Student s Book Jowaheer Consulting and Technologies, R Atkins & E van der Merwe

Transcription

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2 N5 Industrial Electronics Student s Book Jowaheer Consulting and Technologies, R Atkins & E van der Merwe

3 Industrial Electronics N5 Student s Book Jowaheer Consulting and Technologies, R Atkins & E van der Merwe, 2015 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, photocopying, recording, or otherwise, without the prior written permission of the copyright holder or in accordance with the provisions of the Copyright Act, 1978 [as amended]. Any person who does any unauthorised act in relation to this publication may be liable for criminal prosecution and civil claims for damages. First published in 2015 by Troupant Publishers [Pty] Ltd PO Box 4532 Northcliff 2115 Distributed by Macmillan South Africa [Pty] Ltd ISBN: Web PDF ISBN: It is illegal to photocopy any page of this book without written permission from the publishers. While every effort has been made to ensure the information published in this work is accurate, the authors, editors, publisher and printers take no responsibility for any loss or damage suffered by any person as a result of reliance upon the information contained herein. The publisher respectfully advises readers to obtain professional advice concerning the content. While every effort has been made to trace the copyright holders and obtain copyright permission from them, in some cases this has proved impossible due to logistic and time constraints. Any copyright holder who becomes aware of infringement on our side is invited to contact the publisher. To order any of these books, contact Macmillan Customer Services at: Tel: (011) Fax: (011) customerservices@macmillan.co.za

4 Contents Module 1: AC theory 1 Unit 1.1: Response of RLC circuits to sinusoidal voltages...3 Unit 1.2: Resonance...13 Unit 1.3: Response of RC and RL circuits to non-sinusoidal voltages...24 Module 2: Power supplies 39 Unit 2.1: Rectifier circuits and filters...40 Unit 2.2: Voltage multiplication...66 Unit 2.3: Voltage regulators...67 Module 3: Transistor amplifiers 85 Unit 3.1: Basic transistor circuits and operating principles...86 Unit 3.2: Calculation of component values for a common emitter amplifier...91 Unit 3.3: Distortion in transistor amplifiers...95 Unit 3.4: The transistor equivalent circuit and h-parameters...99 Unit 3.5: Using h-parameters to calculate overall circuit performance Module 4: Operational amplifiers 117 Unit 4.1: Basics of operational amplifiers Unit 4.2: Active filters Module 5: Integrated circuits 144 Unit 5.1: Integrated circuits Module 6: Transducers 152 Unit 6.1: Resistive transducers Unit 6.2: Inductive, voltage and current transducers Unit 6.3: Photo electric transducers Module 7: Electronic phase control 180 Unit 7.1: Silicon controlled rectifier (SCR) Unit 7.2: Diac, triac and quadrac Unit 7.3: Control systems Module 8: Test equipment 193 Unit 8.1: Test equipment Module 9: Oscillators 203 Unit 9.1: Basic principles of oscillators Unit 9.2: Types of oscillators Unit 9.3: Multivibrators Unit 9.4: The 555 timer Answers 231 Glossary 233

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6 MODULE AC theory 1 Overview Electrical quantities such as voltage and current are usually regarded as either direct or alternating. Direct current (DC) is constant in amplitude and direction. We get direct current from a battery. Alternating current (AC) or voltage changes in amplitude and direction, usually in a sinusoidal pattern. Some of the most common applications of AC voltages are found in: Power distribution: In power distribution, the power is transmitted at hundreds of kilovolts over long distances. It is brought down in stages by transformers as it approaches built-up areas, until it reaches the familiar 230 V RMS 50 Hz supply at the wall plugs in buildings such as homes and offices. AC voltages in power systems are always sine waves of a constant amplitude and frequency. In most countries, AC voltages are 50 Hz. Electronic instrumentation: An important use of AC voltages is in switch mode power supplies (SMPS). The input DC voltage is converted to a very high frequency AC signal (typically > 100 khz), which can then be transformed up to a much higher voltage. Because of the high frequency, the transformer can be made very small, typically using a ferrite core. A wide range of voltages can be achieved very efficiently, and the SMPS can be designed to tolerate a wide input voltage range without a loss of efficiency. They do, however, generate quite a lot of interference. This can affect radios and other sensitive equipment. Communications: Communications is the field of study in which the AC voltages are called signals and are used to convey information. Starting with the telegraph, electrical communications has evolved through the telephone, radio, TV and we now have cellphones and the Internet. In a different way from when they are used for power applications, AC voltages, though usually sine waves, can be modified when they are used for communications because they need to convey information. These can be AC voltages whose amplitude, frequency or phase is changed (modulated) in order to convey information. These give rise to the terms AM (amplitude modulation), FM (frequency modulation) and PM (phase modulation). Often, we encounter voltages that are basically DC but they have a smaller AC component added. This is called a ripple (seen on the output of rectifier circuits). In contrast we can encounter voltages that are basically AC, but have a DC component. For example, a square wave signal used in a digital circuit typically has a low value of 0 V and a high value of 5 V, with no in-between values, but an average of 2,5 V. Figure 1.1 shows some AC waveforms commonly encountered in electronics. sinusoidal: a quantity whose waveform has a sine or cosine function ferrite: a magnetic material made mainly of iron oxide and ceramic, used in highfrequency electrical components signal: an electrical impulse or wave that carries information ripple: a small AC voltage superimposed on a larger DC voltage 1

