL

X=XL -

(IR)2 +X2

Z=

PF= R Z When working with a parallel circuit you must use the following formulas instead: Ix=IL -IC or IX=IC - IL Iz _ (IR}2 + (I}r)2

U

PF= IR I7

(The impedance of a parallel cir cut is found by the formula Z = E )

z

NOTE: If no value for E is given in a circuit, any value of E can be assumed to find the values of IL, IC, lx, IR, and Iz. The same value of voltage is then used to find impedance. For example, find the value of Z in the circuit shown in Figure 14.12.

Given:

E =300 V R =100 0 XL= 50: X1=150 S

The first step in solving for Z is to calculate the individual branch currents.

14-32

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

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E

Solution,

R=IR =300 V

100 Q

IR =3 A IL = IT

=

E L

300 V

50 S IL =6 A

L

IC = E AC

300 VI I 150 Q IC =2A C

l

E =300V R 100Q

L z-,

IR

Figure 14.12 - Parallel RLC circuit. r1

I

Using the values for IR, IL, and Ic, solve for lx and Iz. IX =I L - IC Ix=6A-2A Ix =4A(inductive)

Iz 11

1

_ (IR)2 + (IX)2

Iz _

(3 A)2 + (4A)2

I z = 25A Iz =5A

Module 3.14 Resistive (R), Capacitive (C) and Inductive (L) Circuits

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Using this value of Iz, solve for Z.

Z= E

Iz Z=

300 V

SA

Z=60:2 If the value for E were not given and you were asked to solve for Z, any value of E could be assumed. If, in the example problem above, you assume a value of 50 volts for E, the solution would be:

Given:

R =100 Q

XL

50 Q

XG =150 0

n

E = 50 V (assumed) First solve for the values of current in the same manner as before.

t-,

2 X3600 = 120

60Hz

A 4-pole, 1800-RPM generator also has a frequency of 60 Hz. A 6-pole, 500-RPM generator has a frequency of 6 x 500 =

120

25Hz

A 12-pole, 4000-RPM generator has a frequency of

12 x 4000 ^ 400Hz 120

L

Module 3.17 AC Generators

17-19

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BOTH ALTGRWATORS ARE ROTATING AT 120 RPM : F=

20

n

Figure 17.12 - Frequency regulation.

17-20

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Li

Voltage Regulation As we have seen before, when the load on a generator is changed, the terminal voltage varies. The amount of variation depends on the design of the generator. The voltage regulation of an generator is the change of voltage from full load to no load, expressed as a percentage of full-load volts, when the speed and DC field current are held

constant. E„L

-EfL x 100 = Percent of regulation

EfL

Assume the no-load voltage of an generator is 250 volts and the full-load voltage is 220 volts.

The percent of regulation is 250 - 220 220

x 100 =13.6%

Remember, the lower the percent of regulation, the better it is in most applications.

Principles of AC Voltage Control

u

Li V

In an AC generator, an generator voltage is induced in the armature windings when magnetic fields of alternating polarity are passed across these windings. The amount of voltage induced in the windings depends mainly on three things: (1) the number of conductors in series per winding, (2) the speed (generator RPM) at which the magnetic field cuts the winding, and (3) the strength of the magnetic field. Any of these three factors could be used to control the amount of voltage induced in the generator windings. The number of windings, of course, is fixed when the generator is manufactured. Also, if the output frequency is required to be of a constant value, then the speed of the rotating field must be held constant. This prevents the use of the generator RPM as a means of controlling the

voltage output. Thus, the only practical method for obtaining voltage control is to control the strength of the rotating magnetic field. The strength of this electromagnetic field may be varied by changing the amount of current flowing through the field coil. This is accomplished by varying the amount of voltage applied across the field coil. U

Parallel Operation of AC Generators AC Generators are connected in parallel to (1) increase the output capacity of a system beyond that of a single unit, (2) serve as additional reserve power for expected demands, or (3) permit shutting down one machine and cutting in a standby machine without interrupting power distribution.

