ENG1021 Electronic Principles

ENG1021 Electronic Principles1 ENG1021 Electronic Principles Learning Package 1 What is electricity? ENG1021 Electronic Principles 2 1 Introduction Welcome to the Electronic Principles module of the course. The course is split up into Learning Packages. At the start of each Learning Package you will be presented with a “Do you know all this already?” section which directs you to some questions on the subject that appear within the Learning Package. If you are 100 % confident with the subject then you may continue with the next Learning Package, but you should be sure that you are able to solve all of the problems that I have selected before moving on. Within each Learning Package you will be given the basic concepts which will be reinforced by problem solving. I will give detailed explanations to the solutions to all of the problems in which I will highlight common pitfalls and suggest alternative solutions and observations. You should always read the solutions and suggestions. However, it is important that you attempt the problems before you study the solutions. This is the only way that you will build up confidence in tackling similar problems in electronics. Often at the end of each Learning Package you will also be given information about where to find opportunities for further reading. This introduction package differs from the others in that there are no set problems. You will familiarise yourself with the concepts of charge, current, voltage and resistance. Do not struggle to understand (or remember) every detail, most of the content will be expanded upon in later chapters, and you can always go back and read them later (when they may make more sense). 2 What is electricity? All substance is made of atoms, and all atoms are made of smaller particles – mainly protons, electrons and neutrons. Each proton has a positive charge, and each electron has the equivalent but opposite negative charge (neutrons do not have a charge). In an atom, the number of protons equals the number of electrons, so that the atom is electronically neutral. In many substances, the atoms can be persuaded to part with one or more electrons temporarily by the application of energy. The free electrons can then flow through the material. The flow of electrons is what is called current, and this is the basis for all of electrical engineering and electronics. The amount of current is measured in amperes or amps for short. When an electrical current flows through a substance such as a metal wire, it produces two effects – heat and magnetism. It is these effects that are used to ENG1021 Electronic Principles 3 create all of the electrical and electronic devices that are used today. For example, when heat is produced this can be exploited to make electric fires, electric cookers, and if the heat is so intense that the wire glows, it can produce electric light. The magnetism that is produced can be exploited to make electro- magnets or electric motors. The energy that has to be provided to move the electrons is usually described as the voltage and is measured in volts. Batteries have a voltage, and when a conductor is placed between the terminals, the electrons flow through the conductor. Similarly, the electric socket that you plug your electrical devices in to has a voltage which in the UK is quoted in volts. There is a close relationship between the voltage and the subsequent amount of current that flows, and this will be discussed in more detail in the next Learning Package. 3 The circuit When describing electricity, the term “circuit” is often used. You may have come across the term “open-circuit” or “short-circuit” before. A simple circuit is shown below. It consists of a battery, which has a positive (+) and a negative (-) terminal. A wire is connected from the positive terminal to a component, which in this case is a light bulb. The other terminal, the negative, is also connected to the other side of the bulb. So now we have a complete circuit – there are no breaks anywhere. Notice that I’ve used standard symbols to represent a battery and a bulb, and I’ve followed the standard convention when showing wiring of using straight lines with right angle bends. This kind of diagram is often referred to as a “circuit diagram” or a “schematic diagram”. So, within the battery is an electronic charge. This would have been created during manufacture, and the way I think of it is that the battery has two compartments. In the one, which would be labelled negative, is a compartment that has millions of free electrons. In the other, labelled positive, there is a lack of electrons. When the negative terminal is connected to the positive terminal, electrons flow out of the negative compartment, through the wire, and into the positive compartment. It will carry on doing this at a fairly constant rate until the number of electrons in each compartment is equal, when the battery would be described as “flat”. The important point here is that the two terminals are joined together. If there was a break in the wire anywhere, then the electrons would not be able to flow and there would be no current. That would be an “open-circuit”. So for current to flow there has to be a closed-circuit, which means no breaks. You can place devices in that circuit, such as light bulbs, as long as the current can flow through them. If you just have a wire between the two terminals and nothing else, that would be described as a “short-circuit” and is usually something that you are advised not to do because it can create very high currents. ENG1021 Electronic Principles 4 Figure 1 A simple electronic circuit Finally, before we leave circuits, you may have come across the terms AC or DC circuits. A circuit which contains a battery is an example of a DC circuit. The voltage that is supplied by the battery is constant, as is the current that is produced. The term DC stands for “direct current” which means that the current doesn’t change. On the other hand a voltage could be supplied by a generator, such as that found in the mains supply, which is not constant. Instead it alternates between a positive and a negative value at a particular rate. In the mains in the UK this change takes place 50 times a second. Therefore this type of circuit would be described as AC, which stands for “alternating current”. We will look at AC circuits in more detail in a later Learning Package. So for now, we will just be dealing with DC circuits. As an aside, it is quite common to come across phrases like “DC current” or “DC voltage” which just means a constant current or constant voltage. 4 Understanding electricity - The water analogy In early dealings with electricity and electronics, students are often confused between the terms voltage, current, resistance and power. Power, current and voltage are often considered (incorrectly) as meaning the same thing when they refer to three completely different physical quantities. It is important for you to be clear of theses distinctions in your own mind, before attempting to understand electronic principles further. Many of the definitions of these electrical quantities are not helpful in your understanding process. Here are some examples: ”One volt is the difference of potential between two points of an electrical conductor when a current of 1 ampere flows between those points and dissipates a power of 1 watt.” ENG1021 Electronic Principles 5 ”Potential Difference (voltage) between two points in an electric field is the work done per unit charge moved from point to point in the field.” To assist understanding, an analogy with water is often made in texts. You may like to consider the charge mentioned above as water in a tank which is sitting on a table. Energy (effort) was used lifting the water tank onto the table (or pumping water from a lower level into the tank), and that energy would be released if the tank were to be emptied. If the table had been higher, more energy would have been required to lift the same quantity of water into it. If we were to connect a pipe to the base of the tank and bring the pipe to the ground then we would have water pressure at the end of the pipe due to the height of the water. The higher the tank, the higher that pressure will be. The height of the tank (or pressure of the water) in this case is analogous to the voltage. In electricity, charge may move from a more negative (or lower) point (or potential) to a more positive (or higher) point (or potential), in our water analogy we have moved water from a lower point to a higher point (or potential). We can continue to use this analogy to give a feel for electrical current. Electrical current is the rate of flow of charge. Charge will only flow between two points if there is a voltage or potential difference between the points. If we connect a flexible pipe to the bottom of our water tank, and move the end of the hose vertically to positions between the ground and the level of the water in the tank, water will flow out of it, providing that there is a difference in height between the end of the hose and the water level in the tank. The flow of water is analogous with the flow of charge (or current), and there is no flow unless there is a difference in height for the water or a voltage exists (potential) for the charge. This brings us to an important property of voltage. Like height, it is a relative measurement, and only meaningful if measured across or between two points. Often one of these points is taken as being “earth” or “ground” which is always regarded as having zero volts. If the pipe is made narrower, then the rate at which water flows out of the pipe will be less. A narrower pipe has more resistance and impedes flow rate of water for a given tank height or pressure. Electrical resistance impedes flow rate of electrons (current) for a given voltage. So there is a potential for the energy to be released, but while the water sits in the tank on the table it is insulated from the floor. Current is the movement or flow of electrical charge. If charge remains static no current flows. Charge will remain static unless there is a difference in potential or voltage. 4 Measuring voltage and current In any electrical circuit the voltage or the current can be measured. Voltage is measured with a voltmeter and, as mentioned in the previous section, has to be ENG1021 Electronic Principles 6 measured between two points. Current is measured with an ammeter where the ammeter has to be placed in an electrical circuit so that the current can flow through it. 5 Where next? The next Learning Package is called “Ohm’s law” which explores to relationships between voltage, current, resistance and power in more detail. However, before leaving the topic of electricity, you may like to have a look at the on-line book that has been supplied with this module. It is entitled “Fundamentals of Electrical Engineering and Electronics” by Tony L Kuphaldt with help from other people. Have a look at the first section entitled “DC”, within that all of the material that comes under the heading of “Basic Concepts of Electricity”, such as: • Static electricity • Conductors, insulators and electron flow • Electric circuits • Voltage and current • Resistance • Voltage and current in a practical circuit • Conventional versus electron flow These will help you to understand the concepts discussed in this Learning Package. ENG1021 Electronic Principles 7 ENG1021 Electronic Principles Learning Package 2 Ohm’s law and power ENG1021 Electronic Principles 8 Ohm’s law and power 1 Do you know all this already? If you are in doubt about Ohm’s law and power, please attempt the self assessment questions that appear later in this Learning Package. If you can answer all of the questions correctly you may omit this Learning Package. If not, please read on. 2 Introduction This section links the electrical quantities current, voltage, resistance and power. It also introduces multiple units, which are prefixes attached to units that are intended to allow us to write very large or very small numbers more concisely. 3 Ohm’s law In the previous Learning Package you were introduced to the concepts of voltage and current and you were told that there was a relationship between these quantities. This relationship is known as Ohm’s law, and states that the voltage divided by the current is a constant (if the temperature remains constant). The constant is given the name “resistance”, and the symbol R. Usually Ohm’s law is expressed as: V = I × R (1) where V is the voltage in volts, I is the current in amperes and R is the resistance in ohms. Example: A battery has a voltage of 9 V. It is connected to a device which has a resistance of 100 Ω. What is the current in the circuit? First, note the units. The voltage is 9 volts, which is written as 9 V. The resistance is 100 ohms, which is written as 100 Ω, where Ω is the upper case Greek character omega. Rearranging Ohm’s law: I = V/R = 9/100 = 0.09 A The solution is 0.09 amperes, or 0.09 amps, which is written as 0.09 A. ENG1021 Electronic Principles 9 4 Power You were also told in the previous Learning Package that when an electric current flows, it generates heat. Heat is a form of energy, and the rate at which heat is generated is the power, measured in watts. You may have come across this when buying devices that use electrical heating devices such as fire or kettle where the manufacturer would be described the device as having 2 kilowatt power, for example. The equation for electrical power is: P = I × V (2) where P is the power in watts, I is the current in amperes and V is the voltage in volts. Example: What is the power in the circuit described earlier, where the battery is 9 V and the device has a resistance of 100 Ω? We calculated the current as 0.09 A, so: P = I × V = 0.09 × 9 = 0.81 W. The power generated is 0.81 watts, which is written as 0.81 W. 4.1 Other forms of the power equation We can also calculate the power in a circuit using just the voltage and resistance, or just the current and resistance by combining the power equation with Ohm’s law. P = I × V V = I × R So, substituting for V is the power equation, we get: P = I × (I × R) = I 2 × R (3) Alternatively, substitute for I in the power equation: P = (V/R) × V = V 2 /R (4) Example: Using the same example again, we have V = 9 v, I = 0.09 A, and R = 100 Ω. ENG1021 Electronic Principles 10 P = I 2 × R = 0.09 x 0.09 x 100 = 0.81 W P = (V/R) × V = V 2 /R = 9 x 9 / 100 = 0.81 W 5 Resistors Resistors are the most common components used in electrical and electronic circuits. Essentially they are components with a fixed resistance value, although usually that value is only approximate. For example, a resistor that states that it has a value of 100 Ω may actually have a value of 99 Ω or 101 Ω. The range of values is usually referred to as the tolerance, and can be ±20%, ±10% ±15% or ±2%. If a resistor is stated as having a value of 100 Ω and a tolerance of ±20%, then its value can be as little as 100 – (20% of 100) = 100 – (0.2 x 100) = 100 -20 = 80 Ω, or as high as 100 + (20% of 100) = 100 + (0.2 x 100) = 100 + 20 = 120 Ω. Resistors have their values written on them as four coloured bands. The colours correspond to the values shown in the following table. Colour 1 st digit 2 nd digit Multiplier Tolerance Silver 10 -2 =0.01 10% Gold 10 -1 =0.1 5% Black 0 0 10 0 =1 20% Brown 1 1 10 1 =10 1% Red 2 2 10 2 =100 2% Orange 3 3 10 3 =1000 Yellow 4 4 10 4 =10000 Green 5 5 10 5 =100000 Blue 6 6 10 6 =1000000 Violet 7 7 10 7 =10000000 Grey 8 8 10 8 =100000000 White 9 9 10 9 =1000000000 The first two bands form a number which is then multiplied by the value of the third band as a power of 10. The fourth band gives the value of the tolerance. ENG1021 Electronic Principles 11 Note that if the fourth band is not present, then the tolerance is taken to be ±20%. For example, a resistor with coloured bands of Brown, Green, Red, Silver would have a resistance of 15 x 10 2 or 1500 Ω and a tolerance of 10%. 10 % of 1500 is 150, so the actual value of the resistor is in the range 1350 to 1650 Ω. Most components, like resistors, are available at various values. Instead of an infinite choice, the values that are available are limited. Not only that, the range of value available depends on the tolerance. There is a wider range of values for lower tolerance resistors. The following Table shows the range of values available for resistors of different tolerance. The values that are used are known as the “preferred values”. Tolerance Preferred Values 5% 10 11 12 13 15 16 18 20 22 24 27 30 10% 10 12 15 18 22 27 20% 10 15 22 Tolerance Preferred Values 5% 33 36 39 43 47 51 56 62 68 75 82 91 10% 33 39 47 56 68 82 20% 33 47 68 These preferred values are then multiplied by factors of 10. So you could have 15 Ω, 150 Ω, 1500 Ω, 15000 Ω etc. So, for example, it is possible to have a resistor with a value of 6800 Ω with a tolerance of 5, 10 or 20%, but you can only have a resistor with a value of 6200 Ω with a tolerance of 5%. Another point to note is the standard symbol for the resistor. British and European circuit diagrams use the following symbol for resistors (Figure 1). ENG1021 Electronic Principles 12 Figure 1 The British Standard resistor symbol However, it is common to find the American standard symbol (which used to be the British standard also). This is shown in Figure 2. Throughout this module the two symbols for resistors will be used. Figure 2 The American Standard resistor symbol In some circuits you may want to use a variable resistor – one where the value can be changed. A device which is simply a variable resistor is called a rheostat. It has two connections, and then either a slider or knob that you turn to vary the resistance. A similar device is a potentiometer. This also has a slider or a knob that turns, but it has three connections, as shown in Figure 3. Figure 3 Rheostat and potentiometer ENG1021 Electronic Principles 13 6 Units You need to understand the conventional way that multiple units are used. They will be used extensively throughout this course without further explanation, so you need to make sure that you understand them now. In a previous table where the colour code of resistors was shown, a multiplier was referred to which used powers of 10. So, for example, 10 was seen to be 10 to the power of 1, written as 10 1 . Similarly 100 was described as 10 2 or 10 to the power of 2, and a thousand, 1000, was described as 10 3 , and so on. These are ways of writing powers of 10 so that you don’t have to write a string of zeroes after every value. You may note that the power is the same as the number of zeroes. Alternatively, the power is the same as the number of times you multiply 10 by itself. So 100 is 10 x 10 and 1000 is 10 x 10 x 10. The following table shows the powers of 10. Value power 10 multiplied by itself metric value symbol 1 10 0 10 10 1 10 100 10 2 10x10 1000 10 3 10x10x10 kilo k 10000 10 4 10x10x10x10 100000 10 5 10x10x10x10x10 1000000 10 6 10x10x10x10x10x10 mega M 1000,000,000 10 9 giga G 1000,000,000,000 10 12 tera T You can see that 10 0 = 1. This is true for any number, X, so that X 0 = 1. I’ve also introduced the metric terms, kilo for a thousand, mega for a million, giga for a thousand million and tera for a million million. The terms may be familiar from kilogram, a thousand grams, megawatt, a million watts. So, for example, a resistor with a value of 120,000 Ω could also be described as having a value of 120 kilo ohms or 120 kΩ. Similarly, a resistor with a value of 1,200 Ω could be said to have a value of 1.2 kΩ. ENG1021 Electronic Principles 14 What about values that are less then 1? We can also use powers of 10, but now the powers are negative. For example, 0.1 is equivalent to 1/10 th which in powers of 10 is written as 10 -1 . Similarly, 0.01 is equivalent to 1/100 th or 1/10 th x 1/10 th and is written as 10 -2 . The following table completes the story. Value power 10 multiplied by itself metric value symbol 1 10 0 0.1 10 -1 1/10 0.01 10 -2 1/10 x 1/10 centi c 0.001 10 -3 1/10 x 1/10 x 1/10 milli m 0.0001 10 -4 1/10 x 1/10 x 1/10 x 1/10 0.00001 10 -5 0.000001 10 -6 micro µ 0.000000001 10 -9 nano n 0.000000000001 10 -12 pico p Some of these terms may be familiar from centimetre, a hundredth of a meter, or millisecond, a thousandth of a second. Note that the choice of multiple unit prefix is up to you and an answer is not incorrect because you have not chosen the correct prefix. For example 470000 Ω may be written as 470 kΩ or 0.47 MΩ. It may be difficult to decide which of the latter two forms to use and either would be suitable. In general, the values that you calculate or measure will be over a wide range, so they might be thousands of millions or millionths. It is difficult to read numbers if they are very long, especially if they have either lots of training or leading zeroes. Therefore a standard called Scientific Notation is often used. In Scientific Notation, all numbers are written as one digit, decimal point, the remaining digits then multiplied by a power of ten. For example, 1.234567 x 10 6 would be how the number 1,234,567 would be written in Scientific Notation. Similarly, 1.234567 x 10 -6 would the way that 0.000001234567 would be written. 7 Additional reading As in the previous Learning Package, you are encouraged to read some of the sections from the on-line book “Fundamentals of Electrical Engineering and ENG1021 Electronic Principles 15 Electronics”. In this Learning Package the relevant section is headed “Ohm’s law”, and the parts that you should read are: • How voltage, current and resistance relate • An analogy for Ohm’s law • Power in electric circuits • Calculating electric power • Resistors The remaining parts of that section can be omitted as they are not relevant, although we may return to some of them later. In addition, you may want to read the section entitled “Scientific Notation and Metric Prefixes”. The relevant parts are: • Scientific notation • Arithmetic with scientific notation • Metric notation • Metric prefix conversion Again, the remainder of the parts can be ignored for now. Finally, tucked away in another section is some information about resistor colour codes in the section entitled “Reference”, and in a section called “Experiments” and sub-section called “DC Circuits” there are a couple of experiments called “Potentiometer as a voltage divider” and “Potentiometer as a rheostat” which you may find useful. 8 Self assessment exercises Now attempt the following Problems. These problems will assess your understanding of Ohm’s law, power, the electrical quantities and you should be able to exercise your use in multiple units. The solutions are given below with full explanations. Problem 1 A 15 V dc source is connected across a 1 kΩ resistance. (a) Draw a schematic diagram. (b) Calculate the current I through the resistance. (c) How much current flows through the voltage source? (d) If the voltage is doubled, how much is the current I in the circuit? ENG1021 Electronic Principles 16 Problem 2 A current of 2 A flows through a 12 Ω resistance connected across a battery. (a) How much is the battery voltage? (b) How much power is dissipated by the resistance? (c) How much power is supplied by the battery? Problem 3 Calculate the resistance, R, in ohms, for each of the following examples: (a) 1 mA drawn from a 12 V source; (b) 4 mA drawn from a 15 V source; (c) 150 mW dissipated with 36 V applied; (d) 16.2 W dissipated with a current of 30 mA. Problem 4 How much will it cost to operate a 1500 W heater for 36 hours if the cost of electricity is 7 cents/kWh? ENG1021 Electronic Principles 17 Solutions Problem 1a See Figure 4. Figure 4 The circuit diagram Problem 1b We use Ohm’s law to find the current I V = IR Firstly we have to rearrange Ohm’s law by dividing both sides of the equation by R: V I R = (5) Substituting our values for V = 15 V dc and R = 1kΩ = 1 × 10 3 we get 3 15 1 10 x = 15 × 10 −3 amperes = 15mA Problem 1c Since current in all parts of the circuit is the same then the same current must also flow through the voltage source, that is, the current through the voltage source is also 15mA. Problem 1d ENG1021 Electronic Principles 18 If the voltage of the source is doubled, then multiplying both sides of Equation 5 by 2 gives: 2 2 V I amperes R × = × and shows that the current will also be doubled. The current I will therefore be 2 × 15 mA = 30 mA An alternative solution is to double the voltage and recalculate I using Equation 5. The old voltage was V = 15 volts so the doubled voltage is V = 30 volts. Substituting into Equation 5 as before gives 3 30 1 10 × = 30 × 10 −3 amperes = 30 mA Problem 2a This first part of this question requires Ohm’s law again. We use the form: V = IR We are told that 2 amps of current flows in a circuit with a resistance of 12Ω. Substituting these values for I and R respectively into Ohm’s law gives: V = 2 × 12 = 24 volts Problem 2b Now that we have found the voltage V we may calculate the power using P = V I. Substituting for I and V (I = 2 amps and V = 24 volts) P = 24 × 2 = 48 watts Problem 2c The power dissipated in the resistor must have been supplied by the battery to heat up the resistor; therefore the power supplied by the battery is also 48 watts. Problem 3 Parts a) and b) of this problem use a rearrangement of Ohm’s law. We require an equation for resistance R alone so we divide both sides of equation 1 by I to give: ENG1021 Electronic Principles 19 V IR I I = Cancelling the I’s and swapping left and right hand sides of the equation gives: V R I = (6) Problem 3a The current I is 1 mA = 1×10 −3 amperes while the voltage V is 12 volts. Substituting these values into Equation 6 gives: 3 3 12 12 10 1 10 R − = = × Ω × Using the k for kilo prefix the answer is: Resistance R = 12 kΩ. Problem 3b Similarly substituting into Equation 6 for I equal to 4 mA = 4 × 10 −3 amperes and the voltage V equal to 15 volts gives: 3 3 3 15 15 10 3.75 10 4 10 4 R − = = × Ω = × Ω × Using the k for kilo prefix the answer is: Resistance R = 3.75 kΩ . Problem 3c We can attempt the solution of this problem in two ways. I prefer the first way because we only have to remember the two equations for voltage and power (Equations 1 and 2). First way (preferred) We have values for the power P and the voltage V and require the resistance R. Ohm’s law will give us a value for R, but we need I. I can be found from Equation 2, by dividing both sides of the equation by V and rearranging. P I V = Substituting for P and V into the equation gives: ENG1021 Electronic Principles 20 3 3 150 10 4.1667 10 36 I amperes − − × = = × Substituting I and V into Equation 6 gives: 3 3 36 8.64 10 8.64 4.1667 10 R k − = = × = Ω × Second way Using the Equation 4 directly: 2 V R P = and substituting for V = 36 volts and P = 150 mW = 150 × 10 −3 gives: 2 3 3 36 1296 10 150 10 150 R − = = × Ω × R = 8.64 × 10 3 = 8.64 kΩ Problem 3d Again we can attempt the solution of this problem in two ways and the first way only requires knowledge of Equations 1 and 2. First way (preferred) Now we have values for the power P and the current I and require the resistance R. Again we can use Ohm’s law to find R, but we need a value V now. V can be found from Equation 2, by dividing both sides by I and rearranging. P V I = Substituting for P and I into the equation gives: 3 3 16.2 0.54 10 30 10 V volts − = = × × Substituting for I and V into Ohm’s law gives: 3 6 3 0.54 10 0.018 10 18 30 10 R k − × = = × Ω = Ω × Second way ENG1021 Electronic Principles 21 Using the Equation 3 directly: 2 P R I = and substituting for P = 16.2 watts and I = 30 mA gives: 6 6 3 2 16.2 16.2 10 0.018 10 18 (30 10 ) 900 R k − = = × = × Ω = Ω × Problem 4a This exercise relates power and energy. Electrical energy is the power multiplied by the time that the power is used for. Energy = power × time (7) Hopefully this is instinctively true from this exercise. When we pay our electricity bills we pay for energy used. It would seem reasonable that the longer the electric heater is on (time) then the more energy we would use and the more we would pay. Before we can calculate the payment we must first determine how much energy has been used. There are two common units for measuring energy, joules and kilowatt hours. They are both in units of power multiplied by time. Joules are watts multiplied by seconds and kilowatt hours (kWh) are kilowatts multiplied by hours. We could calculate the energy used in either joules or kilowatt hours, but since this exercise deals with power in kilowatts (1500W =1.5 kW) and time in hours and the payment details are in terms of kilowatt hours then we will calculate the energy for this question in kilowatt hours. The heater’s power consumption is 1500 W or 1.5 kW. It is used for 36 hours. Let us substitute these values into the energy equation, Equation 7: Energy = 1.5 × 36 = 54 kWh Now that we know how much energy is used we calculate the payment by multiplying the charge per kilowatt hour by amount of energy used (in kilowatt hours) which is 7 cents/kWh. So the answer is: Cost = 54 × 7 = 378 cents = 3.78 dollars 11 Further reading At this point you may like to have a look at the relevant sections of the on-line book “Fundamentals of Electrical Engineering and Electronics”. ENG1021 Electronic Principles 22 Under the section entitled “DC”, you may like to have a look at the part called “Basic Concepts of Electricity”, which reinforces some of the theory given at the beginning of this Learning Package. Following that you could have a look at all the parts under the section heading “Ohm’s Law”. 12 Where next? You are encouraged to study the Learning Package entitled “Series and Parallel circuits” next. ENG1021 Electronic Principles 23 ENG1021 Electronic Principles Learning Package 3 Series and parallel circuits ENG1021 Electronic Principles 24 Series and parallel circuits 1 Do you know all this already? If in doubt, please attempt the self-assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Introduction This section introduces series and parallel circuits. The meaning of the terms “series” and “parallel” are introduced. Then methods for calculating the total resistance of series and parallel combinations of individual resistors are explained. Polarity of voltages and currents are demonstrated. Methods of adding voltages and currents are then shown. We will start with series circuits and then highlight differences and similarities with parallel circuits. 