7 Overview (continued) + one period crest/ peak sinusoidal + time trough sawtooth + square Figure 1.1: Common AC waveforms Units in this module: Unit 1.1: Response of RLC circuits to sinusoidal voltages Unit 1.2: Resonance Unit 1.3: Response of RC and RL circuits to non-sinusoidal voltages Think about it 1. A capacitor and inductor are connected in parallel to an AC voltage source operating at the resonant frequency of the LC circuit. The inductor is suddenly disconnected. What effect does this have on the magnitude and phase of the current flowing out of the voltage source? 2. Our radio receiver cannot tune into a particular station because there is another, stronger station whose frequency is just above the station we want to listen to. What is the problem with our receiver? 3. A DSTV decoder is connected to the dish antenna by a coaxial cable. This cable carries the encoded TV signal down to the decoder, and also conveys DC power up to the receiver module at the dish. How is it possible for the same cable to carry DC in one direction and a wideband digital TV signal in the other direction at the same time without interfering with each other? 2

8 Unit 1.1: Response of RLC circuits to sinusoidal voltages Introduction This unit deals with the way circuits made up of resistors, inductors and capacitors behave or operate when connected to sinusoidal AC voltage sources. An AC voltage is any voltage that changes direction and amplitude repeatedly, at a rate which is called the frequency of the signal. We will only discuss sine or cosine waves in this unit, because they are used in many applications, and are easily handled mathematically. Behaviour of passive components, R, L and C The passive components all limit the amount of AC current that they allow to flow through a circuit connected to an AC voltage, but they behave quite differently. In a resistor, the current is in phase with the applied voltage. However, in an inductor the current lags the voltage by 90 and in a capacitor it leads the voltage by 90. Figure 1.2 shows an AC voltage supply connected in parallel to a resistor, an inductor and a capacitor. The corresponding phasor diagram is shown in Figure 1.3. passive: unable to supply energy or amplify a signal in a circuit; excludes batteries, transistors I T I R I L I C I C V I T R L C I C + I L I R V I L Figure 1.2: Currents in a parallel RLC circuit Figure 1.3: Phasor diagram of parallel RLC circuit The magnitudes of I R, I C and I L are determined by their impedances: IR = V A in phase with V R IL = V XLA lagging V by 90 IC = V A leading V by 90 X C and the reactances X L and X C are frequency dependent, given by: XL = 2πf L Ω XC = 1 2πf C Ω where: f is the frequency in hertz. L is the inductance in henrys. C is the capacitance in farads. 3

9 When using reactances in calculations Note that X L and X C are reactances (in ohms) without any phase information. If they are used in calculations of impedance, voltage or current in a circuit, the complex values must be used. Then, they can be expressed as: XL = j2πf L Ω = 2πf L 90 Ω XC = j 2πf C Ω = 1 2πf C 90 Ω In Figure 1.3 you can see that: I R is in phase with V. I L lags V by 90. I C leads V by 90. I L and I C are 180 out of phase with each other. Their resultant is the phasor I C + I L, which leads V by 90 in this example. If the magnitude (absolute value) of I C is less than the magnitude of I L, then the phasor I C + I L will lag V by 90. The resultant of I C + I L and I R is I T, the total current in the circuit. Think about it We add phasors in a similar way to how we add vectors. However, voltage and current are scalar quantities. Current is just the rate of flow of electric charge. Voltage is a measure of the electric potential at a point in a circuit. It takes one joule to raise the potential of one coulomb by one volt. Because volts and amps do not have direction, they cannot be vectors. imaginary: imaginary part of a complex number, for example, b in the number a + jb real: real part of a complex number, for example, a in the number a + jb rectangular form: representation of a complex number as real + imaginary part, for example, a + jb polar form: representation of a complex number as a magnitude and phase angle, for example, r θ Complex notation In theory, we can determine most currents and voltages in AC circuits by drawing phasor diagrams. But, there is a more simple method: we can express all impedances in terms of complex numbers. Complex arithmetic is used to calculate various quantities in AC networks, as with the phasor graphical method. But, we only apply it to circuits driven by sinusoidal voltage or current sources with a constant frequency. Complex notation allows us to represent quantities that are normally defined in terms of magnitude and phase by a complex quantity, as shown in Figure 1.4. The diagram (which looks like a graph) has a vertical axis, called the imaginary axis and a horizontal axis, called the real axis. On this plane we can plot a point (a; b) representing a complex number a + jb in rectangular form. This also indicates the end-point of a phasor drawn from the origin, O, which then is at an angle, θ to the horizontal. This represents the same number in polar form. 4

10 imaginary axis b (a; b) r O a real axis Figure 1.4: The complex plane with a number represented in rectangular and polar form From the diagram in Figure 1.4, we can derive the following equations: r = a 2 + b2 (modulus or magnitude) θ = tan 1 b a (angle to real axis) a = r cos θ (real part) b = r sin θ (imaginary part) Therefore we can write: a + jb = r θ and use the above formulae to convert from rectangular to polar form, and from polar to rectangular form. Did you know? Most calculators and computer programs use arctan instead of tan 1 for the inverse tangent function. This is to avoid confusion with the reciprocal, which can also be expressed using 1 as an exponent, for example: 1 x = x 1 Moreover, most calculators have special functions to do conversion from polar to rectangular and vice versa, which can make calculations easier. Calculations If we have two complex numbers: Z 1 = a 1 + jb 1 Z 2 = a 2 + jb 2 then we are able to carry out arithmetic operations as follows: Addition and subtraction: Complex numbers must first be converted to rectangular form if they are in polar form. Real parts and imaginary parts are added or subtracted separately: Z 1 + Z2 = ( a 1 + jb1 ) + ( a 2 + jb2 ) = ( a 1 + a2 ) + j( b 1 + b2 ) Note that a or b can be negative. If a complex number is subtracted from another, the normal rules of algebra apply, for example: ( 2 + j4 ) ( 1 j3 ) = ( 2 1 ) + j( ) = 1 + j7 5