Module 3.17 AC Generators

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When generators are of sufficient size, and are operating at different frequencies and terminal voltages, severe damage may result if they are suddenly connected to each other through a common bus. To avoid this, the machines must be synchronized as closely as possible before connecting them together. This may be accomplished by connecting one generator to the bus (referred to as bus generator), and then synchronizing the other (incoming generator) to it before closing the incoming generator's main power contactor. The generators are synchronized when the following conditions are set:

7

• Equal terminal voltages. This is obtained by adjustment of the incoming generator's field strength. • Equal frequency. This is obtained by adjustment of the incoming generator's primemover speed. Phase voltages in proper phase relation. The procedure for synchronizing generators is not discussed in this chapter. At this point, it is enough for you to know that the above must be accomplished to prevent damage to the machines.

iJ

n

t.1

L

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rTh t

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Module 3 Licence Category B1/B2 Electrical Fundamentals

L

3.18 AC Motors

Module 3.18 AC Motors

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Knowledge Levels - Category A, 131, B2 and C Aircraft Maintenance Licence

n

Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows:

1

LEVEL 1 • A familiarisation with the principal elements of the subject. Objectives: • The applicant should be familiar with the basic elements of the subject. • The applicant should be able to give a simple description of the whole subject, using common words and examples. • The applicant should be able to use typical terms.

LEVEL 2 • A general knowledge of the theoretical and practical aspects of the subject. • An ability to apply that knowledge. Objectives: • The applicant should be able to understand the theoretical fundamentals of the subject. • The applicant should be able to give a general description of the subject using, as appropriate, typical examples. • The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. • The applicant should be able to read and understand sketches, drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using detailed procedures.

LEVEL 3 • A detailed knowledge of the theoretical and practical aspects of the subject. • A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: • The applicant should know the theory of the subject and interrelationships with other subjects. • The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. • The applicant should understand and be able to use mathematical formulae related to the subject. • The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. • The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. • The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.

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Table of Contents

Module 3.18 AC Motors Introduction Series AC Motor Rotating Magnetic Fields Rotor Behaviour in a Rotating Field Synchronous Motors Induction Motors

Module 3.18 AC Motors Use and/or disclosure is governed by the statement

5 5 5 7 12 12 14

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Module 3.18 Enabling Objectives Objective

EASA 66 Reference

AC Motors 3.18 Construction, principles of operation and characteristics of: AC synchronous and induction motors both single and polyphase Methods of speed control and direction of rotation Methods of producing a rotating field: capacitor, inductor, shaded or split pole

Level 2

n

n 1

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Module 3.18 AC Motors

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Module 3.18 AC Motors Introduction Most of the power-generating systems on aircraft produce AC. For this reason a majority of the motors used throughout the aircraft are designed to operate on AC. There are other advantages in the use of AC motors besides the wide availability of AC power. In general, AC motors cost less than DC motors. Some types of AC motors do not use brushes and commutators. This eliminates many problems of maintenance and wear. It also eliminates the problem of dangerous sparking. An AC motor is particularly well suited for constant-speed applications. This is because its speed is determined by the frequency of the AC voltage applied to the motor terminals. The DC motor is better suited than an AC motor for some uses, such as those that require variable-speeds. An AC motor can also be made with variable speed characteristics but only within certain limits. Li industry builds AC motors in different sizes, shapes, and ratings for many different types of jobs. These motors are designed for use with either polyphase or single-phase power systems. It is Lnot possible here to cover all aspects of the subject of AC motors. Only the principles of the most commonly used types are dealt with in this chapter. In this chapter, AC motors will be divided into (1) series, (2) synchronous, and (3) induction motors. Single-phase and polyphase motors will be discussed. Synchronous motors, for purposes of this chapter, may be considered as polyphase motors, of constant speed, whose rotors are energized with DC voltage. Induction motors, single-phase or polyphase, whose rotors are energized by induction, are the most commonly used AC motor. The series AC motor, in a sense, is a familiar type of motor. It is very similar to the DC motor

that was covered in chapter 2 and will serve as a bridge between the old and the new.

Series AC Motor A series AC motor is the same electrically as a DC series motor. Refer to Figure 18.1 and use the left-hand rule for the polarity of coils. You can see that the instantaneous magnetic polarities

of the armature and field oppose each other, and motor action results. Now, reverse the current by reversing the polarity of the input. Note that the field magnetic polarity still opposes the armature magnetic polarity. This is because the reversal affects both the armature and the field. The AC input causes these reversals to take place continuously.