3 Series circuits 3.1 Potential difference In the circuits that we’ve seen so far there has only been one resistor. In a circuits with one resistor, the whole of the voltage of the battery as applied across the resistor. In other words, if we put a voltmeter across the resistor which was connected to a 9V battery, it would give a reading of 9V. The voltage across the resistor is called the “voltage drop” or the “potential difference” across the resistor (often abbreviated to “pd”). If the voltage drop across the resistor is V, the current through the resistor is I and the resistance itself is R, then Ohm’s law applies. V = I x R (1) As mentioned before, a voltage has to be measured between two points. It is therefore convenient to think of one point as the reference point, and measure all voltages relative to that reference point. In the case of a battery, the negative terminal would most likely be chosen as the reference point. This means that the positive terminal is 9 V higher than the negative terminal. The voltage of the battery is often called the “electromotive force” and given the symbol E, and is often shortened to the “emf”. Therefore the battery is supplying an electromotive force which is pushing the electrons around the circuit. As the electrons travel around the circuit they lose energy, so that the energy at one side of the resistor will be higher than at the other, and this difference is what we’ve called the potential difference. ENG1021 Electronic Principles 25 So, the energy at the positive terminal is E volts. When we get to the end of the resistor that is connected to the positive terminal the energy is still E (ignoring the very small drop along the wire). Now pushing the electrons through the resistor takes more effort, and uses up the energy, so that by the time we look at the energy at the other end of the resistor it will have dropped by an amount V volts, and this end of the resistor is connected to the negative terminal of the battery. All of the energy that was available, namely E, has to be used up by the time you get around the circuit. So if the energy lost in the resistor is V, we have to be able to say that V = E. So we can say, that: E = V = I x R (2) So, in the case of a circuit with a single resistor, the voltage difference across that resistor equals the electromotive force of the battery. Now let’s look at circuits with more resistors. 3.2 Series resistors and current in series circuit A circuit in which two or more resistors are connected together “in series” means that one end of resistor is connected to one end of another, as shown in Figure 1. Figure 1 Two resistors in series We have already said in an earlier Learning Package that the current flows all through the circuit, and in the circuit in Figure 1 this means that the same current flows through both resistors. From the previous discussion we know that there is a potential difference across R1, which I shall call V1, and a potential difference across R2, which I shall call V2. In Figure 1 the polarity of the voltages has been shown. We know that the battery has a positive and a negative terminal, and that the electron current flows from the negative terminal, around the circuit to the positive terminal. In other words the current flows from the negative to the positive. Similarly, the potential difference across a resistor is shown as having a positive side and a negative where the positive side will have a higher voltage than the negative side, and the current will flow from the negative to the positive again. ENG1021 Electronic Principles 26 Also, we know that the electromotive force, E, of the battery must equal the total potential difference around the circuit (this will be discussed in more detail in a later Learning Package). So, we can say that: E = V1 + V2 (3) For example, if the two resistors are equal, the battery is 9 V, and we have a voltmeter with the negative lead attached to the negative terminal of the battery, then going around the circuit we would find that: At the positive terminal of the battery we would measure 9 V; At the left-hand (+) terminal of R1 we would measure 9 V; At the right-hand (-) terminal of R1 we would measure 4.5 V; At the left-hand (+) terminal of R2 we would measure 4.5 V; At the right-hand (-) terminal of R2 we would measure 0 V; In other words, with equal value resistors, the voltage difference across each resistor would be equal to half the electromotive force of the battery. Returning to a series circuit with two arbitrary resistors, we know that the potential difference across each resistor when added together equal the emf of the battery. Using Ohm’s law we can replace the potential difference across each resistor by the current times the resistor value. V1 = I1R1 (4) V2 = I2R2 (5) E = V1 + V2 = I1R1 + I2R2 However, we already know that the current through each part of the circuit must be the same. So I1 = I2 which we will call I, and the equation becomes: E = IR1 + IR2 = I(R1 + R2) = IR eq We can replace the two resistor values with one equivalent resistor, R eq . The value of this equivalent resistor is: R eq = R1 + R2 (6) So the total resistance of the circuit is the sum of the individual resistors. This is true for any number of series resistors. For any number of resistors this would be written as: RT = R1 + R2 + R3 +… (7) ENG1021 Electronic Principles 27 You should now appreciate what a series circuit is, and calculate the total resistance (or equivalent resistance) of a series string of resistors. You have also learnt why the current in all parts of a series circuit is the same. You should now be able to attempt the following Problem. Do this now without reference to the worked solutions if possible. Problem 1 (Resistor in series) A circuit has 20 V applied across a 10 Ω resistance R 1 . How much is the current in the circuit? How much resistance R 2 must be added in series with R 1 to reduce the current by one half? Show the schematic diagram of the circuit with R 1 and R 2 . Solution The first part of this question is a simple Ohm’s law exercise. The voltage V is 20 volts, the resistance R1 is 10Ω and we are asked to calculate the current I. From the simple re-arrangement of Ohm’s law which we used in the last Learning Package: 20 2 10 V I amperes R = = = We are asked to add some resistance R2 to reduce the current by one half. One half of the above current is 1 ampere, so we must firstly calculate what total (or equivalent) resistance RT is required to cause a 1 ampere current to flow in a circuit with a 20 volt voltage source. Re-arranging Ohm’s law again: 20 20 1 T V R I = = = Ω We are not asked for the total resistance RT, we are asked for the additional resistance R2. Therefore we must use the equation for resistors in series: RT = R1 + R2 + … In this case we only have two resistors so the equation is simply: RT = R1 + R2 Subtracting R1 from both sides of this equation gives: RT − R1 = R2 and then substituting in our values for RT , and R1 gives: R2 = 20 − 10 = 10Ω ENG1021 Electronic Principles 28 You may have arrived at your value for RT using by noting that from Ohm’s law, to half the current we must double the resistance (assuming that the voltage stays the same). So RT = 2 × R1 = 20 Ω Either way is correct. The schematic diagram for the circuit is shown in Figure 2. Figure 2 3.3 Power in series circuit The power dissipated in any resistor is given by the power equation (which was introduced in the previous Learning Package), but the total power in the series circuit is the sum of the powers dissipated in the individual resistors. You should now be able to attempt Problems 2 and 3. Do this now without reference to the worked solutions if possible. Problem 2 (potential difference and power) Draw the schematic diagram of 20, 30 and 40 Ω resistances in series. (a) How much is the total resistance of the entire series string? (b) How much current flows in each resistance, with a voltage of 18 V applied across the series string? (c) Find the voltage drop across each resistance. (d) Find the power dissipated in each resistance. Solution The schematic (or circuit) diagram is shown in Figure 3. Problem 2a ENG1021 Electronic Principles 29 Figure 3 The total resistance is given by the formula for resistors in series, Equation 7 above. In this problem we have three resistors which I will call R1, R2 and R3. I have labelled the resistors so that I can identify the voltage (IR) drops later in the solution. The total resistance is: RT = R1 + R2 + R3 = 20 + 30 + 40 = 90 Ω Problem 2b We should know that the same current flows in all parts of a series circuit and therefore the same current flows in each resistor in this circuit. We calculate the series current using RT and Ohm’s law in the following form. 18 0.2 90 T V I amperes R = = = Problem 2c This part of the problem requires us to calculate the IR drops across the resistors. Let the voltage drop across R1 be V1, the voltage drop across R2 be V2 and the voltage drop across R3 be V3. The IR (or voltage) drops are all given by Ohm’s law. Therefore using our value for I and the three resistances we have: V1 = I × R1 = 0.2 × 20 = 4 volts V2 = I × R2 = 0.2 × 30 = 6 volts V3 = I × R3 = 0.2 × 40 = 8 volts We can check our answers because we know that the total of the IR drops should add up to the source voltage of 18 volts. Let’s do that: ENG1021 Electronic Principles 30 V1 + V2 + V3 = 4 + 6 + 8 = 18 volts which is the source (or applied) voltage. Problem 2d The final part of this question asks us to calculate the power dissipated in each resistor. We can do this using: P = IV P = I 2 R or 2 V P R = I will obtain the solution using P = IV. Let the power dissipated in R1 be P1, the power dissipated in R2 be P2 and the power dissipated in R3 be P3. Using our values for I and the voltage drops across each resistor gives P1 = I × V1 = 0.2 × 4 = 0.8 watts P2 = I × V2 = 0.2 × 6 = 1.2 watts P3 = I × V3 = 0.2 × 8 = 1.6 watts You may want to verify that the same solutions are found if the alternative equations for power are used. Note that if we add up the individual powers, PT = P1 + P2 + P3 = 0.8 + 1.2 + 1.6 = 3.6 watts we get the same result as calculating the power dissipated in the circuit using the applied voltage and the current, P = I × VT = 0.2 × 18 = 3.6 watts Problem 3 Draw a schematic diagram showing two resistances R 1 and R 2 in series across a 100 V source. (a) If the IR voltage drop across R 1 is 60V, how much is the IR voltage drop across R 2 ? (b) Label the polarity of the voltage drops across R 1 and R 2 . (c) If the current is 1A through R 1 , how much is the current through R 2 ? (d) How much is the resistance of R 1 and R 2 ? How much is the total resistance across the voltage source? (e) If the voltage source is disconnected, how much is the voltage across R 1 and across R 2 ? ENG1021 Electronic Principles 31 Solution The circuit diagram is shown in Figure 4. Figure 4 Problem 3a We do not need I to determine the IR drop across R2, because we know that the sum of the IR drops is equal to the applied voltage. If V1 is the voltage drop across R1 and V2 is the voltage drop across R2 the applied voltage VT = V1 + V2. (30) VT = 100 volts and V1 = 60 volts Therefore V2 = VT − V1 = 100 − 60 = 40 volts (31) Problem 3b When I (and you) produced the circuit diagram at the start of this question we decided on the polarity of the voltage source. I decided that the top of the battery is positive. This determines the polarity of the voltage drops. If we assume electron current flow (current flows from negative to positive) then the current flows anti-clockwise around the circuit as shown by the arrow in Figure 5. For current to flow in the anti-clockwise direction, and remembering that the same current flows in all parts of a series circuit, then the left of R1 must be positive with respect to the right of R1 (which in turn will be negative). By a similar argument, the left of R2 will be positive with respect to the right of R2. These polarities are marked on Figure 5. ENG1021 Electronic Principles 32 Figure 5 Problem 3c They cannot catch us with this one, can they? The current flowing in all parts of a series circuit is the same, and since R2 is in the same series circuit as R1 the current must also be the same. That is the current through R2 is 1A. Problem 3d We know that the IR drop across R1 (V1) is 60 V and that the IR drop across R1 (V1) is 40 V Using Ohm’s law again: IR1 = 40 V, since I = 1 A, R1 = 40Ω and IR2 = 60 V, since I = 1 A, R2 = 60Ω. The total resistance across the voltage source RT = R1 + R2 = 40 + 60 = 100 Ω. Problem 3e If the voltage source is disconnected then no current can flow, since there is no longer a potential difference across the resistors. Then I = 0 and the IR drops (or voltage across) IR1 and IR2 must both be zero. 3.4 Series and opposing voltages, tips for analysis You’ve seen circuits with more than one resistor in series, so what happens if there are more batteries in series? Put simply, if the batteries are facing the same way, then the voltages are added up. By the same way I mean that the positive terminal of the first battery is connected to the negative terminal of the second and so on. In other words they are all trying to push the current around ENG1021 Electronic Principles 33 the circuit in the same direction. If the batteries are connected such that some are trying to push the current in the opposite direction, then their voltages are subtracted. Now please attempt the following Problem 4. Try the problem before looking at the solution. Problem 4 (Series adding/opposing voltages) Figure 6 shows the circuit for keeping a 12.6 V car battery charged from a 14.8 V dc generator. Calculate the current I and show the direction of electron flow for current between point A and B. Figure 6 Solution The problem introduces a battery charger as a practical example of opposing voltages. We will work through the problem and then consider the effect of connecting the charger to the battery the wrong way round. Consider the circuit associated with Problem 4 as shown in Figure 6. V1 represents voltage of the battery, while V2 represents the voltage of the charger. These two voltage sources have their positive ends connected together; the voltage therefore oppose. The overall voltage is therefore: V2 − V1 = 14.8 − 12.6 = 2.2 volts We could redraw the circuit replacing both sources with a single 2.2 volt source. We have to consider what polarity this replacement 2.2 volt source will have. This is shown in Figure 7. The polarity will be the same as the battery charger voltage V2 since V2 is the higher voltage. The current is then given by Ohm’s law: 2.2 3.367 0.6 V I amperes R = = = + - - + V 1 =12.6 V V 2 =14.8 V 0.6Ω A B ENG1021 Electronic Principles 34 Figure 7 The opposing voltages are equivalent to a 2.2 volt source Now consider what happens if we connect the battery charger the wrong way round. Figure 8 shows the new circuit. The circuit looks almost identical to Figure 6, but notice that the polarity of the battery has changed. Voltage source V1 now has its negative terminal connected to the positive terminal of voltage source V2. This means that their voltages will add and the overall voltage will be: V2 + V1 = 14.8 + 12.6 = 27.4 volts Figure 8 Now the battery is connected the wrong way round We could redraw the circuit as we did above, but this time replacing both voltage sources with a single 27.4 volt source. This is shown in Figure 9. The polarities of the original voltage sources would have individually caused an electron current to flow in the same direction, that is, anticlockwise around the circuit. The polarity of the replacement voltage source is therefore the same as both of the original sources. The current is again given by Ohm’s law: 27.4 45.67 0.6 V I amperes R = = = ENG1021 Electronic Principles 35 Figure 9 The series aiding voltages are equivalent to a 27.4 volt source This is a very high current and if no other protection was provided would certainly melt the wires of typical battery chargers. The example serves to show a practical application of series aiding and series opposing voltages. 3.5 Open and short circuits in a series path In many practical circuits it is possible to introduce a short-circuit or an open- circuit. A short circuit has zero resistance, and can cause high currents to appear in a circuit. It may be caused by component failure or by contamination. Open circuits, on the other hand, have infinite resistance and prevent current from flowing. This is usually caused by a connection breaking between a component and the board that it is on. Now attempt Problem 5. Problem 5 Three resistors of 100, 200 and 300 Ω are in series across a 24 V source. (a) If the 200 Ω resistor shorts, how much voltage is across the 300 Ω resistor? (b) If the 300 Ω resistor opens, how much voltage is across the 100 and 200 Ω resistors? Solution The problem concerns a series circuit containing 3 resistors 100Ω, 200Ω and 300Ω. I have drawn the circuit in Figure 10. Problem 5a ENG1021 Electronic Principles 36 The first part of the problem asks us to find the voltage across the 300Ω resistor if the 200Ω resistor goes short circuit. The situation is depicted in Figure 11. The short circuit is considered to have no (zero) resistance, and all of the current flows in the short circuit, with no current flowing in the 200Ω resistor. Effectively, it is as if the 200Ω resistor has been removed from the circuit and replaced by a piece of wire. The circuit therefore reduces to a 100Ω resistor and a 300Ω resistor in series. The total series resistance is therefore: RT = 100 + 0 (the short circuit) + 300 = 400 Ω Figure 10 The initial series circuit for Problem 5. Figure 11 The circuit for Problem 5 if the 200Ω resistor goes short circuit The current in the circuit is given by Ohm’s law 24 60 400 T V I mA R = = = The voltage across the 300 mA resistor is also given by Ohms law V300Ω resistor = IR = 60 × 10 −3 300 = 18000 × 10 −3 = 18 V Problem 5b ENG1021 Electronic Principles 37 This part of the Problem considers the situation if the 300 Ω resistor breaks or goes open circuit. The situation is depicted in Figure 12. The open circuit is assumed to have infinite resistance. It is as if the 300 Ω resistor has been removed from the circuit and its connecting wires have been left unconnected. No current can flow in the circuit and since the same current (zero) flows in all parts of a series circuit, no current flows in the 100 Ω and the 200 Ω resistors. Since no current flows in these resistors there is no voltage (IR) drop across the resistors. That is the voltage across the 100 Ω and the 200 Ω resistors is zero in both cases. Figure 12 The circuit for Problem 5 if the 300 Ω resistor goes open circuit. Another way of seeing that the current is zero is by considering the open circuit resistor to have a resistance of ∞ (infinity) and calculate the total resistance. We have RT = 100 + 200 + ∞ = ∞ The current is then the supply voltage divided by infinity which equals zero. 24 0 T V I R = = = ∞ 4 Parallel circuits 4.1 Introduction to parallel circuits and branch currents ENG1021 Electronic Principles 38 Figure 13 Figure 13 shows a simple parallel circuit where we have two resistors, this time joined at both ends to each other. The branches of the circuit with the resistor in are parallel to one another, hence the name. In this instance we can see that the positive terminal of the battery is connected to the left-hand side of both resistors, and that the negative terminal of the battery is connected to the right-hand side of both resistors. If we were to place a voltmeter across either of the resistors it would measure the same value as the electromotive force of the battery, since they are all effectively the same two points. So, in this case the potential difference across each resistor must equal the electromotive force: E = V1 = V2 V1 = I1R1 and V2 = I2R2 So, E = I1R1 and E = I2R2 I1 = E/R1 and I2 = E/R2 This gives us the value of the current in each branch. Although we will discuss this more formally later in the module, we can intuitively see that if the current flowing out of the negative terminal of the battery reaches the point where the two branches are connected, the current has to split so that some of it goes around one branch and some around the other. If we think of the water analogy, then it is clear that the amount of current going into each branch must add up to the total current coming out of the battery, since the current hasn’t got anywhere else to go, and it doesn’t accumulate in puddles at junctions. So, I = I1 + I2 = E/R1 + E/R2 = E(1/R1 + 1/R2) If we introduce an equivalent single resistor again, then we could say that: E = IR eq I = E/R eq Then: E/R eq = E(1/R1 + 1/R2) 1/R eq = 1/R1 + 1/R2 (8) In a circuit with two resistors in parallel, the equivalent resistor is found using this equation. Namely, that the reciprocal of the equivalent resistance equals the sum ENG1021 Electronic Principles 39 of the reciprocals of the two resistances. This can be generalised for any parallel circuit. For example, with many resistors in parallel, the equation would be: 1 2 3 1 1 1 1 .......... eq R R R R = + + + (9) In the special case of a parallel circuit with two resistors in parallel, the equation can be rearranged. 1/R eq = 1/R1 + 1/R2 = (R2 + R1)/R1R2 R eq = R1R2/(R1 + R2) (10) You now know what a parallel circuit is. You will have seen that there are rules which apply to parallel circuits which differ slightly to series circuits. Generally they differ by the transposition of the terms “current” and “voltage”. For example in a series circuit, the current flowing in all parts of the circuit is the same, while in a parallel circuit the applied voltage is the same across all parallel branches. Also, in a series circuit the applied voltage E is the sum of the individual voltage drops, while in a parallel circuit the main current I is the sum of the individual branch currents. Bear these differences and similarities in mind when trying to remember the rules. Now attempt Problem 6. Do this without reference to the worked solutions if possible. Problem 6 (Resistors in parallel) A 6 Ω R 1 and a 12 Ω R 2 are connected in parallel across a 12 V battery. (a) Draw the schematic diagram. (b) How much is the voltage across R 1 and R 2 ? (c) How much is the current in R 1 and R 2 ? (d) How much is the main-line current? (e) Calculate R EQ . Solution Problem 6a The diagram is shown in Figure 14. ENG1021 Electronic Principles 40 Figure 14 Problem 6b Since all parts of the circuit are in parallel and we know that the voltage is the same across all parallel branches, then the voltage across R1 and R2 is the same as the battery voltage which is 12 V. Problem 6c Let us call the current through 6Ω resistor R1, I1 and the current through 12 Ω resistor R2, I2. The current though R1 1 1 12 2 6 T V I amperes R = = = and 2 2 12 1 12 T V I amperes R = = = Problem 6d The total or mainline current IT is the sum of the two branch currents. IT = I1 + I2 = 2 + 1 = 3 amperes Problem 6e We can calculate the equivalent resistance by considering what single resistance would cause the same main-line current of 3A to flow. This is given by Ohms law: ENG1021 Electronic Principles 41 12 4 3 T eq T V R I = = = Ω A handy check to see if you have calculated a parallel resistance properly is that the calculated resistance must always be less than the any of the values of the resistors in the branches. In this case the calculated resistance is 4 Ω and the two parallel resistors are 6 Ω and 12 Ω. If our calculated value had been greater than 6 Ω we must have made a mistake in our calculations. 4.2 More parallel circuits and branch currents and power You will remember the rather uncomfortable formulae for parallel resistors: 1 2 3 1 1 1 1 .......... eq R R R R = + + + 1 2 1 2 eq R R R R R × = + if there are only two resistors Don’t forget that if the resistors all have the same value then the equivalent resistance Req of the parallel combination is simply the value of the resistance divided by the number of resistors in the parallel combination. For example, if there are four resistors with the same value, R, in parallel, the equivalent resistance is: 1 2 3 4 1 1 1 1 1 1 1 1 1 4 4 eq eq R R R R R R R R R R R R = + + + = + + + = = Now attempt Problem 7 and Problem 8. Problem 7 (Resistors in parallel equation and power) For the circuit in Problem 6, how much is the total power supplied by the battery? Solution We are asked to calculate the total power supplied by the battery in the above problem. The total power is the sum of the power dissipated in each branch of the parallel circuit. The current through the first branch I1 is 1 A. The supply voltage is 12 V and this is the same across all parallel branches. Therefore the power in the first branch is, P1 = I1 × V = 1 × 12 = 12 W ENG1021 Electronic Principles 42 The current in the second branch I2 is 2 A. Again the same voltage 12 V exists across this branch, therefore the current in the second branch P2 = I2 × V = 2 × 12 = 24 W The total power P = P1 + P2 = 12 + 24 = 36 W We could have used the main-line current to give the total power supplied by the battery. Since the battery supplies 3 A and its voltage is 12 V, then it must supply: P = I V = 3 12 = 36 W of power. The same result can be arrived at using the equivalent resistance Req and either: 2 2 12 144 36 4 4 eq V P W R = = = = or P = I 2 Req = I 2 × 4 = 3 2 × 4 = 9 × 4 = 36 W All of the above solutions are correct and there is no single correct solution. Do try different solutions so that you become practised in using them. Problem 8 Find the R EQ of the following groups of branch resistances: (a) 10 Ω and 25 Ω; (b) five 10 k Ω resistances; (c) two 500 Ω resistances; (d) 100 Ω, 200 Ω and 300 Ω; (e) two 5 k Ω and two 2 k Ω resistances; (f) four 40 k Ω and two 20 k Ω resistances. Solution Problem 8a Here we have two resistors, with unequal values. We can use Equation 10 with R1 = 10 Ω and R2 = 25 Ω. The equivalent resistance: 10 25 250 7.14 10 25 35 eq R × = = = Ω + Problem 8b ENG1021 Electronic Principles 43 Our luck is in as the resistor values are equal (10 kΩ each), so we simply divide the value of the resistances (10 kΩ) by the number of resistances (5). The equivalent resistance 10 2 5 eq k R k Ω = = Ω Problem 8c Now we only have two resistances of the same value, so the equivalent resistance of the group is half of the value of one resistance. 500 250 2 eq R = = Ω Problem 8d Here we have to use Equation 9. Let the 100 Ω resistor be R1, the 200 Ω resistor be R2 and the 300 Ω resistor be R3. Substitute the values into Equation 9 to find the equivalent resistance: 1 1 1 1 100 200 300 eq R = + + 1 0.01 0.005 0.00333 eq R = + + 1 0.01833 eq R = 1 0.01833 eq R = Req = 54.56 Ω Problem 8e We could do this using Equation 9, but we can take advantage of the fact that we have two sets of resistors of the same value. So we can calculate the equivalent value for each set and then consider that we have a two resistor parallel group containing the equivalent resistors. Firstly take our two 5 kΩ resistors and consider their parallel combination. They have the same value, so the value of the parallel combination is 5kΩ/2 = 2.5 kΩ. Similarly we can treat the next two 2 kΩ resistors as a parallel combination to give 2kΩ/2= 1 kΩ. ENG1021 Electronic Principles 44 Now take each parallel combination and use Equation 10 to give a final equivalent resistance: 2.5 1 2.5 1 eq k k R k k × = + 2.5 1 3.5 eq k k R k × = 2.5 714 3.5 eq k R = = Ω Problem 8f Again we can take advantage of two groups of resistors of the same value. Taking the first group, we have four 40 kΩ resistors, this group’s equivalent resistance 1 40 10 4 eq k R k Ω = = Ω In the second group we have two 20 kΩ resistors and so the equivalent resistance of this group is 2 20 10 2 eq k R k Ω = = Ω Now we have a pair of equal value equivalent resistors (both 10 kΩ), so the overall equivalent resistance is: 10 5 2 eq k R k Ω = = Ω 4.3 Short circuits and open circuits in parallel circuits If the main line of a parallel circuit opens there is no current flow from that point. However, if a branch opens there is no current flow in that branch, but other branches are unaffected, although the total current will drop as the equivalent resistance changes. A short circuit in any branch is an extreme fault condition and will cause the whole circuit to draw excessive current. Now attempt Problem 9. Problem 9 In Figure 15, what is R EQ if R2: (a) opens; (b) shorts? ENG1021 Electronic Principles 45 Figure 15 The circuit diagram for Problem 9 Solution Problem 9a If R2 opens then we can consider it either as being removed from the circuit (leaving only R1 and R3) or as a resistor with infinite resistance. This is shown in Figure 16. Figure 16 Circuit for Problem 9, but R2 is open circuit If we consider it as being totally removed then we can use the simplified parallel resistance formula: 1 3 1 3 60 20 1200 15 60 20 80 eq R R R R R × × = = = = Ω + + Alternatively we can consider the open circuit R2 as having a voltage of infinity. The equivalent resistance of the three resistor parallel combination (REQ) is given by 1 2 3 1 1 1 1 eq R R R R = + + ENG1021 Electronic Principles 46 1 1 1 1 60 20 eq R = + + ∞ Note that: 1 0 = ∞ so the open circuit R2 has no effect in the equation 1 1 1 0 60 20 eq R = + + 1 60 20 60 20 eq R + = × 60 20 60 20 eq R × = + This is the same equation as the equation above. Therefore REQ = 15Ω I have considered the problem this second way, just to demonstrate that we can consider the open circuit as an infinite resistance and arrive at the same result. Normally I would assume that an open circuit resistor in a parallel circuit has effectively been removed. Problem 9b Here we consider the case where R2 is short circuited. The equivalent circuit is shown in Figure 17. We assume the short circuit has zero resistance. We can consider this short circuit as a wire that bypasses the other parallel resistors so that no current can flow in them. The short circuited R2 is equivalent to replacing R2 with a zero ohm resistor. Remember that the addition of parallel resistance can only lower the equivalent resistance REQ. The other parallel resistors cannot lower zero ohms (resistors cannot have negative resistance), therefore the equivalent resistance REQ must be zero. ENG1021 Electronic Principles 47 Figure 17 The circuit of Problem 18 if R2 is short circuited Alternatively we can substitute the value zero into the three resistor parallel equivalent equation to find REQ (the hard way). (REQ) is given by: 1 2 3 1 1 1 1 eq R R R R = + + 1 1 1 1 60 0 20 eq R = + + Note that: 1 0 = ∞ so the short circuit R2 is dominant in the equation: 1 1 1 60 20 eq R = + ∞+ 1 eq R = ∞ 1 eq R = ∞ Therefore REQ = 0Ω Again I have used the second method to show that the mathematics will produce the same answer as the more intuitive approach above. In practice any parallel circuit with a short circuit in any branch will have zero resistance, as the parallel ENG1021 Electronic Principles 48 resistors cannot reduce the zero resistance of the short circuit any lower than zero. 5 Potential and current dividers We’ve seen that in a circuit with two resistors in series, R1 and R2, the electromotive force is divided between the resistors as V1 and V2 and that E = V1 + V2. This is a commonly found circuit in electronics and is called a potential divider. The input voltage is the voltage source, E, and the output voltage is the potential difference across R2, namely V2. To find its value, first apply Ohm’s law to find the current in the circuit. E = I(R1 + R2) I = E/(R1 + R2) Next use Ohm’s law to find the value of the potential difference across R2, namely V2: V2 = IR2 = E/(R1 + R2) x R2 = ER2/(R1 + R2) (11) We also saw earlier that when a circuit contains two resistors in parallel, the equivalent resistance is found using: 1/R = 1/R1 + 1/R2 A quantity that isn’t often used but is useful in parallel circuit is conductivity, which is the reciprocal of resistance and has the symbol G. Taking the parallel circuit and replacing resistances with conductances, the equation for the equivalent conductance is: G = G1 + G2 (12) A circuit with two parallel branches is sometimes referred to as a current divider. The current generated by the battery is divided down each parallel branch. The value through R2 can be found as follows: We saw earlier that the equivalent resistance is R = R1R2/(R1 + R2), so the current from the battery is: E = IR1R2/(R1 + R2) I = E(R1 + R2)/R1R2 = E/R2 + E/R1 This divides into two branches. Since the potential difference across each parallel branch is E, the current in each branch is: I1 = E/R1 ENG1021 Electronic Principles 49 I2 = E/R2 The total current being the sum of these two currents. Thus I2 in terms of I is: I2 = E/R2 = IR1R2/R2(R1 + R2) = IR1/(R1 + R2) If we use conductances instead of resistances, the equation becomes: I2 = IG2/(G1 + G2) 6 Further reading At this point you may like to have a look at the relevant sections of the on-line book “Fundamentals of Electrical Engineering and Electronics”. Under the section entitled “DC”, you may like to have a look at the part called “Series and parallel circuits”. 