RF Circuit Design Chris Bowick, Notas de estudo de Engenharia de Telecomunicações

Baixe RF Circuit Design Chris Bowick e outras Notas de estudo em PDF para Engenharia de Telecomunicações, somente na Docsity! Newnes RF Circuit Design Chris B d k I I- RF CIRCUIT DESIGN Newnes is an imprint of Elsevier Science. Copyright 0 1982 by Chris Bowick 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, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone: (+44) 1865 843830, fax: (+44) 1865 853333, e-mail: permissions@elsevier.co.uk. You may also complete your request on-line via the Elsevier Science homepage (http://www.elsevier.com), by selecting ‘Customer Support’ and then ‘Obtaining Permissions’. @ This book is printed on acid-free paper. Library of Congress Cataloging-in-Publication Data Bowick, Chris. p. cm. RF circuit design / by Chris Bowick Originally published: Indianapolis : H.W. Sams, 1982 Includes bibliographical references and index. ISBN 0-7506-9946-9 (pbk. : alk. paper) 1. Radio circuits Design and construction. 2. Radio Frequency. I. Title. TK6553.B633 1997 96-5 1612 621,384’12-dc20 C P The publisher offers special discounts on bulk orders of this book For information, please contact: Manager of Special Sales Elsevier Science 200 Wheeler Road Burlington, MA 0 1803 Tel: 781-313-4700 Fax: 781-3 13-4802 For information on all Newnes publications available, contact our World Wide Web homepage at http://www.newnespress.corn 15 14 13 12 1110 Printed in the United States of America RF Circuit Design is written for those who desire a practical approach to the design of rf amplifiers, impedance matching networks, and filters. It is totally user oriented. If you are an individual who has little rf circuit design experience, you can use this book as a catalog of circuits, using component values designed for your application. On the other hand, if you are interested in the theory behind the rf circuitry being designed, you can use the more detailed information that is provided for in-depth study. An expert in the rf circuit design field will find this book to be an excellent reference rruznual, containing most of the commonly used circuit-design formulas that are needed. However, an electrical engineering student will find this book to be a valuable bridge between classroom studies and the real world. And, finally, if you are an experimenter or ham, who is interested in designing your own equipment, RF Circuit Design will provide numerous examples to guide you every step of the way. Chapter 1 begins with some basics about components and how they behave at rf frequencies; how capacitors become inductors, inductors become capacitors, and wires become inductors, capacitors, and resistors. Toroids are introduced and toroidal inductor design is covered in detail. Chapter 2 presents a review of resonant circuits and their properties including a discussion of Q, passband ripple, bandwidth, and coupling. You learn how to design single and multiresonator circuits, at the loaded Q you desire. An under- standing of resonant circuits naturally leads to filters and their design. So, Chapter 3 presents complete design procedures for multiple-pole Rutterworth, Chebyshel.. and Bessel filters including low-pass, high-pass, bandpass, and bandstop designs. LVithin minutes after reading Chapter 3, you will be able to design multiple.- pole filters to meet your specifications. Filter design was never easier. Next, Chapter 4 covers impedance matching of both real and complex im- pendances. This is done both numerically and with the aid of the Smith Chart. hlathematics are kept to a bare minimum. Both high-Q and low-Q matching networks are covered in depth. Transistor behavior at rf frequencies is discussed in Chapter 5. Input im- pedance. output impedance, feedback capacitance, and their variation over fre- quency are outlined. Transistor data sheets are explained in detail, and Y and S parameters are introduced. Chapter 6 details complete cookbook design procedures for rf small-signal amplifiers, using both Y and S parameters. Transistor biasing, stability, impedance matching, and neutralization techniques are covered in detail, complete with practical examples. Constant-gain circles and stability circles, as plotted on a Smith Chart, are introduced while rf amplifier design procedures for minimum noise figure are also explained. The subject of Chapter 7 is rf power amplifiers. This chapter describes the differences between small- and large-signal amplifiers, and provides step-by-step procedures for designing the latter. Design sections that discuss coaxial-feedline impedance matching and broadband transformers are included. Appendix A is a math tutorial on complex number manipulation with emphasis on their relationship to complex impedances. This appendix is recommended reading for those who are not familiar with complex number arithmetic. Then, Appendix B presents a systems approach to low-noise design by examining the Noise Figure parameter and its relationship to circuit design and total systems design. Finally, in Appendix C, a bibliography of technical papers and books related to rf circuit design is given so that you, the reader, can further increase your understanding of rf design procedures. CHRIS BOWICK ACKNOWLEDGMENTS The author wishes to gratefully acknowledge the contributions made by various individuals to the completion of this project. First, and foremost, a special thanks goes to my wife, Maureen, who not only typed the entire manuscript at least twice, but also performed duties both as an editor and as the author’s principal source of encouragement throughout the project. Needless to say, without her help, this book would have never been completed. Additional thanks go to the following individuals and companies for their contributions in the form of information and data sheets: Bill Amidon and Jim Cox of Amidon Associates, Dave Stewart of Piezo Technology, Irving Kadesh of Piconics, Brian Price of Indiana General, Richard Parker of Fair-Rite Products, Jack Goodman of Sprague-Goodman Electronics, Phillip Smith of Analog Instru- ments, Lothar Stern of Motorola, and Larry Ward of Microwave Associates. To my wife, Maureen, and daughter, Zoe . . . COMPONENTS Components, those bits and pieces which make up a radio frequency (rf) circuit, seem at times to be taken for granted. A capacitor is, after all, a capacitor -isn’t it? A l-megohm resistor presents an impedance of at least 1 megohm-doesn’t it? The reactance of an inductor always increases with frequency, right? Well, as we shall see later in this discussion, things aren’t always as they seem. Capacitors at certain frequencies may not be capacitors at all, but may look inductive, while inductors may look like capacitors, and resistors may tend to be a little of both. In this chapter, we will discuss the properties of re- sistors, capacitors, and inductors at radio frequencies as they relate to circuit design. But, first, let’s take a look at the most simple component of any system and examine its problems at radio frequencies. WIRE Wire in an rf circuit can take many forms. Wire- wound resistors, inductors, and axial- and radial-leaded capacitors all use a wire of some size and length either in their leads, or in the actual body of the component, or both. Wire is also used in many interconnect appli- cations in the lower rf spectrum. The behavior of a wire in the rf spectrum depends to a large extent on the wire’s diameter and length. Table 1-1 lists, in the American Wire Gauge (AWG) system, each gauge of wire, its corresponding diameter, and other charac- teristics of interest to the rf circuit designer. In the AWG system, the diameter of a wire will roughly double every six wire gauges. Thus, if the last six EXAMPLE 1-1 Given that the diameter of AWG 50 wire is 1.0 mil (0.001 inch), what is the diameter of AWG 14 wire? Solution AWG 50 = 1 mil AWG 44 = 2 x 1 mil = 2 mils AWG 38 = 2 x 2 mils = 4 mils AWG 32 = 2 x 4 mils = 8 mils AWG 26 = 2 x 8 mils = 16 mils AWG 20 = 2 x 16 mils = 32 mils AWG 14 = 2 x 32 mils = 64 mils (0.064 inch) gauges and their corresponding diameters are mem- orized from the chart, all other wire diameters can be determined without the aid of a chart (Example 1-1). Skin Effect A conductor, at low frequencies, utilizes its entire cross-sectional area as a transport medium for charge carriers. As the frequency is increased, an increased magnetic field a t the center of the conductor presents an impedance to the charge carriers, thus decreasing the current density at the center of the conductor and increasing the current density around its perimeter. This increased current density near the edge of the conductor is known as skin effect. It occurs in all con- ductors including resistor leads, capacitor leads, and inductor leads. The depth into the conductor at which the charge- carrier current density falls to l /e, or 37% of its value along the surface, is known as the skin depth and is a function of the frequency and the permeability and conductivity of the medium. Thus, different con- ductors, such as silver, aluminum, and copper, all have different skin depths. The net result of skin effect is an effective decrease in the cross-sectional area of the conductor and, there- fore, a net increase in the ac resistance of the wire as shown in Fig. 1-1. For copper, the skin depth is ap- proximately 0.85 cm at 60 Hz and 0.007 cm at 1 MHz. Or, to state it another way: 63To of the rf current flow- ing in a copper wire will flow within a distance of 0.007 cm of the outer edge of the wire. Straight-Wire Inductors In the medium surrounding any current-carrying conductor, there exists a magnetic field. If the current in the conductor is an alternating current, this mag- netic field is alternately expanding and contracting and, thus, producing a voltage on the wire which op- poses any change in current flow. This opposition to change is called self-inductance and we call anything that possesses this quality an inductor. Straight-wire inductance might seem trivial, but as will be seen later in the chapter, the higher we go in frequency, the more important it becomes. The inductance of a straight wire depends on both its length and its diameter, and is found by: 9 10 RF Cmcurr DFSIGN A, = mlZ A, = TrZZ Skin Depth Area = AZ - A, Fig. 1-1. Skin depth area of a conductor. L = 0.0021[2.3 log ($ - 0.75>] pH (Eq. 1-1) where, L = the inductance in pH, I = the length of the wire in cm, d = the diameter of the wire in cm. This is shown in calculations of Example 1-2. EXAMPLE 1-2 Find the inductance of 5 centimeters of No. 22 copper wire. Solution From Table 1-1, the diameter of No. 22 copper wire is 25.3 