Module 3.18 AC Motors

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FIELD COIL

SERIES FIELD

Figure 18.1 - Series AC motor. The construction of the AC series motor differs slightly from the DC series motor. Special metals, laminations, and windings are used. They reduce losses caused by eddy currents, hysteresis, and high reactance. DC power can be used to drive an AC series motor efficiently, but the opposite is not true.

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The characteristics of a series AC motor are similar to those of a series DC motor. It is a varying-speed machine. It has low speeds for large loads and high speeds for light loads. The starting torque is very high. Series motors are used for driving fans, electric drills, and other small appliances. Since the series AC motor has the same general characteristics as the series DC motor, a series motor has been designed that can operate both on AC and DC. This AC/DC motor is called a universal motor. It finds wide use in small electric appliances. Universal motors operate at lower efficiency than either the AC or DC series motor. They are built in small sizes only. Universal motors do not operate on polyphase AC power. i1 18-6

Module 3.18 AC Motors

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Rotating Magnetic Fields The principle of rotating magnetic fields is the key to the operation of most AC motors. Both synchronous and induction types of motors rely on rotating magnetic fields in their stators to cause their rotors to turn. The idea is simple. A magnetic field in a stator can be made to rotate electrically, around and around. Another magnetic field in the rotor can be made to chase it by being attracted and repelled by the stator field. Because the rotor is free to turn, it follows the rotating magnetic field in the stator. Let's see how it is done. Rotating magnetic fields may be set up in two-phase or three-phase machines. To establish a rotating magnetic field in a motor stator, the number of pole pairs must be the same as (or a multiple of) the number of phases in the applied voltage. The poles must then be displaced from each other by an angle equal to the phase angle between the individual phases of the applied voltage. Two-Phase Rotating Magnetic Field A rotating magnetic field is probably most easily seen in a two-phase stator. The stator of a twophase induction motor is made up of. two windings (or a multiple of two). They are placed at right angles to each other around the stator. The simplified drawing in Figure 18.2 illustrates a

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two-phase stator.

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Figure 18.2 - Two-phase motor stator. If the voltages applied to phases 1-1A and 2-2A are 90°out of phase, the currents that flow in the phases are displaced from each other by 90° Si nce the magnetic fields generated in the coils are in phase with their respective currents, the magnetic fields are also 90°out of phase with each other. These two out-of-phase magnetic fields, whose coil axes are at right angles to each other, add together at every instant during their cycle. They produce a resultant field that rotates one revolution for each cycle of AC. To analyse the rotating magnetic field in a two-phase stator, refer to Figure 18.3. The arrow represents the rotor. For each point set up on the voltage chart, consider that current flows in a direction that will cause the magnetic polarity indicated at each pole piece. Note that from one point to the next, the polarities are rotating from one pole to the next in a clockwise manner. One complete cycle of input voltage produces a 360-degree rotation of the pole polarities. Let's see how this result is obtained.

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Figure 18.3 - Two-phase rotating field.

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The waveforms in Figure 18.3 are of the two input phases, displaced 90°because of the way they were generated in a two-phase alternator. The waveforms are numbered to match their associated phase. Although not shown in this figure, the windings for the poles 1-i A and 2-2A would be as shown in the previous figure. At position 1, the current flow and magnetic field in winding 1-1 A is at maximum (because the phase voltage is maximum). The current flow and magnetic field in winding 2-2A is zero (because the phase voltage is zero). The resultant magnetic field is therefore in the direction of the 1-1 A axis. At the 45-degree point (position 2),

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the resultant magnetic field lies midway between windings 1-1 A and 2-2A. The coil currents and

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magnetic fields are equal in strength. At 90° (position 3), the magnetic field in winding 1-1 A is zero. The magnetic field in winding 2-2A is at maximum.

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Now the resultant magnetic field lies along the axis of the 2-2A winding as shown. The resultant magnetic field has rotated clockwise through 90°to get from position 1 to position 3. When the twophase voltages have completed one full cycle (position 9), the resultant magnetic field has rotated through 360° Thus, by placing two windings at right angles to each other and exciting these windings with voltages 90° out of phase, a rotating magnetic field results.

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Two-phase motors are rarely used except in special-purpose equipment.

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They are discussed here to aid in understanding rotating fields. You will, however, encounter many single-phase and three-phase motors.