7 Where next? You are encouraged to study the Learning Package entitled “Kirchoff’s laws” next. ENG1021 Electronic Principles 50 ENG1021 Electronic Principles Learning Package 4 Kirchoff’s laws ENG1021 Electronic Principles 51 Kirchoff’s laws 1 Do you know all this already? If in doubt, please attempt the self assessments questions in this Learning Package. If you are not sure about any of the terminology, please read on. 2 Introduction Although simple in concept, the application of Kirchoff’s laws can be quite confusing. The main source of the confusion is usually concerned with the direction and polarity of currents and voltages. The situation is not helped by the various methods of solution, which are easy to mix up. Although most books on electronics described many different methods, here I only want you to learn how to use one of the methods, namely branch currents. 3 Kirchoff’s laws In the previous Learning Package I said that the voltage drops around a circuit must equal the electromotive force. I used this to derive the equation for the equivalent resistance in a series circuit. Similarly, in deriving the equivalent resistance of a parallel circuit I used the idea that current at a junction would behave in such a way that the current flowing out of a junction must equal the current flowing into a junction. These two ideas were introduced by Kirchoff and are known as Kirchoff’s laws. Formally stated they are: Kirchoff’s voltage law - The algebraic sum of all voltages in a loop must equal zero; Kirchoff’s current law – The algebraic sum of all currents entering and exiting a node must equal zero Let’s take the current law first, as I believe this is the simpler of the two. ENG1021 Electronic Principles 52 Figure 1 At a node, or a junction, where wires are joined together, the sum of the current flowing into the node must equal the sum of he current leaving the node. If this wasn’t the case there would be a surplus which would form a pool of charge at the node, and this doesn’t happen. If Figure 1 we have two currents flowing into the node, I1 and I2, and two currents flowing out, I3 and I4. The convention is that current flowing into a node should have the opposite sign to current flowing out of a node. It doesn’t matter which, but I will assign current flowing in to a node as positive and current flowing out as negative. Then: I1 + I2 – I3 – I4 = 0 Or put another way: I1 + I2 = I3 + I4 This equation is the mathematical equivalent to Kirchoff’s current law. The sum of all the currents entering and leaving a node is zero. Now let’s try Kirchoff’s voltage law. First of all I need to explain what is meant by a loop. When we were looking series circuits, there was only one loop, and that was the complete circuit. Usually circuit are more complex, such as that shown in Figure 2. Figure 2 In Figure 2 we have a circuit with three resistors. It’s not as simple as a series circuit or a parallel circuit as there is a mixture of the two. There are a number of ways that this circuit could be tackled to analyse what is going on. However, I just want to use it to illustrate what is meant by a “loop” in Kirchoff’s voltage law. ENG1021 Electronic Principles 53 If we start at the point marked A, we could go A – B – E – F – A and that would be a loop. Let’s call it Loop 1. We could also go from A again and this time travel around a different loop: A – B – C – D – E – F – A. Let’s call this one Loop 2. Finally, a third loop starts at B and goes B – C – D – E – B. Let’s call this one Loop 3. So, there are a possible three different loops in this circuit. Kirchoff’s voltage law applies to each one. In Figure 2 I’ve already drawn in the polarity of the potential differences in the circuit, and the currents in each branch. Although I am confident that I‘ve drawn these correctly, it wouldn’t matter if I was wrong. If it turned out that one of the currents actually goes the other way, then when I do the analysis, that current will come out negative. Similarly, if I calculate the potential difference and it comes out negative, that would indicate that I had drawn the polarity of the potential difference the wrong way round. Now, let’s apply Kirchoff’s voltage law. When analysing a loop, you travel around the loop and note down all the potential differences and electromotive forces. Again, it doesn’t matter if you travel clockwise or anticlockwise. Let’s say we go clockwise, and that we note the potential difference as positive if we travel from “- “ to “+”, and as negative if we travel from “+” to “-“. Let the potential difference across R1 be V1, across R2 be V2 and across R3 be V3. Loop1 Starting at A, travel clockwise to B. The potential difference across R3 goes from “+” to “-“ so this would be recorded as negative, -V3. Next we go from B to E and travel through R1, where the potential difference goes from “+” to “-“, so we record a negative voltage again, -V1. Finally we go from E to F to A and pass through the battery, where the voltage goes from “-“ to “+” so we record the voltage as positive, E. Kirchoff’s voltage law states that the sum of the voltages around a loop is zero. So: -V3 – V1 + E = 0 Rearrange to get: E = V1 + V3 ENG1021 Electronic Principles 54 Loop 2 Starting form A and going clockwise to B we find the potential across R3 again, which we record as –V3. From B to C to D we find R2 with a potential difference that goes from “+” to “-“, so we record a negative potential difference, -V2. Finally we go from D to E to F to A and pass through the battery, where the voltage goes from “-“ to “+” so we record the voltage as positive, E. Using Kirchoff’s voltage law we get: -V3 – V2 + E = 0 Rearranging we get: E = V2 + V3 Loop 3 Starting from B and going clockwise to C then D we find R2 and record –V2. From D to E to B we find R1, and this time we find the potential difference going from “-“ to “+” so we record a positive potential difference, V1. Using Kirchoff’s voltage law we get: -V2 + V1 = 0 Rearranging gives: V1 = V2 We end up with three equations. This last one should be no surprise as all it is saying is that the voltage across two parallel branches are equal, which we know already. If we were analysing this circuit we would now go on to substitute using Ohm’s law so that, for example, V1 = I1R1 and so on. I’ll come back to this. Before that I just want to see if I can help to explain Kirchoff’s voltage law using an analogy. In the previous example we’ve seen Kirchoff’s voltage law applied, but it doesn’t quite explain why the voltages should sum to zero. I like to think of it using an analogy, where potential difference is equivalent to a change in height. Think of the battery as an elevator, and the resistors as steps. If I redraw the circuit of Figure 2, you should be able to see what I mean. ENG1021 Electronic Principles 55 Figure 3 In Figure 3, since E, F and D are joined together, they must be at the same height (potential). The elevator (battery) takes you up from F to A. This represents the highest point. Alternatively, you can get to A by going up the stairs from E to B and then form B to A, or you could go up the stairs from D to C and then C to B to A. Either way, you end up at the same height. So if you travel around a loop, you climb to a particular height and then have to come back down by the same amount to end up where you started. That’s why the sum total of the height that you’ve travelled in going around a loop is zero. Even though you have to climb up and down to get around the loop, if you end up where you started then you can’t have gained or lost any height. 3 The method of branch currents If we return to the circuit that was used earlier, as seen in Figure 2, then in our analysis of the loops we ended up with 3 equations, repeated here: E = V1 + V3 E = V2 + V3 V1 = V2 We can’t solve these because we don’t know the values of the potential differences. First, we can replace the potential differences by the current times the resistance using Ohm’s law. ENG1021 Electronic Principles 56 E = I1R1 + I3R3 (1) E = I2R2 + I3R3 (2) I1R1 = I2R2 The next stage in the solution is to use Kirchoff’s current law, which tells us that: I3 = I1 + I2 So now we can get rid of I2 in Equation 2 for example by substituting: I2 = I3 – I1 E = (I3 – I1)R2 + I3R3 E = I3R2 – I1R2 + I3R3 = I3(R2 +R3) – I1R2 (3) We now have equation 1 and Equation 3 which both have two unknowns, namely I1 and I3, and so these can be solved. Rather than solve a theoretical circuit, let’s put some numbers in. Let: E = 10 V R1 = 100 Ω R2 = 200 Ω R3 = 300 Ω What are the values of the currents in the circuit? Starting from the beginning again, we have (using E = 10 V): 10 = V1 + V3 10 = V2 + V3 V1 = V2 We can’t solve this because we don’t know the values of the potential differences. Replace the potential differences by the current times the resistance using Ohm’s law. 10 = I1x100 + I3x300 (4) 10 = I2x200 + I3x300 ENG1021 Electronic Principles 57 I1x100 = I2x200 The next stage in the solution is to use Kirchoff’s current law, which tells us that: I3 = I1 + I2 (5) So now we can get rid of I2 in Equation 2 for example by substituting: I2 = I3 – I1 10 = (I3 – I1)x200 + I3x300 10 = I3x200 – I1x200 + I3x300 = I3x(200 +300) – I1x200 10 = I3x500 -I1x200 (6) Multiply Equation 4 by 2: 20 = I1x200 + I3x600 (7) Add Equations 6 and 7 10 + 20 = I3x500 -I1x200 + I1x200 + I3x600 30 = I3x1100 I3 = 30/1100 = 0.028 A = 28 mA Substitute in Equation 6: 10 = 0.028x500 -I1x200 10 = 14 -I1x200 I1x200 = 14 – 10 = 4 I1 = 4/200 = 0.02 A = 20 mA Finally, using Equation 5: 0.028 = 0.02 + I2 I2 = 0.028 – 0.02 = 0.08 A = 8 mA We have analysed the circuit using the method of branch currents. The steps that you need to take are: Step 1 Identify the currents in each branch of the circuit. If in doubt guess the direction; ENG1021 Electronic Principles 58 Step 2 Label each potential difference with its polarity, using the concept that electron current will flow from “-“ to “+”. Step 3 Apply Kirchoff’s voltage law to at least two loops in the circuit, and derive the equation for each loop Step 4 Apply Kirchoff’s current law to a node in the circuit to get the relationship between currents; Step 5 Manipulate the equations so that you end up with two equations with two unknowns; Step 6 Solve the equations. 4 Additional circuits The circuit that we’ve just analysed had one battery in it. I now want to look at a slightly more complex circuit which has an additional battery in one of the other branches. This is shown in Figure 4. Figure 4 Even before I’ve put any values on the components I’ve made a start. I’ve put in three branch currents, I1, I2 and I3. I’ve guessed the direction of the current flow. If they turn out to be wrong, my current values will be negative. I’ve then put “+” and “-“ across all the potential differences in accordance with the rule that electron current flows from “-“ to “+”. Now to analyse the circuit. ENG1021 Electronic Principles 59 For no particular reason I am going to choose Loop 1 as A - B – E – F – A, and Loop 2 as A – B – C – D – E – F – A. Loop 1 Starting from A and going clockwise, the first potential difference is across R2 and I’m going from “+” to “-“ so it’s –V2. next there’s the battery. Again I’m going from “+” to “-“ so the potential difference is negative, and is recorded as –E2. Next is E to F to A, in which I pass through the other battery from “-“ to “+” so the potential difference is recorded as positive, E1, and then through R1 from “+” to “- “ so it’s negative, -V1. Putting all this together I get: -V2 – E2 + E1 – V1 = 0 (8) Loop 2 From A, going clockwise I eventually pass through R3 going from “+” to “-“, so I record –V3. Then it’s back to F and through to A as in Loop 1. So I get: -V3 + E1 – V1 = 0 (9) Finally, using Kirchoff’s current law I get at Node B: I3 + I2 = I1 (10) Using Ohm’s law I can substitute in Equations 8 and 9: -I2R2 – E2 + E1 – I1R1 = 0 (11) -I3R3 + E1 – I1R1 = 0 (12) Then get rid of I3 in Equation 12 using Equation 10: -(I1 – I2)R3 + E1 – I1R1 = 0 -I1R3 + I2R3 + E1 – I1R1 = 0 I2R3 + E1 – I1(R1 + R3) = 0 (13) Let’s put some values in: E1 = 10V, E2 = 5V R1 = 100 Ω, R2 = 200 Ω, R3 = 300 Ω. Re-writing Equations 11: -I2x200 – 5 + 10 – I1x100 = 0 ENG1021 Electronic Principles 60 I1x100 + I2x200 = 5 (14) Re-writing Equation 13: I2x300 + 10 – I1(100 + 300) = 0 -I1x400 + I2x300 = -10 (15) Multiply Equation 14 by 4: I1x400 + I2x800 = 20 (16) Add Equations 15 and 16: I2x1100 = 10 I2 = 10/1100 = 0.009 A = 9 mA Substitute in Equation 13: -I1x400 + 0.009x300 = -10 -I1x400 + 2.7 = -10 -I1x400 = -12.7 I1 = 12.7/400 = 0.03175 A = 31.75 mA Finally, substitute in Equation 10: I3 + 0.009 = 0.03175 I3 = 0.02275 A = 22.75 mA The three branch currents have been successfully calculated. By luck they all turn out positive, which means that I guessed correctly when I chose the direction of the currents. Had I got any of the directions wrong, the current would have turned out with a negative value. Please now attempt the following Problems. Problem 1 ENG1021 Electronic Principles 61 Figure 5 Circuit for Problem 1 In Figure 5, use the method of branch currents to solve for I 1 , I 2 and I 3 . Assume that V 1 = 30 volts and V 2 = 90 volts. Once the currents have been calculated, determine the values of V R1 , V R2 and V R3 . Solution This problem is an example of branch current analysis. Figure 5 shows the currents I 1 , I 2 and I3 marked on it. Polarities are also marked on the resistors. I have used the convention of electron current flow, which flows from the negative terminal of the voltage sources to the positive terminal. Note: Conventional current flow (from positive to negative) could have been used and the end results would be the same, but I would advise you to stick to one convention to avoid confusion. Since we will use Kirchoff’s law to solve the problem, we need to sum the voltages around two loops. There are three possible loops - one which includes both voltage sources and two which include a voltage source and R3. Any two loops will do. Also the direction in which we sum the voltages (that is, go round the loop) is optional. It is vital, however, to make sure that the polarities of the voltages are correct. I think of voltages here as we normally do, that is the voltage is increasing (going from negative to positive) as we go round the loop then it is a positive voltage. If the voltage is decreasing (going from positive to negative) as we go round the loop then it is a negative voltage. So as we do not introduce too many negative voltages (and increase the risk of mathematical error) we will choose loops and directions which allow us to add voltages across the resistors in a positive sense if we can. This will leave the ENG1021 Electronic Principles 62 voltages across the voltage sources as negative, but when if we take them to the other side of the equals sign (transpose them) they will become positive. If we choose to go around the loop which includes V1 and R3 in an anticlockwise direction, starting at the negative end of R3, and using Kirchoff’s voltage law we get: VR3 + VR1 - V1 = 0 Transposing gives: VR3 + VR1 = V1 By Ohm’s law: I3R3 + I1R1 = V1 Substituting the available values gives: I318 + I1120 = 30 (16) Similarly if we choose to go around the loop which includes V2 and R3 in a clockwise direction, starting at the negative end of R3, and using Kirchoff’s voltage law we get: VR3 + VR2 - V2 = 0 Transposing gives VR3 + VR2 = V2 By Ohm’s law I3R3 + I1R2 = V2 Substituting the available values gives: I318 + I2180 = 90 (17) Using Kirchoff’s current law at the junction of R1, R2 and R3 gives (considering currents into the branch point as positive and those leaving the branch point as negative): I3 - I1 - I2 = 0 Rearranging: I3 = I1 + I2 (18) ENG1021 Electronic Principles 63 If we substitute the right hand side of Equation 63 into Equations 16 and 17 we have: (I1 + I2)18 + I1120 = 30 and (I1 + I2)18 + I2180 = 90 Collecting I1 and I2 terms together: 138I1 + 18 I2 = 30 (19) 198I2 + 18 I1 = 90 (20) To remove I2 divide Equation 19 by 18 and Equation 20 by 198 and subtract the results: 1 2 138 18 30 18 18 18 I I + = 1 2 138 30 18 18 I I + = 1 2 198 18 90 198 198 198 I I + = 1 2 18 90 198 198 I I + = Equation 19 - Equation 20 gives: 1 1 2 2 138 18 30 90 18 198 18 198 I I I I − + − = − 1 138 198 18 18 30 198 90 18 198 18 198 18 I × − × × − × = × × 1 30 198 90 18 138 198 18 18 I × − × = × − × 1 4320 27000 I = 1 0.16 I A = To find I2 we substitute I1 into Equation 19 or 20 giving: ENG1021 Electronic Principles 64 138 × 0.16 + 18 I2 = 30 2 30 138 0.16 18 I − × = I2 = 0.44 A I3 = I1 + I2 I3 = 0.16 + 0.44 I3 = 0.6 A We can now determine the voltages across R1, R2 and R3 using Ohm’s law: VR1 = I1R1 VR1 = 0.16 × 120 VR1 = 19.2 V VR2 = I2R2 VR2 = 0.44 × 180 VR2 = 79.2 V VR3 = I3R3 VR3 = 0.6 × 18 VR3 = 10.8 V Problem 2 In Figure 6, use the method of branch currents to solve for I 1 , I 2 , I 3, V R1 , V R2 , and V R3 . ENG1021 Electronic Principles 65 Figure 6 Circuit diagram for Problem 2 Solution The circuit diagram for Problem 2 is shown in Figure 6, in which the branch currents I1, I2 and I3 are shown. As before, I’ve shown electron current which flows out of the negative terminal of the batteries. In the case of I 3 I’ve just had to guess which direction the current is flowing. If I’ve got the direction wrong then when I find the value for I 3 it will be negative. If we choose to go around the loop which includes V1 and R3 in an anticlockwise direction, starting at the negative end of R3, and using Kirchoff’s voltage law we get: VR3 -VR1 + V1 = 0 Transposing gives: VR1 - VR3 = V1 By Ohm’s law: I1R1 - I3R3 = V1 Substituting the available values gives: I115 - I310 = 20 This can be simplified by dividing both sides of the equation by 5: V 2 =40V 10 V 1 =20V 10 15 R 1 R 2 R 3 - + + + + + - - - - I 1 I 3 I 2 ENG1021 Electronic Principles 66 I13 - I32 = 4 (21) Similarly if we choose to go around the loop which includes V2 and R3 in a clockwise direction, starting at the negative end of R3, and using Kirchoff’s voltage law we get: VR3 + VR2 - V2 = 0 Transposing gives VR3 + VR2 = V2 By Ohm’s law I3R3 + I1R2 = V2 Substituting the available values gives: I310 + I210 = 40 This can be simplified by dividing both sides of the equation by 10: I3 + I2 = 4 (22) Using Kirchoff’s current law at the junction of R1, R2 and R3 gives (considering currents into the branch point as positive and those leaving the branch point as negative): I3 + I1 - I2 = 0 Rearranging: I2 = I1 + I3 (23) If we substitute the right hand side of Equation 23 into Equations 22 we have: I3 + (I1 + I3) = 4 Collecting I1 and I3 terms together: I1 + 2I3 = 4 (24) To remove I 3 add Equation 24 to Equation 21: 4I1 = 8 I1 = 2 A To find I3 we substitute I1 into Equation 24 giving: ENG1021 Electronic Principles 67 2 + 2I3 = 4 2I3 = 2 I3 = 1 A Finally, to find I 2 we substitute I 1 and I 3 into Equation 23: I2 = 2 + 1 = 3A We can now determine the voltages across R1, R2 and R3 using Ohm’s law: VR1 = I1R1 VR1 = 2 × 15 = 30 V VR2 = I2R2 VR2 = 3 × 10 = 30 V VR3 = I3R3 VR3 = 1 × 10 = 10 V 5 Further reading For further information about Kirchoff’s laws and the method of branch currents, you may wish to have a look at the on-line Fundamentals of Electrical Engineering and Electronics. The relevant sections are “Divider circuits and Kirchoff’s laws” all parts, and “DC network analysis”, just the first two parts on “What is network analysis?” and “Branch current method”. 6 Where next? You are encouraged to study the Learning Package entitled “Alternating Voltage and Current” next. ENG1021 Electronic Principles 68 ENG1021 Electronic Principles Learning Package 5 Alternating voltage and current ENG1021 Electronic Principles 69 Alternating voltage and current 1 Do you know all this already? If in doubt, please attempt the self assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Introduction So far in the course we have considered direct voltages and currents (DC); circuits in which the current flows in one direction only. In this section we look at alternating currents (AC). The abbreviation “AC” stands for alternating current, but it is often used to describe an alternating voltage, for example, “an ac voltage” (an alternating current voltage-eh?). This is also true of the abbreviation “DC” which stands for direct current. We are usually dealing with voltages and describe, for example, our mains electrical power as 230V ac (230 volts alternating current). Having spent our studies (up to now) differentiating between voltage and current, it is a pity that this confusing and incorrect use of the language is so universally used that we will have to put up with it. So in the future when we see the abbreviation “AC” we should understand that it refers to alternating current and voltages. AC allows us to transfer information, for example, speech, music and television. Generally in electronics we are usually dealing with AC in one form or another. However, Ohm’s law, the power equations and the rules for combining resistors apply just as well to AC, we just have to take into account that the currents and voltages are alternating (or varying). In this section you we learn how to deal with the varying nature of alternating current, in particular you will learn various ways of describing an AC waveform. 3 AC voltages and currents 3.1 AC, the sine wave and amplitude values The term AC refers to an alternating current or voltage. The form that the current and voltage take over time is sinusoidal, as shown in Figure 1. Figure 1 shows the value of sin(x) against x in degrees. You will see that it has a maximum value of 1, a minimum value of -1, and that one whole cycle runs every 360 degrees. ENG1021 Electronic Principles 70 -1.5 -1 -0.5 0 0.5 1 1.5 0 200 400 600 800 degrees Figure 1 Sin(x) versus x in degrees This shape is found often in nature. In electronics it comes about as the result of creating voltage using a generator or dynamo. In these devices a coil is rotated within a magnetic field so that the voltage generated oscillates between a maximum and a minimum value. The maximum value is called the amplitude. The number of times per second that the waveform repeats is called the frequency, f, and is usually measured in hertz (Hz). We generally write the equation for the waveform as: v = VSin(ωt) = VSin(2πft) (1) We use lower case v to indicate the instantaneous value of voltage. The sine function is a function of an angle. To use this function as a waveform over time, the terms ω is used (Greek letter omega) to represent angular velocity, which is measured in radians per second. Then when you multiply the angular velocity by the time, you end up with an angle. Alternatively, instead of the angular velocity you can use the frequency, f, where: ω = 2πf (2) You may recall that 2π in radians is equivalent to 360 degrees. Another value which is used is the period, T, which is the length of time for one cycle to occur. It equals the reciprocal of the frequency, so we have: T = 1/f (3) A typical AC waveform is shown in Figure 2. ENG1021 Electronic Principles 71 -300 -200 -100 0 100 200 300 0 0.02 0.04 0.06 0.08 time/s V / v o l t s Figure 2 AC voltage This voltage has a peak voltage of 250 V and the frequency is 50 Hz. With a frequency of 50 Hz, the period would be 1/50 = 0.02 seconds or 20 ms. A single sinewave has two properties which are amplitude and frequency. If we are comparing two sinewaves, then a third property is used which is called the relative phase or just phase. Figure 3 shows two sinewaves of the same frequency. -300 -200 -100 0 100 200 300 0 0.02 0.04 0.06 0.08 time/s v o l t a g e / V Figure 3: The first waveform that we’ve seen already has the form: v1 = 250sin(100πt) ENG1021 Electronic Principles 72 The second waveform is: v2 = 220sin(100πt + 1) The first waveform has an amplitude of 250 V, and the second an amplitude of 220 V. The relative phase is 1 radian. It’s as if the second wave started before the first wave. 3.2 Power in an AC circuit We saw in an earlier Learning Package that the power in a DC circuit is equal to the voltage times the current. It is very similar in AC circuits except that the power is a bit less than the voltage times the current. To explain I need to introduce some terms. We have already seen that a sinewave has an amplitude which corresponds to its peak value. Another value which might be quoted is the peak-to-peak value. This is the size of the gap between the maximum and minimum values. Since the waveform is symmetrical, the minimum value is minus the amplitude. So if a waveform has an amplitude of A, the peak-to peak value is 2A. In the waveform that we used earlier the peak value or amplitude was 250 V. The peak-to-peak value is therefore 500 V. Now we said that when current flows it produces heat, which is measured as the power. An AC circuit also produces heat, since the electrons are flowing in the circuit, even if they are only oscillating back and forth inside the conductor. We use the term RMS, which stands for root-mean-squared, in AC circuits to give an equivalence to the power in a DC circuit. If a battery has a voltage of 10 V and delivers a current of 2 amps, the power would be 10 x 2 = 20 watts. An equivalent AC circuit could have a generator that produced 10 V rms, and delivers 2 amps rms, and the result would also be a power of 20 watts. The rms value is found in a sinewave by squaring the sinewave, finding the average value, and then taking the square root of the average. The relationship between the rms voltage and the amplitude, V, in a sinewave is: 0.707 2 rms V V V = = (4) Similarly, for current: 0.707 2 rms I I I = = (5) ENG1021 Electronic Principles 73 So, the power in an AC circuit would be: 0.5 2 2 2 rms rms V I VI Power V I VI = = × = = (6) In other words, the power is half the value of the amplitude of the voltage times the amplitude of the current. The value 0.5 that multiplies VI is called the “power factor”, and varies for different waveforms. A similar but less useful value is the average value. Now a sinewave has an average of zero, since it is symmetrical about the x-axis. However, we can find the average of half a cycle. For sinewaves this turns out to be 0.637V. The ratio of the rms value to the average value is called the “form factor” and again is different for different waveform. For a sinewave the form factor is: form factor = rms value/average value = (V × 0.707)/(V × 0.637) = 0.707/0.637 = 1.1107 (7) Please now attempt the following Problems. Problem 1 Give the angle in degrees and radians for each of the following: one cycle, one half cycle, one quarter cycle, three quarter cycles. Solution This question exercises your understanding of a waveform’s phase in both degrees and radians. The phase angles calculated in this exercise are so common that we should eventually be able to quote them (in degrees and radians) without thinking. One cycle One whole cycle of ac waveform is equivalent to the waveform produced by a single loop ac generator travelling through one complete circle. (It is assumed that the rate of rotation is constant.) One whole cycle in degrees is therefore equal to the number of degrees in a circle or 360 degrees. There are 2π radians in a circle and so one cycle of waveform is also equivalent to 2π radians. Radians are used in ac calculations so it is worth becoming comfortable with using them. Half a cycle The angle which represents one half cycle is simply 360 divided by 2 which equals 180 (in degrees) or 2π divided by 2 which equals π in radians. ENG1021 Electronic Principles 74 One quarter and three quarters of a cycle Similarly we obtain one quarter of a cycle to be 360/4 = 90 degrees or 2π/4 radians and three quarters of a cycle to be 3 x 360/4 = 270 degrees or 3 x 2π/4 = 1.5π radians. Beware it is an easy mistake to think of π/2 as half a cycle when in fact it is a quarter of a cycle. This is also true of other fractions of π. Problem 2 A 5 Ω resistor in a circuit connected to the ac power line has an rms current of 1.17 Amps. Calculate the power dissipated in the resistor. Solution This is a simple Ohm’s law/Power equation problem applied to ac circuit. The current I is measured in amperes rms. Since we are using rms values we can apply Ohm’s law and the power equation as we did for dc circuits. It is important to realise that we cannot use Ohm’s law directly if our current and voltage values are not rms values. We can either use the equation. P = I 2 /R (8) or calculate V using Ohm’s relationship and find the power using: P = I × V (9) I will use the second method. The voltage (rms value) is: Vrms = I R = 1.17 × 5 = 5.85 V rms Now we use the power equation to obtain the power P = I × V = 1.17 × 5.85 = 6.845 watts We could not have used the power equation if either the current or voltage (or both) were not rms values. Problem 3 Convert to RMS voltage the following peak-to-peak values of sine-wave ac signal voltage: (a) 462.5 µV; (b) 9.84 mV; (c) 35.19 mV. Solution We are asked to convert peak to peak values to rms. Note that these values are peak to peak, not just peak. The rms value of a sine-wave is 0.707 or 2 2 multiplied by the peak value. The peak value is half of the peak-to-peak value. ENG1021 Electronic Principles 75 Problem 3a Our first value to convert is 462.5 µV p - p. This is equal to: 462.5/2 = 231.25 × 10 -6 V peak To find the rms value we need to multiply the peak value by 2 2 . Vrms = 231.25 × 10 -6 × 2 2 = 163.5 × 10 -6 = 163 µV rms Problems 3b, 3c These problems are solved in the same way. The value 9.84 mV p - p is equal to: 9.84/2 × 10 -3 = 4.92 × 10 -3 = 4.92 mV peak To find the rms value we need to multiply the peak value by 2 2 . Vrms = 4.92 × 10 -3 × 2 2 = 3.97 × 10 -3 = 3.97 mV rms And finally, the value 35.19 mV p - p is equal to: 35.19/2 × 10 -3 = 17.595 × 10 -3 = 17.595 mV peak To find the rms value we need to multiply the peak value by 2 2 . Vrms = 17.595 × 10 -3 × 2 2 = 12.44 × 10 -3 = 12.44 mV rms The important point to note here is while is acceptable to divide any vertically symmetrical waveform by two to convert the peak to peak value to the peak value, the conversion factor for converting from peak to rms (or average) is not fixed and is specific to the waveform. Generally, the broader the waveform the larger the conversion factor will be. Problem 4 A sinewave ac voltage has an rms value of 19.2 V. (a) Find the peak value. (b) What is the instantaneous value at 50 o of the cycle? ENG1021 Electronic Principles 76 Solution This problem involves calculating the instantaneous value of a sinusoidal voltage given the phase, but first we must find the peak value. To find the peak value we must multiply the rms value by 2 or 1.414. The peak value of the waveform: Vpeak = 19.2 × 2 = 27.15 V peak The peak value, by definition is the largest value that the sine wave reaches in it cycle. It reaches its positive peak at 90 degrees (when the sine function equals 1) and the negative peak at 270 degrees (when the sine function equals -1). At 50 degrees it must be less than the peak value. We need to look up what is the