mils. Since 1 mil equals 2.54 x 10-3 cm, this equals 0.0643 cm. Substituting into Equation 1-1 gives L = (0.002) ( 5 ) [ 2.3 log (a - 0.75)] = 57 nanohenries The concept of inductance is important because any and all conductors at radio frequencies (including hookup wire, capacitor leads, etc.) tend to exhibit the property of inductance. Inductors will be discussed in greater detail later in this chapter. RESISTORS Resistance is the property of a material that de- termines the rate at which electrical energy is con- verted into heat energy for a given electric current. By definition : 1 volt across 1 ohm = 1 coulomb per second = 1 ampere The thermal dissipation in this circumstance is 1 watt. P = E I = 1 volt X 1 ampere = 1 watt Fig. 1-2. Resistor equivalent circuit. Resistors are used everywhere in circuits, as tran- sistor bias networks, pads, and signal combiners. How- ever, very rarely is there any thought given to how a resistor actually behaves once we depart from the world of direct current (dc). In some instances, such as in transistor biasing networks, the resistor will still perform its dc circuit function, but it may also disrupt the circuit’s rf operating point. Resistor Equivalent Circuit The equivalent circuit of a resistor at radio frequen- cies is shown in Fig. 1-2. R is the resistor value itself, L is the lead inductance, and C is a combination of parasitic capacitances which varies from resistor to resistor depending on the resistor’s structure. Carbon- composition resistors are notoriously poor high-fre- quency performers. A carbon-composition resistor con- sists of densely packed dielectric particulates or carbon granules. Between each pair of carbon granules is a very small parasitic capacitor. These parasitics, in aggregate, are not insignificant, however, and are the major component of the device’s equivalent circuit. Wirewound resistors have problems at radio fre- quencies too. As may be expected, these resistors tend to exhibit widely varying impedances over various frequencies. This is particularly true of the low re- sistance values in the frequency range of 10 MHz to 200 MHz. The inductor L, shown in the equivalent cir- cuit of Fig. 1-2, is much larger for a wirewound resistor than for a carbon-composition resistor. Its value can be calculated using the single-layer air-core inductance approximation formula. This formula is discussed later in this chapter. Because wirewound resistors look like inductors, their impedances will first increase as the frequency increases. At some frequency ( Fr) , however, the inductance ( L ) will resonate with the shunt capaci- I I Frequency (F) Fig. 1-3. Impedance characteristic of a wirewound resistor. COMPONENTS 11 120 3 loo B 5 2 4 80 - e @ H P) 8 40 -8 - 20 0 1.0 10 100 loo0 Frequency (MHz) Fig. 1-4. Frequency characteristics of metal-film vs. carbon-composition resistors. (Adapted from Handbook of Components for Electronics, McGraw-Hill ) . tance (C), producing an impedance peak. Any further increase in frequency will cause the resistor's im- pedance to decrease as shown in Fig. 1-3. A metal-film resistor seems to exhibit the best char- acteristics over frequency. Its equivalent circuit is the same as the carbon-composition and wirewound resistor, but the values of the individual parasitic elements in the equivalent circuit decrease. The impedance of a metal-film resistor tends to de- crease with frequency above about 10 MHz, as shown in Fig. 1-4. This is due to the shunt capacitance in the equivalent circuit. At very high frequencies, and with low-value resistors (under 50 ohms), lead inductance and skin effect may become noticeable. The lead in- ductance produces a resonance peak, as shown for the 5-ohm resistance in Fig. 1-4, and skin effect decreases the slope of the curve as it falls off with frequency. Many manufacturers will supply data on resistor be- havior at radio frequencies but it can often be mislead- ing. Once you understand the mechanisms involved in resistor behavior, however, it will not matter in what form the data is supplied. Example 1-3 illustrates that fact. The recent trend in resistor technology has been to eliminate or greatly reduce the stray reactances as- sociated with resistors. This has led to the development of thin-film chip resistors, such as those shown in Fig. 1-6. They are typically produced on alumina or beryl- lia substrates and offer very little parasitic reactance at frequencies from dc to 2 GHz. Fig. 1-6. Thin-film chip resistors. ( Courtesy Piconics, Inc. ) EXAMPLE 1-3 In Fig. 1-2, the lead lengths on the metal-film resistor are 1.27 cm (0.5 inch), and are made up of No. 14 wire. The total stray shunt capacitance ( C ) is 0.3 pF. If the resistor value is 10,OOO ohms, what is its equivalent d im- pedance at 200 MHz? Sotution From Table 1-1, the diameter of No. 14 AWG wire is 64.1 mils (0.1628 cm). Therefore, using Equation 1-1: L = 0.002( 1.27) [ 2.3 log (- - 0.75)] = 8.7 nanohenries This presents an equivalent reactance at 200 MHz of: X L = OL = 2?r( 200 x 106) (8.7 x 10-0) = 10.93 ohms The capacitor ( C ) presents an equivalent reactance of: X e = x 1 1 - - 2 4 200 x 106) (0.3 x 10-12) = 2653 ohms The combined equivalent circuit for this resistor, at 200 MHz, is shown in Fig. 1-5. From this sketch, we can see that, in this case, the lead inductance is insignificant when compared with the 10K series resistance and it may be j10.93 0 i10.93 0 4- - j2563 0 Fig. 1-5. Equivalent circuit values for Example 1-3. neglected. The parasitic capacitance, on the other hand, cannot be neglected. What we now have, in effect, is a 2563-ohm reactance in parallel with a 10,000-ohm re- sistance. The magnitude of the combined impedance is: RX. dR2 + X.2 Z= - (10K)(2583) - <( 10K)z + (2563)z = 1890.5 ohms Thus, our 10K resistor looks like 1890 ohms at 200 MHz. 14 RF CIRCUIT DESIGN 30 20 f 10 V a o 9 -10 2 -20 2 -30 10 5 f 4 0 ; - 5 9 -10 8 -15 20 $ 0 2 -20 -40 9 -60 s -80 -55 -35 -15 + 5 +25 +45 +65 +85 +105+125 Temperature, "C Fig. 1-1 1. Temperature characteristics for ceramic dielectric capacitors. million-per-degree-Celsius ) with tolerances as small as 215 ppm/ "C. Because of their excellent temperature stability, NPO ceramics are well suited for oscillator, resonant circuit, or filter applications. Moderately stable ceramic capacitors (Fig. 1-11) typically vary +1570 of their rated capacitance over their temperature range. This variation is typically nonlinear, however, and care should be taken in their use in resonant circuits or filters where stability is im- portant. These ceramics are generally used in switching circuits. Their main advantage is that they are gener- ally smaller than the NPO ceramic capacitors and, of course, cost less. High-K ceramic capacitors are typically termed general-purpose capacitors. Their temperature char- acteristics are very poor and their capacitance may vary as much as 80% over various temperature ranges (Fig. 1-11). They are commonly used only in bypass applications at radio frequencies. There are ceramic capacitors available on the market which are specifically intended for rf applications. These capacitors are typically high-Q (low ESR) de- vices with flat ribbon leads or with no leads at all. The lead material is usually solid silver or silver plated and, thus, contains very low resistive losses. At vhf frequencies and above, these capacitors exhibit very low lead inductance due to the flat ribbon leads. These devices are, of course, more expensive and require spe- cial printed-circuit board areas for mounting. The capacitors that have no leads are called chip capaci- tors. These capacitors are typically used above 500 MHz where lead inductance cannot be tolerated. Chip capacitors and flat ribbon capacitors are shown in Fig. 1-12. Fig. 1-12. Chip and ribbon capacitors. Mica Capacitors-Mica capacitors typically have a dielectric constant of about 6, which indicates that for a particular capacitance value, mica capacitors are typically large. Their low K, however, also produces an extremely good temperature characteristic. Thus, mica capacitors are used extensively in resonant circuits and in filters where pc board area is of no concern. Silvered mica capacitors are even more stable. Ordi- nary mica capacitors have plates of foil pressed against the mica dielectric. In silvered micas, the silver plates are applied by a process called vacuum evaporation which is a much more exacting process. This produces an even better stability with very tight and reproduc- ible tolerances of typically +20 ppm/"C over a range The problem with micas, however, is that they are becoming increasingly less cost effective than ceramic types. Therefore, if you have an application in which a mica capacitor would seem to work well, chances are you can find a less expensive NPO ceramic capaci- tor that will work just as well. Metalized-Film Capacitors-"Metalized-film" is a -60 "C to +89 "C. Fig. 1-13. A simple microwave air-core inductor. ( Courtesy Piconics, Inc. ) COMPONENTS 15 broad category of capacitor encompassing most of the other capacitors listed previously and which we have not yet discussed. This includes teflon, polystyrene, polycarbonate, and paper dielectrics. Metalized-film capacitors are used in a number of applications, including filtering, bypassing, and coup- ling. Most of the polycarbonate, polystyrene, and teflon styles are available in very tight ( &2%) capacitance tolerances over their entire temperature range. Poly- styrene, however, typically cannot be used over +85 “C as it is very temperature sensitive above this point. Most of the capacitors in this category are typically larger than the equivalent-value ceramic types and are used in applications where space is not a con- straint. INDUCTORS An inductor is nothing more than a wire wound or coiled in such a manner as to increase the magnetic flux linkage between the turns of the coil (see Fig. 1-13). This increased flux linkage increases the wire’s self-inductance ( or just plain inductance ) beyond that which it would otherwise have been. Inductors are used extensively in rf design in resonant circuits, filters, phase shift and delay networks, and as rf chokes used to prevent, or at least reduce, the flow of rf en- ergy along a certain path. Real-World Inductors As we have discovered in previous sections of this chapter, there is no “perfect” component, and inductors are certainly no exception. As a matter of fact, of the components we have discussed, the inductor is prob- ably the component most prone to very drastic changes over frequency. Fig. 1-14 shows what an inductor really looks like Fig. 1-14. Distributed capacitance and series resistance in an