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Use and/or disclosure is governed by the statement

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Three-Phase Rotating Fields The three-phase induction motor also operates on the principle of a rotating magnetic field. The following discussion shows how the stator windings can be connected to a three-phase AC input and has a resultant magnetic field that rotates.

Figure 18.4, views A-C show the individual windings for each phase. Figure 18.4, view D, shows how the three phases are tied together in a Y-connected stator. The dot in each diagram indicates the common point of the Y-connection. You can see that the individual phase windings are equally spaced around the stator. This places the windings 120°apart.

Figure 18.4 - Three-phase, Y-connected stator. The three-phase input voltage to the stator of Figure 18.4 is shown in the graph of Figure 18.5. Use the left-hand rule for determining the electromagnetic polarity of the poles at any given instant. In applying the rule to the coils in Figure 18.4, consider that current flows toward the terminal numbers for positive voltages, and away from the terminal numbers for negative voltages. n

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Figure 18.5 - Three-phase rotating-field polarities and input voltages. The results of this analysis are shown for voltage points 1 through 7 in Figure 18.5. At point 1, the magnetic field in coils 1-1 A is maximum with polarities as shown. At the same time, negative voltages are being felt in the 2-2A and 3-3A windings. These create weaker magnetic fields, which tend to aid the 1-1 A field. At point 2, maximum negative voltage is being felt in the 3-3A windings. This creates a strong magnetic field which, in turn, is aided by the weaker fields in 11A and 2-2A. As each point on the voltage graph is analysed, it can be seen that the resultant magnetic field is rotating in a clockwise direction. When the three-phase voltage completes one full cycle (point 7), the magnetic field has rotated through 3600

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Li Module 3.18 AC Motors Use and/or disclosure is

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Rotor Behaviour in a Rotating Field For purposes of explaining rotor movement, let's assume that we can place a bar magnet in the centre of the stator diagrams of Figure 18.5. We'll mount this magnet so that it is free to rotate in this area. Let's also assume that the bar magnet is aligned so that at point 1 its south pole is

opposite the large N of the stator field.

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You can see that this alignment is natural. Unlike poles attract, and the two fields are aligned so that they are attracting. Now, go from point 1 through point 7. As before, the stator field rotates clockwise. The bar magnet, free to move, will follow the stator field, because the attraction between the two fields continues to exist. A shaft running through the pivot point of the bar magnet would rotate at the same speed as the rotating field.

This speed is known as synchronous speed. The shaft represents the shaft of an operating motor to which the load is attached. Remember, this explanation is an oversimplification. It is meant to show how a rotating field can cause mechanical rotation of a shaft. Such an arrangement would work, but it is not used. There are limitations to a permanent magnet rotor. Practical motors use other methods, as we shall see in the next paragraphs.

Synchronous Motors The construction of the synchronous motors is essentially the same as the construction of the salient-pole alternator. In fact, such an alternator may be run as an AC motor. It is similar to the drawing in Figure 18.6. Synchronous motors have the characteristic of constant speed between no load and full load. They are capable of correcting the low power factor of an inductive load when they are operated under certain conditions.

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They are often used to drive DC generators. Synchronous motors are designed in sizes up to thousands of horsepower. They may be designed as either single-phase or multiphase machines. The discussion that follows is based on a three-phase design.

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n 18-12 TTS Integrated Training System © Copyright 2010

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Figure 18.6 - Revolving-field synchronous motor. To understand how the synchronous motor works, assume that the application of three-phase AC power to the stator causes a rotating magnetic field to be set up around the rotor. The rotor is energized with DC (it acts like a bar magnet). The strong rotating magnetic field attracts the strong rotor field activated by the DC. This results in a strong turning force on the rotor shaft. The rotor is therefore able to turn a load as it rotates in step with the rotating magnetic field. It works this way once it's started. However, one of the disadvantages of a synchronous motor is that it cannot be started from a standstill by applying three-phase AC power to the stator. When AC is applied to the stator, a high-speed rotating magnetic field appears immediately. This rotating field rushes past the rotor poles so quickly that the rotor does not have a chance to get started. In effect, the rotor is repelled first in one direction and then the other. A synchronous motor in its purest form has no starting torque. It has torque only when it is running at synchronous speed.