value of the sine function when the phase angle is 50 degrees. Using my calculator it is 0.766. The instantaneous voltage is the peak voltage multiplied by the sine function at that instant. Vinst = 27.15 × 0.766 = 20.80 V Note that instantaneous voltages are not normally useful quantities. They tell us nothing about the ac waveform unless we are given the phase angle also. Problem 5 A sine wave of voltage has an average value of 38.22 V. Calculate the waveforms: (a) rms value; (b) peak value; (c) peak to peak value. Solution In the last problem of this section we will have to deal with average voltages. It is easier to measure an average voltage than an rms voltage, although some digital voltmeters have special circuitry to do it. Voltmeters commonly measure average voltages, but their scale reads the rms equivalent assuming that the waveform is sinusoidal. To do this the average values are multiplied by the “form factor” for the scale values. The form factor for a sinusoidal waveform form factor = rms value/average value = Vpeak × 0.707/Vpeak × 0.637 = 0.707/0.637 = 1.1107 Different shapes of waveforms will have different form factors. In the case of measuring a non-sinusoidal waveform with such an instrument, the reading is meaningless (because it does not even tell us the average value). In this problem we are initially going to convert from average to rms given that the waveform is sinusoidal. (If we did not know the waveform’s shape we could not solve this problem.) Problem 5a ENG1021 Electronic Principles 77 The easiest way to convert average values to sinusoidal values is to multiply by the form factor for the ac waveform, but we will do it the hard way first and then show that the form factor method works. We will firstly convert the average value to peak and then convert the peak value to rms. We convert average values to peak values by dividing by 0.637 or 2/π. Vpeak = Vave/0.637 = 38.22/0.637 = 60.04 V peak Now we will convert the peak value to rms Vrms = Vpeak × 2 2 = 60.04 × 2 2 = 42.45 V rms Now the form factor method. Vrms = Vaverage × 1.1107 = 38.22 × 1.1107 = 42.45 V rms Problem 5b The peak value has already been found in Part (a). Vpeak = 60.04 V peak Problem 5c The peak-to-peak value is twice the peak value. Vp-p = 2 × Vpeak = 2 × 60.04 = 120.08 V peak - to - peak Problem 6 The 60 Hz power line voltage of 120 V is applied across a resistance of 20 Ω. (a) How much is the rms current in the circuit? (b) What is the frequency of the current? (c) What is the phase angle between the current and the voltage? (d) How much dc applied voltage would be necessary for the same heating effect in the resistance? Solution This problem serves as revision from the last section with a reminder of the meaning of rms values and as a gentle introduction to frequency and phase. Problem 6a The power line voltage is 120 V and since we are not told otherwise we should assume that this is an rms value. The rms current is given by Ohm’s relationship ENG1021 Electronic Principles 78 I = V/R = 120/20 = 60 A rms Problem 6b The alternating current will vary in sympathy with the alternating voltage according to Ohm’s relationship. The current will vary at the same frequency as the power-line voltage which is stated as being 60 Hz Problem 6c As the voltage alternates from zero up through its positive peak, before completing the rest of the cycle, the current starts from zero at exactly the same time and reaches its positive and negative peaks at the same time as the voltage. The current is then “in phase” with the voltage and the phase angle is zero. Problem 6d This part of the question tests if you understand the meaning of rms. If you remember the rms value is defined as that value of an ac waveform (not necessarily sinusoidal) that will produce the same heating effect in a resistor as the dc value. The dc voltage that would produce that same heating effect in a resistance is therefore the same as the rms voltage, that is 120 volts. Problem 7 What is the frequency for the following ac variations? (a) 10 cycles in 1 s; (b) 1 cycle in 1/10 second; (c) 50 cycles in 1 s; (d) 50 cycles in ½ s; (e) 50 cycles in 5 s. Solution Here we will find the frequency of an ac waveform by considering how many cycles the waveform goes through in a given time and then calculate how many cycles would have occurred in one second. This will give us our answer in cycles per second (cps) or Hertz (Hz). The units cps have been replaced by the Hertz as an international standard, but the term cycles per second or cps it still used, probably because it is more descriptive. Problem 7a The first example is easy. We have 10 cycles occurring in one second. That is, the frequency: f = 10 cycles per second or 10Hz ENG1021 Electronic Principles 79 Problem 7b Here we have 1 cycle every 1/10th of a second. We ask ourselves, “How many cycles in one second”. I hope it is clear that 10 cycles will occur in one second. This gives us the answer for the frequency as: f = 10 Hz Alternatively, since we have one cycle occurring in 1/10 th second, we can say that the waveform has a period T of 1/10 th second and we can take the reciprocal of T to find the frequency. f = 1/T = 1/(1/10) = 10 Hz Problem 7c This part is similar to Part (a). We have 50 cycles occurring in one second, so the frequency is: f = 50Hz Problem 7d We have 50 cycles occurring in ½ second, so 100 cycles would occur in one second, giving f = 100 Hz Alternatively, if there are 50 cycles in ½ second, each cycle has a period of ½ second divided by 50, which is: T = (1/2)/50 = 1/100 The frequency is: f = 1/T = 1/(1/100) = 100 Hz Problem 7e If we have 50 cycles in 5 seconds then in one second 10 cycles will occur. The frequency is: f = 10Hz You should also try to calculate the period of the waveform and take its reciprocal to arrive at the frequency. ENG1021 Electronic Principles 80 Problem 8 Calculate the time delay for a phase angle of 45 o at the frequency of (a) 500 Hz; (b) 2 MHz. Solution A 45 degree phase angle will mean that one waveform starts 1/8 th of a cycle (45/360=1/8) after the other. The time delay will therefore be 1/8th of the waveforms period later. Therefore in the following exercises, we must first calculate the period and then divide the period by 8. Problem 8a The period of a waveform is the reciprocal of its frequency. In this case the frequency is 500 Hz, the period is: T = 1/f = 1/500 = 0.002 s The time delay for 45 degree phase angle td = 1/8 × 0.002 = 0.00025 = 0.25 ms Problem 8b As in Part (b) we find the period and divide by 8. The period is: T = 1/f = 1/(2 × 10 6 )= 0.5 × 10 -6 s The time delay for 45 degree phase angle td = 1/8 × 0.5 × 10 -6 = 0.0625 × 10 -6 = 62.5 ns Problem 9 Calculate the period T of a radio wave whose wavelength λ is 2 m. Solution So far our problems have dealt with phase, frequency and period. We need to exercise our understanding of the wavelength of a waveform. The wavelength of an ac waveform is related to the frequency (and period) through the velocity of the wave. In other words, if we know the wavelength and the velocity we can calculate the frequency (and/or period) and vice versa. Different waves have different velocities, but we are told that this is a radio wave and so it travels at the velocity of light. I prefer to work in SI units, using metres rather than centimetres, so the speed of light is 3 × 10 8 metres/second. In this problem we are given the wavelength and asked to calculate the period. The wavelength is: ENG1021 Electronic Principles 81 λ = velocity/frequency (10) We are going to find the frequency first and then find the period by taking the reciprocal. Rearranging Equation 10 for the frequency f = velocity/wavelength = (3 × 10 8 )/2 = 1.5 × 10 8 Hz = 150 MHz This frequency has a period: T = 1/f = 1/(1.5 × 10 8 ) = 6.667 ns 3.4 Electromagnetic waves Another type of wave that you may encounter in electronics is an electromagnetic wave. Examples are radio waves, microwaves, visible light, infra-red and gamma radiation. These waves, rather than just being oscillations in a conductor, travel through space, even through a vacuum. Since they are travelling they must have a velocity, and this turns out to be a constant generally known as the speed of light. These waves also have a frequency, and a wavelength. The relationship between the frequency, velocity and wavelength is: v = fλ (11) where v is the speed of light, f is the frequency and λ (Greek letter lambda) is the wavelength. For example, a radio wave could have a frequency of 3 MHz. If the speed of light is approximately 300,000,000 m/s, what is the wavelength of the wave? λ = v/f = 300,000,000/3,000,000 = 100 m 3.4 Non-sinusoidal waveforms Very briefly, sometimes in electronics waveforms other than sinewaves are encountered. Probably the most common is the square wave which is found in digital electronics. There is a whole wealth of theory about waveforms other than sinewaves, usually coming under the heading of Fourier transforms. In essence, what Fourier showed was that any repetitive waveform could be constructed from sinewaves with varying amplitudes and frequencies which are multiples of the fundamental frequency. In other words, a square which repeats every 50 ms has a fundamental frequency of 1/50 ms = 20 Hz. We can create a square wave by adding up sine waves with a frequency of 20 Hz, 40 Hz, 60 Hz, 80 Hz and so on. These secondary waveforms with frequencies that are multiples of the fundamental frequency are called harmonics. In a square wave, the values switch from the maximum for half a period to the minimum for the other half of the period. If the amplitude is V, then the peak value is V, the peak-to-peak value is 2V, the average value is V/2, and the rms ENG1021 Electronic Principles 82 value is V/2. Therefore, the power in a square wave is VI/4 giving the power factor of 0.25, and the form factor is rms/average = 1 since the average and the rms value are both the same. 3.5 Harmonics Harmonics are at frequencies which are multiples of the fundamental frequency. In music the term “octave” is used to mean the same note only an octave higher or lower. In physical terms an octave is a doubling of a frequency. So if a note has a frequency of 256 Hz, say, then an octave higher would be 512 Hz. Two octaves higher would be 1024 Hz and so on. The alternative to octaves are “decades”. If a frequency is increased by a decade then the new higher frequency is 10 times the lower frequency. If note of frequency 256 Hz is increased by a decade, its new frequency is 2560 Hz. Now attempt the following Problems. They are problems that reinforce your understanding of harmonics and multiples of frequencies. Problem 10 List the first four harmonics of 7.5 MHz. Also, identify each harmonic as being either an even or odd multiple of the fundamental frequency. Solution The harmonics are integer (whole number) multiples of the fundamental frequency. In this case the fundamental frequency is 7.5 MHz. The first harmonic is “one” times the fundamental frequency, that is, the first harmonic is the fundamental which has a frequency of 7.5 KHz. It is odd, because we have multiplied the fundamental by “one” and one is an odd number. In general, we don’t usually talk about the first harmonic, because it can be confusing. (It becomes confusing because the first “interesting” harmonic, interesting in the sense that it is in anyway different to the fundamental, is the second harmonic) The second harmonic is “two” times the fundamental frequency, that is 7.5 × 2 = 15 MHz. Two is an even number and so this is an even harmonic. The third harmonic is “three” times the fundamental frequency or 22.5 MHz. Three is an odd number and so this harmonic is odd. The fourth harmonic is “four” times the fundamental frequency or 30 MHz. Four is an even number and so this harmonic is even. Problem 11 ENG1021 Electronic Principles 83 List the frequency three decades above 100 Hz. Solution Decades are multiples of ten. Every time we multiply a frequency by ten we have increased the frequency by one decade. Every time we divide a frequency by ten, we have decreased the frequency by one decade. In this problem the frequency of concern is 100Hz and we are asked to find the frequency three decades above it. The first decade will be 100 × 10 = 1000 Hz or 1 kHz. To find the third decade above 100Hz, we multiply our frequency of concern by 10, three times, or by 1000. So three decades above 100 Hz is: 100 × 10 × 10 × 10 = 100000 = 100 kHz. Another way of doing this is to multiply our frequency by 10 3 . 100 × 10 3 = 100 kHz. In general, multiplying by the “nth” power of ten is equivalent to raising the frequency by n decades, and dividing by the “nth” power of ten is equivalent to decreasing the frequency by n decades. Problem 12 Raising the frequency of 400 Hz by two octaves corresponds to what frequency? Solution Octaves are multiples of two. Every time we multiply a frequency by two we have increased the frequency by one octave. Every time we divide a frequency by two, we have decreased the frequency by one octave. In this problem we are asked to find the frequency two octaves above 40 kHz. We can do this by multiplying 40 kHz by two, twice, this gives 40000 × 2 × 2 = 160 kHz. Alternatively, we can multiply by 2 2 . So, 40000 × 2 2 = 160 KHz. In general, multiplying by the nth power of 2 is equivalent to raising the frequency by n octaves, and dividing by the “nth” power of two is equivalent to decreasing the frequency by n decades. Problem 13 What is the frequency three octaves below 40 kHz? Solution ENG1021 Electronic Principles 84 Here we are asked for 3 octaves below 40 kHz. We can do this by dividing by two, three times. ((40000/2)/2)/2 = (20000/2)/2 = 10000/2 = 5000 or 5 kHz Or by dividing by 2 3 : 40000/2 3 = 40 000/8 = 5000 Hz or 5kHz Note that dividing by 2 3 is equivalent to multiplying by 2 -3 = 1/8, giving us another method of solution. 4 Further reading If you want to read more about the topics discussed in this Learning Package you could have a look at the appropriate sections of the on-line book “Fundamentals of Electrical Engineering and Electronics”. The relevant sections are “Basic AC theory”, all parts except perhaps “principles of radio” which isn’t particularly relevant. 6 Where next? The next suggested Learning Package in entitled “Electromagnetism”. ENG1021 Electronic Principles 85 ENG1021 Electronic Principles Learning Package 6 Electromagnetism ENG1021 Electronic Principles 86 Electromagnetism 1 Do you know all this already? If in doubt, please attempt the self assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Electromagnetism The relationship between magnetism and electricity is very important. Electric motors, generators, computer hard disks, tape recorders, microphones and loudspeakers are a small selection of devices that depend on the electromagnetic relationship. In this section we will explore this relationship. We will start with magnetism and then introduce laws which connect electricity and magnetism. You will learn that any wire carrying a current will generate a “magnetic field” and that a loop of wire in a changing magnetic field will have a voltage induced in it. 2.1 Magnets What is magnetism? I am sure that you are familiar with bar magnets. These are usually pieces of iron which have a north and a south pole, as shown in Figure 1. Figure 1 Bar magnet If another magnet is brought close to the first one, then you find that north poles are attracted to south poles. Alternatively, similar poles, such as two north poles, repel each other. Magnets can therefore produce a force. In Figure 1 I have drawn in the “lines of force”. These lines run from the north poles to the south ENG1021 Electronic Principles 87 pole outside of the magnet, and from south to north inside the magnet. If another magnet is brought near the first, a force will act on in the direction shown by the lines of force. In a real system, the lines of force can be found by either putting a sheet of paper over a bar magnet and sprinkling iron filings on the paper. If the paper is tapped gently a number of times, the iron filings will each become small bar magnets, and align themselves along the lines of force. Alternatively, using a small compass, the lines of force can be traced on the paper. Position the compass anywhere on the paper and draw a small arrow in the in the direction the needle of the compass is pointing. By repeating this over the paper you will get a set of small arrows pointing in different directions. If these are then joined up by extending the line of the arrows, you will find you trace out the lines of force. The area around a magnet is therefore able to exert a force. We therefore call this a magnetic field (think of a force field in science fiction). The strength of the magnetic force increases with the number of lines. In Figure 1 there were a total of 6 lines of force, so this magnetic field is said to have a magnetic flux equal to 6 maxwells. Magnetic flux is defined as the total number of magnetic lines, and is given the Greek letter phi, φ, and is measured in maxwells, which is abbreviated to Mx. So in Figure 1: φ = 6 Mx The magnetic field can be thought of as imaginary lines of force. It is important to realise that these lines are imaginary and that a magnetic field exists at all points around the magnet (not just on the lines). In other words, when I drew Figure 1 I was drawing a representation of a bar magnet which has a flux of 6 Mx, so I drew 6 lines. These lines are imaginary, but represent the magnetic field and its strength. It is something of a coincidence, that if you trace the lines of force using either the iron filings method or the compass method that you get something that looks the same. However, the difference is that the number of lines you would get when tracing them would not correspond to the flux, but would still have some of the same properties such as direction. The units of maxwells have now been superseded by the unit called the weber, which is abbreviated to Wb. One weber is equivalent to 100,000,000 lines of force. Since this is quite a large unit, we often use the micro weber (µWb) which is equivalent to 100 Mx or 100 lines of force. The total number of lines doesn’t really convey the idea that the magnetic force varies according to the position relative to the magnet. Magnetic flux density goes some way to address this. Magnetic flux density is the magnetic flux per unit area or often described as the number of lines of force passing perpendicularly through a unit area. In Figure 1 the lines are closer at the poles and get more diffuse as you move further away from the magnet. Thus the flux ENG1021 Electronic Principles 88 density would be highest at the poles and get smaller as you move away from the poles. The magnetic flux density is given the symbol, B, and is measured in Gauss, abbreviated to G, if you are using the units of maxwells. That is: B = φ /A = maxwells/cm 2 = gauss (1) Alternatively, B is measured in tesla, abbreviated to T, and in this case: B = φ /A = weber/m 2 = tesla This has hopefully explained what a magnetic field is and introduced the terms flux and flux density. Flux is given the symbol φ and is the total number of the imaginary lines of the magnetic field, while flux density is given the symbol B and is the flux (or number of imaginary lines) in a given area. We will now exercise our understanding of the new terms. Please attempt Problems 1, 2, 3. Problem 1 (Magnetism) A magnet produces 5000 field lines. Find φ in maxwells and webers. Solution We are told that the magnet produces 5000 field lines. This is a measurement of the magnetic flux φ. One maxwell is one field line, therefore 5000 field lines is 5000 maxwells. There are 10 8 maxwells in a weber, therefore this same flux measured in webers is: 5000/10 8 webers or 10 -5 webers. To reinforce the idea that the magnetic field is continuous and that field lines are imaginary consider the nanoweber, which is 10 -9 webers. This is a practical unit for the measurement for weak magnetic fields. If we consider this unit in maxwells it is equal to 10 -9 /10 -8 = 10 -1 maxwell This unit then measures a tenth of a maxwell (or a tenth of a field line). If magnetic lines of force were real then we could not have a tenth of one and the smallest magnetic field that could exist would be one line or 1 maxwell. Problem 2 If the area of the pole in Problem 1 is 5 cm 2 , calculate B in gauss units. Solution ENG1021 Electronic Principles 89 This problem introduces the area of the pole of the magnet and we are asked to calculate B (the flux density) in gauss units. Flux density is measured in flux per unit area. The gauss unit measures how many maxwells pass through one square centimetre. We are told that the total number of lines is 5000 and this flux flows out of a 5cm 2 area. The number lines flowing through one centimetre is therefore the flux density: B = φ/Area = 5000/5 = 1000 gauss Problem 3 Calculate B in tesla units for a 200 µWb flux through an area of 5 × 10 -4 m 2 . Solution This problem is similar to the previous problem except that we are now working with tesla units. Teslas are a measure of the number of webers (10 8 maxwells) flowing through a one square metre area. We are told that 200 µWb flow through 5 × 10 -4 m 2 . Given this information, we have to calculate how many webers will flow through a one square metre area. The number lines flowing through one square metre is therefore the flux density in teslas B = φ /Area = (200 × 10 -6 )/(5 × 10 -4 ) = 200/5 × 10 -2 = 0.4 tesla 2.2 Electromagnets The type of magnet that we’ve looked at so far is a permanent magnet, usually made of iron. An alternative is an electromagnet. When electricity flows through a conductor it produces a magnetic field, as shown in Figure 2. Figure 2 In Figure 2, current refers to electron current. The current produces a magnetic field which radiates from the wire. In Figures 2b and 2c the wire is in the centre of the circle and the magnetic line of force is the outer circle. The direction of the current in the wire is supposedly represented by a dart. If the wire flows into the ENG1021 Electronic Principles 90 page then you would see the flights at the back of the dart, whereas if the current is coming out of the page you would see the point of the dart. The direction of the magnetic field is described by the left-hand grip rule. If you coiled your fingers of your left hand up, leaving the thumb sticking out, rather like the hitch-hiking gesture, then you thumb represents the direction of the current, and your fingers represent the direction of the magnetic field. Note: if, like me, you studied electronics some time ago, and are used to using conventional current, which flows from the positive terminal to the negative, as opposed to electron current as we’ve used in this module, then you would know this rule as the right-hand grip rule. If we coil a wire up, as in Figure 3, the lines of force come together and the result is just like a bar magnet, with a north pole and a south pole. This magnet exists while current if flowing, but stops being a magnet when the current is switched off. Figure 3 Electromagnet A coil, like that in Figure 3 is called a solenoid. It usually contains a core, with the wire coiled around the core. Typically that core would be made of soft iron which is easily magnetised and demagnetised. Within electrical circuits you may find an electromagnet like this one in electric bells, and the big electromagnets used in scrap yards. One device, known as a relay, uses an electromagnet to open or close a circuit. 2.3 Magnetic units We’ve just seen that we can create a magnetic field using electricity, particularly using coils of wire, but I haven’t said anything about the strength of the magnetic field produced by such a device. What do you think would happen to the strength of the magnetic field if: ENG1021 Electronic Principles 91 a) you increase the current; b) increase the number of turns in the coil? It turns out that both of these would increase the magnetic field strength or flux. The value of the current and the number of turns is so important in electromagnetism that it’s own term, namely the magnetomotive force or mmf. The mmf is equal to the current, I, times the number of turns, N. Mmf = Ix N (2) Strictly speaking the mmf is measured in ampere-turns, but since “turns” is a unitless number, the units should be amperes. This would get confusing as you wouldn’t know if someone is referring to the current or the mmf, so we will use the unit A.t for ampere-turns. Now if you imagine a coil with a fixed number of turns and a constant current flowing through it, mmf would be constant. This mmf creates a magnetic field. If you stretch the coil along its length, the number of coils is still the same and the current is still the same, so the mmf is still the same, but the strength of the magnetic field is weakened. To reflect this phenomenon, the field intensity H, is defined as the mmf divided by the length of the coil. H = mmf/L (3) The units for H are ampere-turns per unit length, A.t/m. Finally, the current in the coils produces a magnetic field which has an intensity H. This then creates a magnetic flux with a flux density, B. The relationship between H and B depends on the material that the core of the electromagnet is made from and is called the permeability and given the Greek character mu, µ. B = µ × H (4) The permeability of air is 4π × 10 -7 or 1.26 x 10 -6 T/A.t/m and is given the special symbol µ o . Most material are then quoted as having a relative permeability, µr, which is its permeability relative to air. To find the absolute permeability you have to multiply the relative permeability by µ o . There is one other term which is often used in electromagnetism, which is reluctance. In comparison with electric circuits, the magnetic flux corresponds to current. The flux φ is produced by ampere-turns I × N of magnetomotive force. Therefore the mmf corresponds to voltage. Opposition to the production of flux in a material is called the reluctance, comparable with resistance. The symbol for reluctance is ℜ. Reluctance is inversely proportional to permeability. Iron has a high permeability and low reluctance. Air or a vacuum has low permeability and high reluctance. ENG1021 Electronic Principles 92 The three factors – flux, ampere-turns or mmf, and reluctance – are related as follows: φ = mmf/ℜ (5) which is known as Ohm’s law for magnetic circuits. Please attempt Problem 4. Problem 4 (Magnetic units) A battery is connected across a coil of 100 turns and a resistance of 20 Ω, with an iron core 0.2 m long. (a) How much battery voltage is needed for 200 A.t? (b) Calculate H in the iron core in Ampere-turns per meter. (c) Calculate B in teslas in the iron core if its µr is 300. (d) Calculate φ in webers at each pole with an area of 8 × 10 -4 m 2 . (e) How much is the reluctance ℜ of the iron core, in ampere-turns per weber? Solution This problem is very useful, because it mixes concepts you have learnt about electrical voltage and current and resistance, with magnetic potential, magnetic field intensity, magnetic flux, magnetic flux density, permeability and reluctance. Problem 4a The first part of the problem requires us to calculate the battery voltage having been given the mmf of the coil as 200 ampere turns. We are given the resistance of the coil as 20 Ω which is the total resistance of the electrical circuit. Before we can calculate the battery voltage we need to calculate the electrical current I. Since: mmf = NI where N is the number of turns on the solenoid and is given as 100, I = mmf/N = 200/100 = 2 A We can now calculate the battery voltage using Ohm’s law. The battery voltage is: V = IR = 2 × 20 = 40 volts Problem 4b We now have to calculate the magnetic field intensity H. The magnetic field intensity H is dependant on the mmf and the length of the coil L, which we are given as 200 ampere turns and 0.2 metres respectively. ENG1021 Electronic Principles 93 H = mmf/L = 200/0.2 = 1000 A.t/m Problem 4c We are now required to calculate B, the magnetic flux density. You may have noticed that the answer is required in teslas which are SI units and that all other units (such as length) are also SI units. When we use SI units in the equation: B = µ × H and we have a figure for the relative permeability µr we must introduce the permeability for air (or a vacuum) µ0 = 4π × 10 -7 to find the absolute permeability µ. That is: µ = µr × µ0 We are given the relative permeability µr of the iron core as 300 therefore the absolute permeability is: µ = µr × µ0 = 300 × 4 π × 10 -7 = 3.77 × 10 -4 T/A.t/m Having found the absolute permeability, the flux density is found using Equation 13 above: B = µ × H = 3.77 × 10 -4 × 1000 = 0.377 tesla Problem 4d The total flux at the pole ends is given by: φ = B × A The area A is 8 × 10 -4 , therefore the flux is: φ = 0.377 × 8 × 10 -4 = 3.02 × 10 -4 webers Problem 4e Here we have to find the reluctance ℜ of the iron core. The reluctance ℜ of a magnetic circuit is analogous to the resistance in an electrical circuit. In an electrical circuit more current flows, for a given emf (voltage), as the resistance is decreased (Ohm’s law). In the magnetic circuit there is more flux, for a given mmf, as the reluctance is decreased. The equation for φ in terms of reluctance and mmf shows this. φ = mmf/ℜ Rearranging and substituting our values for φ and mmf we arrive at a value for the reluctance: ENG1021 Electronic Principles 94 ℜ = mmf/φ = 200/(3.02 × 10 -4 ) = 66 × 10 4 A.t/W 2.4 Electro-Magnetic Induction You have seen that current flowing in a conductor creates a magnetic field. Also, that a forces act on magnets when they close top each other. So, we can conclude that if a conductor is placed within a magnetic field and a current is passed through it, then a force is applied to the conductor. This is basically how motors work. Figure 4 illustrates this principle. Figure 4 shows a conductor which has been placed within a magnetic field. The magnetic field is produced by the horseshoe magnet, and the direction of the magnetic field is from the North pole to the South pole. The conductor is connected to a battery, which produces a current flowing from the negative terminal to the positive terminal. The result would be that the conductor moves upwards. Figure 4 Motor effect The direction that the conductor moves is described by Lenz’s law. It is often memorised using Fleming’s motor rule, or Fleming’s right hand rule. If you arrange your left hand such that your thumb is sticking up, and your index finger is pointing straight ahead, then if you move your middle finger so that it’s pointing ENG1021 Electronic Principles 95 to the side, you have it. The thumb, index finger and middle finger should all be pointing in different directions. The thumb represents the motion, the index of first finger represents the field, and the middle or second finger represents current. Thumb motion First finger field Second finger current In a similar way, if you move a conductor in a magnetic field then a current is created. This is illustrated in Figure 5. In this case you can memorise the directions using Fleming’s dynamo or left hand rule. The fingers of the left hand are arranged as before, and have the same meaning. Therefore, moving a conductor in a magnetic field generates electricity. This is the basis of a dynamo or a generator. ENG1021 Electronic Principles 96 Figure 5 Dynamo effect In the dynamo effect, I said that a current was produced. In order for a