inductor. at rf frequencies. As previously discussed, whenever we bring two conductors into close proximity but separated by a dielectric, and place a voltage differen- tial between the two, we form a capacitor. Thus, if any wire resistance at all exists, a voltage drop (even though very minute) will occur between the windings, and small capacitors will be formed. This effect is shown in Fig. 1-14 and is called distributed capaci- tance ( C ~ ) . Then, in Fig. 1-15, the capacitance (C,) is an aggregate of the individual parasitic distributed capacitances of the coil shown in Fig. 1-14. Fig. 1-15. Inductor equivalent circuit / / / J r Capacitive Frequency Fig. 1-16. Impedance characterishc vs. frequency for a practical and an ideal inductor The effect of Cd upon the reactance of an inductor is shown in Fig. 1-16. Initially, at lower frequencies, the inductor’s reactance parallels that of an ideal in- ductor. Soon, however, its reactance departs from the ideal curve and increases at a much faster rate until it reaches a peak at the inductor’s parallel resonant frequency ( F, ) , Above F,, the inductor’s reactance begins to decrease with frequency and, thus, the in- ductor begins to look like a capacitor. Theoretically, the resonance peak would occur at infinite reactance (see Example 1-4). However, due to the series re- sistance of the coil, some finite impedance is seen at resonance. Recent advances in inductor technology have led to the development of microminiature fixed-chip induc- tors. One type is shown in Fig. 1-17. These inductors feature a ceramic substrate with gold-plated solder- able wrap-around bottom connections. They come in values from 0.01 p H to 1.0 mH, with typical Qs that range from 40 to 60 at 200 MHz. It was mentioned earlier that the series resistance of a coil is the mechanism that keeps the impedance of the coil finite at resonance. Another effect it has is 16 RF Cmcurr DFSIGN Fig. 1-17. Microminiature chip inductor. ( Courtesy Piconics, Inc. ) ~~~~ EXAMPLE 1-4 To show that the impedance of a lossless inductor at resonance is infinite, we can write the following: (Eq. 1-3) XLXC Z=- XL + xc where,c Z = the impedance of the parallel circuit, XL = the inductive reactance ( joL ) , Xc = the capacitive reactance Therefore, Multiplying numerator and denominator by jmC, we get: From algebra, j2 = -1; then, rearranging: joL 1 - o2LC Z = (Eq. 1-5) (Eq. 1-6) If the term oZLC, in Equation 1-6, should ever become equal to 1, then the denominator will be equal to zero and impedance Z will become infinite. The frequency at which o2LC becomes equal to 1 is: WZLC = 1 1 L C = - o2 C C = $ 2TdE&T 1 1 2 % - m = (Eq. 1-7) which is the familiar equation for the resonant frequency of a tuned circuit. to broaden the resonance peak of the impedance curve of the coil. This characteristic of resonant circuits is an important one and will be discussed in detail in Chapter 3. The ratio of an inductor’s reactance to its series re- sistance is often used as a measure of the quality of the inductor. The larger the ratio, the better is the inductor. This quality factor is referred to as the Q of the inductor. If the inductor were wound with a perfect conductor, its Q would be infinite and we would have a lossless inductor. Of course, there is no perfect conductor and, thus, an inductor always has some finite Q. At low frequencies, the Q of an inductor is very good because the only resistance in the windings is the dc resistance of the wire-which is very small. But as the frequency increases, skin effect and winding capacitance begin to degrade the quality of the in- ductor. This is shown in the graph of Fig. 1-18. At low frequencies, Q will increase directly with fre- quency because its reactance is increasing and skin effect has not yet become noticeable. Soon, however, skin effect does become a factor. The Q still rises, but at a lesser rate, and we get a gradually decreasing slope in the curve. The flat portion of the curve in Fig. 1-18 occurs as the series resistance and the reactance are changing at the same rate. Above this point, the shunt capacitance and skin effect of the windings combine to decrease the Q of the inductor to zero at its resonant frequency. Some methods of increasing the Q of an inductor and extending its useful frequency range are: 1. Use a larger diameter wire. This decreases the ac and dc resistance of the windings. 2. Spread the windings apart. Air has a lower dielectric constant than most insulators. Thus, an air gap be- tween the windings decreases the interwinding capacitance. 3. Increase the permeability of the flux linkage path. This is most often done by winding the inductor around a magnetic-core material, such as iron or Frequency Fig. 1-18. The Q variation of an inductor vs. frequency. COMPONENTS 19 Q, would take 90 turns of a very small wire (in order to fit all turns within a %-inch length) to reach 35 pH; however, the toroidal inductor would only need 8 turns to reach the design goal. Obviously, this is an ex- treme case but it serves a useful purpose and illustrates the point. The toroidal core does require fewer turns for a given inductance than does an air-core design. Thus, there is less ac resistance and the Q can be increased dramatically. B Fig. 1-23. Magnetization curve for a typical core. ( A ) Typicol inductor. Magnetic Flux ( B ) Toroidal inductor. Fig. 1-22. Shielding effect of a toroidal inductor. The self-shielding properties of a toroid become evident when Fig. 1-22 is examined. In a typical air- core inductor, the magnetic-flux lines linking the turns of the inductor take the shape shown in Fig. 1-22A. The sketch clearly indicates that the air surrounding the inductor is definitely part of the magnetic-flux path. Thus, this inductor tends to radiate the rf signals flow- ing within. A toroid, on the other hand (Fig. 1-22B), completely contains the magnetic flux within the ma- terial itself; thus, no radiation occurs. In actual prac- tice, of course, some radiation will occur but it is min- imized. This characteristic of toroids eliminates the need for bulky shields surrounding the inductor. The shields not only tend to reduce available space, but they also reduce the Q of the inductor that they are shielding. Core Characteristics Earlier, we discussed, in general terms, the relative advantages and disadvantages of using magnetic cores. The following discussion of typical toroidal-core char- acteristics will aid you in specifying the core that you need for your particular application. Fig. 1-23 is a typical magnetization curve for a magnetic core. The curve simply indicates the mag- netic-flux density ( B ) that occurs in the inductor with a specific magnetic-field intensity ( H ) applied. As the magnetic-field intensity is increased from zero (by in- creasing the applied signal voltage ), the magnetic- flux density that links the turns of the inductor in- creases quite linearly. The ratio of the magnetic-flux density to the magnetic-field intensity is called the permeability of the material. This has already been mentioned on numerous occasions. p = B/H ( WebersIampere-turn) (Eq. 1-9) Thus, the permeability of a material is simply a mea- sure of how well it transforms an electrical excitation into a magnetic flux. The better it is at this transforma- tion, the higher is its permeability. As mentioned previously, initially the magnetiza- tion curve is linear. It is during this linear portion of the curve that permeability is usually specified and, thus, it is sometimes called initial permeability (h) in various core literature. As the electrical excitation increases, however, a point is reached at which the magnetic-flux intensity does not continue to increase at the same rate as the excitation and the slope of the curve begins to decrease. Any further increase in ex- citation may cause saturation to occur. HBnt is the ex- citation point above which no further increase in magnetic-flux density occurs (B,,,) . The incremental permeability above this point is the same as air. Typi- cally, in rf circuit applications, we keep the excitation small enough to maintain linear operation. Bsat varies substantially from core to core, depend- ing upon the size and shape of the material. Thus, it is necessary to read and understand the manufacturer’s literature that describes the particular core you are using. Once BBat is known for the core, it is a very simple matter to determine whether or not its use in a particular circuit application will cause it to saturate. The in-circuit operational flux density (B,,,,) of the core is given by the formula: (Eq. 1-10) E x lox Bop = (4.44)fNL 20 RF Cmcurr DESIGN where, Bo, = the magnetic-flux density in gauss, E = the maximum rms voltage across the inductor f = the frequency in hertz, N = the number of turns, A, = the effective cross-sectional area of the core Thus, if the calculated Bo, for a particular application is less than the published specification for BBat, then the core will not saturate and its operation will be somewhat linear. Another characteristic of magnetic cores that is very important to understand is that of internal loss. It has previously been mentioned that the careless addition of a magnetic core to an air-core inductor could possibly reduce the Q of the inductor. This con- cept might seem contrary to what we have studied so far, so let’s examine it a bit more closely. The equivalent circuit of an air-core inductor (Fig. 1-15) is reproduced in Fig. 1-24A for your convenience. The Q of this inductor is in volts, in cm2. Q = & (Eq. 1-11) R, where, X L = OL, R, = the resistance of the windings. If we add a magnetic core to the inductor, the equivalent circuit becomes like that shown in Fig. 1-24B. We have added resistance R, to represent the losses which take place in the core itseIf. These losses are in the form of hysteresis. Hysteresis is the power lost in the core due to the realignment of the magnetic particles within the material with changes in excita- tion, and the eddy currents that flow in the core due to the voltages induced within. These two types of internal loss, which are inherent to some degree in every magnetic core and are thus unavoidable, com- bine to reduce the efficiency of the inductor and, thus, increase its loss. But what about the new Q for the magnetic-core inductor? This question isn’t as easily answered. Remember, when a magnetic core is in- serted into an existing inductor, the value of the in- ductance is increased. Therefore, at any given fre- quency, its reactance increases proportionally. The question that must be answered then, in order to de- ( A ) Air core. ( B ) Magnetic core. Fig. 1-24. Equivalent circuits for air-core