11 Multiplication and division: Complex numbers must first be converted to polar form if they are in rectangular form. Multiplication: Multiply the moduli (or value) and add the angles, for example: Z 1 Z2 = (r 1 θ 1 ) (r2 θ 2 ) = ( r 1 r2 )(θ 1 + θ 2 ) Division: Divide the moduli and subtract the angles, for example: Z 1 Z2 = (r 1 θ 1 ) (r2 θ 2 ) = ( r 1 r2 )(θ 1 θ 2 ) Example 1.1 Find 3 + j4 1 j2 + 3 j8 Solution 3 + j4 1 j2 + 3 j8 = ( tan 1 ( 4 3 ) ) = (5 53,13 ) (2,24 63,43 ) + 3 j8 = 5 53, , j8 2, tan 1 ( -2 1 ) + 3 j8 = 2,23 cos 116,56 + j2,23 sin 116, j8 = 1,00 + j1, j8 = 2,00 j6,01 RLC circuits Now that we can perform complex arithmetic operations, we can solve any series or parallel circuit if it is driven by sinusoidal voltage source. Example 1.2 In the parallel RL circuit shown in Figure 1.5, calculate the following in polar form: 1. The circuit impedance, Z T. 2. The total current, I T. 3. The current in the inductor, I L. 4. The current in the resistor, I R. V T R L 25 Ω 100 mh 50 Hz 230 V Figure 1.5: Parallel RL circuit 6

12 Example 1.2 (continued) Solution 1. First calculate the inductive reactance, X L : XL = 2π = 31,42 Ω Then calculate the impedance of R in parallel with Z L : R = 25 + j0 Ω = 250 Ω Z L = 0 + j31,42 Ω = 31,42 90 Ω Zparl = Z 1 Z2 Z 1 + Z2 ZT = (25 0 ) 31, j j31,42 2. IT = V T Z T = 25 31, j31,42 = 785, ,15 51,5 = 19,56 38,5 Ω = ,56 38,5 3. IL = V T Z L = 11,76 38,5 A = ,42 90 = 7,32 90 A 4. IR = = 9,20 0 A Note that Z T can also be calculated from I L and I R, adding them to get I T, and then calculating Z T from Z T = V T I T. 7

13 Example The circuit of a series RLC circuit is shown in Figure 1.6. Calculate the impedance of the circuit and express it in rectangular and polar form if the frequency (f) of V T is: a) 15 Hz. b) 50 Hz. 2. Calculate the current in polar form if f = 50 Hz. 3. Draw a phasor diagram showing V T, I T, V L, V R and V C when f = 50 Hz. R L C 75 Ω 200 mh 150 µf V T 250 V Figure 1.6: Series RLC circuit Solution 1. Since the components are in series, their impedances are added in rectangular form. a) Z T = R + j( X L XC ) = 75 + j ( 2π π ) = 75 + j( 18,85 70,74 ) = 75 j51, ( 51,892 ) 1 tan ( 51,89 75 ) = = 91,20 34,68 Ω b) ZT = R + j ( X L XC ) = 75 + j ( 2π π ) = 75 + j( 62,83 21,22 ) = 75 + j41,61 Ω (41,612 ) 1 tan ( 41,61 75 ) = 85,7729,02 Ω = 8

14 Example 1.3 (continued) 2. Use the polar form of Z T so that the required division can be calculated: IT = V T Z T = V/85,77 29,02 Ω = 2,91 29,02 A 3. Calculate the magnitudes and phase angles of the required phasors: From 2. above: IT = 2,91 29,02 A V L = IT X L = 2,91 29,02 2π = 182,84 29, = 182,84 60,98 V VR = I T R = 2,91 29, = 218,25 29, = 218,25 29,02 V VC = I T X C = 2,91 29, π = 2,91 29,02 21,22 90 = 61,75 119,02 V V L = V V C = V T = V R = V I T = 2,91 29 A Figure 1.7: Phasor diagram 9

15 Example Draw the circuit diagram including component values of the network shown in Figure 1.8. Terminals A and B are connected to an AC voltage source of 100 V with a frequency of 50 Hz. 15 j15 A 20 + j25 B Figure 1.8: Impedance network 10 + j10 2. Calculate the impedance (at 50 Hz) between A and B in complex polar form. 3. Draw a phasor diagram of the current flowing from A to B. Solution 1. Name the components (R 1, L 1, etc.) as they are calculated. Z 1 = 20 + j25 Ω, which consists of R 1 = 20 Ω in series with an inductor L 1 whose reactance at 50 Hz is 25 Ω. Calculate L 1 : XL = 2πf L L1 = X L 2πf = 25 = 79,6 mh 2 π 50 Z 2 = 15 j15 Ω, which consists of R 2 = 15 Ω in series with a capacitor C 1 whose reactance is 15 Ω at 50 Hz. Calculate C 1 : XC = 1 2πf C Ω C1 = 1 2πf XC = 1 = 212 μf 2 π Z 3 = 10 + j10 Ω, which consists of R 3 = 10 Ω in series with an inductor, L 2 whose reactance at 50 Hz is also 10 Ω. Calculate L 2 : XL = 2πf L L2 = X L 2πf = 10 = 31,8 mh 2 π 50 The circuit diagram is shown in Figure 1.9. C 1 R 2 R 1 L µf 15 Ω 20 Ω 79,6 mh L 2 R 3 31,8 mh 10 Ω V T Figure 1.9: Circuit diagram derived from impedances 10

16 Example 1.4 (continued) 2. Let Z T be the impedance between A and B at 50 Hz. Z T = Z 1 in series with Z 2 and Z 3 connected in parallel. ZT = Z 1 + Z 2 Z3 Z 2 + Z3 = 20 + j25 + ( 15 j15 ) ( 10 + j10 ) 15 j j10 ( tan 1 [15/15] ) ( = 20 + j tan 1 [10/10] ) 25 j5 = 20 + j25 + (21,21 45 ) (14,14 45 ) ( 5)2 tan 1 ( 5/25 ) = 20 + j ,50 11,31 = 20 + j ( 11,31 ) 25,50 = 20 + j ,76 11,31 = 20 + j ,53 + j2,31 = 31,53 + j27,31 = , ,31 tan ( 27,31 31,53 ) = 41,71 40,90 Ω 3. Before we can draw the phasor diagram (Figure 1.10), we must first calculate I T : IT = V T Z T = ,71 40,90 = 2,40 40,90 A V T = V Figure 1:10: Phasor diagram I T = 2,40 40,90 A 11