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A squirrel-cage type of winding is added to the rotor of a synchronous motor to cause it to start. The squirrel cage is shown as the outer part of the rotor in Figure 18.7. It is so named because it is shaped and looks something like a turnable squirrel cage. Simply, the windings are heavy copper bars shorted together by copper rings. A low voltage is induced in these shorted windings by the rotating three-phase stator field. Because of the short circuit, a relatively large current flows in the squirrel cage. This causes a magnetic field that interacts with the rotating field of the stator. Because of the interaction, the rotor begins to turn, following the stator field; the motor starts. We will run into squirrel cages again in other applications, where they will be covered in more detail.

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SQUIRREL-CAGE WINDING OVER SALIENT-POLE WINDINGS Figure 18.7 - Self-starting synchronous AC motor. To start a practical synchronous motor, the stator is energized, but the DC supply to the rotor field is not energized. The squirrel-cage windings bring the rotor to near synchronous speed. At that point, the DC field is energized. This locks the rotor in step with the rotating stator field. Full torque is developed, and the load is driven. A mechanical switching device that operates on centrifugal force is often used to apply DC to the rotor as synchronous speed is reached. The practical synchronous motor has the disadvantage of requiring a DC exciter voltage for the rotor. This voltage may be obtained either externally or internally, depending on the design of the motor.

Induction Motors The induction motor is the most commonly used type of AC motor. Its simple, rugged construction costs relatively little to manufacture. The induction motor has a rotor that is not connected to an external source of voltage. The induction motor derives its name from the fact that AC voltages are induced in the rotor circuit by the rotating magnetic field of the stator. In many ways, induction in this motor is similar to the induction between the primary and secondary windings of a transformer. Large motors and permanently mounted motors that drive loads at fairly constant speed are often induction motors. Examples are found in washing machines, refrigerator compressors, bench grinders, and table saws. The stator construction of the three-phase induction motor and the three-phase synchronous motor are almost identical. However, their rotors are completely different (see Figure 18.8). The induction rotor is made of a laminated cylinder with slots in its surface. The windings in these slots are one of two types (shown in Figure 18.9). The most common is the squirrel-cage winding. This entire winding is made up of heavy copper bars connected together at each end

18-14 Use and/or disclosure is Governed by the statement

Module 3.18 AC Motors Module Motors

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by a metal ring made of copper or brass. No insulation is required between the core and the bars. This is because of the very low voltages generated in the rotor bars. The other type of winding contains actual coils placed in the rotor slots. The rotor is then called a wound rotor.

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Figure 18.8 - Induction motor.

r1 METAL RING

SQUIRREL-CAGE ROTOR U

WOUND ROTOR Figure 18.9 - Types of AC induction motor rotors.

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Module 3.18 AC Motors

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Regardless of the type of rotor used, the basic principle is the same. The rotating magnetic field generated in the stator induces a magnetic field in the rotor. The two fields interact and cause the rotor to turn. To obtain maximum interaction between the fields, the air gap between the rotor and stator is very small. As you know from Lenz's law, any induced EMF tries to oppose the changing field that induces it. In the case of an induction motor, the changing field is the motion of the resultant stator field. A force is exerted on the rotor by the induced EMF and the resultant magnetic field. This force tends to cancel the relative motion between the rotor and the stator field. The rotor, as a result, moves in the same direction as the rotating stator field. It is impossible for the rotor of an induction motor to turn at the same speed as the rotating

magnetic field. If the speeds were the same, there would be no relative motion between the stator and rotor fields; without relative motion there would be no induced voltage in the rotor. In order for relative motion to exist between the two, the rotor must rotate at a speed slower than that of the rotating magnetic field. The difference between the speed of the rotating stator field and the rotor speed is called slip. The smaller the slip, the closer the rotor speed approaches the stator field speed. The speed of the rotor depends upon the torque requirements of the load. The bigger the load, the stronger the turning force needed to rotate the rotor. The turning force can increase only if the rotor-induced EMF increases. This EMF can increase only if the magnetic field cuts through the rotor at a faster rate. To increase the relative speed between the field and rotor, the rotor must slow down. Therefore, for heavier loads the induction motor turns slower than for lighter loads. You can see from the previous statement that slip is directly proportional to the load on the motor. Actually only a slight change in speed is necessary to produce the usual current changes required for normal changes in load. This is because the rotor windings have such a low resistance. As a result, induction motors are called constant-speed motors. Single-Phase Induction Motors

There are probably more single-phase AC induction motors in use today than the total of all the other types put together.