and magnetic-core inductors. termine the new Q of the inductor, is: By what factors did the inductance and loss increase? Obviously, if by adding a toroidal core, the inductance were in- creased by a factor of two and its total loss was also increased by a factor of two, the Q would remain unchanged. If, however, the total coil loss were in- creased to four times its previous value while only doubling the inductance, the Q of the inductor would be reduced by a factor of two. Now, as if all of this isn’t confusing enough, we must also keep in mind that the additional loss intro- duced by the core is not constant, but varies (usually increases) with frequency. Therefore, the designer must have a complete set of manufacturer’s data sheets for every core he is working with. Toroid manufacturers typically publish data sheets which contain all the information needed to design inductors and transformers with a particular core. (Some typical specification and data sheets are given in Figs. 1-25 and 1-26.) In most cases, however, each manufacturer presents the information in a unique manner and care must be taken in order to extract the information that is needed without error, and in a form that can be used in the ensuing design process. This is not always as simple as it sounds. Later in this chapter, we will use the data presented in Figs. 1-25 and 1-26 to design a couple of toroidal inductors so that we may see some of those differences. Table 1-2 lists some of the commonly used terms along with their symbols and units. Powdered Iron Vs. Ferrite In general, there are no hard and fast rules govern- ing the use of ferrite cores versus powdered-iron cores in rf circuit-design applications. In many instances, given the same permeability and type, either core could be used without much change in performance of the actual circuit. There are, however, special appli- cations in which one core might out-perform another, and it is those applications which we will address here. Powdered-iron cores, for instance, can typically handle more rf power without saturation or damage than the same size ferrite core. For example, ferrite, if driven with a large amount of rf power, tends to retain its magnetism permanently. This ruins the core by changing its permeability permanently. Powdered iron, on the other hand, if overdriven will eventually return to its initial permeability (pi). Thus, in any application where high rf power levels are involved, iron cores might seem to be the best choice. In general, powdered-iron cores tend to yield higher- Q inductors, at higher frequencies, than an equivalent size ferrite core. This is due to the inherent core char- acteristics of powdered iron which produce much less internal loss than ferrite cores. This characteristic of powdered iron makes it very useful in narrow-band or tuned-circuit applications. Table 1-3 lists a few of the common powdered-iron core materials along with their typical applications. COMPONENTS 21 d, d2 h Values measured a t 100 KHz. T = 25OC. PART NUMBER MR-7401 BER.7402 MR.7YI3 MR.7- TOL UNITS 0.135 0.156 0.230 0.100 t 0 . a in. 0.065 0.088 0.120 0.060 t0.m in. 0.055 0.061 0.060 0.060 t0.m in. 7400 Series Toroids Nom. pi 2500 m Temperature Coefficient (TC) = 0 to +0.75% P C max., 40 to +7OoC. rn Disaccornrnodation (D) = 3.0% max., 10-100 rnin., 25OC. m Hysteresis Core Constant (vi) measured at 20 KHz to 30 gauss (3 milli Tesla). For mm dimensions and core constants, see pegem. MECHANICAL SPECIFICATIONS ELECTRICAL SPECIFICATIONS MR-7- 440 0.276 8.9 54 ~ 5.1 2,150 TO1 t2096 tm min. rnin. rnax. rnax. nH/turn2 ohm/turn2 VSA.2 ~ - 3 1 2 TYPICAL CHARACTERISTIC CURVES - Part Numbers 7401,7402,7403 a d 7404 Inductance Factor VI. RMS VoltsIArra Turn lnduarna Factor It. Temp.r8tur8 1nduct.na Factor VI. DC Polarization 100 100 loo 80 80 80 80 80 80 40 40 40 20 10 10 0 0 0 20 -20 -20 40 40 40 -60 60 60 80 -Y) -80 ,100 ,100 -100 0 0.89 1.8 2.6 3.5 4.4 5.2 6.2 7.0 8.0 8.9 40 0 40 80 120 160 ,02 , 103 , ,04 TEMPERATURE OC D.C. MILLIAMP TURNS VrmJA,N (X lO’I Volts mm.’ Cont. on next page Fig. 1-25. Data sheet for ferrite toroidal cores. (Courtesy Indiana General) 24 RF Cmcum DFSIGN IRON-POWDER TOROIDAL CORES FOR R E S O N A N T CIRCUITS MATERIAL # 0 permeability 1 50 MHz to 300 MHz Tan Core number T- 130-0 T- 106-0 T- 94-0 T- 80-0 T- 68-0 T- 50-0 T- 44-0 T- 37-0 T- 30-0 T- 25-0 T- 20-0 T- 16-0 T- 12-0 Outer diarn. ( in . ) 1.300 1.060 .942 .795 .690 .500 .440 ,375 .307 .255 .200 .160 .125 Inner diam. .780 . S O .SO .495 .370 .303 .229 .205 .151 .120 .088 .W8 .062 ( in . Height ( in . ) .437 A37 ,312 .250 .190 190 .159 .128 ,128 .096 .067 .060 .050 AL value uh / 100 t 15.0 19.2 10.6 8.5 7.5 6.4 6.5 4.9 6 .O 4.5 3.5 3.0 3.0 MATERIAL # 12 permeability 3 20 MHz to 200 MHz Green 8 White Core number Outer diarn. inner diarn. Height AL value ( i n . ) ( in . ) ( in . ) uh / 100 t T-80-12 T-68- 12 T-50-12 T-44- 12 T-37-12 T-30-12 T-25- 12 T-20- 12 T- 16- 12 T-12-12 .795 .690 .500 ,440 .375 .307 .255 .200 .160 .125 .495 .370 .300 .229 .205 .151 ,120 .088 .078 .062 .250 .190 .190 .159 .128 .128 ,096 .067 .060 .050 22 21 18 15 16 12 10 7 la a T -- 200 -------- IRON Key to POWDER part numbers TOROIDAL for : CORES 7 - - - e Outer diameter Material Tom i d Cont. on next page Fig. 1-26-cont. Data sheet for powdered-iron toroidal cores. (Courtesy Amidon Associates) COMPONENTS 25 IRON-POWDER TOROIDAL CORES FOR R E S O N A N T C I R C U I T S MATERIAL # 10 permeability 6 10 MHz to 100 MHz Core number Outer diam. Inner diam. Height ( in. ) ( in . ) ( in . ) T-94- 10 T-80-10 T-68- 10 T-50- 10 T-44- 10 T-37- 10 T-30-10 T-25- 10 T-20- 10 T-16-10 T- 12- 10 e 942 .795 .690 .500 .440 ,375 .307 .255 .200 .160 .125 .560 .495 3 0 .303 .229 .205 .151 .120 .088 .078 .062 .312 .250 .190 .190 .159 ,128 ,128 .W6 .067 .060 .050 Black AL value uh /100t 32 32 31 33 25 25 19 16 13 12 58 Core Size T-12 T-16 1-20 T-25 T-30 1-37 T-44 T-50 T-68 T-80 T-94 T- 106 T- 130 T- 157 T-I84 T-200 T-225 ~~ NUMBER OF TURNS vs. WIRE SIZE and CORE SIZE Approximate number of turns of wire - single layer wound - single insulation w i r e s i z e 4 0 3 8 3 6 34 32 3 0 2 8 2 6 2 4 2 2 2 0 18 16 14 12 I O 47 63 72 101 129 177 199 265 325 438 496 496 693 846 046 1115 1250 37 49 56 79 101 140 157 210 2 57 347 393 393 550 672 672 886 993 29 38 43 62 79 110 124 166 205 276 313 313 439 536 536 707 793 21 29 33 48 62 87 97 131 162 219 248 248 348 426 426 562 63 1 15 21 25 37 48 67 76 103 127 172 195 195 275 336 336 445 499 11 16 18 28 37 53 60 81 101 137 156 156 220 270 270 357 400 8 11 14 21 28 41 46 63 79 108 123 123 173 213 213 282 317 5 8 9 15 21 31 35 49 61 84 96 96 137 168 168 223 250 4 2 1 1 1 0 0 1 5 3 3 1 1 1 0 0 6 5 4 3 1 1 1 1 1 1 7 5 4 3 1 1 1 1 5 1 1 7 5 4 3 1 2 3 1 7 1 2 9 7 5 3 ’ 27 20 15 IO 7 6 5 3 37 28 21 16 1 1 8 6 i 47 36 28 21 15 1 1 9 - 66 51 39 30 23 17 12 6 75 58 45 35 27 20 14 10 75 58 45 35 27 20 14 10 107 83 66 51 40 30 23 17 132 104 82 64 50 38 29 22 132 104 82 64 50 38 29 22 176 139 109 86 68 53 4 1 31 198 156 123 98 77 60 46 3 Cont. on next page Fig. 1-26-cont. Data sheet for powdered-iron toroidal cores. (Courtesy Amidon Associates) 26 RF C m m DESIGN Q-CURVES IRON-POWDER TOROIDAL CORES . , Y Cont. on next page Fig. 1-M-cont. Data sheet for powdered-iron toroidal cores. ( Courtesy Amidon Associates) COMPONENTS 29 EXAMPLE 1-6 Using the data given in Fig. 1-25, design a toroidal in- ductor with an inductancc of 50 pH. What is the largest 4WG wire that we could possibly use while still maintain- ing a single-layer winding? What is the inductor’s Q at 100 MHz? Solution There are numerous possibilities in this particular design since no constraints were placed on us. Fig. 1-25 is a data sheet for the Indiana General 7400 Series of ferrite toroidal cores. This type of core would normally be used in broad- band or low-Q transformer applications rather than in narrow-band tuned circuits. This exercise will reveal why. The mechanical specifications for this series of cores in- dicate a fairly typical size for toroids used in small-signal rf circuit design. The largest core for this series is just under a quarter of an inch in diameter. Since no size con- straints were placed on us in the problem statement, we will use the BBR-7403 which has an outside diameter of 0.0230 inch. This will allow us to use a larger diameter wire to wind the inductor. The published value for AL for the given core is 495 nH/turn2. [:sing Equation 1-14, the number of turns re- quired for this core is: = 10 turns Note that the inductance of 50 /AH was replaced with its equivalent of 50,000 nH. The next step is to determine the largest diameter wire that can be used to wind the trans- former while still maintaining a single-layer winding. In ~ o m e cases, the data supplied by the manufacturer will include this type of winding information. Thus, in those cases, the designer need only look in a table to determine the maximum wire size that can he used. In our case, this infomiation was not given, so a simple calculation must he made. Fig. 1-27 illustrates the geometry of the problem. It is obvious from the diagram that the inner radius ( rl) of Wire Radius I< = d 2 Fig. 1-27. Toroid coil winding geonietiy the toroid is the limiting factor in determining the maxi- mum number of turns for a given wire diameter. The exact maximum diameter wire for a given number of turns can be found by: (Eq. 1-15) where, d = the diameter of the wire in inches, rl = the inner radius of the core in inches. N the number of turns. For this example, we obtain the value of rl from Fig. 1-25 (d, = 0.120 inch). 0.120 2T __ 2 d=- 10 + T = 28.69 x lo-” inches = 28.69 mils As a practical rule of thumb, however, taking into ac- count the insulation thickness variation among manufac- turers, it is best to add a “fudge factor” and take 90% of the calculated value, or 25.82 mils. Thus, the largest diameter wire used would be the next size below 25.82 mils, which is AWG No. 22 wire. EXAMPLE 1-7 Using thc information provided in the data sheet of Fig. 1-26, design a high-Q ( Q > 80), 300 nH, toroidal inductor For use at 100 MH;. Due to pc board space available, the toroid may not hr any larger than 0.3 inch in diameter. Coltition Fig. 1-26 is 311 rxcerpt from an Aniidon Associates iron- powder toroidal-core data sheet. The recommended oper- ating frequencies f o r various materials are shown in the Iron-Powder Material vs. Frequency Range graph. Either material No. 12 or material NO. 10 seems to be well suited for operation at 100 MHz. Elsewhere on the data sheet, ma- terial No. 12 is listed as IRN-8. (IRN-8 is described in Table 1-3.) Material No. 10 is not described, so choose material No. 12. Then, under a heading of Iron-Powder Toroidal Cores, the data sheet lists the physical dimensions of the toroids i h n g with the value of ,4r, for each. Note, however, that this particular company chooses to specify AL in pH/100 turns rather than pH/100 t u r d . The conversion factor between their value of AL and AL in nH/turnZ is to divide their value of AL by 10. Thus, the T-80-12 core with an A L of 22 pH/100 turns is equal to 2.2 nH/turn2. Next, the data sheet lists a set of Q-curves for the cores listed in the preceding charts. Note that all of the curves shown indicate Qs that are greater than 80 at 100 MHz. Choose the largest core available that will fit in the allotted pc hoard area. The core you should have chosen is the number T-25-12, with an outer diameter of 0.255 inch. A, = 12 /.