17 Assessment activity Complete the following table of complex numbers: Rectangular form: z = a + jb Polar form: z = r θ 2. Calculate: a b r è a) (2 + j3)(3 j4) b) ( 5 j8 ) 1 + j6 1 + j3 c) (leave answer in rectangular form) d) (5 + j10)2 3. Refer to the circuit in Figure 1.11 and calculate: a) The total impedance in polar form. b) The current value in each of the four components. C µf V T L 1 0,15 H R 2 20 V, 50 Hz R 1 5 Ω 80 Ω Figure 1.11: Circuit diagram 12

18 Assessment activity 1.1 (continued) 4. The phasor diagram of a series RLC circuit has the following values: VT = V VC = 64,9 160,2 V VL = 160,1 17,8 V VR = 30,6 72,2 V IT = 2,0 72,2 V a) Draw the phasor diagram. b) Calculate the following, given that the frequency is 50 Hz: (i) The value of L. (ii) The value of C. (iii) The value of R. Unit 1.2: Resonance Introduction Resonance is the tendency of a system to oscillate or vibrate with greater amplitude at a particular frequency than at other frequencies. This is sometimes called sensitivity to that particular frequency. Selectivity is the degree to which a resonant circuit can respond to one frequency out of many. Resonance occurs when the system is able to transfer energy from one form to another. For example, in a pendulum the potential energy at the end of a swing is transferred into kinetic energy as it swings through the vertical position. In a resonant circuit the energy is transferred from the coil to the capacitor and back again as the circuit oscillates. In a series or parallel RLC circuit, resonance will occur when the capacitive and inductive reactances are equal. The reactances X L (inductive) and X C (capacitive) are dependent on frequency as we can see in the following formulae: sensitivity: the degree to which a resonant circuit responds more to one particular frequency than to all other frequencies; also related to selectivity selectivity: the degree to which a resonant circuit can respond to one frequency out of many XL = 2πf L Ω XC = 1 2πf C Ω In the graph shown in Figure 1.12, the reactance values for L = 1 mh and C = 1 μf are plotted. Resonance will occur at a frequency of about 5 khz where the curves intersect. To determine the resonant frequency by calculation, we must find the frequency for which X L = X C : XL = X C 2π f L = 1 2π f C (2π f)2 = 1 L C 2π f = 1 L C f = 1 2π LC 13

19 60 50 X L 40 reactance in ohms f r X c frequency in Hz Figure 1.12: Graph of capacitive and inductive reactance versus frequency The above expression shows that only the values of the inductor and capacitor determine the resonant frequency. There are two ways that an inductor and a capacitor can be connected to form a resonant circuit, namely in series (Figure 1.13) and in parallel (Figure 1.14). The resonant frequency is the same for both circuits, but the voltage, current and impedance are different. A L C L C A Figure 1.13: Series resonant circuit B B Figure 1.14: Parallel resonant circuit parasitic: a circuit element (for example, resistance) that is possessed by a circuit, but which is not desirable because it reduces the effectiveness of the circuit Series resonant circuits Look at Figure No circuit is completely without some parasitic resistance (unless it uses super-conductor wires) and at the resonant frequency X L = X C, so we can write: Zseries = R + j ( X L XC ) = R where R is the resistance of the wiring including the inductor. 14

20 From this information, we can list the following characteristics of a series resonant circuit (assuming a voltage V T applied across terminals A and B results in a current I T in the circuit): f r = 1 2π LC X L = X C = X V L = V C = I T X and is a maximum at resonance. Note that V L and V C can be greater than V T when X > R. Z series = R and is a minimum at resonance. I T = V T and is a maximum at resonance. R I T and V T are in phase, i.e. θ = 0. A graph of Z series versus frequency is shown in Figure The impedance is capacitive below resonance as indicated by 90 < θ < 0. It becomes resistive at resonance (θ = 0 ) and becomes inductive at frequencies above resonance (0 < θ < 90 ) º impedance (Z series ) magnitude of impedance resonant frequency phase angle of impedance 60º 20º 20º phase angle ( ) 20 60º 0 100º 1,80 1,84 1,88 1,92 1,96 2,00 2,04 2,08 2,12 2,16 2,20 frequency (KHz) Figure 1.15: Impedance magnitude and phase angle versus frequency for a series resonant circuit Parallel resonant circuits Look at Figure As with the series resonant circuit, we assume that there is some resistance in the circuit, mainly in the inductor, which is a coil of wire. The impedance at resonance is not as easily calculated as for the series resonant circuit. It is referred to as Z D, the dynamic impedance (also sometimes called Z R, dynamic resistance). Z D is given by: ZD = L CR where: Z D is the equivalent resistance in ohms, of the parallel LC circuit. C is the capacitance in farads. L is the inductance in henrys. R is the parasitic circuit resistance in ohms. dynamic impedance: also called dynamic resistance; the resistance of a parallel resonant circuit at its resonant frequency 15