It is logical that the least expensive, lowest maintenance type of AC motor should be used most often. The single-phase AC induction motor fits that description. Unlike polyphase induction motors, the stator field in the single-phase motor does not rotate. Instead it simply alternates polarity between poles as the AC voltage changes polarity. Voltage is induced in the rotor as a result of magnetic induction, and a magnetic field is produced around the rotor. This field will always be in opposition to the stator field (Lenz's law applies). The interaction between the rotor and stator fields will not produce rotation, however. The interaction is shown by the double-ended arrow in Figure 18.10, view A. Because this force is across the rotor and through the pole pieces, there is no rotary motion, just a push and/or pull along this line.

18-16 TTS Integrated Training System Use andlor disclosure Is governed by the statement

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on page 2 of this Chapter.

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integrated Training System Designed in association with the club66pro.co.uk question practice aid

NR, SR = ROTOR FIELD Ns, Ss = STATOR FIELD

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A. STATIONARY

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B. ROTATING

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Figure 18.10 - Rotor currents in a single-phase AC induction motor. Now, if the rotor is rotated by some outside force (a twist of your hand, or something), the pushpull along the line in Figure 18.10, view A, is disturbed. Look at the fields as shown in Figure 18.10, view B. At this instant the south pole on the rotor is being attracted by the left-hand pole. The north rotor pole is being attracted to the right-hand pole. All of this is a result of the rotor being rotated 90°by the outside force. The pull th at now exists between the two fields becomes a rotary force, turning the rotor toward magnetic correspondence with the stator. Because the two fields continuously alternate, they will never actually line up, and the rotor will continue to turn once started. It remains for us to learn practical methods of getting the rotor to start. There are several types of single-phase induction motors in use today. Basically they are identical except for the means of starting. In this chapter we will discuss the split-phase and shaded-pole motors; so named because of the methods employed to get them started. Once

they are up to operating speed, all single-phase induction motors operate the same.

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Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Split-Phase Induction Motors One type of induction motor, which incorporates a starting device, is called a split-phase induction motor. Split-phase motors are designed to use inductance, capacitance, or resistance to develop a starting torque. The principles are those that you learned in your study of alternating current. Capacitor-Start - The first type of split-phase induction motor that will be covered is the capacitor-start type. Figure 18.11 shows a simplified schematic of a typical capacitor-start motor. The stator consists of the main winding and a starting winding (auxiliary). The starting winding is connected in parallel with the main winding and is placed physically at right angles to it. A 90-degree electrical phase difference between the two windings is obtained by connecting the auxiliary winding in series with a capacitor and starting switch. When the motor is first energized, the starting switch is closed. This places the capacitor in series with the auxiliary

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winding. The capacitor is of such value that the auxiliary circuit is effectively a resistive-

capacitive circuit (referred to as capacitive reactance and expressed as Xc). In this circuit the current leads the line voltage by about 45°(becaus a Xc about equals R). The main winding has enough resistance-inductance (referred to as inductive reactance and expressed as XL) to cause the current to lag the line voltage by about 45° (because X, about equals R). The currents in each winding are therefore 90°out of p hase - so are the magnetic fields that are generated. The effect is that the two windings act like a two-phase stator and produce the rotating field required to start the motor.

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MAIN WINDING fl

AC SINGLE- AUXILIARY PHASE WINDING SUPPLY

Figure 18.11 - Capacitor-start, AC induction motor. When nearly full speed is obtained, a centrifugal device (the starting switch) cuts out the starting winding. The motor then runs as a plain single-phase induction motor. Since the auxiliary winding is only a light winding, the motor does not develop sufficient torque to start heavy loads. Split-phase motors, therefore, come only in small sizes.