~H/100 t = 1.2 nH/turnz Therefore, tising Equation 1-14. the niimher of turns re- quired is = 15.81 = 16 turn.; Finally, the chart of Number of Turns vs. Wire Size and Core Size on the data sheet clearly indicates that, for a T-25 size core, the largest size wire we can use to wind this particular toroid is No. 28 AWC wire. 30 RF Cmcurr DESIGN while reactance X, increases. At about 3 MHz, X, equals R, and the Q becomes unity. The Q then falls below unity until about 100 MHz where resistance R, begins to increase dramatically and causes the Q to again pass through unity. Thus, due to losses in the core itself, the Q of the coil at 100 MHz is probably very close to 1. Since the Q is so low, this coil would not be a very good choice for use in a narrow-band tuned circuit. See Example 1-7. PRACTICAL WINDING HINTS Fig. 1-28 depicts the correct method for winding a toroid. Using the technique of Fig. 1-28A, the inter- winding capacitance is minimized, a good portion of the available winding area is utilized, and the resonant frequency of the inductor is increased, thus extending the useful frequency range of the device. Note that by using the methods shown in Figs. 1-28B and 1-28C, both lead capacitance and interwinding capacitance will affect the toroid. 40' ( A ) Correct. ( B ) Incorrect. Interwinding Capacitance ( C ) Incorrect. Fig. 1-28. Practical winding hints. RESONANT CIRCUITS In this chapter, we will explore the parallel resonant circuit and its characteristics at radio frequencies. We will examine the concept of loaded-Q and how it re- lates to source and load impedances. We will also see the effects of component losses and how they affect circuit operation. Finally, we will investigate some methods of coupling resonant circuits to increase their selectivity. SOME DEFINITIONS The resonant circuit is certainly nothing new in rf circuitry. It is used in practically every transmitter, receiver,, or piece of test equipment in existence, to selectively pass a certain frequency or group of fre- quencies from a source to a load while attenuating all other frequencies outside of this passband. The perfect resonant-circuit passband would appear as shown in Fig. 2-1. Here we have a perfect rectangular-shaped passband with infinite attenuation above and below the frequency band of interest, while allowing the desired signal to pass undisturbed. The realization of this filter is, of course, impossible due to the physical characteristics of the components that make up a filter. As we learned in Chapter 1, there is no perfect component and, thus, there can be no perfect filter. If we understand the mechanics of resonant circuits, however, we can certainly tailor an imperfect circuit to suit our needs just perfectly. Fig. 2-2 is a diagram of what a practical filter re- 1 Passband sponse might resemble. Appropriate definitions are presented below : 1. Bandwidth-The bandwidth of any resonant circuit is most commonly defined as being the difference between the upper and lower frequency (f, - f l ) of the circuit at which its amplitude response is 3 dB below the passband response. It is often called the half-power bandwidth. 2. Q-The ratio of the center frequency of the res- onant circuit to its bandwidth is defined as the circuit Q. (Eq. 2-1) This Q should not be confused with component Q which was defined in Chapter 1. Component Q does have an effect on circuit Q, but the reverse is not true. Circuit Q is a measure of the selectivity of a resonant circuit. The higher its Q, the narrower its bandwidth, the higher is the selectivity of a resonant circuit. 3. Shape Factor-The shape factor of a resonant cir- cuit is typically defined as being the ratio of the 60-dB bandwidth to the 3-dB bandwidth of the resonant circuit. Thus, if the 60-dB bandwidth ( fq - f3) were 3 MHz and the 3-dB bandwidth ( f 2 - f l ) were 1.5 MHz, then the shape factor would be : 3 MHz SF = 1.5 MHz = 2 1 0 dB Ripple Insertion Loss 1 ... I Ultimati I \ -6ddB . . Frequency Fig. 2-1. The perfect filter response. 31 Fig. 2-2. A practical filter response. 34 RF Cmcurr DESIGN Multiply the numerator and the denominator by joC. (Remember that j2 = -1.) Thus, substituting and transposing in Equation 2-5, we have : ioL Vi, 1 - W2LC joL + 1 - W2LC Multiplying the numerator and the denominator through by 1 - w2LC yields: Vout - jwL Vi, -- (R, - w2R,LC) + joL Thus, the loss at any frequency may be calculated from the above equation or, if needed, in dB. where 1 [ represents the magnitude of the quantity within the brackets. Notice, in Fig. 2-8, that as we near the resonant fre- quency of the tuned circuit, the slope of the resonance curve increases to 12 dB/octave. This is due to the fact that we now have two significant reactances present and each one is changing at the rate of 6 dB/ octave and sloping in opposite directions. As we move away from resonance in either direction, however, the curve again settles to a 6-dBIoctave slope because, again, only one reactance becomes significant. The other reactance presents a very high impedance to the circuit at these frequencies and the circuit behaves as if the reactance were no longer there. LOADED Q The Q of a resonant circuit was defined earlier to be equal to the ratio of the center frequency of the cir- cuit to its 3-dB bandwidth (Equation 2-1). This “cir- cuit Q,” as it was called, is often given the label loaded Q because it describes the passband character- istics of the resonant circuit under actual in-circuit or loaded conditions. The loaded Q of a resonant circuit is dependent upon three main factors. (These are il- lustrated in Fig. 2-9.) 1. The source resistance ( Ra) . 2. The load resistance ( R L ) . 3. The component Q as defined in Chapter 1. Effect of R, and RL on the Loaded Q Let’s discuss briefly the role that source and load impedances play in determining the loaded Q of a resonant circuit. This role is probably best illustrated Fig. 2-9. Circuit for loaded-Q calculations. through an example. In Fig. 2-8, we plotted a resonance curve for a circuit consisting of a 50-ohm source, a 0.05-pH lossless inductor, and a 25-pF lossless capaci- tor. The loaded Q of this circuit, as defined by Equa- tion 2-1 and determined from the graph, is approxi- mately 1.1. Obviously, this is not a very narrow-band or high-Q design. But now, let’s replace the 50-ohm source with a 1000-ohm source and again plot our results using the equation derived in Fig. 2-7 (Equa- tion 2-5). This new plot is shown in Fig. 2-10. (The resonance curve for the 50-ohm source circuit is shown with dashed lines for comparison purposes.) Notice that the Q, or selectivity of the resonant circuit, has been increased dramatically to about 22. Thus, by rais- ing the source impedance, we have increased the Q of our resonant circuit. Neither of these plots addresses the effect of a load impedance on the resonance curve. If an external load of some sort were attached to the resonant circuit, as shown in Fig. 2-llA, the effect would be to broaden or “de-Q the response curve to a degree that depends on the value of the load resistance. The equivalent circuit, for resonance calculations, is shown in Fig. 2-llB. The resonant circuit sees an equivalent resis- tance of R, in parallel with R L as its true load. This total external resistance is, by definition, smaller in value than either R, or RI,, and the loaded Q must de- crease. If we put this observation in equation form, it becomes (assuming lossless components) : 0 -5 -10 5 P 3 5 -20 z fi -15 .- ., - 25 - 30 - 35 10 20 50 100 200 500 1000 Frequency (MHz) Fig. 2-10. The effect of Rs and RL on loaded Q. 35 (.A ) Resonant circuit with an external load. _ _ _ - Frequent\ Fig. 2-13 Plot of loaded-Q cui\cs for ~ i i c i i i t s $ 1 1 Fie 2-12 f 1. 1 ( B ) Eqiticalent circuit for Q calculations. Fig 2-1 1. The equivalent parallel impedance across a resonant circuit. (Eq. 2-6) where, R,, = the equivalent parallel resistance of R, and RL, Y,, = either the inductive or capacitive reactance. Equation 2-6 illustrates that a decrease in R,, will decrease the Q of the resonant circuit and an increase in R,, will increase the circuit Q, and it also illustrates another very important point. The same effect can be obtained by keeping R, constant and varying X,. Thus, for a given source and load impedance, the optimum Q of a resonant circuit is obtained when the inductor is a $mall value and the capacitor is a large value. Therefore, in either case. XI, is decreased. This effect is shown using the circuits in Fig. 2-12 and the character- istics curves i n Fig 2-13. The circuit designer, therefore, has two approaches he can follow in designing a resonant circuit with a particular Q ( Example 2-1 ) . 1. He can select an optimum value of source and load impedance. 2 . He can select component values of L and C which optimize Q. Often there is no real choice in the matter because, in many instances, the source and load are defined and we have no control over them. When this occurs, X,, ( They are equal at resonance. ) (2 = 1 I t = 142 3 7 XlHz Q = 22 4 f = 142 35 MHz (it) Large zndrrctor, ( E ) Small inductor, ymall capacitor large capacitor. Fig. 2-12. Effect of Q vs. X, at 142.35 MHz. is automatically defined for a given 4 and \e usually end up with component values that ilre irnpractical at best. Later in this chapter. we will study some methods of eliminating this problem. EXAMPLE 2-1 wurcc resistance of 150 ohms and a load iesistance of 1000 ohin5 The loaded Q must be equal to 20 at the icwnant ficquencl of 50 MHz Assume lossless componriit\ and no impcdanc e matching. Design a reson,int circuit to operate bet\\tvw Solution The effective parallel resistance acioss the resonmt CII - cuit is 150 ohms in parallel with 1000 ohms ( 1 1 R,. = 130 ohms Thus, using Equation 2 4 . R s, =L Q 130 20 -- = fi.S ohms m t l , 1 x,, = WT, = -. (0 ( Therefore, L = 20 7 nH. and C' x 489 7 pl; The Effect of Component Q on Loaded Q Thus far in this chapter, we have assumed that the components used in the resonant circuits are lossless and, thus, produce no degradation in loaded Q. In reality, however, such is not the case and the individual component Q s must be taken into account. In a lossless resonant circuit, the impedance seen across the cir- cuit's terminals at resonance is infinite. In a practical circuit, however, due to component losses, there evists some finite equivalent parallel resistance. This is il- lustrated in Fig. 2-14. The resistance ( R,,) and its associated shunt reactance ( X I , ) ran be found from the following transformation equations: R, = (Q2 + l ) R R (E¶ . 