21 impedance (Z parallel ) 66 kω 60 kω 54 kω 48 kω 42 kω 36 kω 30 kω 24 kω 18 kω 12 kω 6 kω The following are characteristics of a parallel resonant circuit (assuming a voltage V T applied across terminals A and B results in a current I T in the circuit): f r = 1 2π LC X L = X C = X I L = IC = V T X Z parallel is maximum at resonance, and is equal to Z D = L CR. I T = V T and is a minimum at resonance. Z D I T and V T are in phase, i.e. θ = 0. A graph of Z parallel versus frequency is shown in Figure The impedance is inductive below resonance, as indicated by 0 < θ < 90. It becomes resistive at resonance (θ = 0 ) and becomes capacitive at frequencies above resonance ( 90 < θ < 0 ). Limitation on accuracy of resonant frequency formula f r = 1 2π LC is correct in a perfect circuit without resistance or where R < X L 10. However, if the inductor resistance is R > X L, then use the following formula: 10 fr = 1 2π 1 LC R 2 L 2 magnitude of impedance phase angle of impedance 0 kω 1,80 1,84 1,88 1,92 100º 20º 60º resonant frequency 80º 100º 1,96 2,00 2,04 2,08 2,12 2,16 2,20 frequency (KHz) Figure 1.16: Impedance magnitude and phase angle versus frequency for a parallel resonant circuit 80º 60º 40º 20º 0º 40º phase angle ( ) 16

22 Did you know? There are at least three names for the parallel resonant circuit (as shown in Figure 1.17): Parallel resonant circuit. Tuned circuit. Tank circuit. I + L V C Figure 1.17: Parallel LC circuit Tuned circuit and tank circuit are often used in descriptions of radio circuits, where these circuits are frequently used as high frequency band-pass filters. tuned circuit: a parallel resonant circuit which resonates freely at its natural (resonant) frequency, and not at any other frequency; this circuit is used for tuning a radio to a signal at a particular frequency tank circuit: another term for tuned circuit band-pass filter: a circuit which allows only a limited continuous range of frequencies to pass through from its input to its output Table 1.1: Characteristics of series and parallel resonant circuits Type of resonance f < f r f = fr = 1 2π LC Series Parallel Z: capacitive X C > X L θ < 0 Z: inductive X L > X C θ > 0 Z = R Z: minimum θ = 0 X C = X L = X I T : maximum IT = V T R V L = V C = I T X I T and V T are in phase, i.e. θ = 0 Z = ZD = L CR Z: maximum θ = 0 X C = X L = X I T : minimum IT = V T Z D IL = I C = VT X I T and V T are in phase, i.e. θ = 0 f > f r Z: inductive X L > X C θ > 0 Z: capacitive X C > X L θ > 0 17

23 Example 1.5 Draw phasor diagrams for the following resonant circuits: 1. An AC current I T = 1 A at the resonant frequency passes from A to B through the circuit as shown in Figure This results in voltages V L and V C across the inductor and capacitor. A L C B Figure 1.18: Ideal series resonant circuit 2. An AC voltage of 1,0 V at the resonant frequency is connected across terminals A and B, as shown in Figure This results in currents I L and I C flowing in the inductor and capacitor. A L C B Figure 1.19: Ideal parallel resonant circuit Solution 1. V L = X L 1,0 90 V 2. I C = 1/X C 90 A I T = 1,0 0 A V T = 1,0 0 V bandwidth: the range of frequencies which a circuit allows to pass through from its input to its output; it is defined as the range of frequencies where the output voltage is above 0,7071 the maximum voltage V C = X C 1,0 90 V Figure 1.20: Phasors of series a) and parallel b) resonant circuits Q of a resonant circuit I L = 1/X L 90 A Q stands for quality factor. It is a measure of how well the circuit can operate as a band-pass filter. Q is directly dependent on the amount of resistance in the circuit, especially the inductor which is a coil of copper wire. A high value of Q corresponds to a narrow bandwidth, which is desirable in many applications. 18

24 Calculation of Q In these circuits the resistor represents all the losses in the circuit. Note that the formula for the Q of a parallel resonant circuit depends on how the resistor R is connected. Series resonant circuit R L A Figure 1.21: Q of a series RLC resonant circuit C B Q = X L R = X C R = 1 R L C Parallel resonant circuit There are two ways that the resistor can be connected: R in series with the inductor, as shown in Figure The same formulae as for the series circuit may be used. R in parallel with L and C, as shown in Figure A A C R L Q = X L R = X C R = 1 R L R C L C Q = R X L = R X C B Figure 1.22: Q of parallel RLC circuit, R in series with L B Figure 1.23: Q of parallel RLC circuit, R in parallel with L and C Think about it Some textbooks incorrectly use the same formula for the series RLC circuit (for example, Q = X L ) as for the circuit in Figure 1.23, where R is in parallel with L and C. R This cannot be correct, because a perfect coil has zero resistance, so the R in parallel would be a short-circuit across L. Relation between Q and bandwidth Figure 1.24 shows the voltage across a parallel resonant circuit as a function of frequency. f 1 and f 2 are known as the half-power or 3 db points, and are the frequencies at which the output voltage has dropped to 0,7071 of its mid-band value. This means that it is 3 db below the maximum value. 19

25 V max (0 db) 0,7071 V max ( 3 db) f r bandwidth f 1 f 2 Figure 1.24: Bandwidth of a resonant circuit About decibels (db) The decibel is a logarithmic unit used to express the ratio between two values of a physical quantity. This means that it represents one value divided by another. The basic formula for a power ratio is: db value = 10 log10 ( P 1 P 2 ) where P 1 and P 2 are two power values. Because power is proportional to the square of the voltage, the equation for a voltage ratio becomes: db value = 20 log10 ( V 1 V 2 ) Other forms of decibel Sometimes values of physical quantities can be expressed directly in a decibel related quantity. For example, RF power is often expressed in dbm, where the m means that the db value expresses the ratio of the RF power to 1 mw. Decibels are also often used to express losses in signal amplitude. Examples 1. Convert 1,0 W to dbm. db value = 10 log10 ( P 1 P 2 ) = 10 log 10 ( 1, ) = 30 dbm 20