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Module 3.18 AC Motors Module 3.18 AC Motors

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Resistance-Start - Another type of split-phase induction motor is the resistance-start motor, This motor also has a starting winding (shown in fig. 4-12) in addition to the main winding. It is switched in and out of the circuit just as it was in the capacitor-start motor. The starting winding is positioned at right angles to the main winding. The electrical phase shift between the currents in the two windings is obtained by making the impedance of the windings unequal. The main winding has a high inductance and a low resistance. The current, therefore, lags the voltage by a large angle. The starting winding is designed to have a fairly low inductance and a high resistance. Here the current lags the voltage by a smaller angle, For example, suppose the current in the main winding lags the voltage by 70°. The current in the auxiliary winding lags the voltage by 40°. The currents are, therefore, out of phase by 30°. The magnetic fields are out of phase by the same amount. Although the ideal angular phase difference is 90°for maximum starting torque, the 30-degree phase difference still generates a rotating field. This supplies enough torque to start the motor. When the motor comes up to speed, a speed-controlled switch disconnects the starting winding from the line, and the motor continues to run as an induction motor. The starting torque is not as great as it is in the capacitor-start.

MAIN NG AC

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SINGLE- AUXILIARY PHASE WINDING SU P PLY

Figure 18.12 - Resistance-start AC induction motor. L

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Shaded-Pole Induction Motors The shaded-pole induction motor is another single-phase motor. It uses a unique method to start the rotor turning. The effect of a moving magnetic field is produced by constructing the stator in a special way. This motor has projecting pole pieces just like some DC motors. In addition, portions of the pole piece surfaces are surrounded by a copper strap called a shading coil. A pole piece with the strap in place is shown in Figure 18.13. The strap causes the field to move back and forth across the face of the pole piece. Note the numbered sequence and points on the magnetization curve in the figure. As the alternating stator field starts increasing from zero (1), the lines of force expand across the face of the pole piece and cut through the strap. A voltage is induced in the strap. The current that results generates a field that opposes the cutting action (and decreases the strength) of the main field. This produces the following actions: As the field increases from zero to a maximum at 90° a large portion of the magnetic lines of force are concentrated in the unshaded portion of the pole (1). At 90°the field reaches its maximum value. Si nce the lines of force have stopped expanding, no EMF is induced in the strap, and no opposing magnetic field is generated. As a result, the main field is uniformly distributed across the pole (2). From 90°to 180° the main field starts decreasing or collapsing inward. The field generated in the strap opposes the collapsing field. The effect is to concentrate the lines of force in the shaded portion of the pole face (3). You can see that from 0°to 180° the main field ha s shifted across the pole face from the unshaded to the shaded portion. From 180°to 360° the main field goes through the same change as it did from 0°to 180° however, it is no w in the opposite direction (4). The direction of the field does not affect the way the shaded pole works. The motion of the field is the same during the second half-cycle as it was during the first half of the cycle.

18-20 ITS Integrated Training System r ,

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Module 3. I o me Motors

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on page 2 of this Chapter.

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Figure 18.13 - Shaded poles as used in shaded-pole AC induction motors. The motion of the field back and forth between shaded and unshaded portions produces a weak torque to start the motor. Because of the weak starting torque, shaded-pole motors are built only in small sizes. They drive such devices as fans, clocks, blowers, and electric razors. Speed of Single-Phase Induction Motors The speed of induction motors is dependent on motor design. The synchronous speed (the speed at which the stator field rotates) is determined by the frequency of the input AC power

and the number of poles in the stator. The greater the number of poles, the slower the synchronous speed. The higher the frequency of applied voltage, the higher the synchronous speed. Remember, however, that neither frequency nor number of poles are variables. They are both fixed by the manufacturer. The relationship between poles, frequency, and synchronous speed is as follows:

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n (rpm) =

120f

where `n' is the synchronous speed in RPM, f is the frequency of applied voltage in hertz, and p is the number of poles in the stator. L

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

Let's use an example of a 4-pole motor, built to operate on 60 hertz. The synchronous speed is determined as follows:

n = 120f p

n

n= 120x60 4

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n

n =1800 rpm

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Common synchronous speeds for 60-hertz motors are 3600, 1800, 1200, and 900 RPM, depending on the number of poles in the original design. As we have seen before, the rotor I s never able to reach synchronous speed. If it did, there would be no voltage induced in the rotor. No torque would be developed. The motor would not operate. The difference between rotor speed and synchronous speed is called slip. The difference between these two speeds is not great. For example, a rotor speed of 3400 to 3500 RPM can be expected from a synchronous speed of 3600 RPM.