2-7) where, R,, = the equivalent parallel resistance. R, = the series resistance of the component. 36 RF CIRCUIT DESIGN OT c3 Or XS -i& Rs = Component Losses Fig. 2-14. A series-to-parallel transformation. Q = QB which equals Qp which equals the Q of the component. and, (Eq. 2-8) If the Q of the component is greater than 10, then, R p ~9 Q2RB (Eq. 2-9) and, These transformations are valid at only one frequency because they involve the component reactance which is frequency dependent (Example 2-2). Example 2-2 vividly illustrates the potential drastic effects that can occur if poor-quality (low Q) com- ponents are used in highly selective resonant circuit designs. The net result of this action is that we effec- tively place a low-value shunt resistor directly across the circuit. As was shown earlier, any low-value resis- tance that shunts a resonant circuit drastically reduces its loaded Q and, thus, increases its bandwidth. In most cases, we only need to involve the Q of the inductor in loaded-Q calculations. The Q of most capacitors is quite high over their useful frequency range, and the equivalent shunt resistance they pre- sent to the circuit is also quite high and can usually be neglected. Care must be taken, however, to ensure that this is indeed the case. x, - x, (Eq. 2-10) INSERTION LOSS Insertion loss (defined earlier in this chapter) is another direct effect of component Q. If inductors and capacitors were perfect and contained no internal re- sistive losses, then insertion loss for LC resonant cir- cuits and filters would not exist. This is, of course, not the case and, as it turns out, insertion loss is a very critical parameter in the specification of any resonant circuit, Fig. 2-16 illustrates the effect of inserting a resonant circuit between a source and its load. In Fig. 2-16A, the source is connected directly to the load. Using the voltage division rule, we find that: Vi = 0.5 Vi, ~~ ~ EXAMPLE 2 2 Given a 50-nanohenry coil as shown in Fig. 2-15A, com- pute its Q at 100 MHz. Then, transform the series circuit of Fig. 2-15A into the equivalent parallel inductance and resistance circuit of Fig. 2-15B. P 55.1 nH 108.7 0 Q ( A ) Series circuit. ( B ) Equivalent parallel circuit. Fig. 2-15. Example of a series-to-parallel transformation. Solution The Q of this coil at 100 MHz is, from Chapter 1, X. Q = x - 2a( 100 x 1Oe)(50 x 10-9) 10 - = 3.14 Then, since the Q is less than 10, use Equation 2-7 to find RP. R, = ( 4 2 + 1)R. = [(3.14)2 + 11 10 = 108.7 ohms Next, we find X, using Equation 2-8: R P X, =- QP 108.7 - 3.14 = 34.62 -- Thus, the parallel inductance becomes: L , Z W X, - 34.62 - 2 4 loo x 106) = 55.1 nH These values are shown in the equivalent circuit of Fig. 2-15B. Fig. 2-16B shows that a resonant circuit has been placed between the source and the load. Then, Fig. 2-16C illustrates the equivalent circuit at resonance. Notice that the use of an inductor with a Q of 10 at the resonant frequency creates an effective shunt re- sistance of 4500 ohms at resonance. This resistance, combined with RI,, produces an 0.9-dB voltage loss at VI when compared to the equivalent point in the cir- cuit of Fig. 2-16A. An insertion loss of 0.9 dB doesn’t sound like much, but it can add up very quickly if we cascade several RESONANT CIRCUITS 39 EXAMPLE 2-4 Design a resonant circuit with a loaded Q of 20 at a cen- ter frequency of 100 MHz that will operate between a source resistance of 50 ohms and a load resistance of 2000 ohms. Use the tapped-C approach and assume that induc- tor Q is 100 at 100 MHz. Solution We will use the tapped-C transformer to step the source resistance up to 2000 ohms to match the load resistance for optimum power transfer. (Impedance matching will be covered in detail in Chapter 4.) Thus, R,' = 2000 ohms and from Equation 2-13, we have: = 5.3 or, c, = 5.3G (Eq. 2-16) Proceeding as we did in Example 2-3, we know that for the inductor: Therefore, R, = 100 X, (Eq. 2-17) We also know that the loaded Q of the resonant circuit is equal to: where, RtOt. i -= the total equivalent shunt resistance, =Re ' 1 1 R p I / RL = 1000 I / R, and, where we have taken R.' and RL to each be 2000 ohms, in parallel. Hence, the loaded Q is (Eq. 2-18) 1000R, = ( 1000 + R,)X, Substituting Equation 2-17 (and the value of the desired loaded Q ) into Equation 2-18, and solving for X,, yields: X, = 40 ohms And, substituting this result back into Equation 2-17 gives Rp = 4000 ohms and, X L = a . w = 63.6 nH 1 G=- X P W We now know what the total capacitance must be to res- onate with the inductor. We also know from Equation 2-16 that C, is 5.3 times larger than C2. Thus, if we substitute Equation 2-16 into Equation 2-14, a n d solve the equations simultaneously, we get: = 39.78 pF C, = 47.3 pF Ci = 250.6 pF The final circuit is shown in Fig. 2-18D. Critical Coupling Under Coupling dB/octave 6 dB/octave Fig. 2-20. The effects of various values of capacitive coupling on passband response. tnain purpose of the two-resonator passively coupled filter is not to provide a narrower 3-dB bandwidth, but to increase the steepness of the stopband skirts and, thus, to reach an ultimate attenuation much faster than a single resonator could. This characteristic is shown in Fig. 2-21. Notice that the shape factor has decreased for the two-resonator design. Perhaps one way to get an intuitive feel for how this occurs is to consider that each resonator is itself a load for the other resonator, and each decreases the loaded Q of the other. But as we move away from the passband and into the stopband. the response tends to fall much ~ Frequency Fig. 2-21. Selectivity of single- and h q o-resonator d e ~ i g n ~ more quickly due to the combined response of each resonator. The value of the capacitor used to couple two identi- cal resonant circuits is given by C c12 = g where, CI2 = the coupling capacitance. C = the resonant circuit capacitance, Q = the loaded Q of a single resonator. (Eq. 2-19) 40 OdB- C .B E 2 RF Cmcurr DESIGN Critical Coupling 6 dB/octave ( A ) Below resonance. ( B ) Above resonance. Fig. 2-22. Equivalent circuit of capacitively coupled resonant circuits. One other important characteristic of a capacitively coupled resonant circuit can be seen if we take another look at Fig. 2-20. Notice that even for the critically coupled case, the response curve is not symmetric around the center frequency but is skewed somewhat. The lower frequency portion of the response plum- mets down at the rate of 18 dB per octave while the up- per slope decreases at only 6 dB per octave. This can be explained if we take a look at the equivalent circuit both above and below resonance. Below resonance, we have the circuit of Fig. 2-22A. The reactance of the two resonant-circuit capacitors ( Fig. 2-19) has in- creased, and the reactance of the two inductors has decreased to the point that only the inductor is seen as a shunt element and the capacitors can be ignored. This leaves three reactive components and each con- tributes 6 dB per octave to the response. On the high side of resonance, the equivalent cir- cuit approaches the configuration of Fig. 2-22B. Here the inductive reactance has increased above the ca- pacitive reactance to the point where the inductive reactance can be ignored as a shunt element. We now have an arrangement of three capacitors that effectively looks like a single shunt capacitor and yields a slope of 6 dB per octave. Inductive Coupling Two types of inductively coupled resonant circuits are shown in Fig. 2-23. One type (Fig, 2-23A) uses ( A ) Series inductor. Magnetic Coupling n ( B ) Transformer. Fig. 2-23. Inductive coupling. Frequency ( A ) Inductive coupling. ,Over Coupling YY, Critical Coupling 1 ,Under Coupling I Frequency ( B ) Transformer coupling. Fig. 2-24. The effects of various values of inductive coupling on passband response. a series inductor or coil to transfer energy from the first resonator to the next, and the other type (Fig. 2-23B) uses transformer coupling for the same purpose. In either case, the frequency response curves will resem- ble those of Fig. 2-24 depending on the amount of coupling. If we compare Fig. 2-24A with Fig. 2-20, we see that the two are actually mirror images of each other. The response of the inductively coupled resonator is skewed toward the higher end of the frequency spectrum, while the capacitively coupled re- sponse is skewed toward the low frequency side. An examination of the equivalent circuit reveals why. Fig. 2-25A indicates that below resonance, the capaci- ( B ) Above resonance. Fig. 2-25. Equivalent circuit of inductively coupled resonant circuits. RESONANT CIRCUITS 41 tors drop out of the equivalent circuit very quickly because their reactance becomes much greater than the. shunt inductive reactance. This leaves an arrange- ment of three inductors which can be thought of as a single tapped inductor and which produces a 6-dB per octave rolloff. Above resonance, the shunt inductors can be ignored for the same reasons, and you have the circuit of Fig. 2-2%. We now have three effective ele- ments in the equivalent circuit with each contributing 6 dR per octave to the response for a combined slope of 18 dB per octave. The mirror-image characteristic of inductively and capacitively coupled resonant circuits is a very useful concept. This is especially true in applications that require 4ymmetrical response curves. For example, suppose that a capacitively coupled design exhibited too much skew for your application. One very simple way to correct the problem would be to add a “top-L” coupled section to the existing network. The top-L coupling would attempt to skew the response in the opposite direction and would, therefore, tend to coun- teract any skew caused by the capacitive coupling. The net result is a more symmetric response shape. The value of the inductor used to couple two identi- cal resonant circuits can be found by LIZ = QL (Eq. 2-20) where, = the inductance of the coupling inductor, 0 = the loaded Q of a single resonator, 1, = the resonant circuit inductance. 