26 2. A filter introduces a loss of 25% of the input signal voltage. Calculate the voltage loss at the output in db. (The negative value shows that the output is smaller than the input. The loss is therefore +2,50 db.) db value = 20 log10 ( V out V in ) = 20 log10 ( 0,75 Vin V in ) = 2,50 db Think about it Why is the half-power point at the frequency where V = 0,7071 V max? If the graph shown in Figure 1.24 is the frequency response of an amplifier connected to a load of resistance R, then the output power is given by: P = V 2 R Therefore, we can calculate the voltage at the half-power points as follows: P half-power P max = V half-power 2 2 V max = 1 2 V half-power V max = 1 2 = 0,7071 Vhalf-power = 0,7071 V max Formulae for Q and bandwidth BW = f2 f 1 Q = X L R = X C R = 1 R L C Q = f r BW = f r f 2 f1 where: BW is the bandwidth in Hz. f 2 is the higher half-power frequency in Hz. f 1 is the lower half-power frequency in Hz. Did you know? In radio receivers, a process called frequency mixing is used to convert a received signal to a fixed intermediate frequency (IF) which is much lower in frequency. The IF filter has a much lower bandwidth than the resonant circuit; this circuit operates at the original radio carrier frequency. This allows improved selectivity to be achieved. 21

27 Example 1.6 The following components are connected in series to an AC voltage source: R = 22 Ω L = 0,25 H C = 63 μf Calculate the Q, bandwidth, upper and lower half-power frequencies. Solution Calculate the resonant frequency, then Q and the other required values. f r = 1 2π LC = 1 2π 0, = 40,1 Hz X Q = L R 2π 40,1 0,25 = 22 = 2,86 f BW = r Q = 40,1 2,86 = 14,0 Hz Lower half-power frequency: f1 = f r BW 2 = 40,1 14,0 2 = 33,1 Hz Upper half-power frequency: f2 = f r + BW 2 = 40,1 + 14,0 2 = 47,1 Hz Example 1.7 A voltage source supplies an AC voltage of 20 volts to a parallel resonant circuit consisting of an inductor of 56 mh with an internal resistance of 15 Ω and a capacitor of value 150 nf. The frequency of the AC voltage is the same as the resonant frequency of the circuit. Calculate: 1. Resonant frequency, f r. 2. Bandwidth of circuit. 3. Lower and upper half-power frequencies, f 1 and f The supply current at resonance, I t. 22

28 Example 1.7 (continued) Solution 1. fr = 1 2π LC 3. Then calculate f 1 and f 2 : = 1 2π = 1,737 khz 2. First calculate Q: Q = X L R = 2π = 40,7 Q = fr BW BW = f r Q = ,7 = 42,7 Hz f1 = f r BW 2 42,7 = = Hz f2 = f r + BW 2 = ,7 2 = Hz 4. Calculate dynamic impedance: ZD = L CR = = Ω Then calculate the supply current (V t = supply voltage): I t = V t Z D = = 0,804 ma Assessment activity In a series or parallel RLC circuit, what is the condition that determines the resonant frequency? 2. Supply the missing words in the sentences below: a) As the frequency increases, the inductive reactance. b) At a lower frequency, the capacitive reactance is. c) When the inductive and reactances are equal, the circuit is. d) In a series resonant circuit, the current is at resonance. e) In a series resonant circuit, X C > X L, if the supply frequency is than the resonant frequency. f) In a parallel resonant circuit, the current in the inductor and capacitor are and at resonance. g) At resonance the of a series RLC circuit is equal to and is at its value. h) At resonance the impedance of a parallel resonant circuit is called the. 23

29 Assessment activity 1.2 (continued) i) The impedances of both kinds of resonant circuits are at resonance. j) At resonance the supply current I T is with the, V T for both series and parallel resonant circuits. 3. a) Draw the amplitude versus frequency curve of a parallel resonant circuit, and show the following features: (i) Resonant frequency. (ii) Half-power frequencies. (iii) Voltage at the frequencies identified in (i) and (ii). (iv) Bandwidth. b) Give the range of frequencies for which X L > X C. c) What is the impedance of the circuit at the resonant frequency (give term and formula)? 4. Consider a series resonant circuit consisting of a resistor of R ohms, an inductor of L henrys, and a capacitor of C farads. a) What is the impedance of the circuit at resonance? Explain. b) If the resonant frequency is f r, for which range of frequencies is X L < X C? c) In a series resonant circuit, the voltages developed across the inductor and capacitor are and are degrees out of phase with each other. 5. Make a sketch graph of X L and X C versus frequency and mark the resonant frequency f r. a) Does this graph apply to series or parallel resonant circuits, or both? b) Name the units of the vertical and horizontal axes. 6. The following components are connected in series to an AC voltage source: R = 27 Ω L = 0,39 H C = 56 μf Calculate the Q, bandwidth and upper and lower half-power frequencies. 7. A voltage source supplies an AC voltage of 20 volts to a parallel resonant circuit consisting of an inductor of 12 mh with an internal resistance of 5 Ω and a capacitor of value 12 μf. The frequency of the AC voltage is the same as the resonant frequency of the circuit. Calculate: a) Resonant frequency, f r. b) Bandwidth of the circuit. c) Lower and upper half-power frequencies, f 1 and f 2. d) The supply current at resonance, I t. Unit 1.3: Response of RC and RL circuits to non-sinusoidal voltages Introduction In this unit we will consider how RC and RL circuits respond to: Step functions, as shown in Figure Square waves, as shown in Figure