current to flow there must be a potential difference. Therefore, the dynamo or generator could be said to be generating a voltage. Figure 6 Figure 6 shows a solenoid again, where a coil is wrapped around a core. If that core is a magnet, then either moving the core or moving the coil would generate a voltage. The value of this voltage depends on three factors: • The magnetic flux – a stronger the magnetic field would produce a higher voltage; • The rate at which the core (or the coil) is moving – the faster it moves the more voltage is generated; • The number of coils or turns – the more turns the higher the voltage. These are summarised in Faraday’s law, which is: v ind = N dφ/dt (21) Where N is the number of turns, φ is the magnetic flux, and dφ/dt means “the rate of change” of flux. This is interpreted as dφ is a small change in flux, and dt is a small change in time, t. For example, if a coil has 100 turns, and the flux changes from 2 Wb to 5 Wb in 2 seconds, the induced voltage would be: v ind = N dφ/dt = 100 x (5 – 2)/2 = 150 volts Similarly, if the flux changed from 5 Wb back to 2 Wb, the induced voltage would be: v ind = N dφ/dt = 100 x (2 – 5)/2 = -150 volts ENG1021 Electronic Principles 97 You now know the fundamental relationships between electricity and magnetism. That is, you will know how to produce magnetism from electricity and how to produce electricity from magnetism. The principles behind motors and generators have been described. That is, a current carrying conductor in a magnetic field is subject to a force, which will cause it to move (the motor effect) and a conductor moving such that it cuts across a magnetic field will have a voltage generated across it (the generator effect). Two more important laws have been introduced (Faraday’s law and Lenz’s law). Now attempt Problem 5. Problem 5 (Electro-Magnetic Induction) A magnetic flux of 800 Mx cuts across a coil of 1000 turns in 1 µs. How much is the voltage induced in the coil? (1 Mx = 10 -8 Wb) Solution This problem involves the use of Faraday’s law. If you are not familiar with mathematical differentiation, the dφ/dt expression may appear a little daunting. It is described as the “rate of change”, where the “d” in “dφ” means a small change in φ. When you see this notation, don’t be tempted to cancel out the “d” on the top and bottom of the equation. Just as φ stands for flux and t for time, so dφ stands for a small change in flux and dt stands for a small change in time The expression dφ/dt is simply the gradient or slope of a graph at any point. What is wrong with using the term “gradient” then? We could, if the graph was a straight line, but often the slope of a graph varies along its length so that it is not straight but curved. You can imagine a real hill on a road; no matter how steep it is, it will not be the same steepness at the bottom of the hill or at the top of the hill as it is in the middle. A gradient sign near to the road may indicate the gradient, but does not say at which point this applies. The dφ/dt allows us to see the gradient at any point, not just an average gradient. This will become more important later when we study AC. and transformers. In these problems the graph will always be a straight line. In this problem we have a magnetic flux of 800 maxwells cutting across a coil of wire in 1 µs. Firstly, let us find dφ/dt. We should assume that the coil cuts across the field at a steady rate giving a straight line graph for the magnetic field φ plotted against time t. The gradient is: dφ/dt = The change flux/The change in time for that change in flux = 800 × 10 -8 Wb/1 × 10 -6 seconds = 8 Wb/s We know the number of turns N is 1000, so we can substitute for N and dφ/dt to determine the voltage. ENG1021 Electronic Principles 98 V = N dφ/dt = 1000 × 8 = 8000 V or 8 kV Now attempt Problems 6 and 7. Problem 6 A circuit has a 20 V battery connected to a 100 Ω coil of 400 turns with an iron core 0.2 m long. Using SI magnetic units, calculate (a) I; (b) ampere-turns of mmf; (c) field intensity H; (d) flux density B in a core with µr of 500; (e) total flux φ at each pole with an area of 6 × 10 -4 m 2 . Solution The first part of this problem relies on our knowledge of the relationships between the current in the coil, the flux φ, the flux density B, the mmf and the magnetic field intensity H. Problem 6a The current is given by Ohm’s law. We have a 100 Ω coil connected to a 20 volt battery, therefore: I = V/R = 20/100 = 0.2 A Problem 6b We have 400 turns through which flows a current of 0.2A, therefore the mmf in ampere-turns is given by: mmf = N I = 400 × 0.2 = 80 ampere - turns Problem 6c We need the length of the coil to calculate the field intensity H which is the mmf per unit length. The length of the coil is given as 0.2 metres, therefore the field intensity H = mmf/length = 80/0.2 = 400 ampere.turns/metre Problem 6d We want to calculate the flux density B. We are given the relative permeability µr so, if we want to work in SI units, then we have to calculate the absolute permeability: µ = µ0 × µr = 4π × 10 -7 × 500 = 6.283 × 10 -5 ENG1021 Electronic Principles 99 We can then find the flux density B using B = µ × H = 6.283 × 10 -5 × 400 = 0.251 tesla Problem 6e Having found the flux density B the amount of flux φ is dependant upon the area of the coil. The area of the coil is 6 × 10 -4 m -2 , therefore the flux φ = B × A = 0.251 × 6 × 10 -4 = 1.508 × 10 -4 Wb Problem 7 For the coil in Problem 6: (a) If the iron core is removed, how much will the flux be in the air core coil? (b) How much induced voltage would be produced by this change in flux while the core is being moved out in 1 s? (c) How much is the induced voltage after the core is removed? Solution The main reason that you were advised to attempt Problem 6 is that it revises the previous work and sets you up for this rather interesting problem. It is interesting because we are changing the flux by removal of the iron core of the coil. Note that in previous examples and problems we have moved the coil, moved the magnet or subject a coil (somehow) to a changing magnetic field, in order to generate an emf in the coil. The question again exercises your understanding of Faraday’s law. Problem 7a The first thing that we are asked to calculate, is the flux if the iron core is removed. The flux will change, because the permeability has changed. The permeability is now just µ0 = 4π × 10 -7 , the permeability of air. Although the field intensity H has not changed, the flux density is now B = µ × H = 4π × 10 -7 × 400 = 5.027 × 10 -4 tesla The flux in the area of the coil is then φ = B × A = 5.027 × 10 -4 × 6 × 10 -4 = 3.0162 × 10 -7 Wb Problem 7b The next part of the question requires us to calculate dφ/dt. The change in flux is equal to the value before the removal of the iron core minus the value of the flux after the removal of the iron core which is: 1.508 × 10 -4 - 3.0162 × 10 -7 = 1.505 × 10 -4 Wb ENG1021 Electronic Principles 100 We are told that this change takes place in 1 second, therefore: dφ/dt = 1.505 × 10 -4 Wb/1 second = 1.505 × 10 -4 Wb/s The induced voltage is N times the change in flux, where N is the number of turns which equals 400. So the induced voltage is: V = 400 × 1.505 × 10 -4 = 0.0602 V Problem 7c After the core is removed there is no more change so the induced voltage is 0 V. 2.5 Summary Read the Review at the end of the Chapter. You have studied new concepts and learnt some new terms. It is easy to confuse the terms: • magnetic flux φ, • magnetic field intensity H, • magnetic flux density B, and • magnetic field which is a concept described in terms of the three other physical quantities φ, H, B. 3 Further reading If you would like to read some more about electromagnetism, the on-line book “Fundamentals of Electrical Engineering and Electronics” has some relevant sections. I suggest you look under DC where there is a sub-section called magnetism and electromagnetism. Look at all of the parts under this heading except for the last part on mutual inductance. 4 Where next? You are advised to study the learning package entitled “Capacitance” next. ENG1021 Electronic Principles 101 ENG1021 Electronic Principles Learning Package 7 Capacitance ENG1021 Electronic Principles 102 Capacitance 1 Do you know all this already? If in doubt, please attempt the self assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Introduction Nearly all electronic circuits contain capacitors. Capacitors have the ability to store charge. Although they generally do not hold as much charge as a rechargeable battery (and they store charge in a very different way), you may like to think of them as small rechargeable batteries. You will also learn about reactance. Capacitors have reactance in AC circuits. Reactance has similarities with resistance, and is measured in ohms, but has some surprising differences. Finally this section introduces simple capacitive circuits, often called RC circuits which consist of a capacitor in a series or parallel combination with a resistor. You will learn how to combine resistive and reactance values (which cannot simply be added together) to form an impedance value. To do this we have to take into account the relative phase between the voltage and current in a capacitive circuit which leads to the concept of phase angle. 3 Capacitance 3.1 Capacitance, dielectrics, and charge A capacitor is a device that can store electrical charge. It consists of two plates, and between the plates is a material called a dielectric. This dielectric material is an insulator, the there is no way that electrons can flow through the capacitor. Figure 1 shows a capacitor connected to a battery. Figure 1 ENG1021 Electronic Principles 103 When a battery is connected to a capacitor, as in Figure 1, a potential difference is created across the capacitor, so that one plate (the one connected to the positive terminal) has a higher potential that the other plate (connected to the negative terminal of the battery). We know that continuous current can’t flow round the circuit because of the insulator. However, when the battery is first connected, the potential difference created causes electrons to accumulate in the plate connected to the negative terminal of the battery. In the other plate, electrons are lost to the positive terminal of the battery. The end result is that there is a difference in charge on either side of the capacitor, and that this difference creates its own potential difference. The system settles when this potential difference across the capacitor equals the potential difference created by the battery. Now if the battery is disconnected, the charge difference within the capacitor remains. The capacitor has been fully charged. We can discharge the capacitor by connecting a wire between the plates, so that the charge can flow from the negatively charged side until both side have equal charge, Then the capacitor is said to be fully discharged. The amount of charge stored in a capacitor is given the symbol Q, and is measured in coulombs, C. One coulomb corresponds to 6.25 x 10 18 electrons. If a voltage, V, is applied to a capacitor, then the ratio of the charge produced to the voltage applied is called the capacitance of the capacitor. Capacitance is also given the symbol C, and is measured in farads, F. C = Q/V (1) Q = CV The unit of a farad is very large, and so capacitance is often measured in microfarads, µF, which is 10 -6 Farads or even picofarads, pF, which is 10 -12 farads. For example, a 2 µF capacitor is connected to a 9 V battery. How much charge is stored? Q = CV = 2 x 10 -6 x 9 = 18 x 10 -6 C When a battery is connected to a capacitor, we have said that the capacitor becomes charged. In an ideal circuit, where there is no resistance, this would be instantaneous. In reality, there is always some resistance in a circuit, either in the wires, the battery or even in the capacitor itself. So, charging takes a finite time. Exactly how much is a subject of a later Learning Package. However, if a constant current can be applied, rather than a battery, then that constant current will flow which will be continuously delivering charge to one plate of the capacitor, which the other side loses charge. If the constant current has a value of I amps, and is applied for a time t, then the charge introduced to the capacitor is: ENG1021 Electronic Principles 104 Q = I x t (2) Now attempt Problems 1, 2 and 3. Problem 1 How much charge in coulombs is in a 4 µF capacitor charged to 100 V? Solution All of these problems require the use of equation for charge Q = CV Where C is the capacitance in Farads, V is the voltage across the capacitor, in volts, and Q is the charge in coulombs. So to find the charge we simply substitute the values for capacitance and voltage into the equation Q = CV = 4 × 10 −6 × 100 = 4 × 10 −4 = 0.4 mC or 400 µC Problem 2 A 4 µF capacitor has 400 µC of charge. (a) How much voltage is across the capacitor? (b) How much is the voltage across an 8 µF capacitor with the same 400 µC charge? Solution Again we use Equation 1 in a rearranged form for the voltage V = Q/C Problem 2a The first part of the question asks for the voltage across a 4 µF capacitor which is holding 400 µC of charge. Substituting these values into Equation 1 gives: V = 400 × 10 −6 /4 × 10 −6 = 100 volts Problem 2b If the same charge is stored in a 8 µF capacitor then the voltage V = 400 × 10 −6 /8 × 10 −6 = 50 volts Problem 3 ENG1021 Electronic Principles 105 A 2µF capacitor is charged by a constant 3 µA charging current for 6 s. (a) How much charge is stored in the capacitor? (b) How much is the voltage across the capacitor? Solution Constant current charging of capacitors is implemented in many electronic circuits, because the voltage rises linearly with time. Timing circuits and slope generators use this principle. Problem 3a In this problem a 2 µF capacitor is charged with a constant current of 3 µA for 6 s. We are asked to find the charge in the capacitor. Although it is not stated, we must assume that the capacitor is fully discharged. If it were not constant current charging would add to any charge already in the capacitor. The charge is found by substituting the values for the current and time into the equation: Q = I × t = 3 × 10 −6 × 6 = 18 µC Problem 3b The voltage across the capacitor is given by Equation 1: V = Q/C = 18 × 10 −6 /2 × 10 −6 = 9 volts 3.2 Practical capacitors, energy stored and trouble shooting 3.2.1 Construction of capacitors A practical capacitor is made up of plates and a dielectric. One example is the “paper” capacitor in which layers of tin foil surround a layer of paper, and then these layers are rolled up. The symbol used in electronics is shown in Figure 2. Figure 2 Symbols used for a capacitor ENG1021 Electronic Principles 106 The symbol on the left is most common, but you can also often see the symbol in the middle being used. This symbol represents an electrolytic capacitor which can only be connected one way round. The curved side always represents the negative connection. Finally the symbol on the right indicates a variable capacitor. Often the value of the capacitance is written on the capacitor, usually in either microfarads or picofarads. In some cases they are colour coded using a similar scheme as resistors. In either case, capacitors are only available in preferred values, with the same range of values as resistors. In general, the area of the plates influences the value of the capacitance, with bigger plates creating bigger capacitance. Another factor is the size of the gap between the plates. The smaller this gap, the bigger the capacitance. Finally, the material used as the dielectric has a major influence. When we looked at magnetism we found a quality called permeability, which influenced how much magnetic flux is produced. Similarly, in capacitance we have permittivity, ε, the Greek character epsilon. As with permeability, we usually quote a relative permittivity of a material, ε r , which then has to be multiplied by the absolute permittivity of air, ε o , which equals 8.854 x 10 -12 F/m. The relative permittivity is also given the symbol K ε . For example, air has a relative permittivity of 1, paper is 2 to 6, ceramic is 80 to 1200. Given all of the above, we can now say that the capacitance is given by the following equation, where A is the area of the plates and d is the size of the gap between the plates. C = K ε x A/d x 8.85 x 10 -12 F (3) Now please attempt Problems 4, 5 and 6. Problem 4 Calculate C for a mica capacitor, with Kє = 8, a thickness of 0.02 cm, plates of 6 cm 2 , and five sections in parallel. (Hint: 1 cm = 10 -2 m and 1 cm 2 = 10 -4 m 2 .) Solution The construction of the mica capacitor is such that it many single layer capacitors in parallel. In this problem we have 5 layers which is equivalent to having 5 single layer capacitors in parallel. So firstly we must calculate how much capacitance there is in a single layer. The capacitance of a single layer capacitor is given by the equation C = Kє × A/d × 8.85 × 10 −12 ENG1021 Electronic Principles 107 where A is the area of the plates in metre 2 , d is the distance between the plates in metres, K є is the dielectric constant alternatively called the relative permittivity є r . The figure 8.85 × 10 −12 is the absolute permittivity of air or a vacuum, which is given the symbol є 0 . Given the above information I prefer to write Equation 3 as C = A/d × єrє0 (4) The equation is more concise in this form and it reminds us that the capacitance is proportional to the area of the plates and inversely proportional to the distance between the plates. Substituting our values for A, d, є 0 , and є r into Equation 4 we have: C = 6 × 10 −4 /(0.02 × 10 −2 ) × 8 × 8.85 × 10 −12 = 212.4 pf This is the capacitance between a single plate. Capacitors in parallel add together, therefore the capacitance across 5 sections will be: 5 × 212.4 = 1.062 nF 3.2.2 Energy stored Charge stored in a capacitor has energy. The amount of energy is found using the following equation: E = ½ CV 2 (5) It is important to realise that capacitors store this charge even when disconnected. This means that electrical equipment that contain capacitors can often hold on to large amounts of charge even when the equipment is switched off. For example, a television picture tube will can hold enough charge to generate twenty five thousand volts even when disconnected form the wall. The energy stored is small, but this voltage can still give an electric shock and cause burns. Other capacitors within electronic equipment may also store hazardous voltages when equipment is disconnected from the mains. Please bear this in mind when handling equipment that is supposed to be ”dead” (disconnected from power). Problem 5 Calculate the energy in joules stored in (a) a 500 pF capacitor charged to 10 kV; (b) a 1 µF capacitor charged to 5 kV; (c) a 40 µF capacitor charged to 400 V. Solution Here we are asked to calculate the energy stored in capacitors. We use the equation: E =½CV 2 ENG1021 Electronic Principles 108 The solutions are found simply by substituting the vales of C and V into the equation for energy. Problem 5a Here C = 500 pf and V = 10 kV. Substituting the values into the equation gives: E = ½(500 × 10 −12 )(10 ×10 3 ) 2 = 250 × 10 −12 ×10 8 = 2.5 × 10 −2 joules Problem 5b Here C = 1 µf and V = 5 kV. Substituting the values into the equation gives: E = ½ (1 × 10 −6 )(5 × 10 3 ) 2 =½ × 1 × 10 −6 × 25 × 10 6 = 12.5 joules Problem 5c Here C = 40 µf and V = 400 V. Substituting the values into the equation Gives: E = ½ (40 × 10 −6 )(400) 2 = 20 × 10 −6 ×16 × 10 4 = 320 × 10 −2 = 3.2 joules 3.2.4 Series and parallel capacitors When capacitors are connected in parallel it is equivalent to increasing the plate area. Therefore the capacitance of two capacitors in parallel is the sum of the individual capacitances. C eq = C1 + C2 (6) Connecting capacitors in series is equivalent to increasing the dielectric thickness or the gap between the plates. Therefore the reciprocal of total capacitance of two capacitors in series is the sum of the reciprocal of the individual capacitances. 1/C eq = 1/C1 + 1/C2 (7) Problem 6 Calculate C T for the series-parallel combination of capacitors shown in Figures 3a and 3b. ENG1021 Electronic Principles 109 Figure 3a Figure 3b Solution This problem exercises our familiarity with parallel and series capacitor combinations. Remember that the rules for parallel and series capacitor combinations are the reverse of the rules for combinations of series and parallel resistors. Problem 6a The problem can be though of as a parallel combination of two capacitors where one of the capacitors is a series combination of two capacitors. Therefore we work out the equivalent capacitance of the series capacitors first. Let us call the series combination of C1 and C2, C12. Then using the equation for capacitors in series: 1/C12 = 1/C1 + 1/C2 = 1/(0.02 × 10 −6 ) + 1/(0.04 × 10 −6 ) = (50 + 25) × 10 6 = 75 × 10 6 Taking reciprocals: C12 = 0.0133 × 10 −6 F C T is the parallel combination of C 12 and C 3 and parallel capacitors add in value. So: CT = C3 + C12 = 0.047 × 10 −6 + 0.0133 × 10 −6 = 0.06033 × 10 −6 or 0.0633 µF C T C T C 1 =0.02µF C 2 =0.04µF C 3 =0.047µF C 1 =150pF C 1 =100pF C 1 =47pF ENG1021 Electronic Principles 110 Problem 6b This part of the problem should be thought of as a series combination of two capacitors, but one of those capacitors is a parallel combination of two capacitors. Therefore we work out the value of the parallel combination C 1 and C 2 first. Let use call the equivalent capacitance of the parallel capacitors C 12 , then C12 = C1 + C2 = 47 pF + 100 pF = 147 pF C 12 is in series with C 3 , a 150 pF capacitor. The total capacitance is given by: CT = C12 × C3/(C12 + C3) = 147 × 150/(147 + 150) = 22050/297 = 74.2 pF Note that in the later part of this problem I have used the formula for two series capacitors and worked entirely in picofarads in order to show the alternative methods. 3.2.3 Trouble shooting Leaky and open capacitors can be detected with an ohmmeter. If this technique works, all well and good, but often open and leaky capacitors are not that obvious to an ohmmeter and since we must disconnect one end of the capacitor from the circuit anyhow, it is usually more efficient to trouble shoot leaky and open capacitors by replacement. 3.3 Capacitive reactance So far we’ve looked at DC circuits. A capacitor is essentially a break in the circuit, so a continuous current cannot flow. When a battery is connected there is an initial current that builds up charge on the capacitor plates, but then reaches a stable point where the potential difference are equal and no more current flows. An analogy that is sometimes used is that of a hydraulic system, where a pump represents the battery. A capacitor is represented by two chambers separated by a rubber diaphragm, and all connected together with pipes. Initially the pipes and the chambers are full of water. When the pump is switched on, it forces more water into one of the chamber which pushes against the rubber diaphragm. The diaphragm stretches which forces water out of the other chamber. This continues until the pressure produced by the pump equals the pressure produced by the ENG1021 Electronic Principles 111 rubber diaphragm, at which point no more water can move. This is analogous to the charge building up on one of the plates of the capacitor, and the lack of negative charge on the other plate. Clearly, in this analogy, you cannot have a continuous flow of water because the circuit is blocked by the diaphragm. Now let’s think of an AC circuit. In this scenario the pump keeps changing direction. Initially it will pump water in one direction, gradually increasing the pressure. Then the pressure eases of and the pump reverses direction and repeats the cycle. Now, although there is still a diaphragm, the water all through the circuit can move back and forth under pressure from the pump. Similarly, in a capacitor the charge can move back and forth in the circuit under the emf of the AC voltage generator. If we have a sinusoidal voltage applied, then the current will also be sinusoidal, with the same frequency. The amplitude of the voltage sinewave divided by the amplitude of the current sinewave is similar to resistance, and is called reactance, measured in ohms. Reactance is given the symbol X, and as this is capacitive reactance it is referred to as Xc. It obeys Ohm’s law as follows: V = I Xc (8) The main difference between reactance and resistance is that reactance is dependent on frequency. In the case of capacitive reactance, when the frequency is zero (DC) then the reactance is infinite, i.e. no current flows. At higher frequencies the reactance gets less. This is summarised in the following equation for the reactance of a capacitor: Xc = 1/2πfC (9) In an AC circuit, if two capacitors are connected in series, the total reactance is the sum of the individual reactances. We see this by starting with the equation we found earlier for the total capacitance of two series capacitors. 1/Ceq = 1/C1 + 1/C2 Xeq = 1/2πfCeq = 1/2πf C1 + 1/2πf C2 = Xc1 + Xc2 Similarly, the reciprocal of the total reactance of two capacitors in parallel is found as the sum of the reciprocal of the individual reactances. 1/Xeq = 1/Xc1 + 1/Xc2 (10) Please attempt Problems 7 and 8. Problem 7 Give the values of C needed for 2000 Ω of XC at the following four frequencies – 1 MHz, 0.5 MHz, 0.2 MHz and 0.1 MHz. ENG1021 Electronic Principles 112 Solution We are asked to find the value of a capacitance which has a reactance of 2000 Ω at four frequencies 1 MHz, 0.5 MHz, 0.2 MHz and 0.1 MHz. We take the equation for capacitive reactance: XC = 1/2πfC and rearrange it to give: C = 1/2πfXC The four solutions to this question are obtained by substitution into the rearranged equation. When frequency we are told that X C = 2000 Ω. Therefore: C = 1/2π1 × 10 6 2000 = 1/12566.37061436 × 10 6 = 79.6 pF At f = 0.5 MHz: C = 1/2π0.5 × 10 6 2000 = 1/6283 × 10 6 = 159 pF At f = 0.2 MHz: C = 1/2π0.2 × 10 6 2000 = 1/2513.274122872 × 10 6 = 399 pF At f = 0.1 MHz: C = 1/2π0.1 × 10 6 2000 = 1/1256.637061436 × 10 6 = 796 pF Problem 8 Four capacitive reactances of 100, 200, 300 and 400 Ω each are connected in series across a 40 V rms source. (a) Draw the schematic diagram. (b) How much is the total X CT ? (c) Calculate I. (d) Calculate the voltages across each capacitance. (e) If the frequency of the applied voltage is 1600 kHz, calculate the required value of each capacitance. Solution This problem introduces series combinations of capacitors and the calculation of their reactance. Problem 8a The circuit diagram is shown in Figure 4. ENG1021 Electronic Principles 113 Figure 4 Circuit diagram for Problem 8a Problem 8b The equivalent (or total) reactance of a series combination of reactances is found using the same rules as for series resistors. Note, that this is not true for finding the equivalent capacitance for a series combination of capacitors. We have four reactances of 100, 200, 300 and 400Ω in series, the total reactance is: XT = X1 + X2 + X3 + X4 = 100 + 200 + 300 + 400 = 1000 Ω Problem 8c The current is given by Ohm’s law: I = V/XT = 40/1000 = 40 mA rms Problem 8d To find the voltage across each capacitor we use Ohm’s law and the fact that the same current I flows in all parts of a series circuit. Therefore, the voltage across the first capacitor (X C1 = 100 Ω) is: VC1 = 40 × 10 −3 × 100 = 4 volts The voltage across the next capacitor (XC2 = 200 Ω) VC2 = 40 × 10 −3 × 200 = 8 volts The voltage across the next capacitor (XC3 = 300 Ω) is VC3 = 40 × 10 −3 × 300 = 12 volts The voltage across the last capacitor (XC4 = 200 Ω) is 40 V 100Ω 200Ω 300Ω 400Ω ENG1021 Electronic Principles 114 VC4 = 40 × 10 −3 × 400 = 16 volts We can check our results because for purely capacitive reactance the individual voltage drops should add up to the supply voltage. So, 4 + 8 + 12 + 16 = 40 volts = the supply voltage of 40 volts Problem 8e If the frequency of the AC source is 1600 kHz = 1.6 MHz we use the rearranged equation for reactance Equation 25 substituting for reactance in each case. For the first capacitor (XC1 = 100 Ω). The capacitance is: C 1 = 1/(2π1.6 × 10 6 × 100) = 0.995 × 10 −9 = 0.995 nF For the next capacitor (XC2 = 200 Ω) the capacitance is: C 2 = 1/(2π1.6 × 10 6 × 200) = 0.497 × 10 −9 = 0.497 nF For the next capacitor (XC3 = 300 Ω) the capacitance is: C 3 = 1/(2π1.6 × 10 6 × 300) = 0.332 × 10 −9 = 0.332 nF For the last capacitor (XC4 = 400 Ω) the capacitance is: C 4 = 1/(2π1.6 × 10 6 × 400) = 0.249 × 10 −9 = 0.249 nF 3.4 Relative phase Unlike in a resistor, the alternating current in a capacitor is not in phase with the alternating voltage across it. In other words the alternating voltage across a capacitor rises at falls at different times from the alternating current through the capacitor. The phase angle between the voltage and current in a capacitor is 90º and is the same as the difference between a cosine wave and a sine wave. The current rises before the voltage and the current is said to “lead” the voltage, as shown in Figure 5. ENG1021 Electronic Principles 115 -300 -200 -100 0 100 200 300 0 0.02 0.04 0.06 0.08 time/s v o l t a g e / V c u r r e n t / I Figure 5 Voltage and current in a capacitor (current is dashed) The rate of voltage change with time operator is introduced and related to the current dv i C dt = (11) Here we see the notation for differentiation again, where dv means a small change in voltage, and dt means a small change in time. So, dv/dt represents the rate at which the voltage is changing with time. Please attempt Problem 9. Problem 9 A capacitor has a discharge current i c of 15 mA when the voltage across its plates decreases at the rate of 150 V/µs. Calculate C. Solution This problem exercises your understanding of rates of voltage change and the capacitor charge/discharge cycle. We are told that a capacitor is subject to a discharge current of 15 mA and that the rate of change of voltage is 150 V/µs. Let us rearrange Equation 11: I C dV dt = ENG1021 Electronic Principles 116 and substitute our values into it to find C: 3 3 9 10 6 6 15 10 1 10 10 10 100 150 10 10 10 10 C pF − − − − − − − × × = = = = = − Note that I have used minus signs in front of the current and the decreasing voltage, because the current is a discharge current and the voltage is decreasing. Although the minus signs cancel out, it is good practice to include them, because in larger problems they become important. 