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TTS Integrated Training System L LI

Module 3 Licence Category B1/B2 Electrical Fundamentals Appendix

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Module 3 Appendix

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2 TTS Integrated Training System © Copyright 2010

Module 3 Appendix

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Module 3 Appendix

Li Colour Diagrams The following diagrams from the main chapters of these notes have been reproduced here in full colour due to the essential nature of the colour-code information. Type

K

Temperature range 9C (continuous)

BS Colour codeANSI Colour code

0 to +1100 NMI

iL

L

Brown

Yellow

Blue

Red

J

0 to +700

Yellow

White

Bl ue

Red

N

0 to +1100

Orange

Orange Red

R

0 to +1600

White Blue

Not defined.

S

0 to 1600

White Blue

Not defined.

B

+20Q to +1700

T

-185 to +300

E

0 to +800

Whit e

No standard use copper wire White Blue

Brown Blue

Not defined. Blue N=I Red

Blue NMI Red

Table 5.1 - Thermocouple comparison and wire identification

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Module 3 Appendix

3 TTS Integrated Training System © Copyright 2010

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Tolerance Multiplier 2nd digit '1st digit Figure 7.3 -- A common 4-band resistor YEL

1st&2nd bands

0

L 3rd bandx1

1

2

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x100

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5

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7

8

x10M n/an/a

Table 7.2 - Standard Colour Code for Resistors

SLV

Tolerance Band

+11%

+12%

+10.5%

+10.25%

0.1%

0.06%

5%

10%

Table 7.4 - 5th Band Colour Codes (Tolerance Band)

Tolerance �.. rv111ltil}tier arc! digit

2nd digit 'Ist digit Figure 7.5 - A modern 5-band resistor

4

1

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Integrated Training System Designed in association with the club66pro,co.uk question practice aid

U

r,

L

BRN-iDOpprn

TempCo Tolerance

REL -6IO:ppm

Multiplier

3rd digit 2nd digit

1st digit Table 7.5 - Temperature Coefficients

Figure 7.6 - A 6-band resistor

6-band color cod

47.5 K Ohms±1%

6-band color code

276 Ohms ± 5%

r-I

Tolerance

L

SLV 0.01

1-1

U

i

L

BLK-0

BLK-0

BLK-0

BLK-1

BRN•I

BRN-1

BRIv-1

BRN-10

BRN + 1.t

RED-2

RED-2

RE 0.2

RED-100

RED

YEL.-4 BLU-S VIU-7

YEL-4 GRN-S RLU-6 VlO-7

GRY4

GRY-8

GRY-8

WHT-9

WHT-9

WHT-9

GRN-5 i

SLY ± 10%

YEL-a,

GRN BLU-6 VIU-7

21.,.

YiEL- IDK" .

SRN-100ppm R E D-50ppm

`Y�L=gym,

GRN-100K BLU-1 M vIO-10M

Figure 7.7 - Combined 4-Band, 5-Band and 6-Band Chart

L Module 3 Appendix

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© Copyright 2010

Integrated Training System Designed in association with the club66pro.co.uk question practice aid

n

1OnF,20%

100V

47nF, 10%

240V

n

Figure 9.24 - Ceramic capacitor colour bands

Colour • '-•

Digit A 0 1 2

Digit B 0 1 2

• -• Yellow

Multiplier Tolerance Tolerance D

T > 10pf

T < 10pf

x1 x10 x100

± 20% ±1% ± 2%

± 2.OpF ± 0.1 pF ± 0.25pF

3 3 x1000 4 4 x1 0k 5 5 x100k Blue 6 6 x1 m 7 7 Grey 8 8 xO.01 White 9 9 xO.1 Table 9.2 - Colour code for capacitors.

± 3% +100%,-0% ± 5%

Temperature Working Coefficient

voltage

TC

V

-33x10-6 -75x1 0-6 -150x10-6 -220x10-6

t 0.5pF

-330x10"6 -470x10"6

250v

tJ

400v 100v 630v

-750x10-6 +80%,-20% ± 10%

r

Figure 9.26 - Mica capacitors

Figure 13.26 - Clockwise rotation phase sequence: 1-2-3

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Module 3 Appendix