30 V V V1 V1 V0 Figure 1.25: Step function, for example, closing a switch t V0 Figure 1.26: Square wave function t There are many applications where RL and RC circuits are used for specific purposes, mainly in the areas of: Signal processing. Timers. Communication. Both inductors and capacitors can store energy, but in different ways: Inductors store energy in the magnetic field that forms in and around them when current flows through them. Inductors consist of coils of wire, sometimes wound around a core or rod of magnetic material which increases their inductance. An example of a coil inductor is shown in Figure Inductance is not easy to calculate from the physical dimensions of the coil. Furthermore, inductors are much more bulky and expensive than capacitors. For this reason, capacitors are preferred for most signal processing or timing purposes. Figure 1.27: A coil inductor dielectric: or dielectric material; an insulator with a high dielectric constant free space: a vacuum permittivity: the ability of a substance to store electrical energy in an electric field dielectric constant: the ratio of the permittivity of a substance to the permittivity of free space; the higher the dielectric constant, the higher the capacitance if the material is used in a capacitor Capacitors store energy in the electric field formed between their plates when a voltage is applied between their terminals. A capacitor consists of two conducting plates or flat sheets separated by a small gap, as shown in Figure The space between the plates is made of an insulator or dielectric, such as ceramic or polyester. All insulators including free space (or a vacuum) have a property called permittivity (for which the Greek letter ε is used). This is a measure of how much energy the substance can store when it contains an electric field. A more common term is dielectric constant (k). This is the ratio of the permittivity of the substance to the permittivity of free space. ε k = substance ε 0 where ε 0 is the permittivity of free space. conductive plates A d dielectric Figure 1:28: A plate capacitor 25

31 Calculation of capacitance The general formula for a capacitor which has one pair of plates is: C = A ε d where: C is the capacitance (in farads). A is the area of each plate in m 2. ε is the permittivity of the material (in farads/metre). d is the separation distance in metres. If N is the number of plate pairs in the capacitor, and A is the area of each plate, then the formula becomes: C = N A ε d From the above equations we can see that capacity is increased when: The area of the plates is increased. The number of plates is increased. The separation distance is reduced. The permittivity or dielectric constant of the dielectric material is increased. Properties of RC and RL circuits Energy stored in inductors and capacitors Both inductors and capacitors can store energy, and it takes time for the energy to flow into them. The stored energy is given by the following equations: Inductor: E stored = 1 LI2 2 where: L is the inductance in henrys. I is the current. Capacitor: E stored = 1 CV2 2 where: C is the capacitance in farads. V is the voltage. time constant: a measure of how quickly a system responds to change; the time taken for 63,2% of the eventual change to take place Time constants The time constant of a circuit is an important property. It will be used in many calculations involved in timers, oscillators and signal processing circuits. The time constant is a measure of how long it takes for energy to flow into the component, whether it is L or C. RC circuit capacitor charging and discharging We will look in detail at the RC circuit first. Consider the circuit shown in Figure

32 switch V + R C Figure 1.29: Capacitor charging circuit The graph in Figure 1.30 shows the capacitor voltage as a function of time, the switch closing at Time = 0. In this example, V = 1 V, C = 1 F and R = 1 Ω. The time constant is the time taken for the voltage to reach 63,2% of the final value. In this example it is 1 second. The capacitor voltage rises to 0,632 V in 1 second and finally reaches 1 V. 1,2 1 V max capacitor charging voltage 0,8 0,6 0,4 time const. 63,2% V max 0, time Figure 1.30: Charging of a capacitor through a resistor RL circuit current build-up in inductor The RL circuit is shown in Figure The current build-up graph is the same shape as the capacitor voltage build-up graph, except that the final current in the inductor is V R A. switch + V1 V R L Figure 1.31: Inductor current build-up when switch is closed 27

33 Formulae for time constant Capacitor/resistor: τ = RC where τ is in seconds, R in ohms, C in farads. Inductor/resistor: τ = L R where τ is in seconds, R in ohms, L in henrys. Example Calculate the energy stored in: a) A capacitor of 100 μf charged to a voltage of 100 V DC. b) An inductor of 100 mh carrying a current of 10 A DC. 2. Calculate i) the time constant and ii) final values of voltage and current in the following circuits: a) A capacitor of 1,0 μf is charged through a resistor of 150 kω from a voltage source which supplies a step function of 6,0 V DC. b) An inductor of 120 mh is connected through a resistor of 1,0 kω to a voltage generator which gives a step function of 10 V DC. Solution 1. a) E stored = 1 2 CV 2 = 0, = 0,50 J b) Estored = 1 2 LI 2 = 0, = 5,0 J 2. a) (i) τ = RC = , = 0,15 sec (ii) V final = 6,0 V I final = 0 b) (i) τ = L R = = 120 μ sec (ii) V final = 0 I final = V R = = 10 ma 28

34 Pulse shaping and filtering circuits The C-R differentiator circuit C output input + R Figure 1.32: C-R differentiator circuit As a capacitor charges up, its voltage rises at a rate proportional to the current. This means that the current is a measure of the rate of change of the voltage across the capacitor. Q = C V = I t I t = C V Now differentiate both sides: I = C dv dt The current through the capacitor produces a voltage equal to dv across the resistor. dt Note that this circuit gives only an approximate value for dv unless the output dt voltage across the resistor is very small relative to the voltage developed across the capacitor. Figure 1.33 shows the input square wave, the voltage across the capacitor and the output voltage across the resistor. Notice that the output gives spikes corresponding to the rising and falling edges of the input signal. These spikes approximate the derivative of V in, and the DC level of the input is lost, because it is being blocked by the capacitor. The peak-to-peak voltage across the capacitor is the same as the input maximum voltage swing, V m. Spikes of amplitude V m are generated at the output. These spikes are positive when the input voltage rises, and negative when it falls. 29

35 V in V m 0 t V cap V m 0 t V m V out 0 t V m Figure 1.33: Waveforms of the C-R differentiator with a square wave input The C-R differentiator is characterised by a short time constant. It is frequently used for the following purposes: High-pass filter. To produce short pulses required as inputs into digital circuits, such as counters and timers. To remove the DC component from a signal. The R-C integrator circuit This circuit is shown in Figure As with the C-R differentiator, the output of the R-C integrator is only approximate unless the output voltage is relatively small compared with the input, or the time constant of the RC circuit is much larger than the period of the input signal. In R Figure 1.35, the input is a square output wave which is centred around zero volts. The output voltage rises linearly when the input is positive, and falls at the same rate when the + C input goes negative. input Figure 1.34: R-C integrator circuit 30