3.4 Capacitive (RC) circuits We will now have a look at circuits containing a resistor and a capacitor in either series or in parallel. Figure 6 shows both cases, and the associated “phasor triangle”. Figure 6 a) series circuit; b) parallel circuit In Figure 6a), a resistor is in series with a capacitor. We know from earlier discussions that in a series circuit the current is the same through each component, so the current flowing through the capacitor must be the same as the current flowing through the resistor, I. We also know that the current through the capacitor leads the voltage across the capacitor by 90 degrees. If the reactance of the capacitor is Xc, then by Ohm’s law the voltage across the capacitor is Vc = IXc. Similarly, the voltage across the resistor is Vr = IR. We can represent these two voltages using a phasor triangle, where the horizontal arrow shows the voltage across the resistor which is in phase with the current through the resistor. The vertical arrow shows the voltage across the capacitor which is 90 degrees behind the current through the capacitor. ENG1021 Electronic Principles 117 You think of a phasor diagram as a rotating arrow, pivoted at the centre of a circle, then when it is pointing to the right, that is 0 degrees. Now rotate it anticlockwise. When it’s point upwards, that is +90 degrees. Keep going until it is pointing to the right, and that’s +180 degrees. Keep going until it is pointing downwards and that corresponds to +270 degrees. Alternatively, if you rotate clockwise, then from zero degrees, rotate until the arrow is pointing downwards and this corresponds to -90 degrees. Figure 7 illustrates this. Figure 7 Phasor diagrams In Figure 7, a phasor is shown pointing to the left. The other phasor is shown relative to this first one, either as a clockwise (negative) rotation or an anticlockwise (positive) rotation. When we add two phasors we have to take into account not only their magnitude but also their direction. For example, if I asked you to start at your house, walk east for 3 miles, then north for 4 miles. How far have you walked? That’s easy, you just add the miles, 3 plus 4 equals 7 miles. If I asked you how far would be as the crow flies from your house, then you have to take the direction into account. Figure 8Vector distance ENG1021 Electronic Principles 118 The distance from your house is shown as the diagonal line from the home to the journey’s end. This distance, d, is calculated using Pythagoras’s theorem. 2 2 2 3 4 25 25 5 d d = + = = = So the distance is 5 miles. In our phasor triangle, the resistive voltage is at 0 degrees, and the capacitive voltage is at -90 degrees. We can either say that the current leads the voltage in a capacitor by 90 degrees, or that the voltage lags behind the current by -90 degrees. It amounts to the same thing. Now we know that the supply voltage should equal the sum of the resistive voltage and the capacitive voltage. But because they are out of phase, when represented in a phasor diagram they point in different directions. We there fore have to use Pythagoras’s theorem to find the sum. This is shown as the hypotenuse of the triangle. The sum is therefore: 2 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) r c c c c c E V V E IR IX E I R I X E I R X E I R X E IZ = + = + = + = + = + = (12) In the last line, the term Z is the total impedance of the circuit. Impedance is the combination of resistance and reactance, and is also measured in ohms. You can see that to find the value of Z you have to take the square root of the resistance squared plus the reactance squared. The other value that we can calculate is the relative phase, θ, between the voltage and the current. In Figure 6a) this is shown as the angle between the supply voltage, E, and the resistive voltage Vr. This is because Vr is in phase with the current, therefore the angle between these two voltages is the same as the angle between the voltage and the current. From trigonometry we can see that: c c IX X Tan IR R θ = − = − (13) ENG1021 Electronic Principles 119 So the phase is the ratio of the reactance and the resistance. Since the capacitive reactance is pointing downwards, it is negative. You have to bear this in mind when calculating the phase. Using the values in the circuit of Figure 6, the reactance is: Xc = 1/2πfC = 1/2x 3.142 x 40 x 100 x 10 -6 = 39.79 Ω ≈ 40 Ω. This means that the total impedance is: 2 2 2 2 100 40 10000 1600 11600 107.7 c Z R X Z = + = + = + = = Ω The phase is: 40 0.4 100 21.8 c o X Tan R θ θ = − = − = − = − Now for parallel circuits. As we know from previous Learning Packages the voltage across each parallel branch is the same. So in the parallel circuit of Figure 6b) the voltage across the resistor and the capacitor is equal to E. The current in the resistive branch is therefore Ir = E/R and is in phase with the voltage. The current in the capacitive branch is Ic = E/Xc and is 90 degrees ahead of the voltage. If the resistive current is drawn horizontally, then the capacitive current leads by 90 degrees, as in the phasor triangle of Figure 6b). The total current must equal the sum of the two individual branch currents. As they are phasor values with direction as well as size, we use the triangle again and Pythagoras’s theorem. This gives the total current as: 1 1 1 1 1 r c c c c E E I I I R X I E R X I E R X Z = + = + | | = + | \ ¹ | | = + = | \ ¹ So this is the same as any parallel circuit, where the reciprocal of the total impedance equals the sum of the reciprocal of the resistance and the reciprocal of the reactance. Similarly, from the phasor triangle of Figure 6b) it can be seen that: ENG1021 Electronic Principles 120 c r c c c I E E Tan I X R E R R Tan X E X θ θ = = ÷ = × = Using the values in Figure 6, we can find the impedance and relative phase. 1 1 1 1 1 140 100 40 4000 4000 28.57 140 c Z R X Z = + = + = = = Ω 100 2.5 40 68.2 c o R Tan X θ θ = = = = You should now have a good idea about the phase relationship between the current and voltage in a capacitive circuit and be able to construct a phasor triangle (often called a phasor diagram). Please attempt Problem 10, 11, and 12. Problem 10 This problem exercises your understanding of impedance and phase relationships in a series capacitive circuit. Problem 10a A 40 Ω R in series with a 30 Ω X c across a 100 V sinewave AC source. (a) Draw the schematic diagram. (b) Calculate Z T . (c) Calculate I. (d) Calculate the voltage across R and C. (e) What is the phase angle of the circuit? Solution The schematic diagram is shown in Figure 9. ENG1021 Electronic Principles 121 Figure 9 Diagram for Problem 10a Problem 10b To calculate Z T it is useful (but not essential) to draw a phasor diagram. The voltages across the resistor and the capacitor must add up to the source voltage (at any given time), but because these voltages are out of phase we must add the voltage across the capacitor phasor (V C = IX C ) to the voltage across the resistor phasor (V R = IR). Doing this forms the phasor triangle shown in Figure 10. Consider the phasor triangle, the length of the hypotenuse is equal to the source voltage (100 volts), which is equal to the current I multiplied by the total impedance Z T . By Pythagoras’ theorem: Figure 10 Phasor diagram for Problem 10 2 2 100 ( ) ( ) T C IZ IX IR = = + The I’s cancel giving: ENG1021 Electronic Principles 122 2 2 ( ) ( ) T C Z X R = + Problem 10c The current I may be found from Ohm’s law by using the previously obtained value for the total impedance Z T and the knowledge that the source voltage is 100V. I = V/ZT = 100/50 = 2 amperes Problem 10d The individual voltages across R and C are, again, given by Ohm’s law. The voltage across the capacitor is: VC = IXC = 2 × 30 = 60 volts The voltage across the resistor VR = I × 40 = 80 volts Note that the algebraic addition of the voltages across the resistor and the capacitor is greater than the supply voltage. This is because the two voltages are out of phase. If we looked at the voltages at any instant in time, one voltage will be rising when the other is at its peak. If you were to add these instantaneous voltages they would add up to the supply voltage at that instant. Problem 10e The phase angle θ is the angle between the generator voltage and the series current or equivalently the generator voltage and current. Although this phasor diagram does not show the current, we know that the current and voltage in a resistor are always in phase. The phase angle θ in this circuit is the angle between the generator voltage and the voltage across the resistor. From the phasor diagram shown in Figure 10: tan θ = IXC/IR = XC/R = 3/4 = 0.75 θ = 36.86 o Problem 11 A 40 Ω R and a 30 Ω X c are in parallel across a 100 V sinewave AC source. (a) Draw the schematic diagram. (b) Calculate each branch current. (c) How much is I T ? (d) Calculate Z EQ . (e) What is the phase angle of the circuit? (f) Compare the phase angle of the voltage across R and X c . Solution ENG1021 Electronic Principles 123 This problem is similar to Problem 10, but we are dealing with a parallel circuit. Problem 11a The circuit diagram is shown in Figure 11. Figure 11 Circuit diagram for Problem 11a Problem 11b Since the supply voltage appears across capacitor C and resistor R the current flowing though each of them are the branch currents and are given by Ohm’s law. The current through the resistor is: IR = V/R = 100/40 = 2.5 amperes The current through the resistor is: IC = V/XC = 100/30 = 3.33 A Problem 11c The total current I T is found by adding the branch currents, but because the currents are out of phase we must add their phasors. A phasor diagram would be helpful now, as in Figure 12: ENG1021 Electronic Principles 124 Figure 12 Phasor diagram for Problem 11c From the diagram we can see that I T is the phasor addition of I T R and IR, again using Pythagoras’ theorem: 2 2 2 2 ( ) ( ) 2.5 3.33 4.14 T R C I I I amperes = + = + = Problem 11d The equivalent impedance ZEQ is given by Ohm’s law: ZEQ = V/I = 100/4.14 = 24.15 Ω Problem 11e The phase angle can be found from the phasor diagram. It is marked as current on Figure 12. The phase angle is the angle between the voltage and current as “seen” by the voltage supply. The phase of the voltage is the same as the phase of the current through the resistor vector IR, since the resistor is directly connected across the voltage supply. The phase of the current is in the direction of vector I T , since this is the current we calculated as the total current supplied to the circuit. On Figure 12 the angle between these two vectors is marked as θ. tan θ = IC/IR = 3.33/2.5 = 1.332 θ = 53.1 o If you are wondering if this angle should be positive or negative, it depends only on whether you take the current or the voltage as a reference. The voltage will always lag the current in an RC circuit and so the phase angle of the voltage with ENG1021 Electronic Principles 125 respect to the current will be negative. Alternatively, we can say that the current leads the voltage in an RC circuit and so the phase angle of the current with respect to the voltage will be positive. Verify this to yourself using the phasor diagram and try turning the phasor diagram around to that the current is the reference phasor and checking that the phase of the voltage (which is the same as the current through the resistor IR) now lags the current. Remember that positive angles are in the anticlockwise direction, while negative angles are the clockwise direction. Problem 11f The phase of the voltage is the same in all parts of the parallel circuit, so the phase angle across the resistor is equal to the phase angle across the capacitor. Problem 12 Draw a schematic diagram of a capacitor in series with a 20 kΩ resistance across a 10 V ac source. What size C is needed for equal voltages across R and X c at frequencies of 100 Hz and 100 kHz? Solution The circuit diagram for the first part of this problem is given in Figure 13. We are then asked what value of capacitor will result in equal voltages existing across the resistor and the capacitor. This is a series circuit so the same current I flows in all parts of the circuit. The voltage across the resistor (by Ohm’s law) is: VR = IR The voltage across the capacitor is: VC = IXC Figure 13 Circuit diagram for Problem 12 We are told that: VC = VR ENG1021 Electronic Principles 126 So: IR = IXC we can cancel the I’s, so that the voltages across the resistor and the capacitor are equal when: R = XC XC = 1/2πfC = R We can rearrange this equation to get a value for C. C = 1/2πfR We are asked to find the value of C for two different values of frequency f, so let’s substitute our values of R = 20 kΩ and f = 100 Hz into our equation for C first. C = 1/2πfR = 1/2π × 100 × 20 × 103 = 0.08 µF Now for R = 20 kΩ and f = 100 kHz C = 1/2πfR = 1/2π × 100 × 10 3 × 20 × 10 3 = 80 pF 4 Further reading For further reading you may want to look at Fundamentals of Electrical Engineering and Electronics. The relevant sections are “DC” and the sub-section called “Capacitors””, where all parts could be read. Also, under the section “AC”, sub-section “Reactance and impedance – capacitive” there is relevant information. Have a look at all the parts, but be warned. This book starts using both complex notation (or j notation) and polar notation rather than using a phasor diagram. Where next? The next suggested learning package in entitled “Inductance and Transformers”. ENG1021 Electronic Principles 127 ENG1021 Electronic Principles Learning Package 8 Inductance and Transformers ENG1021 Electronic Principles 128 Inductance 1 Do you know all this already? If in doubt, please attempt the self assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Introduction In this Learning Package you will learn about inductors and the property inductance. In particular we will see how inductors produce an electromagnetic field and that two inductors can interact to produce mutual inductance. This property of mutual inductance is put to use in transformers where AC voltages can be “stepped-up” or “stepped-down” to other voltages. 3 What is inductance? When a magnetic flux cuts across a conductor a voltage is induced, as you saw in Learning Package 6. The voltage is induced if the conductor is moving or if the magnetic field is moving or changing. In an AC circuit, a conductor produces a magnetic flux which alternates in direction as the current alternates, and also changes size in proportion to the size of the current. This changing flux induces a voltage. So, it must be remembered that it is the change in current that is inducing a voltage. A current that is large but doesn’t change would not induce a voltage. The frequency of the alternating current also affects the size of the induced voltage, as a higher frequency means that the current is changing ore quickly. Figure 1 shows a simple AC circuit with a magnetic coil. Figure 1 Circuit with magnetic coil The inductance, L, is defined as: ENG1021 Electronic Principles 129 / v L di dt = (1) In this equation, L is the inductance, measured in Henries, v is the voltage that is induced in the coil and di/dt is the rate of change of current in amperes per second. Rearranging we get: di v L dt = (2) In AC circuits the current is continuously changing. However, a voltage can be induced by any change in current, as the following problem indicates. Now attempt Problem 1. Problem 1 Calculate the values of v L across a 5 mH inductance for the following current variations. (a) zero to 3 A in 2 s; (b) zero to 50 mA in 5 µs; (c) 100 to 150 mA in 5 µs; (d) 150 to 100 mA in 5 µs. Solution Problem 1a We are asked for the voltage across a 5 mH inductor subject to different current variations. For all parts of this problem we will use the Equation 2. Returning to our problem, we need to calculate the change of current with respect to time, dI/dt. In this part we have a current which changes form 0 to 3 A in 2 seconds. Therefore: Change in current 3 - 0 = 3A Change in time = 2 seconds. dI/dt = 3/2 = 1.5 As -1 The voltage vL = LdI/dt = 5 × 10 -3 × 1.5 = 7.5 mV Problem 1b In this part we have a current which changes form 0 to 50 mA in 5 µs seconds. Therefore: ENG1021 Electronic Principles 130 Change in current 50 - 0 mA = 50 mA Change in time = 5 µs seconds. dI/dt = (50 × 10 -3 )/(5 × 10 -6 ) = 10 4 As -1 The voltage is: vL = LdI/dt = 5 × 10 -3 × 10 4 = 50 V Problem 1c In this part we have a current which changes from 100 to 150 mA in 5 µs. Therefore: Change in current 150 - 100 = 50 mA Change in time = 5 µs seconds. dI/dt = (50 × 10 -3 )/(5 × 10 -6 ) = 10 4 As -1 This is the same value for dI/dt as in part (b), even though the current is higher, because it is the change in current that is important and not the absolute value. The voltage is: vL = LdI/dt = 5 × 10 -3 × 10 4 = 50 V Problem 1d In this part we have a current which changes form 150 to 100 mA in 5 µs Therefore: Change in current 100 - 150 = -50 mA. The minus sign signifies the decrease in current. Change in time = 5 µs seconds. dI/dt = (-50 × 10 -3 )/(5 × 10 -6 ) = -10 4 As -1 The voltage is: vL = LdI/dt = 5 × 10 -3 × -10 4 = -50 V 3.1 Inductance in coils The value of the inductance of a coil depends on a number of factors. These include: ENG1021 Electronic Principles 131 The number of turns, N; The area enclosed by the coil, A; The permeability of the core, µ r ; The length of the coil, l. The inductance can be calculated as: 2 6 1.26 10 H r N A L l µ − × = × × × (3) Now try Problem 2. Problem 2 Calculate the inductance L for the following long coils: (a) air core, 20 turns, area 3.14 cm 2 , length 25 cm; (b) same coil as (a) with a ferrite core having µr of 5000; (c) air core, 200 turns, area 3.14 cm 2 , length 25 cm; (d) air core, 20 turns, area 3.14 cm 2 , length 50 cm; (e) air core, 20 turns, diameter 4 cm, length 50 cm, (Note 1 cm = 10 -2 m, and 1 cm 2 = 10 -4 m 2 .) Solution Here we are asked to calculate the inductance for coils of wire of differing dimensions and cores. The problem illustrates the impact of changing each of the values in Equation 3. Problem 2a This inductor has an air core. This tells us that the relative permeability µr = 1. All the other values are given for substitution into Equation 3, but remember that the equation requires that area A is in square metres and that length l is in metres. We are given: The number of turns N = 20 The area of the coil A = 3.14 cm 2 = 3.14 × 10 -4 m 2 The length of the coil l = 25 cm = 25 × 10 -2 m Substituting these values into Equation 3 with µr = 1: L = µr(N 2 × A)/l × 1.26 × 10 -6 = 1 × (20 2 × 3.14 × 10 -4 )/(25 × 10 -2 ) × 1.26 × 10 -6 = 6.33 × 10 -7 = 0.663 µH ENG1021 Electronic Principles 132 Problem 2b This inductor has a ferrite iron core. We are told that that the relative permeability of the ferrite core µr = 5000. We are given: The number of turns N = 20 The area of the coil A = 3.14 cm 2 = 3.14 × 10 -4 m 2 The length of the coil l = 25 cm = 25 × 10 -2 m The same values are used as in Part (a). Substituting these values into Equation 3 with µr = 5000: L = µr(N 2 × A)/l × 1.26 × 10 -6 = 5000 × (20 2 × 3.14 × 10 -4 )/(25 × 10 -2 ) × 1.26 × 10 -6 = 3.165 × 10 -3 = 3.165 mH Problem 2c This inductor has an air core so µr = 1. The other values given are: The number of turns N = 200 The area of the coil A = 3.14 cm 2 = 3.14 × 10 -4 m 2 The length of the coil l = 25 cm = 25 × 10 -2 m Substituting these values into Equation 3.with µr = 1: L = µr(N 2 × A)/l × 1.26 × 10 -6 = 1 × (200 2 × 3.14 × 10 -4 )/(25 × 10 -2 ) × 1.26 × 10 -6 = 6.33 × 10 -5 = 66.3 µH Problem 2d This inductor has an air core so µr = 1. The other values given are: The number of turns N = 20 The area of the coil A = 3.14 cm 2 = 3.14 × 10 -4 m 2 The length of the coil l = 50 cm = 50 × 10 -2 m Substituting these values into Equation with µr = 1: ENG1021 Electronic Principles 133 L = µr(N 2 × A)/l × 1.26 × 10 -6 = 1 × (20 2 × 3.14 × 10 -4 )/(50 × 10 -2 ) × 1.26 × 10 -6 = 3.165 × 10 -7 = 0.3165 µH Problem 2e This inductor has an air core so µr = 1. The other values given are: The number of turns N = 20 We are not given the area, but we are given the diameter d = 4 cm, therefore the radius of the coil r = 2 cm and assuming that the coil is circular, the area of the coil A = π.r 2 = 3.14 × 2 2 cm 2 = 12.57 × 10 -4 m 2 The length of the coil l = 50 cm = 50 × 10 -2 m Substituting these values into Equation with µr = 1: L = µr(N 2 × A)/l × 1.26 × 10 -6 = 1 × (20 2 × 12.57 × 10 -4 )/(50 × 10 -2 ) × 1.26 × 10 -6 = 1.267 × 10 -6 = 1.267 µH 4 Mutual inductance When the current in an inductor changes it creates a magnetic flux which may cut across another inductor and so induce a voltage in it. The voltage induced in the second inductor also produces a current which in turn produces a magnetic flux which will also cut across the first inductor and so produce a voltage in that one. This process is called mutual induction, and the effect has to be included in any calculations. The fraction of flux produced by one coil, L 1 , that cuts across another coil, L 2 , is called the coupling and given the letter k. The value of the coupling is defined as: k = flux linkage between L 1 and L 2 /flux produced by L 1 This overall effect is known as mutual inductance, L m . The additional inductance is measured as: 1 2 m L k L L = (4) When two inductors are in series in a circuit, their inductances are added, just like resistances. However, if there is any coupling and therefore any mutual inductance, this has to be included. ENG1021 Electronic Principles 134 There is a complication. If there is no coupling, the inductances are simply added. However, the two inductors or coils could be wound in opposite directions. This means that for a current flowing through the coils in one direction, the magnetic flux produced in one coil may be in the opposite direction to the magnetic flux in the other. If the magnetic field is in the same direction the coils are said to be “series–aiding”. If they produce magnetic fluxes in opposite directions then they are called “series-opposing”. The total inductance of a pair of series coils would be: L T = L 1 + L 2 + L m for series aiding (5a) L T = L 1 + L 2 – L m for series opposing (5b) Finally, inductors in parallel also follow the same pattern as resistors. For two inductors in parallel (without mutual inductance) the total inductance is: 1/L T = 1/L 1 + 1/L 2 (6) Please now attempt Problems 3 and 4 Problem 3 For a 100 µH inductance L 1 and a 200 µH inductance L 2 , calculate: (a) the total inductance L T of L 1 and L 2 in series without mutual coupling; (b) the combined inductance of L 1 and L 2 in parallel without mutual inductance; (c) the L T of L 1 and L 2 series aiding, and series opposing, with 10 µH mutual inductance; (d) the value of the coefficient of coupling k. Solution This is an exercise in combining inductances, with and without mutual inductance. The equations for combinations of inductors, without considering mutual inductance, are the same as for resistors. Problem 3a The total inductance for inductors in series (in the absence of mutual inductance) is then: LT = L1 + L2 Substituting the values given: LT = 100 + 200 µH = 300 µH Problem 3b Inductors in parallel combine in a similar way to resistors in parallel (without mutual inductance). So the total inductance of the given inductors in parallel is: ENG1021 Electronic Principles 135 1/LT = 1/L1 + 1/L2 Or LT = L1 × L2/(L1 + L2) = 100 × 200/(100 + 200) = 20000/300 = 200/3 = 66.7 µH Notice that the equations for series and parallel inductors are “swapped around” forms of the equations for series and parallel capacitors. Problem 3c If mutual coupling exists between inductors then Equation 5 is used. For series aiding: LT = L1 + L2 + 2LM and for series opposing: LT = L1 + L2 - 2LM Substituting the given values for both cases gives: LT = 100 + 200 + 2 × 10 = 320 µH for series aiding and: LT = 100 + 200 - 2 × 10 = 280 µH for series opposing. Problem 3d The mutual coupling k is a measure of the amount of flux in one inductor links another (the coupled) inductor. It can be found in this case by using the Equation 4 for mutual inductance. 1 2 m L k L L = Rearranging this equation for k and substituting our values into the equation gives: 1 2 10 10 0.07 100 200 20000 m L k L L = = = = × 5 Transformers ENG1021 Electronic Principles 136 Transformers are a practical application of mutual inductance. As shown in Figure 2 there is a primary circuit which has a coil and a secondary circuit which has a coil. When the primary coil is connected to an AC voltage source, the transformer produces an AC voltage in the secondary circuit which causes a current to flow through the load resistor. Figure 2 A transformer The size of the voltage in the secondary circuit depends on the number of turns in the secondary coil, Ns, relative to the number of turns in the primary coil, Np. The turns ratio is the number of turns in the primary coil divided by the number of turns in the secondary coil. Turns ratio = Np/Ns (7) For example, if there are 500 turns in the primary coil and 50 turns in the secondary, the turns ratio is 500/50 = 10, so the transformer could be described as a 10:1 transformer. It is usually assumed in transformers that the coupling between the primary and secondary coils is perfect, so has a value of 1. This means that the voltage induced in the secondary coil is created by the same magnetic field that produces the induced voltage in the primary coil. The difference between the secondary and primary voltages, therefore, is in proportion to the number of turns in the coils. This means that: Vp/Vs = Np/Ns (8) The power in the primary circuit is VpIp, and in the secondary circuit is VsIs. If the transformer is 100% efficient, all of that power is transferred, so that: VpIp = VsIs (9) This means that Vp/Vs = Is/Ip ENG1021 Electronic Principles 137 But earlier we showed that Vp/Vs = Np/Ns, which means that: Is/Ip = Np/Ns = turns ratio. (10) If the transformer isn’t 100% efficient, the not all of the power is transmitted. In this case the efficiency would be defined as: % Efficiency = Ps/Pp x 100 (11) The load resistor in the secondary has a value RL. When looking into the circuit from the primary end, the resistance (or more accurately impedance in the general case) appears at a different value, known is the reflected impedance. Figure 3 shows this, where Zp is the reflected impedance and Zs is the secondary impedance such as the load resistor. Figure 3 Reflected impedance By manipulating the equations that we have already it is found that the reflected impedance is: Zp = (Np/Ns) 2 x Zs (12) One of the ways that reflected impedance can be used is for impedance matching. In any circuit the power delivered to the load is maximised if the impedance of the load is equal to the impedance of the source. For example, Figure 4 shows a voltage source which delivers a voltage but has its own internal resistance r of 200 Ω. This could be an amplifier connected to a speaker, for example. The load resistor that it is delivering power to has a resistance of 8 Ω, which means that if you connected the load directly to the source, it would receive 1.85 W of power, as I shall show. ENG1021 Electronic Principles 138 Figure 4 Impedance matching circuit To match the impedance we want to choose a turns ratio such that Zp = 200Ω. We start with the equation for the reflected impedance: Zp = (Np/Ns) 2 x Zs (13) Since it the turns ratio we require we can rearrange the equation: (Np/Ns) 2 = Zp/Zs = 200/8 = 25 Np/Ns = 5 So a turns ratio of 5:1 would give the required value for the reflected impedance. With this value, the power delivered to the primary coil would be: Vp = Vr/(r+Zp) = 100x200/(200+200)= 20000/400 = 50 V So half the supply voltage gets to the primary coil. The power at the primary coil would be: Pp = Vp 2 /R = 50 2 /200 = 2500/200 = 12.5 watts. Assuming 100% efficiency for the transformer, the power reaching the load would be 12.5 watts – a big improvement on 1.85 watts. Alternatively we could show that the secondary voltage is: Vp/Vs = Np/Ns = 5 Vs = Vp/5 = 10 V Is/Ip = Np/Ns = 5 Is = 5Ip Ip = V/(r + Zp ) = 100/(200+200) = 100/400 = 0.25 Amps So Is = 5 x 0.25 = 1.25 Amps. Power across RL = Vs x Is = 10 x 1.25 = 12.5 watts. Problem 4 ENG1021 Electronic Principles 139 A power transformer with a 1:8 turns ratio has 50Hz 120 V across the primary. (a) What is the frequency of the secondary voltage? (b) How much is the secondary voltage? (c) With a load resistance of 10,000 Ω across the secondary, how much is the secondary current? (d) How much is the primary current? Assume 100% efficiency. (Note: The ratio of N p to N s is 1:8.) Solution This problem involves a power transformer and gives us a feel for the current and voltage relationships in the primary and secondary of transformers. Problem 4a This part of the problem is trivial. The alternating voltage across the primary produces a current in the primary windings of the transformer of the same frequency. Since the alternating current in the primary of the transformer produces an alternating flux (which alternates at the same rate as the primary’s current), the voltage induced in the secondary winding (by the coupled alternating flux) must also alternate at the same rate. So, the frequency of the secondary voltage is the same as the primary voltage, that is, 50 Hz. Problem 4b The ratio between the primary and secondary voltages is equal to the turns ratio (assuming that the transformer is 100 % efficient. We are asked for the value of the secondary voltage. We are told that the turns ratio is 1:8. Therefore the voltage ratio must be 1:8. That is: VP/VS = 1/8 Rearranging and substituting in the values gives: VS = VP/1/8 = VP × 8 = 120 × 8 = 960 volts Problem 4c A load resistance of 10,000 Ω is connected across the secondary winding of the transformer. Since we have just calculated the secondary voltage, the secondary current is given by Ohm’s law: IS = VS/R = 960/10,000 = 96 mA Problem 4d The primary current may be found by using the inverse relationship between the primary and secondary current and the turns ratio: IS/IP = turns ratio = 1/8 ENG1021 Electronic Principles 140 Alternatively, it may be found by calculating the power dissipated in the secondary and the knowledge that if the transformer is 100 percent efficient that the same power will be dissipated in the primary. We will do it the second way. Power in primary = PP = IPVP Power in Secondary = PS = ISVS That is: IPVP = ISVS Rearranging and substituting in our values gives the primary current IP = ISVS/VS = 96 × 960/120 = 768 mA Problem 5 (a) A transformer delivers 400 W out with 500 W in. Calculate the efficiency in percent. (b) A transformer with an 80 percent efficiency delivers 400 W total secondary power. Calculate the primary power. Solution Real transformers are not 100% efficient. Problem 5a Efficiency, generally, relates how much power we can get out of a device compared to how much power we put in. In the case of this transformer example we are told that if we put 500 watts in to it we can only get 400 watts out of it, therefore its efficiency is: power in/power out × 100% = 400/500 × 100% = 80% Problem 5b We know from part (a) that an 80% efficient transformer that delivers 400 watts takes 500 watts, but let’s do it the hard way using the equation for efficiency. Efficiency = power in/power out × 100% We rearrange the equation to leave power out (the secondary power) on the right-hand side of the equation and substitute our values in: power out = 100 × power in/Efficiency = 100 × 400/80 = 500 W ENG1021 Electronic Principles 141 Problem 6 An amplifier can be modelled as an AC voltage source of V G = 100 V, and an internal resistance r i of 500 Ω. If the secondary load resistance is R L = 4 Ω what turns ratio N p /N s will provide the maximum transfer of power from the amplifier to R L ? Solution This is a problem in the use of impedance matching. As stated in Grob the maximum power transfer between a voltage generator and a resistive load occurs when the source resistance is equal to the load resistance. We are given a voltage generator with a 500 Ω resistance and a 4 Ω load. We can solve the matching problem by either, making the load look like 500 - to the generator or making the generator look like 4 Ω to the load. We will do it the first way, but you are encouraged to try the second way and convince yourself that you arrive at the same result. So we want the load to look like 500 Ω, by Ohm’s law this means that we want the relationship between the primary voltage and the primary current in the matching transformer. VP/IP = RP = 500 Ω We know that the turns ratio: NP/NS = VP/VS So rearranging for VP gives: VP = NP VS /NS For current the turns ratio: NP/NS = IS/IP and rearranging for IP gives: IP = NS IS/NP Now if we substitute these expressions for VP and IP we get: VP/IP = (N P /N S )V S /(N S /N P )I S = (NP/NS) 2 × VS/IS = RP = 500 The secondary voltages and currents are also given by Ohm’s law: VS/IS = RS = 4 Ω ENG1021 Electronic Principles 142 Therefore: 500 = (NP/NS) 2 ¶ × 4 The turns ratio is: NP/NS = 500/4 = 125 = 11.18 : 1 6 Stored energy and practical problems The energy that is stored in a coil is stored in the magnetic field. The amount can be calculated using the following equation: Energy = ½ LI 2 (14) For example, a 10 H inductor with a 3 A current stores ½ x 10 x 3 2 = 45 Joules. Now attempt Problem 7. Problem 7 Calculate the energy in joules stored in the magnetic field of a 60 mH inductor with a current of 90 mA. Solution Remember the equation for the energy stored in a inductor is: E = ½ LI 2 The solution to this problem is a simple matter of substitution of the given values into the equation. E = ½ LI 2 = ½ × 60 × 10 -3 × 90 × 10 -3 = 2.7 mJ A word of caution. The energy in an inductor is released by removal of the current through it as (by Lenz’s law) a back emf (voltage) is produced to oppose the change in current. So if an inductor is connected across a small battery and then disconnected, the change in current with respect to time, dI/dt, is very large and the voltage produced as the energy is released will be very high. Motor vehicle ignition systems and “switched mode” power supplies work on this principle; generating tens of thousands of volts from low voltage batteries. 