1 little manipulation of Equations 2-19 and 2-20 will reveal a very interesting point. The reactance of CI, calculated with Equation 2-19 will equal the reactance of LI2 calculated with Equation 2-20 for the same operating Q and resonant frequency. The de- signer now has the option of changing any “top-C” coupled resonator to a top-L design simply by re- placing the coupling capacitor with an inductor of equal reactance at the resonant frequency. When this is done, the degree of coupling, Q, and resonant fre- quency of the design will remain unchanged while the slope of the stopband skirts will flip-flop from one side to the other. For obvious reasons, top-L coupled designs work best in applications where the primary objective i5 a certain ultimate attenuation that must be met above the passband. Likewise, top-C designs arp best for meeting ultimate attenuation specifications below the p‘issband Transformer coupling does not lend itself well to an exact design procedure because there are so many factors which influence the degree of coupling. The geometry of the coils, the spacing between them, the core materials used, and the shielding, all have a pro- noiinced effect 0 1 1 the degree of coupling attained in any design. Probably the best way to design your own trans- former is to use the old trial-and-error method. But do it in an orderly fashion and be consistent. It’s a very sad day when one forgets how he sot from point A to point B, especially if point B i? an improvement in the design. Remember: 1. Decreasing the spacing between the primary and secondary increases the coupling. 2. Increasing the permeability of the magnetic path increases the coupling. 3. Shielding a transformer decreases its loaded Q and has the effect of increasing the coupling Begin the design by setting the loaded Q of each resonator to about twice what will be needed in the actual design. Then, slowly decrease the spacing be- tween the primary and secondary until the response broadens to the loaded Q that is actually needed. If that response can’t be met, try changing the geometry of the windings or the permeability of the magnetic path. Then, vary the spacing again. Use this a? an iterative process to zero-in on the response that is needed. Granted, this is not an exact process, but it works and, if documented, can be reproduced. There are literally thousands of commerciall! avail- able transformers on the market that just might suit your needs perfectly. So before the trial-and-error method is put into practice, try a little research-it just might save a lot of time and money. Active Coupling It is possible to achieve very narrow 3-dR band- widths in cascaded resonant circuits through the use of active coupling. Active coupling, for this purpose, is defined as being either a transistor or vacuum tube which, at least theoretically, allows signal flow in only one direction (Fig. 2-26). If each of the tuned circuits is the same and if each has the same loaded Q, the total loaded Q of the cascaded circuit is approuimatel? equal to (Eq. 2-21) where, Qtlltal = the total Q of the cascaded circuit, Q = the Q of each individual resonant circuit, 11 = the number of resonant circuits. The first step in any design procedure must be to Bt Fig. 2-26. Active coiipling FILTER DESIGN Filters occur so frequently in the instrumentation and communications industries that no book covering the field of rf circuit design could be complete without at least one chapter devoted to the subject. Indeed, entire books have been written on the art of filter de- sign alone, so this single chapter cannot possibly cover all aspects of all types of filters. But it will familiarize you with the characteristics of four of the most com- monly used filters and will enable you to design very quickly and easily a filter that will meet, or exceed, most of the common filter requirements that you will encounter. We will cover Butterworth, Chebyshev, and Bes- sei filters in all of their common configurations: low- pass, high-pass, bandpass, and bandstop. We will learn how to take advantage of the attenuation char- acteristics unique to each type of filter. Finally, we will learn how to design some very powerful filters in as little as 5 minutes by merely looking through a catalog to choose a design to suit your needs. BACKGROUND In Chapter 2, the concept of resonance was ex- plored and we determined the effects that component value changes had on resonant circuit operation. You should now be somewhat familiar with the methods that are used in analyzing passive resonant circuits to find quantities, such as loaded Q, insertion loss, and bandwidth. You should also be capable of designing one- or two-resonator circuits for any loaded Q desired (or, at least, determine why you cannot). Quite a few of the filter applications that you will encounter, how- ever, cannot be satisfied with the simple bandpass arrangement given in Chapter 2. There are occasions when, instead of passing a certain band of frequencies while rejecting frequencies above and below (band- pass), we would like to attenuate a small band of fre- quencies while passing all others. This type of filter is called, appropriately enough, a bandstop filter. Still other requirements call for a low-pass or high-pass response. The characteristic curves for these responses are shown in Fig, 3-1. The low-pass filter will allow all signals below a certain cutoff frequency to pass while attenuating all others. A high-pass filter’s re- sponse is the mirror-image of the low-pass response and attenuates all signals below a certain cutoff fre- quency while allowing those above cutoff to pass. These types of response simply cannot be handled very well with the two-resonator bandpass designs of Chapter 2. In this chapter, we will use the low-pass filter as our workhorse, as all other responses will be derived from it. So let’s take a quick look at a simple low-pass filter and examine its characteristics, Fig. 3-2 is an example of a very simple two-pok, or second-order low-pass filter. The order of a filter is determined by the slope of the attenuation curve it presents in the stopband. A second-order filter is one whose rolloff is a function of the frequency squared, or 12 dB per octave. A third-order filter causes a rolloff that is pro- portional to frequency cubed, or 18 dB per octave. Thus, the order of a filter can be equated with the number of significant reactive elements that it pre- sents to the source as the signal deviates from the passband. The circuit of Fig. 3-2 can be analyzed in much the same manner as was done in Chapter 2. For instance, an examination of the effects of loaded Q on the re- sponse would yield the family of curves shown in Fig. 3-3. Surprisingly, even this circuit configuration can cause a peak in the response. This is due to the fact that at some frequency, the inductor and capacitor will become resonant and, thus, peak the response if the loaded Q is high enough. The resonant frequency can be determined from 1 Fr = 2& ( Eq. 3-1 ) For low values of loaded Q, however, no response peak will be noticed. The loaded Q of this filter is dependent upon the individual Q s of the series leg and the shunt leg where, assuming perfect components, and, Qz = x, R L (Eq. 3-3) 44 FILTER DESIGN 45 and the total Q is: ( Eq. 3-4 ) If the total Q of the circuit is greater than about 0.5, then for optimum transfer of power from the source to the load, Q1 should equal Q z . In this case, at the Frequency ( A ) Low-pass. peak frequency, the response will approach 0-dB in- sertion loss. If the total Q of the network is less than about 0.5, there will be no peak in the response and, for optimum transfer of power, R, should equal RL. The peaking of the filter’s response is commonly called ripple (defined in Chapter 2) and can vary consider- 0 l o f + - - - 2 g 20 2 Y m 3 30 40 50 5 0 10 0 1 0 2 0 5 1 0 L O Frequency ( I f,i Fig. 3-3. Typical two-pole filter response curves. Fig. 3-4. Three-element low-pass filter. 8 .- Y Y tt Frequency ( B ) High-pass. Frequency Fig. 3-5. Typical response of a three-element low-pass filter. Frequency ( C ) Bandstop. Fig. 3-1. Typical filter response curves. Fig. 3-2. A simple low-pass filter. 50 60 0 10 5 20 0 1 0 2 0 5 1 0 2 0 50 10 Frequency ( f / f c ) Fig. 3-6. Curves showing frequency response vs. loaded Q for three-element low-pass filters. 46 RF CIRC~UIT DFSICN ably from one filter design to the next depending on the application. As shown, the two-element filter ex- hibits only one response peak at the edge of the pass- band. It can be shown that the number of peaks within the passband is directly related to the number of ele- ments in the filter by: Number of Peaks = N - 1 where, Thus, the three-element low-pass filter of Fig. 3-4 should exhibit two response peaks as shown in Fig. 3-5. This is true only if the loaded Q is greater than one. Typical response curves for various values of loaded Q for the circuit given in Fig. 3-4 are shown in Fig. 3-6. For all odd-order networks, the response at dc and at the upper edge of the passband ap- proaches 0 dB with dips in the response between the two frequencies. All even-order networks will pro- duce an insertion loss at dc equal to the amount of passband ripple in dB. Keep in mind, however, that either of these two networks, if designed for low values of loaded Q, can be made to exhibit little or no pass- band ripple. But, as you can see from Figs. 3-3 and 3-6, the elimination of passband ripple can be made only at the expense of bandwidth. The smaller the ripple that is allowed, the wider the bandwidth be- comes and, therefore, selectivity suffers. Optimum flatness in the passband occurs when the loaded Q of the three-element circuit is equal to one ( 1 ) . Any value of loaded Q that is less than one will cause the response to roll off noticeably even at very low frequencies, within the defined passband. Thus, not only is the selectivity poorer but the passband inser- tion loss is too. In an application where there is not much signal to begin with, an even further decrease in signal strength could be disastrous. Now that we have taken a quick look at two repre- sentative low-pass filters and their associated responses, let’s discuss filters in general: N = the number of elements, High-Q filters tend to exhibit a far greater initial slope toward the stopband than their low-Q coun- terparts with the same number of elements. Thus, at any frequency in the stopband, the attenuation will be greater for a high-Q filter than for one with a lower Q. The penalty for this improvement is the increase in passband ripple that must occur