36 V V out 0 t V in Figure 1.35: Input and output waveforms of an R-C integrator The R-C integrator is characterised by a long time constant. It is frequently used for the following purposes: Low-pass filter. To smooth out unwanted very short pulses or gaps in a low frequency signal or DC. To remove the unwanted AC noise and ripple components from a signal. Timing circuits (for example, the 555 timer IC, a standard component used for a variety of timer, pulse generation, and oscillator applications). Coupling circuits A coupling circuit is used when we need to connect the output signal from one device into the input of another, and a direct connection is not possible. This can be because the output has a DC component, or the input impedance of the second device is too low at high frequencies (too capacitive). Common coupling applications are: From one stage of an amplifier to the next. From the output of an audio device to a loudspeaker or earphone. To prevent the capacitance of a telephone from blocking the high frequency signals in a DSL Digital Subscriber Line (DSL or ADSL). coupling circuit: a circuit that functions as an interface between two other circuits, designed to prevent the transmission of some unwanted component of the input signal, such as DC voltage C-R coupling circuit This circuit is shown in Figure It is the same circuit as the differentiator, but the component values will be chosen according to the particular application. The main function of this circuit is as a DC blocker. This is shown in Figure 1.37, where the input is a 2 V sine wave with a 6 V DC component. In the output, the DC has been completely removed with no loss of quality of the input signal. C output input + R Figure 1.36: C-R coupling circuit 31

37 V V in 6,0 V V out 0 t Figure 1.37: Input and output waveforms of a C-R coupling circuit L-R coupling circuit L output input + R Figure 1.38: L-R coupling circuit attenuate: reduce the amplitude of something The L-R coupling circuit is shown in Figure It is used where DC current needs to be transferred to another circuit, but the AC component needs to be isolated without too much loading of the input. A typical application is the filter connected between an ADSL line and a telephone. The filter presents low impedance to the DC current which allows it through. But, the filter presents a high impedance to the high frequency signal (X L is directly proportional to frequency of the applied signal). Therefore the line can supply DC current to the telephone. But, it does not place a capacitive load on the line, which would seriously attenuate the high frequency signal conveying digital information on the line. L-R coupling circuits are used when: DC current needs to be transferred to a circuit. The AC component of the input signal is to be blocked. The first circuit should not be loaded by the other circuit. Think about it If a coupling circuit blocks high frequencies from a generator entering a load, then the same coupling circuit will prevent the same load from affecting the generator. Sometimes people say The generator doesn t see the load. This is another way of saying, The load doesn t have any effect on the generator. 32

38 Square waves used for testing analogue circuits In order to test the performance of a circuit, such as an amplifier, we need to observe the output under different input conditions. Ideally we should observe the output at a range of frequencies, including using step functions on the input voltage. A quick way to achieve these goals is to use a square wave. This is because a square wave contains many different frequencies all added together. Harmonics Mathematical theory has shown that any repeating wave of any shape can be represented by an infinite series of sine and cosine waves added together. In this example, we are dealing with a square wave. In Figure 1.39, some of the harmonics of a square wave are shown. harmonic: a wave with a frequency equal to a multiple of the fundamental The largest is a sine wave (called the fundamental or first harmonic). This has the same frequency as the square wave, to which are added a number of higher frequencies with smaller amplitudes. These are the 2nd, 3rd, 4th, and higher harmonics. When added together in the right proportions, they make up the ordinal square wave. The more harmonics that are combined, the more accurate the result is. In Figure 1.39, the first seven harmonics of a square wave are shown. Notice that there are no even numbered harmonics shown. This is because they all have zero amplitude. You can clearly see that the resultant is beginning to look like a square wave, and will improve if more harmonics are added. resultant 5 th harmonic 7 th harmonic 3 rd harmonic fundamental Figure 1.39: Harmonics of a square wave Testing analogue systems using square waves This method is used for many different systems, not just electronic. We will look at how square waves are used for testing an amplifier. input square wave overshoot DC error output slow rising edge Figure 1.40: Square wave testing of an amplifier 33

39 Figure 1.40 shows the input and output waveforms of a system (in this example, an amplifier) being tested using square waves. Information about the performance of the amplifier is shown by how well it handles the vertical and horizontal parts of the wave: Rising and falling edges contain information about the high frequency behaviour of the system. In this example: Rising and falling edges indicate limited bandwidth (rather slow transitions). The overshoot indicates a measure of instability, which could lead to oscillation of the amplifier. The high level and low levels of the square wave contain information about the low frequency behaviour of the system. In this example: There is a DC error, which seems to get worse over time. This means the low frequency response does not extend down to DC. The larger the amount of sagging the worse the low frequency response of the amplifier. Assessment activity Fill in the missing words: a) Inductors store energy in the that forms in and around them when flows through them. They consist of of wire, sometimes wound around a or rod of material which increases their. b) Capacitors store energy in the formed between their plates when a is applied between their terminals. They consist of two separated by a layer of an insulating material called a. 2. Name four factors which determine the capacity of a capacitor. State whether the factors should be increased or reduced to increase the capacitance. 3. Calculate the energy stored in: a) A capacitor of 220 μf charged to 150 V. b) An inductor of 150 mh carrying a current of 50 ma. 4. The time constant is a measure of how quickly a system responds to change. What percentage of the eventual change will have taken place when one time constant has elapsed? 5. Calculate the time constant and final values of voltage and current in the following circuit: A capacitor of 2,2 μf is charged through a resistor of 470 kω from a voltage source which supplies a step function of 10,0 V DC. 6. An inductor of 180 mh is connected through a resistor of 1,8 kω and a switch to a 12 V DC power supply. When the switch is closed the current starts to build up in the inductor. Calculate the time constant and final values of voltage and current in the inductor. 7. Make a sketch of: a) A C-R differentiator circuit. b) An R-C integrator circuit. 34