7 Further reading For further reading you may want to look at Fundamentals of Electrical Engineering and Electronics. The relevant sections are “DC” and the sub-section called “Inductors””, where all parts could be read. ENG1021 Electronic Principles 143 8 Where next? You are advised to study the learning package “Inductors and Reactance” next. ENG1021 Electronic Principles 144 ENG1021 Electronic Principles Learning Package 9 Inductance and Reactance ENG1021 Electronic Principles 145 Inductance 1 Do you know all this already? If in doubt, please attempt the self assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Introduction Inductors, like capacitors, have reactance and produce phase shifts between current and voltage in AC circuits. We also use phasor diagrams to add currents and voltages in RL circuits. However, in many senses inductors can be considered as ”opposites” to capacitors - their phase shift is opposite to that of a capacitor and their reactance increases with frequency rather than decreases. You will see that the voltage and current properties for capacitors ”swap places” in inductors. We can therefore use what we have learnt about capacitors to assist in our understanding of inductors and find a quick route through inductors and inductive properties. 3 Inductive reactance Just as with capacitors, all inductors have reactance, X L , in an AC circuits. The reactance is measured in ohms, and is frequency dependent. The equation for reactance is: X L = 2πfL (1) For example, an inductor with an inductance of 2.65 H at a frequency of 60 Hz has a reactance of: X L = 2πfL = 2 x 3.142 x 60 x 2.65 = 1000 ohms Similarly, if we know the value of the reactance in ohms and the frequency we can find the inductance. L = X L /2πfL (2) Reactance is dealt with in the same way as resistance when it comes to series and parallel circuits. In series reactances are added, whereas in parallel it the reciprocal of reactance gets added. Series: X L = X L1 + X L2 + X L3 … etc (3) ENG1021 Electronic Principles 146 Parallel: 1/X L = 1/X L1 + 1/X L2 + 1/X L3 … etc (4) Finally, one of the main differences between reactance and resistance is that the voltage and current are not in phase. In the case of inductive reactance the voltage leads the current, as shown in Figure 1. -300 -200 -100 0 100 200 300 0 0.02 0.04 0.06 0.08 time/s v o l t a g e / V c u r r e n t / I Figure 1 Voltage and current in a capacitor (voltage is dashed) This means that in a purely inductive reactance the voltage and current are both sine waves, but the current lags behind the voltage by 90 degrees or π/2 radians. They can be written as: v = Vsin(2πft) (5) i = Isin(2πft - π/2) (6) In these equations v and i are the instantaneous values of voltage and current, and V and I are the amplitudes. This is the correct form of these equations as the term 2πft is a measure of the angular frequency times time, so it is saying that with a frequency of f, 2π radians are swept out every 1/f seconds. When we talk about relative phase, we usually use degrees. We know that 2π radians are equivalent to 360 degrees, and that 90 degrees would be π/2 radians. So when the phase is -90 degrees, we should show it as -π/2 radians, so that both terms in the sin function are in radians. However, it is very common to find the phase in degrees, and the equations look like: v = Vsin(2πft) (7) i = Isin(2πft – 90) (8) ENG1021 Electronic Principles 147 Now attempt Problems 1, 2 and 3. Problem 1 Calculate the XL of a 0.5 H inductance at 100, 200 and 1000 Hz. Solution This problem involves the substitution of the inductor and frequency values into the equation for inductive reactance: XL = 2πfL The problem asks for the reactance of a 0.5 H inductor at three different frequencies. When f = 100 Hz XL = 2π × 100 × 0.5 = π × 100 = 314 Ω When f = 200 Hz XL = 2π × 200 × 0.5 = π × 200 = 628 Ω When f = 100 Hz XL = 2π × 100 × 0.5 = π × 1000 = 3140 Ω = 3.14 kΩ In Problems 2 and 3 we consider inductors in parallel and series. These problems are very similar to their capacitive equivalents, but we will attempt them to ensure that we understand that similar rules apply to inductors as they do to resistors and capacitors. Problem 2 A 1000 Ω X L1 and a 4000 Ω X L2 are in series across 10 V 60 Hz source. Draw the schematic diagram and calculate the following: (a) total X L ; (b) current in X L1 and X L2 ; (c) voltage across X L1 and across X L2 ; (d) L 1 and L 2 . Solution Problem 2a The circuit diagram is shown in Figure 2. ENG1021 Electronic Principles 148 Figure 2 The circuit diagram for Problem 2 The total reactance is: XT = XL1 + XL2 Substituting the given values: XT = 1000 + 4000 = 5000 Ω Problem 2b The current through XL1 and XL2 is the same since they are in series. By Ohm’s law the current is: I = VT/XT = 10/5000 = 2 mA Problem 2c The current through the two inductors produces a voltage drop across the inductors. The voltage across XL1 is: V1 = IXL1 = 2 × 10 -3 × 1000 = 2 V The voltage across XL2 is: V2 = IXL2 = 2 × 10 -3 × 4000 = 8 V Problem 2d The inductance is found by rearranging XL = 2πfL to give: ENG1021 Electronic Principles 149 L = XL/2πf and substituting our values into it. For L1: L1 = 1000/(2π × 60) = 2.65H For L2: L1 = 4000/(2π × 60) = 10.61H Problem 3 The same 1000 Ω X L1 and 4000 Ω X L2 are in parallel across the 10 V 60 Hz source. Draw the schematic diagram and calculate the following: (a) branch currents in X L1 and X L2 ; (b) total current in the generator; (c) voltage across X L1 and X L2 ; (d) inductance of L 1 and L 2 . Solution Figure 3 The circuit diagram for Problem 3 The circuit diagram is shown in Figure 3. The supply voltage of 10 V appears across both inductors since it is a parallel circuit. The branch currents are given by Ohm’s law: IL1 = V/XL1 = 10/1000 = 10mA And: IL2 = V/XL2 = 10/4000 = 2.5mA The total current in the generator is the sum of these currents: IT = I1 + I2 = 10 + 2.5 = 12.5 mA ENG1021 Electronic Principles 150 The voltage across both inductors is the same and is equal to the supply voltage of 10V. The inductance of L1 and L2 as calculated in Problem 2 is 2.65 H and 10.61 H respectively. 4 Inductive circuits You have seen most of these concepts before in the RC circuit module. We will now have a look at circuits containing a resistor and an inductor in either series or in parallel. Figure 4 shows both cases, and the associated “phasor triangle”. In Figure 4a), a resistor is in series with an inductor. We know from earlier discussions that in a series circuit the current is the same through each component, so the current flowing through the inductor must be the same as the current flowing through the resistor, I. We also know that the current through the inductor lags the voltage across the inductor by 90 degrees. Figure 4 a) series circuit; b) parallel circuit If the reactance of the inductor is X L , then by Ohm’s law the voltage across the inductor is V L = I X L . Similarly, the voltage across the resistor is Vr = IR. We can represent these two voltages using a phasor triangle, where the horizontal arrow shows the voltage across the resistor which is in phase with the current through the resistor. The vertical arrow shows the voltage across the inductor which is 90 degrees ahead of the current through the inductor. ENG1021 Electronic Principles 151 In this phasor triangle, the resistive voltage is at 0 degrees, and the inductive voltage is at +90 degrees. We can either say that the current lags the voltage in an inductor by 90 degrees, or that the voltage leads the current by 90 degrees. It amounts to the same thing. Now we know that the supply voltage should equal the sum of the resistive voltage and the inductive voltage. But because they are out of phase, when represented in a phasor diagram they point in different directions. We therefore have to use Pythagoras’s theorem to find the sum. This is shown as the hypotenuse of the triangle. The sum is therefore: 2 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) r L L L L L E V V E IR IX E I R I X E I R X E I R X E IZ = + = + = + = + = + = In the last line, the term Z is the total impedance of the circuit. Impedance is the combination of resistance and reactance, and is also measured in ohms. You can see that to find the value of Z you have to take the square root of the resistance squared plus the reactance squared. The other value that we can calculate is the relative phase, θ, between the voltage and the current. In Figure 4a) this is shown as the angle between the supply voltage, E, and the resistive voltage Vr. This is because Vr is in phase with the current, therefore the angle between these two voltages is the same as the angle between the voltage and the current. From trigonometry we can see that: L L IX X Tan IR R θ = = (9) So the phase is the ratio of the reactance and the resistance. Since the inductive reactance is pointing upwards, it is positive. You have to bear this in mind when calculating the phase. Using the values in the circuit of Figure 4, the reactance is: X L = 2πfL = 2x 3.142 x 40 x 100 x 10 -3 = 25.13 Ω ≈ 25 Ω. This means that the total impedance is: ENG1021 Electronic Principles 152 2 2 2 2 100 25 10000 625 1625 40.3 L Z R X Z = + = + = + = = Ω The phase is: 25 0.25 100 14 L o X Tan R θ θ = = = = Now for parallel circuits. As we know from previous Learning Packages the voltage across each parallel branch is the same. So in the parallel circuit of Figure 4b) the voltage across the resistor and the inductor is equal to E. The current in the resistive branch is therefore Ir = E/R and is in phase with the voltage. The current in the inductive branch is I L = E/X L and is 90 degrees behind the voltage. If the resistive current is drawn horizontally, then the capacitive current lags by 90 degrees, as in the phasor triangle of Figure 4b). The total current must equal the sum of the two individual branch currents. As they are phasor values with direction as well as size, we use the triangle again and Pythagoras’s theorem. This gives the total current as: 1 1 1 1 1 r L L L L E E I I I R X I E R X I E R X Z = + = + | | = + | \ ¹ | | = + = | \ ¹ So this is the same as any parallel circuit, where the reciprocal of the total impedance equals the sum of the reciprocal of the resistance and the reciprocal of the reactance. Similarly, from the phasor triangle of Figure 4b) it can be seen that: L r L L L I E E Tan I X R E R R Tan X E X θ θ = − = − ÷ = − × = − (10) Using the values in Figure 4, we can find the impedance and relative phase. ENG1021 Electronic Principles 153 1 1 1 1 1 125 100 25 2500 2500 20 125 L Z R X Z = + = + = = = Ω 100 4 25 76 L o R Tan X θ θ = − = = = − You should now have a good idea about the phase relationship between the current and voltage in a capacitive circuit and be able to construct a phasor triangle (often called a phasor diagram). As far as phase angle and phasor diagrams are concerned, remember that the current and voltage relationships are “swapped around” so that a sine wave current in an inductor lags the voltage by 90 o , while a sine wave voltage in an inductor lags the current by 90 o . The simple mnemonic “civil” may help you remember this: Figure 5 “CIVIL” One new idea that I want to introduce in this Learning Package is the concept of “Q” or quality of an inductor. We only have “Q” for inductors because it is difficult to make a “good” inductor, that is, without some significant resistance. We do not have an equivalent problem with capacitors. The Q of a coil is defined as: Q = X L /r i = 2πfL/r i (11) In this equation L is the inductance of the coil and r i is the resistance of the coil. For example, a coil with a reactance of 500 ohm has an internal resistance of 5 ohms. Its quality is then: Q = 500/5 = 100 Now attempt the Problems 4, 5 and 6. Problem 4 Draw the schematic diagram of a circuit with X L and R in series across a 100 V source. Calculate Z T , I, IR, IX L , and θ for these values: (a) 100 Ω R, 1 Ω X L ; (v) 1 Ω R, 100 Ω X L ; (c) 50 Ω R, 50 Ω X L . ENG1021 Electronic Principles 154 Solution Whenever there is a need to examine different phase angles, it is worthwhile drawing a phasor diagram. It will help you to deal with the concept of vector addition (adding of phasors of different angles) more easily. The phasor diagram for inductors is very similar to the phasor diagram for capacitors, except that the voltage leads the current for inductors. The phasor diagram is shown in Figure 7. The circuit diagram is shown in Figure 6. Problem 4a The problem is to find ZT. From the phasor diagram: (IZT ) 2 = (IR) 2 + (IXL) 2 Figure 6 The circuit diagram for Problem 4 We can cancel the current term I to leave: ZT 2 = R 2 + XL 2 Substituting our values of R = 100 Ω and XL = 1 Ω gives: ZT 2 = R 2 + XL 2 = 100 2 + 1 2 = 10001 Taking the square roots: ZT = 10001 = 100.004 Ω Note that this is very close to 100 Ω. The current is given by: I = V/ZT = 100/100.004 ≈ 1 A IR = 1 × 100 = 100 volts ENG1021 Electronic Principles 155 IXL = 1 × 1 = 1 volt Figure 7 The phasor diagram for Problem 4 The phase angle θ is given by: tan -1 θ = IXL/IR = 1/100, θ = 0.057 o ≈ 0 o This low value of phase angle tells us that the impedance ZT is nearly purely resistive. Problem 4b The problem is to find ZT. From the same phasor diagram (IZT ) 2 = (IR) 2 + (IXL) 2 We can cancel the current term I to leave ZT 2 = R 2 + XL 2 Substituting our values of R = 1 Ω and XL = 100 Ω gives: ZT 2 = R 2 + XL 2 = 1 2 + 100 2 = 10001 Taking the square roots: ZT = 10001 = 100.004 Ω Note that this is very close to 100 Ω. The current is given by: I = V/ZT = 100/100.004 ≈ 1 A IXL = 1 × 100 = 100 volt IR = 1 × 1 = 1 volts The phase angle θ is given by ENG1021 Electronic Principles 156 tan -1 θ = IXL/IR = 100/1, θ = 89.4 o ≈ 90 o This high value of phase angle tells us that the impedance ZT is nearly purely inductive. Problem 4c The problem is to find ZT. From the same phasor diagram: (IZT ) 2 = (IR) 2 + (IXL) 2 We can cancel the current term I to leave: ZT 2 = R 2 + XL 2 Substituting our values of R = 50 Ω and XL = 50 Ω in gives: ZT 2 = R 2 + XL 2 = 50 2 + 50 2 = 5000 Taking the square roots: ZT = 5000 = 70.7 Ω The current is given by: I = V/ZT = 100/70.7 = 1.414 A IR = 1.414 × 35.36 = 1 volts IXL = 1.414 × 50 = 35.36 volts The phase angle θ is given by tan -1 θ = IXL/IR = 50/50, θ = 90 o Problem 5 A 200 Ω R is in series with L across a 141 V 60 Hz generator V T . The V R is 100V. Find L. (Hint: V L 2 = V T 2 – V R 2 .) Solution Again, I would suggest that the first step in solving this problem is to draw a circuit diagram and phasor diagram. Then we can see the relationships between the voltages and currents in the circuit. They are shown in Figure 1 To find L we must first find XL, the phasor diagram is helpful here. We know that the VR phasor and the VL phasor must add up (by vector addition) to the total ENG1021 Electronic Principles 157 voltage or supply voltage phasor VT. If we find VL, we can find the current I from the voltage in the resistor, then we can find XL by Ohm’s law. Examination of the phasor diagram shows that VT is the vector addition of VL and VR. That is VT 2 = VL 2 + VR 2 Figure 8 The circuit and phasor diagrams for Problem 5 Rearranging for VL and taking square roots gives: 2 2 R T L V V V − = Substituting in our values gives: VL = 1002 - 1412 = 10000 - 19881 = 9881 = 99.4 volts The same current, I, flows in all part of the series circuit. The current in the inductor is then equal to the current flowing in the resistor. I = VR/R Substituting our values in gives I = 100/200 = 0.5 By Ohms law XL = V l /I so we substitute in our values for I and VL to give: XL = 99.4/0.5 = 199Ω ENG1021 Electronic Principles 158 Now XL = 2πfL. Rearranging for L and substituting in our values gives: L = XL/2πf = 199/(2 × 3.14 × 60) = 199/376.8 = 0.528 H Problem 6 A 350 µH L has a Q of 35 at 1.5 MHz. calculate the internal resistance R i of the coil. Solution This problem introduces “Q”, the quality of the coil (or inductor). The more resistance the coil has, the less it behaves as a pure inductor, and the lower its Q will be. Q = XL/Ri Note that since XL is dependant on frequency (for any given L) that Q must also be dependant on frequency. In other words we cannot determine or state a value of Q without knowing or stating the frequency. We are given a value for the frequency in this problem. Rearranging the equation for Ri: Ri = XL/Q XL = 2πfL, so we can substitute this expression for XL in the equation for Ri and then enter our values to gives: Ri = 2πfL/Q = 2 × 3.14 × 1.5 × 10 6 × 350 × 10 -6 /35 = 30 × 3.14 = 94.2Ω 7 Further reading For further reading you may want to look at Fundamentals of Electrical Engineering and Electronics. The relevant sections are “AC”, sub-section “Reactance and impedance – inductive” there is relevant information. Have a look at all the parts, but be warned. This book uses both complex notation (or j notation) and polar notation rather than using a phasor diagram. 8 Where next? You are advised to study the learning package “Time constants and LCR circuits” next. ENG1021 Electronic Principles 159 ENG1021 Electronic Principles Learning Package 10 Time constants and LCR circuits ENG1021 Electronic Principles 160 Time constants and LCR circuits 1 Do you know all this already? If in doubt, please attempt the self assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Introduction The first section of this Learning Package takes a closer look at the time dependant behaviour of capacitors and inductors, in particular the time taken for the voltage to rise and fall in a capacitor (current in an inductor) and the shape of the rise and fall curves. The second section introduces the combinations of inductors, capacitors, and resistors together in series and parallel AC circuits. Two very significant points are made here: • Capacitive and inductive reactance cancel out. • The real power consumed in any reactive circuit differs from the apparent power, which is the current drawn from the power source multiplied by the voltage of the power source. 3 Study Guide 3.1 Time constants So far we’ve considered circuits to be either DC with a fixed voltage supply like a battery, or AC with a sinusoidal voltage supply as you get from a generator. What we haven’t discussed are the effects that are seen when a voltage is switched in and out of a circuit. When this happens we get what are known as transient effects. ENG1021 Electronic Principles 161 Figure 1 Transient response when the switch is closed For example, in the circuit shown in Figure 1, when the switch is open, no current flows, so there is no voltage across the coil. When the switch closes, suddenly there is a voltage across the resistor and the coil. The current starts to increase as the coil sets up a magnetic field. Initially the coil produces a “back emf” which is a voltage that opposes the voltage that is producing the magnetic field, and hence the current is small. As the field collapses, the back emf falls and the current increases until it reaches a steady value. In this final steady state, all of the voltage is dropped across the resistor and the coil is effectively a short circuit. The graph in Figure 1 shows the value of the current. At this point I only want you to look at the shape of the curve and not worry too much about the values. The important points are that it rises sharply at the beginning, and flattens off towards the end. In this example I have given it a steady state value of the current, I, as 1 amp. Also, it reaches a value of 0.63 amps after 0.5 seconds. In general, this sort of curve reaches a value that is 63% of the final value after one time constant. For a circuit with a resistor and an inductor, the time constant, T, is equal to: T = L/R (1) Generally, it is assumed that the current reaches its steady state value after 5 time constants. In the diagram of Figure 1, if the battery is 10V, the resistance is 10 ohms and the inductance is 5 H, the time constant is: T = L/R = 5/10 = 0.5 seconds. The steady state current is V/R = 10/10 = 1 amp. For the same circuit as in Figure 1, Figure 2 shows the current when the switch is opened. It is assumed that the circuit is in the steady state before the switch is opened. ENG1021 Electronic Principles 162 0 0.2 0.4 0.6 0.8 1 1.2 0 1 2 3 4 Figure 2 Current as switch opens Initially the current is at 1 amp. When the switch opens the current drops to zero. In this instance the current drops by 63% in one time constant. So, starting at 1 amp, after one time constant the current has dropped to 1 – 0.63 = 0.37 amps. Figure 3 shows a capacitive circuit, or an RC circuit, with a switch. This time the voltage is shown as the switch is closed. Figure 3 RC circuit This time the voltage across the capacitor increases to a steady state value over a period of time. In the steady state the voltage across the capacitor equals the battery voltage and no current flows. Again, the voltage rises to 63% of its steady state value in one time constant. However, the time constant this time is: T = CR (2) As before, when the switch is opened again, the voltage will fall just as in Figure 2. Please attempt Problems 1, 2 and 3. Problem 1 ENG1021 Electronic Principles 163 Calculate the time constants of the following inductive circuits: (a) L is 20 H and R is 400 Ω; (b) L is 20 µH and R is 400 Ω; (c) L is 50 mH and R is 50 Ω; (d) L is 40 µH and R is 2 Ω. Solution Here is a set of simple time constant calculations for inductor, to check that you can use remember the formula and to give you some idea of what sorts of values of L and R components, give you certain time constants. The equation to use is: T = L/R Using this equation the time constants are: Problem 1a L = 20 H, R = 400 Ω T = L/R = 20/400 = 0.05 s Problem 1b L = 20 µH, R = 400 Ω T = L/R = 20 × 10 -6 /400 = 0.05 µs Problem 1c L = 50 mH, R = 50 Ω t =L/R = 50 × 10 -3 /50 = 1 ms Problem 1d L = 40 µH, R = 2 Ω T = L/R = 40 × 10 -6 /2 = 20 µs Problem 2 Calculate the time constants of the following capacitive circuits: (a) C is 0.001 µF and R is 1 MΩ; (b) C is 1 µF and R is 1000 Ω; (c) C is 0.05 µF and R is 250 kΩ; (d) C is 100 pF and R is 10 kΩ. Solution This problem is similar to Problem 1, but for capacitors. The equation to use is T = CR ENG1021 Electronic Principles 164 Using this equation the time constants are calculated as follows. Problem 2a C = 0.001 µF, R = 1 M Ω The M and the µ cancel out when multiplied together therefore: T = CR = 0.001 × 1 = 0.001 s Problem 2b C = 1 µF, R = 1000 Ω T = CR = 1 × 10 -6 × 1000 = 0.001 s Problem 2c C = 0.05 µF, R = 250 k Ω T = CR = 0.05 × 10 -6 × 250 × 10 3 = 0.0125 s Problem 2d C = 100 pF, R = 10 k Ω T = CR = 100 × 10 -12 × 10 × 103 = 1 µs Problem 3 A 100 V source is in series with a 2 MΩ R and a 2 µF C. (a) How much time is required for v c to be 63 V? (b) How much is v c after 20 s? Solution This problem allows us to determine charging times and voltages without the use of extensive calculation. Problem 3a 63 V is 63% of 100 V and we have learnt that an RC circuit charges to 63% of the supply voltage in the time of one time constant. So we calculate the time constant (C = 2 µF, R = 2 M Ω) and this is our answer. T = CR = 2 × 2 = 4 s (The M and the µ cancel out when multiplied together.) Problem 3b ENG1021 Electronic Principles 165 The “rule of thumb” for RC and L/R time constants is that they reach their final voltage (or current for inductors) after 5 times their time constants time. 20 s is 5 times the time constant of 4 s and so the voltage after 20 s is the final voltage (supply voltage), that is 100 V. 3.2 Calculation of charging voltages, effect on square waves If we want to know the value of the rising or falling voltages or currents in circuits then we have to use the equation that defines the way that they change. For a rising value (current in L/R circuits, voltage in CR circuits) the equation is: v = V s (1-e -t/CR ) or (3a) i = I s (1-e -t/L/R ) (3b) In these equations V s and I s are the steady state values, and v and I are the instantaneous values. The “e” term is the exponential, which has a value of 2.718. On a calculator it may appear as “e” or else you use the inverse of the natural logarithm which is usually shown as “ln”. The natural logarithm is the logarithm to the base of “e”. Figure 4 Calculator showing the value if “e” In Figure 4 I put in the value 1, then selected “inverse” and clicked on “ln”. The display shows the value of e 1 = 2.718. ENG1021 Electronic Principles 166 For example, if the value of the inductor is L = 5 H and the resistor is 10 ohms, and the steady state voltage is 10 volts, then: T = L/R = 0.5 seconds i = I s (1-e -t/L/R ) = 10/10 (1 - e -t/0.5 ) = 1 - e -t/0.5 After 0.1 seconds, the current would be: i = I s (1-e -t/L/R ) = 1 - e -t/0.5 = 1 – e -0.1/0.5 = 1 – e -0.2 = 1 – 0.82 = 0.18 amps Similarly, the equations for the change when the switch is open, so the value decreases to zero, is: v = V s (e -t/CR ) or (4a) i = I s (e -t/L/R ) (4b) For example, if the voltage in a RC circuit is 10 V, the resistance is 5 kΩ and the capacitance is 100 µF, what is the value of the voltage after 0.1 seconds when the switch is opened? T = CR = 5 x 1000 x 100 x 10 -6 = 0.5 s v = V s (e -t/CR ) = 10(e -t/0.5 ) = 10(e -0.1/0.5 ) = 10(e -0.2 ) = 10 x 0.82 = 8.2 V Figure 5 Square wave input to an RC circuit Figure 5 shows a square wave as an input to an RC circuit. You can imagine this as a switch being quickly and repeatedly opened and then closed. In this V i R = 10 kΩ C = 0.01 µF +10 V 25 ms 0 V 30 ms V i ENG1021 Electronic Principles 167 instance the switch is closed for 25 ms so that a voltage of 10 V is applied, and then opened for 5 ms so that the voltage drops to zero again. With the values given, the time constant is T = 10 x 10 3 x 0.01 x 10 -6 = 0.1 ms, so the time constant is short or small compared to the times of the waveform. This means that the output voltage will rise to 10 V in around 0.5 ms and later will fall in around 0.5 ms. The output voltage therefore looks like the lower waveform in Figure 5. Even in this waveform I’ve exaggerated the time it takes for the voltage to rise and fall so that you can see these effects. Now attempt Problems 4, 5 and 6. Problem 4 An RC circuit consists of a 0.01 µF capacitor in series with 10 kΩ resistor. The circuit is supplied with a 15 V battery and a switch. What will the value of the voltage across the capacitor be 150 µs after the switch has been closed, assuming that the capacitor is completely discharged before at the start Solution The time constant T is: T = CR = 10 x 10 3 x 0.01 x 10 -6 = 10 -4 s or 0.1 ms or 100 µs. After 150 µs, the voltage will be: v = V s (1-e -t/CR ) = 15 x (1 - e -150/100 ) = 15 x (1 - e -1.5 ) = 15 x 0.223 = 3.345 V Problem 5 A 0.05 µF C is charged to 264 V through a 40 kΩ resistor. How much is the time for v c to charge to 132 V? Solution In this problem we are asked for voltages after charge times which are not either equal to the time constant or five times the time constant, so we need to use the charge and discharge equations (or the time constant graph) to arrive at our answer. The equation for the voltage after t seconds of charge is: v C = V (1 - e -t/RC ) Where V is the maximum voltage to which the capacitor is charged, which in this case is 264 V , e is the base of natural logarithms and is equal to 2.718. Substitute the values we are into the equation: 132 = 264(1 - e -t/40 × 10 3 × 0.05 × 10 -6 ) ENG1021 Electronic Principles 168 Divide both sides of the equation by 264 and then subtract 1 from both sides: -0.5 = -e -t/40 × 10 3 × 0.05 × 10 -6 Multiply both sides by -1: 0.5 = e -t/40 × 10 3 × 0.05 × 10 -6 If we take natural logarithms (“ln” on our calculator) of both sides of the above equation, noting that since e is the base of natural logarithms ln(e x ) = x, we have: ln(0.5) = -t/(40 × 10 3 × 0.05 × 10 -6 ) ln(0.5) = - 0.693. Therefore: -0.693 = -t/(40 × 10 3 × 0.05 × 10 -6 ) Rearranging: t = 0.693 × 40 × 10 3 × 0.05 × 10 -6 = 0.001386 s = 1.386 ms Problem 6 In Figure 6, draw the waveform you would expect to measure across the 0.01 µF capacitor. Indicate the capacitor voltage V c at the beginning and end of each 25 ms pulse interval. Draw the V c waveform in the proper time relationship with respect to V in . (C is initially uncharged.) Figure 6 Circuit for Problem 6 Solution Before we can progress we need to calculate the time constant. The values of R and C 10 kΩ and 0.1 µF, and so the time constant T = RC is 1 ms. This means V i R = 10 kΩ C = 0.1 µF +10 V 25 ms 0 V 35 ms V i ENG1021 Electronic Principles 169 that the voltage will rise to 10 V in 5 ms, which is shorter than the 25 ms of the wide part of the square wave, and then fall in 5 ms which exactly equals the short part of the square wave. The resulting waveform is shown below. Figure 7 Resulting waveform across the capacitor in Figure 6 3.3 LCR circuits We’ve already seen AC circuits with resistances and capacitors (RC circuits) and AC circuits with inductors and resistors (LR circuits). In this section we will look at circuits with all three components in – the so called LCR circuits. When an inductor and a capacitor are in series, the same current flows through them, but the voltage across the inductor leads the current by 90 degrees, whereas the voltage across the capacitor lags the current by 90 degrees. The consequence of this is that the voltage across the inductor is 180 degrees ahead of the voltage across the capacitor, and 180 degrees is equivalent to a change in sign. Thus the total voltage across the inductor and the capacitor is: V tot = V L - V c = IX L – IX c = I(X L – X c ) = IX eq This means that the equivalent reactance to an inductor and a capacitor in series is: X eq = X L – X c (5) Similarly, the reactance of an inductor in parallel with a capacitor is: X eq = X L . X c /( X L – X c ) (6) So for pure reactance, the inductive and capacitive inductance are combined using Equations 5 and 6. When resistance is introduced we have a combination of resistance and reactance, which combine to give a total impedance. If we call the reactance X, which may be inductive, capacitive or both, then a series combination of resistance and reactance would result in an impedance, Z, where the magnitude is: 2 2 Z R X = + (7) And the phase is: ENG1021 Electronic Principles 170 1 tan z X R φ − = (8) For example, a circuit which contains a resistance of 40 ohms, an inductor with a reactance of 60 ohms and a capacitor with a reactance of 90 ohms would produce a total reactance of 60 – 90 = -30 ohms. The impedance would be: 2 2 2 2 40 ( 30) 1600 900 2500 500 Z R X ohms = + = + − = + = = And the phase is: 1 1 1 30 tan tan tan ( 0.75) 37 40 z X R φ − − − − = = = − = − o In all of these circuits, Ohm’s law still applies. So, for example, having found the total impedance of a circuit, if we know the value of the voltage source, then we can calculate the current using: V = IZ (9) If, in the example above, the voltage source had been 100 V AC, then the current would be: I = V/Z = 100/500 = 0.2 A We know that the impedance is capacitive (because of the minus sign) which means that the voltage lags behind the current by 37 degrees. This means that the current leads the voltage by 37 degrees. So if we take the supply voltage as the zero phase reference point, then the current leads this by 37 degrees. So the phase of the current is: φ = +37 degrees Now attempt Problem 7. Problem 7 For the circuit shown in Figure 8, calculate Z T , I, φ Z and φ. ENG1021 Electronic Principles 171 Figure 8 Circuit for Problem 7 Solution The first part of the problem asked for the total impedance ZT. We need to break the circuit into equivalent reactances and resistances. I would start this problem by finding the equivalent reactance of the parallel combination of components in the branches. That is the XC2 XC3 combination and the XL1 XL2 combination. I will call the equivalent combination of XC2 XC3 Xcpar, and the equivalent combination of XL1 XL2 Xlpar Xcpar = 500 × 400/(500 + 400) = 200000/900 = 222.2 Ω Xlpar = 300 × 200/(300 + 200) = 60000/500 = 120 Ω Xlpar is an inductive reactance and will cancel out 120 Ω of the capacitive reactance in the same branch, that is, XC1. The total reactance in the left hand branch (which we can call Xleft is 120 - 400 = -280 Ω. Similarly Xcpar is cancelled out by 100 Ω of XL3 giving the total reactance of the right hand branch Xright to be 100 - 222.2 = -122.2 Ω. Now we need to calculate the parallel impedance of the left and right hand side branches together, luckily they are both capacitive and can be treated as two capacitive reactances in parallel. I will call the equivalent reactance XCT, then: XCT = -280 × -122.2/(-280 - 122.2) = 34216/-402.2 = -85.07 Ω This equivalent capacitor is connected to the voltage source by two resistors in series, so in order to find the total impedance ZT we need to add the values of these resistances to give a total resistance RT. RT = R1 + R2 = 47 + 68 = 115 Ω R 1 =47Ω R 2 =68Ω X c1 =400Ω X c3 =400Ω X c2 =500Ω X L3 =100Ω X L2 =200Ω X L1 =300Ω V T =100V ENG1021 Electronic Principles 172 Then find the total impedance from 2 2 CT T T X R Z + = . So, substituting our values into the equation we get: Ω = = + = + = 143 20462 07 . 85 115 2 2 2 2 CT T T X R Z The current I is now easily found by using Ohm’s law with the total impedance, ZT. I = Vs/ZT = 100/143 = 0.699 A Using Ohm’s law as follows: V = IZ Just as we can think of Ohm’s law as saying that we have to multiply the value I by the value Z to give the value of V , we can also think of Ohm’s law as saying that we have twist the current I through the angle φZ to give the angle of V. That is φZ is the angle of the voltage phasor V with respect to the current phasor, I. Since the circuit phase angle is defined as being the reverse of this, that is, the angle of the current phasor I with respect to the voltage phasor V, φZ = -φ. Figure 9 Phasor diagram showing voltage with respect to current We can now proceed to find φZ by asking, “What is the angle of the voltage applied to impedance ZT with respect to the current in flowing into impedance ZT ?”. The phasor diagram is shown in Figure 9. The current in the resistors is in the same phase as the voltage drop IR across the resistors and this is the current which flows into ZT since the equivalent circuit breaks down into a series circuit. ENG1021 Electronic Principles 173 The voltage across ZT is VT The angle φZ is given by tan -1 IXCT/IRT. The I’s cancel and we get: tan -1 XCT/RT = -85.07/115 = -36.5 degrees Similarly we find φ the circuit phase by asking “What is the angle of the current flowing in impedance ZT with respect to the voltage applied to impedance ZT?”. This can be seen by rotating the phasor diagram as shown in Figure 10. Figure 10 Rotated phasor diagram showing current with respect to the voltage. As we move from the voltage phasor towards the current phasor (or the IR phasor since the voltage in the resistance in phase with the current through it) we move in an anticlockwise or positive direction. The angle can be found as above to be: tan -1 XCT/RT = 85.07/115 = 36.5 degrees 3.3.1 Real Power In an AC circuit with reactance there is a relative phase angle between the voltage and the current. This means that the product of VI does not represent the real power in the circuit since V and I represent the RMS values of the voltage and current which may at any one time be in a different part of the sine wave. In fact the value of VI is termed the “apparent power”. The real power can be calculated using the following equation as it only contains one variable, the current: P = I 2 R (10) Alternatively, if we want to use voltage and current we have to take the phase into account, and can write the real power as: ENG1021 Electronic Principles 174 P = VICosφ (11) The cosine term in the equation is often referred to as the “power factor”. The real power is the apparent power multiplied by the power factor. In series circuits the power factor is: PF = Cosφ = R/Z (12) In parallel circuits the power factor is: PF = cosφ = I R /I T (13) Where I T is the total current and I R is the current flowing through the resistive part of the circuit. Now try Problem 8. Problem 8 In Figure 11, calculate I L , I C , I R , I T , Z EQ , θ I , real power, apparent power, and power factor. Figure 11 Solution This is another problem involving phasors, but this time power factor is introduced. The individual branch currents are all simply given by Ohm’s law, since they are all connected across the 12 volt source. IL = VT/XL = 12/200 = 60 mA IC = VT/XC = 12/100 = 120 mA IR = VT/R = 12/150 = 80 mA The total reactive current (current due to capacitors and inductors) is: R=150Ω X c =100Ω X L =200Ω V T =12Vac ENG1021 Electronic Principles 175 IC - IL = 120 - 60 = 60 mA The capacitive and inductive currents are in opposite phase and therefore subtract from one other. The capacitive current is greater therefore the resultant reactive current is capacitive and will lead the applied voltage VT by 90 0 . Figure 12 The phasor diagram The resistive current is 80 mA which is in phase with the applied voltage VT. We can draw a phasor diagram, see Figure 12. From the diagram: mA I I I I L C R T 100 10000 60 80 ) ( 2 2 2 2 = = + = − + = The equivalent impedance of the circuit is given by Ohm’s law: ZT = VT/IT = 12/(100 × 10 -3 ) = 120 Ω The phase angle between the current in the resistor (or the applied voltage) and the total current IT is given by: φI = tan -1 60/80 = tan -1 0.75 = 36.87 o The angle is positive because we move anticlockwise as we move from the resistive current phasor to the total current phasor. The real power in the circuit is the power dissipated in the resistor: P = I 2 R = (80 × 10 -3 ) 2 .150 = 0.96W The apparent power VI V I = VT IT = 12(100 × 10 -3 ) = 1.2 VAR The power factor is: Realpower/apparentpower = 0.96/1.2 = 0.8 We could have arrived at this figure by taking cos φI (cos 36.87 = 0.8). This will work for reactive circuits. However, there are other electronic circuits in which the ENG1021 Electronic Principles 176 apparent power is not equal to the real power times the cosine of the phase, because phase angle is meaningless. The power factor in these cases is still realpower/apparentpower. Therefore, if you always think of power factor as realpower/apparentpower you will not be “caught out”. 4 Further reading For further reading you may want to look at Fundamentals of Electrical Engineering and Electronics. The relevant sections are “DC” and the sub-section called “RC and L/R time constants””, where all parts could be read. Also, under the section “AC”, sub-section “Reactance and impedance – R, L and C” there is relevant information. Have a look at all the parts, but again be warned. This book uses both complex notation (or j notation) and polar notation rather than using a phasor diagram. 5 Where next? You are advised to study the learning package entitled “Resonance” next. ENG1021 Electronic Principles 177 ENG1021 Electronic Principles Learning Package 11 Resonance ENG1021 Electronic Principles 178 Resonance 1 Do you know all this already? If in doubt, please attempt the self assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Introduction This section introduces electrical resonance and shows that when inductors and capacitors have the same reactance value, resonance occurs in both series and parallel circuits. 2.1 Series and parallel resonance In a circuit with capacitors and inductors we have already seen that the reactances are subtracted, as they are 180 degrees out of phase. Since both capacitive and inductive reactance are functions of frequency, it is sometimes possible to find a frequency at which the capacitive and inductive reactance have the same value and therefore cancel out. This phenomenon is called resonance, and a circuit resonates when the reactance is zero. Assuming there is some resistance in the circuit, at resonance the total impedance would equal the total resistance only. A circuit in which there is a resistor, capacitor and inductor in series would have a resistance R, and a reactance which is XL – Xc. We know that the reactances are: 2 1 2 L c X fL X fC π π = = If these are equal and opposite in sign then: 0 1 2 2 L c L r r X X X X f L f C π π − = = = Rearranging this equation we can find the value of f. ENG1021 Electronic Principles 179 2 2 1 4 1 2 r r f LC f LC π π = = (1) This is the resonant frequency. The impedance at this frequency is just equal to the resistance, so the current is a maximum value, and is in phase with the supply voltage. If we were measuring the voltage across the resistance, then at other frequencies it would be some fraction of the supply voltage. At resonance, it equals the supply voltage, so is the maximum possible value. In an ideal parallel circuit, an inductance would be in parallel with a capacitor. At resonance, the reactances are equal and opposite in sign. This means that the total reactance is: 1 1 1 1 1 1 1 0 1 0 L c L L L L X X X X X X X X = + = + = − = − = = ∞ At resonance, the total reactance is infinite, like an open circuit. So no current flows, and the voltage just equals the supply. In reality every inductor has a small internal resistance, so the impedance would be some very large value. Please now attempt Problems 1, 2 and 3. Problem 1 Calculate f r for a series LC circuit with L = 5 µH and C = 202.64 pF. Solution We are asked to calculate the resonant frequency f r of a series LC circuit when L = 5 µH and C = 202.64 pF. We use the equation: LC f r π 2 1 = Simply substituting our values for L and C gives 12 6 10 64 . 202 10 5 2 1 − − × × × = π r f ENG1021 Electronic Principles 180 15 10 0132 . 1 2 1 − × = π = 1/(2π × 3.18 × 10 -8 ) = 5 MHz Problem 2 Calculate fr for a parallel LC circuit with L = 25.8 µH and C = 500 pF. Solution Here we have L = 25.48 µH and C = 500 pF. This is a parallel circuit, but the above equation still applies. We simply substitute in our values for L and C. 12 6 10 500 10 48 . 25 2 1 − − × × × = π r f 14 10 274 . 1 2 1 − × = π = 1/(2π × 1.129 × 10 -7 ) = 1.41 MHz Although we used Equation 1 directly to solve this problem, it is important to remember that resonance only occurs when XC = XL. This is true of both series and parallel circuits. Problem 3 In Figure 1, calculate the following values at fr: XL, XC, ZT, I, V I , V C and φ Z . ENG1021 Electronic Principles 181 Figure 1 Solution This problem highlights the features of series resonance. In order to calculate XL and XC at the resonant frequency we must find the resonant frequency fr first. LC f r π 2 1 = Substitute our values for L and C into it. 12 6 10 67 . 50 10 20 2 1 − − × × × = π r f 15 10 0134 . 1 2 1 − × = π = 1/(2π × 3.18 × 10 -8 ) = 5 MHz We find XL by using XL = 2πfL and substitution of our values: XL = 2π × 5 × 10 6 × 20 × 10 -6 = 628 Ω At resonance XC will have the same magnitude as X L , but with the opposite sign. Let’s prove that to ourselves. XC = 1/2πfC: XC = 1/(2π × 5 × 10 6 × 50.57 × 10 -12 ) = 628 Ω At resonance the reactive components cancel out, and so ZT = rS = 12.56 Ω. Rs =12.56Ω C=50.67 pF (40-400 pF) V T =1mV L=20µH ENG1021 Electronic Principles 182 Notice the use of rS rather than RS to signify that this resistance is the AC resistance. At high frequencies current flows only near to the surface of a conductor (this is called the skin effect). This means that the whole conductor is not used to carry the current and the AC resistance rS is higher than the dc resistance R S where the whole conductor carries the current. The current I = V T /R T = 1 × 10 -3 /12.56 = 79.6 × 10 -6 = 79.6 µA The voltage across the inductor is: VL = IXL = 79.6 × 10 -6 × 628 = 50 000µV = 50 mV The voltage across the capacitor will be equal and opposite. Let us prove it. The voltage across the capacitor VC = IXC = 79.6 × 10 -6 × -628 = -50 000µV = -50 mV The impedance phase angle is the angle between the resistive component of the impedance and the total impedance, that is φZ = tan -1 X T /R. In a resonant circuit the inductive and capacitive reactances have cancelled out. Therefore, XT = 0, X T /R = 0 and φZ = tan -1 0 = 0 2.2 Q of resonant circuits, tuning, damping and bandwidth In a resonant circuit we define the quality or “figure of merit” of the circuit as the Q magnification factor. It is a measure of the sharpness of the resonance i.e. if we measure the voltage across a resistor at the resonance frequency f r , how quickly does that voltage fall away as the frequency is increased or decreased. Q is defined as follows: Q = X L /r s (2) Remember that at resonance, X L = -X c , so that absolute value of either the inductive or capacitive reactance can be used. For example, if the inductive reactance at resonance is 1500 ohms, and the AC resistance is 10 ohms, then the Q magnification factor is 1500/10 = 150. Although we have defined resonance as a single frequency, fr, at which the reactances of the capacitive and inductive parts of the circuit are equal, it is usual to talk of a band of frequencies centred on the resonance frequency as producing resonance. At frequencies close to the resonant frequency, the circuit still produces a relatively high voltage response. We define the band of frequencies as the bandwidth, and it is centred on the resonance frequency. We define the edge frequency f1 as the frequency below the resonance frequency at which the response is 70.7% of the maximum response, and an edge frequency f2 above the resonance frequency at which the response is 70.7% of the maximum. This is shown in Figure 2. This shows the current for a series resonance circuit with a ENG1021 Electronic Principles 183 capacitor of 1 µF, an inductor of 1 H, a resistance of 10 ohms, and a supply voltage of 10 V. I 0 0.2 0.4 0.6 0.8 1 1.2 0 100 200 300 400 0 0.2 0.4 0.6 0.8 1 1.2 154 156 158 160 162 164 166 Figure 2 (a) Resonance peak in current; (b) detail of the resonance peak The symbol we use to denote the bandwidth is ∆f = f2 – f1. The triangle is the Greek character capital delta and usually means “the difference between”. In this case it means the difference between the edge frequencies f2 and f1. The bandwidth is found to be: r f f Q ∆ = (3) The resonant frequency is 159 Hz. At resonance the current is 1 amp. The Q magnification factor is 100, and the bandwidth is 1.59 Hz. This means that at 0.5 x 1.59 = 0.795 Hz either side of 159 Hz the value of the current has dropped to 0.707 amps, as shown in the detail of Figure 2(b). So f1 = 159 – 0.795= 158.205 Hz and f2 = 159.795 Hz. Now attempt Problems 4, 5, 6, 7, 8 and 9. Problem 4 In Figure 1 again, calculate Q, ∆f, and the edge frequencies f 1 and f 2 . Solution The Q of the circuit is given by : Q = XL/rS Substituting in our values for XL (from Problem 3 above) and rS gives: Q = 628/12.56 = 50 To find the bandwidth we use: ENG1021 Electronic Principles 184 ∆f = fr/Q = 5 × 10 6 /50 = 100 kHz The edge frequencies are situated either side of the resonance frequency; spaced by the bandwidth ∆f. That is f1 = fr - ½∆f and f2 = fr + ½∆f. f1 = 5 × 10 6 - ½ 10 5 = 4.95 MHz f2 = 5 × 10 6 + ½ 10 5 = 5.05 MHz The following problems examine a tank (parallel) resonant circuit and consider bandwidth, Q, and damping. Problem 5 Figure 3 Circuit for problems 5 to 7 Calculate f r in Figure 3. Solution The resonance frequency is given by substitution into the equation: LC f r π 2 1 = L = 100 µH and C = 162.11 pF so: 12 6 10 11 . 162 10 100 2 1 − − × × × = π r f = 1/(2π × 1.27 × 10 -7 ) = 1.25 MHz Problem 6 In Figure 3, calculate the following at fr: X L , X C and Q. rs =7.85Ω C=162.11 pF (100-1000 pF) V A =1Vpp L=100µH ENG1021 Electronic Principles 185 Solution Now we know f r we can calculate X L using XL = 2πfL = 2π × 1.25 × 10 6 × 100 × 10 -6 = 785.4 Ω Similarly we calculate X C using: XC = 1/2πfC = 1/(2π × 1.25 × 10 6 × 162.11 × 10 -12 ) = 785.4 Ω We could have stated that XC = XL at resonance, but doing it the long way checks that we have not made any mistakes. Q is given by the equation: Q = XL/rS = 785.4/7.85 = 100 Problem 7 In Figure 3, calculate ∆f, f i and f 2 . Solution We have a resonance frequency of 1.25 MHz and a Q of 100. As before (Problem 3): ∆f = fr/Q = 1.25 × 10 6 /100 = 12.5kHz The edge frequencies are given by f1 = fr - ½∆f and f2 = fr + ½∆f. f1 = 1.25 × 10 6 – ½ × 12.5 × 10 3 = 1.243750 MHz f2 = 1.25 × 10 6 + ½ × 12.5 × 10 3 = 1.256250 MHz Problem 8 In Figure 4, what value of series resistance R S must be added to double the bandwidth ∆f when C = 56.29 pF? ENG1021 Electronic Principles 186 Figure 4 Circuit for Problem 8 Solution Since Q determines the bandwidth: ∆f = fr/Q To double the bandwidth we must halve Q. Q/2 = XL/2rS The equation shows that a series resistance of 2r S will halve Q and hence double the bandwidth. Therefore we must add another r S = 18.85 Ω to double the bandwidth. Note that we did not have to consider the resonant frequency or the value of C in this calculation. Doubling the series resistance in any series resonant circuit will double the bandwidth. 3 Further reading For further reading you may want to look at Fundamentals of Electrical Engineering and Electronics. The relevant sections are “AC” and the sub-section called “Resonance”, where all parts could be read. Have a look at all the parts, but again be warned. This book uses both complex notation (or j notation) and polar notation rather than using a phasor diagram. 4 Where next? You are advised to study the last learning package entitled “Filters” next. rs =18.85Ω C=56.29 pF (30-300 pF) V T =50µV L=50µH ENG1021 Electronic Principles 187 ENG1021 Electronic Principles Learning Package 12 Filters ENG1021 Electronic Principles 188 Filters 1 Do you know all this already? If in doubt, please attempt the self assessment questions in this Learning Package. If you can answer all of the questions correctly you may omit this section. If not, please read on. 2 Introduction This section introduces the idea that signals can contain components at different frequencies, and that filter circuits can remove some of these components. 2.1 Filtering AC and DC signals As we saw in the previous Learning Package, the reactance of a circuit varies with frequency. This means that at some frequencies the voltage or current in the circuit is larger than at other frequencies. This property can be used to “filter out” certain frequencies or bands of frequencies. Most signals in electronics will consist of a range of frequencies. For example, an audio signal will contain signals in the range of 20 Hz to 20 kHz. Radio uses frequencies in bands such as VHF (very high frequency) is in the range 30 – 300 MHz. In order to let these frequencies through and get rid of any signals at other frequencies you use electronic filter circuits. The most common are: Low pass filter – this allows signals with frequencies from 0 up to some cut-off frequency through, and blocks any signals with frequencies higher than this; High pass filter – this allows all signals with frequencies above the cut-off frequency through and blocks signals with low frequencies; Band-pass filter – this allows signals through with frequencies that lie between a lower and an upper cut-off frequency, and blocks all other signals; Band-stop filter – this blocks signals with frequencies between a lower and an upper cut-off frequency and allows all other signals through. This is sometimes referred to as a “notch” filter. Many signals found in electronics will consist of a DC voltage, or a bias, in addition to an AC signal. Very often we want to remove the DC component, and to do this we either use a transformer or a coupling capacitor as a high-pass filter. Figure 1 shows an input voltage which is a combination of a DC voltage of E2 plus an AC voltage with an amplitude of E1. The coupling capacitor block the ENG1021 Electronic Principles 189 DC voltage from getting through, so that the output voltage is just the AC component. Figure 1 RC coupling circuit We know that the reactance of the capacitor is Xc, and therefore the output voltage is: 2 2 i o c v R v R X = + (1) When the frequency is zero, the reactance of the capacitor is 1/0 = infinity, which is an open circuit. So, from the point of view of a signal with a frequency of zero, which is how we might describe a DC signal, the capacitor is an open circuit and no voltage gets through. If we write the reactance as 1/2πfC then the equation becomes: 2 2 2 2 (2 ) 1 1 2 i i o v R v fRC v fRC R fC π π π = = + ( + ( ¸ ¸ This circuit has a cut-off frequency, fc, which equals 1/2πRC. fc = 1/2πRC (2) At this frequency the equation becomes: ENG1021 Electronic Principles 190 2 2 (1) 1 i i o v v v = = + This is the “half-power” point or -3 dB point that has been mentioned before, where the output has fallen to 70.7% of it’s maximum value. If we use an RC coupled circuit to block a DC component, we would usually choose the values of R and C such that the cut-off frequency is 1/10 th of the frequency of the AC component. This achieves the goal of having the reactance equal to 1/10 th of the resistance at the frequency of the AC component. For example, if the AC component is 100 Hz, then the cut-off frequency would be chosen as 10 Hz. Assuming that a 1 uF capacitor is used, the resistor value should be: fc = 1/2πRC R = 1/2πfcC = 1/(2 x 3.142 x 10 x 0.000001) = 15.915 kΩ = 16 kΩ (approx) Please now attempt Problem 1. Problem 1 Calculate the minimum coupling capacitance C c in series with a 10 kΩ R if the frequency of the applied voltage ranges from (a) 100 Hz to 10 kHz; (b) 15 kHz to 300 kHz. Solution Generally, for coupling capacitors, we require that the reactance X C of the coupling capacitor is less than a tenth of the value of the resistance at the frequencies that we want to allow through. In this problem we are given ranges of frequencies. The capacitor will have its highest reactance at the lower of these frequencies, therefore if we ensure that the reactance of the capacitor is less than a tenth of the resistance at the lower frequency of the range, then the capacitor must have a reactance which is less than a tenth of the input resistance for any of the higher frequencies. Problem 1a Let’s take the first range 100 Hz to 10 KHz we need to find the minimum coupling capacitance. So we make sure that at 100 Hz the reactance of the capacitor C c is less than or equal to a tenth of the input resistance of 10 kΩ. XC = 1/2πfCc ·= 10 kΩ/10 = 1 kΩ Rearranging: 1/2πf ≤ Cc × 10 3 ENG1021 Electronic Principles 191 Dividing both sides by 10 3 and substituting for f gives: 1/(2π × 100 × 10 3 ) = 1.59 µF ≤ Cc So providing that Cc ≥ 1.59 µF it will be a suitable coupling capacitor for this frequency range. Problem 1b Similarly we take the lower end of the frequency range 15 kHz to 300 kHz and make sure that at 15 kHz that the reactance of the capacitor C c is less than or equal to a tenth of the input resistance of 10 kΩ. XC = 1/2πfCc ·= 10 kΩ/10 = 1 kΩ Rearranging: 1/2πf ≤ Cc × 10 3 Dividing both sides by 10 3 and substituting for f gives: 1/(2π × 15 × 10 3 × 10 3 ) = 10.6 nF ≤ Cc So providing that Cc ≥ 10.6 nF it will be a suitable coupling capacitor for this frequency range. 2.2 Filter circuits ENG1021 Electronic Principles 192 Figure 2 (a) RC low-pass (b) RL low-pass (c) RL high-pass (d) RC high-pass The RC coupling is an example of a high pass filter. These sections describe how you can design filter circuits by choosing appropriate values for resistors and capacitors and inductors. Figure 2 shows the simplest examples of low-pass and high-pass filters made out of resistors, inductors and capacitors. In each case the circuit represents a potential divider, where the output voltage is divided between a resistance and a reactance. For all of these circuits the cut-off frequency is given by either: fc = 1/2πRC (3) or fc = R/2πL (4) We can make more sophisticated filters with more components and in particular with the inclusion of amplifiers. However, that goes beyond the scope of this module. In order to build a band-pass or a band-stop filter, you effectively combine a low- pass and a high-pass filter. Figure 3 shows an example of an RC band-pass filter which is effectively a high-pass filter followed by a low-pass filter. Figure 3 An RC band-pass filter A band-pass filter allows signals through which have frequencies that lie between the low cut-off frequency and the high cut-off frequency. It is fairly clear from this circuit that the low cut-off frequency is 1/2πR1C1 an d the high cut-off frequency is 1/2πR2C2. ENG1021 Electronic Principles 193 Figure 4 An RC notch filter An RC band-stop filter could be constructed the same way. However, an alternative is the notch filter shown in Figure 4. It still only uses a combination of resistors and capacitors, but it is a bit more complicated that simply taking a low pass filter followed by a high pass filter. For this circuit the rather than the cut-off frequency, the notch frequency is defined, which is the frequency which you want to remove. This frequency is defined as: f N = 1/4πRC (5) Now attempt Problems 2 and 3. Problem 2 Figure 5(a) Circuit for problem 2(a); (b) Circuit for problem 2(b) Figure 5(c) Circuit for problem 2(c); (d) Circuit for problem 2(d) Calculate the cut-off frequency, fc, for the filter in: (a) The RC low-pass filter in Figure 5(a); (b) the RL low-pass filter in Figure 5(b); R =2.2kΩ C=0.022 µF R =1.5kΩ L=100 R =1kΩ L=30 mH R =1.8kΩ C=0.047 µF ENG1021 Electronic Principles 194 (c) the RC high-pass filter in Figure 5(c); (d) the RL high-pass filter in Figure 5(d). Solution Problem 2a The cut-off frequency is found using the formula: 1 2 c f RC π = Substituting the value for R and C: 3 6 1 1 2 2 2.2 10 0.022 10 3.3 c c f RC f kHz π π − = = × × × × = Problem 2b The cut-off frequency is found using the formula: 2 c R f L π = Substituting the value for R and L: 3 3 1 10 2 2 30 10 5.3 c c R f L f kHz π π − × = = × × = Problem 2c The cut-off frequency is found using the formula: 1 2 c f RC π = Substituting the value for R and C: 3 6 1 1 2 2 1.8 10 0.047 10 1.9 c c f RC f kHz π π − = = × × × × = Problem 2d ENG1021 Electronic Principles 195 The cut-off frequency is found using the formula: 2 c R f L π = Substituting the value for R and L: 3 3 1.5 10 2 2 100 10 2.4 c c R f L f kHz π π − × = = × × = Problem 3 In Figure 6, calculate the notch frequency f N if R = 18kΩ and C = 0.001 µF. Figure 6 Circuit for Problem 3 Solution The notch frequency, f N , is given by the formula: 1 4 N f RC π = Substituting fro R 1 and C 1 gives: V in R 2C 2R C C 2R R L V out ENG1021 Electronic Principles 196 3 6 1 1 4 4 18 10 0.001 10 4.4 N N f RC f kHz π π − = = × × × × = 2.3 Filter circuit gain We often define the gain of a filter circuit as the value of the output voltage divided by the value of the input voltage. For all the filters that we’ve looked at so far, that gain will always be less than 1, because the circuits are effectively voltage dividers. In more complex filters may find amplifiers included. These are called “active filters” as they contian “active” components suchg as transisters of operational amplifiers which have their own power supply, and are therefore “active”. In tese circuits the output voltage can be greater than the input voltage. The gain of the filter is usually stated in decibels. To explain this, we have to define the unit of powwer ratio, the bel. Essentially, if you measure the power of the output, and divide by the poweer of the input, then take logarithms to the base 10, you get the power ratio in bels. N = log(Pout/Pin) in bels It was found tha the bel is actually very large, so it is divided into tenths of a bel, called decibels. So 1 bel = 10 decibels. To state th egain in decibels you multiply the logarithm by 10. The decibel is abbreviated to dB. N = 10log(Pout/Pin) in decibels or dBs (6) Finally, in electronic circuits the power is defined in a number of ways, but one way is as: P = V 2 /R Therefore the power is proportional to the square of the voltage. It is therefore convenient to substitute the voltage into the equation to get: N = 10log(V 2 out/V 2 in) dB However, one of the properties of logarithms is that log (x 2 ) = 2log(x). So the equation becomes: N = 20log(Vout/Vin) dB (7) Problem 4 The input power to an amplifier is 1 watt. Calculate the gain in decibels for the values of the output power shown in the following table: ENG1021 Electronic Principles 197 Input Power Output Power Gain in dB 1 W 2 W 1 W 10 W 1 W 20 W 1 W 100 W 1 W 1 kW 1 W 2 kW Table 1: Table for Problem 4 Solution The gain in dB is given by the following equation: 10log out dB in P N P = If the input power is 1 W and the output power is 2 W, the gain is: 2 10log 3.01 1 dB N dB = = Similarly, when the input power is 1 W and the output power is 10 W, the gain is: 10 10log 10 1 dB N dB = = The complete table looks like this: Input Power Output Power Gain in dB 1 W 2 W 3 1 W 10 W 10 1 W 20 W 13 1 W 100 W 20 ENG1021 Electronic Principles 198 1 W 1 kW 30 1 W 2 kW 33 Table 2: Solution for Problem 4 3 Further reading For further reading you may want to look at Fundamentals of Electrical Engineering and Electronics. The relevant sections are “AC” and the sub-section called “Filters”, where all parts could be read. In addition, in the section called “Semiconductors”, in the sub-section called “Amplifiers and Active Devices”. I don’t want you to read most of this section as it goes beyond this module. However, there is a section called “Decibels” which is worth a read. 4 Where next? That’s the end of the module. Make sure you’ve handed in all of the assignments and then have a well-earned rest!