as a result. Low-Q filters tend to have the flattest passband response but their initial attenuation slope at the band edge is small. Thus, the penalty for the re- duced passband ripple is a decrease in the initial stopband attenuation. As with the resonant circuits discussed in Chapter 2, the source and load resistors loading a filter will have a profound effect on the Q of the filter and, therefore, on the passband ripple and shape factor 4. of the filter. If a filter is inserted between two re- sistance values for which it was not designed, the performance will suffer to an extent, depending upon the degree of error in the terminating im- pedance values. The final attenuation slope of the response is de- pendent upon the order of the network. The order of the network is equal to the number of reactive elements in the low-pass filter. Thus, a second-order network ( 2 elements) falls off at a final attenuation slope of 12 dB per octave, a third-order network ( 3 elements) at the rate of 18 dB per octave, and so on, with the addition of 6 dB per octave per ele- ment. MODERN FILTER DESIGN Modern filter design has evolved through the years from a subject known only to specialists-in the ~ field (because of the advanced mathematics involved) to a practical well-organized catalog of ready-to-use cir- cuits available to anyone with a knowledge of eighth grade level math. In fact, an average individual with absolutely no prior practical filter design experience should be able to sit down, read this chapter, and within 30 minutes be able to design a practical high- pass, low-pass, bandpass, or bandstop filter to his specifications. It sounds simple and it is-once a few basic rules are memorized. The approach we will take in all of the designs in this chapter will be to make use of the myriad of normalized low-pass prototypes that are now avail- able to the designer. The actual design procedure is, therefore, nothing more than determining your re- quirements and, then, finding a filter in a catalog which satisfies these requirements. Each normalized element value is then scaled to the frequency and im- pedance you desire and, then, transformed to the type of response ( bandpass, high-pass, bandstop) that you wish. With practice, the procedure becomes very sim- ple and soon you will be defining and designing filters. The concept of normalization may at first seem foreign to the person who is a newcomer to the field of filter design, and the idea of transforming a low-pass filter into one that will give one of the other three types of responses might seem absurd. The best advice I can give (to anyone not familiar with these prac- tices and who might feel a bit skeptical at this point) is to press on. The only way to truly realize the beauty and simplicity of this approach is to try a few actual designs, Once you try a few, you will be hooked, and any other approach to filter design will suddenly seem tedious and unnecessarily complicated. NORMALIZATION AND THE LOW-PASS PROTOTYPE In order to offer a catalog of useful filter circuits to the electronic filter designer, it became necessary to FILTER DESIGN 49 ized for a load resistance of 1 ohm, while taking what we get for the source resistance. Dividing both the load and source resistor by 10 will yield a load re- sistance of 1 ohm and a source resistance of 5 ohms as shown in Fig. 3-12. We can use the normalized terminating resistors to help us find a low-pass proto- type circuit. Table 3-2 is a list of Butterworth low-pass proto- type values for various ratios of source to load im- pedance ( Rs/RL). The schematic shown above the table is used when R,/R, is calculated, and the ele- ment values are read down from the top of the table. Table 3-2A. Butterworth Low-Pass Prototype Element Values 1.111 1.250 1.429 1.667 2.000 2.500 3.333 5.000 10.000 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 1.111 1.250 1.429 1.667 2.000 2.500 3.333 5.000 10.000 rJ 30 1) 1.035 0.849 0.697 0.566 0.448 0.342 0.215 0.156 0.074 1.414 0.808 0.844 0.915 1.023 1.181 1.425 1.838 2,669 5.167 1.500 0.466 0.388 0.325 0.269 0.218 0.169 0.124 0.080 0.039 1.531 1.835 2.121 2.439 2.828 3.316 4.095 5.313 7.707 14.814 0.707 1.633 1.384 1.165 0.965 0.779 0.604 0.440 0.284 0.138 1.333 1.592 1.695 1.862 2.103 2.452 2.986 3.883 5.684 11.094 1.577 1.599 1.926 2.277 2.702 3.261 4.064 5.363 7.910 15.455 0.500 1.744 1.511 1.291 1.082 0.883 0.691 0.507 0.331 0.162 I .469 1.811 2.175 2.613 3.187 4.009 5.338 7.940 15.642 1.082 0.383 1 I Alternately, when RI,/ Rs is calculated, the schematic below the table is used while reading up from the bottom of the table to get the element values (Ex- ample 3-2). EXAMPLE 3-2 Find the low-pass prototype value for an n = 4 Butter- worth filter with unequal terminations: RP = 50 ohms, RL = 100 ohms. Solution Normalizing the two terminations for RI, = 1 ohm will yield a value of RS = 0.5. Reading down from the top of Table 3-2, for an n = 4 low-pass prototype value, we see that there is no R ~ R L = 0.5 ratio listed. Our second choice, then, is to take the value of RL/Rs = 2, and read up from the bottom of the table while using the schematic below the table as the form for the low-pass prototype values. This approach results in the low-pass prototype circuit of Fig. 3-13. Fig. 3-13. Low-pass prototype circuit for Example 3-2. Obviously, all possible ratios of source to load re- sistance could not possibly fit on a chart of this size. This, of course, leaves the potential problem of not being able to find the ratio that you need for a par- ticular design task. The solution to this dilemma is to simply choose a ratio which most closely matches the ratio you need to complete the design. For ratios of 1OO:l or so, the best results are obtained if you assume this value to be so high for practical purposes as to be infinite. Since, in these instances, you are only approximating the ratio of source to load resis- tance, the filter derived will only approximate the re- sponse that was originally intended. This is usually not too much of a problem. The Chebyshev Response The Chebyshev filter is a high-Q filter that is used when: ( 1 ) a steeper initial descent into the stopband is required, and (2) the passband response is no longer required to be flat. With this type of require- ment, ripple can be allowed in the passband. As more ripple is introduced, the initial slope at the beginning of the stopband is increased and produces a more rectangular attenuation curve when compared to the rounded Butterworth response. This comparison is made in Fig. 3-14. Both curves are for n = 3 filters. The Chebyshev response shown has 3 dB of passband ripple and produces a 10 dB improvement in stopband attenuation over the Butterworth filter. 50 Table 3-2B. Butterworth Low-Pass Prototype Element Values RF Cmcurr DESIGN n R d R L 5 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 6 1.111 1.250 1.429 1.667 2.000 2.500 3.333 5.000 10.000 7 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 W W W Cl L2 0.412 1.027 0.470 0.866 0.517 0.731 0.586 0.609 0.686 0.496 0.838 0.388 1.094 0.285 1.608 0.186 3.512 0.091 1.545 1.694 0.289 1.040 0.245 1.116 0.207 1.236 0.173 1.407 0.141 1.653 0.111 2.028 0.082 2.656 0.054 3.917 0.026 7.705 1.553 1.759 0.299 0.711 0.322 0.606 0.357 0.515 0.408 0.432 0.480 0.354 0.590 0.278 0.775 0.206 1.145 0.135 2.257 0.067 1.558 1.799 1.910 2.061 2.285 2.600 3.051 3.736 4.884 7.185 14.095 1.382 1.322 1.126 0.957 0.801 0.654 0.514 0.379 0.248 0.122 1.553 1.404 1.517 1.688 1.928 2.273 2.795 3.671 5.427 10.700 1.659 1.756 1.544 1.333 1.126 0.924 0.727 0.537 0.352 0.173 0.894 2.054 2.239 2.499 2.858 3.369 4.141 5.433 8.020 15.786 1.202 1.489 1.278 1.091 0.917 0.751 0.592 0.437 0.287 0.142 1.397 c5 1.389 1.738 2.108 2.552 3.133 3.965 5.307 7.935 15.710 0.309 1.744 1.550 1.346 1.143 0.942 0.745 0.552 0.363 0.179 0.758 2.125 2.334 2.618 3.005 3.553 4.380 5.761 8.526 16.822 1.055 1.335 1.688 2.062 2.509 3.094 3.931 5.280 7.922 15.738 0.259 1.727 1.546 1.350 1.150 0.951 0.754 0.560 0.369 0.182 0.656 c7 1.296 1.652 2.028 2.477 3.064 3.904 5.258 7.908 15.748 0.223 The attenuation of a Chebyshev filter can be found by making a few simple but tiresome calculations, and can be expressed as: are given in Table 3-3. The parameter E is given by: (Eq. 3-8) E = ~ l O R a , / 1 0 - 1 where, A d B = 10 log [ 1 + @Cn2 (E) ’1 (Eq. 3-7) RdB is the passband ripple in decibels. Note that - ’ is not the same as - . The quan- (3 (3 where, Cn2 (:)’ is the Chebyshev polynomial to the order can be found by defining another parameter: n evaluated at (2)’. \O,I The Chebyshev polynomials for the first seven orders 1 B = - n cosh-1 (i) (Eq. 3-9) FILTER DESIGN 51 4(1 Chebyshev Respons \\ . L I 1 2 3 4 Frequency I f f,i Fig 3- 14 Comparison of three-element Chebyshev ,tnd Buttern orth responses Table 3-3. Chebyshev Polynomials to the Order n n Chebvshev Polunomial w w , - where, t i = the order of the filter, E = the parameter defined in Equation 3-8, cosh-I = the inverse hyperbolic cosine of the quan- tity in parentheses. Finally. we have: (E)' = (:) cosh B (Eq. 3-10) where, (2) = the ratio of the frequency of interest to the cutoff fre uency, cosh = the hyper B olic cosine. If your calculator does not have hyperbolic and in- verse hyperbolic functions, they can be manually de- termined from the following relations: cosh x = 0.5( e x + e-I) cosh - l y = ln (x 2 d x z - 1 ) The preceding equations yield families of attenua- tion curves, each classified according to the amount of and ripple allowed in the passband. Several of these fami- lies of curves are shown in Figs. 3-15 through 3-18, and include 0.01-dB, 0.1-dB, 0.5-dB, and 1.0-dB ripple. Each curve begins at o / w , = 1, which is the normal- ized cutoff, or 3-dB frequency. The passband ripple is. therefore, not shown. If other families of attenuation curves are needed with different values of passband ripple, the preceding Chebyshev equations can be used to derive them. The problem in Example 3-3 illustrates this. Obviously, performing the calculations of Example 3-3 for various values of w/w, , ripple, and filter order is a very time-consuming chore unless a programmable caIculator or computer is available to do most of the work for you, The low-pass prototype element values correspond- ing to the Chebyshev responses of Figs. 3-15 through 3-18 are given in Tables 3-4 through 3-7. Note that the Chebyshev prototype values could not be separated into two distinct sets of tables covering the equal and 0 12 24 36 m 48 c ;;j 60 E 72 d 84 96 - 108 - + + I I20 I O 1 5 2 0 2 5 3 0 3 5 . 1 : Q 3 I O Frequency Ratio 11 t Fig. 3-15 Attenuation charncteristic, for a Chthyshr \ filter with 0.01-dR ripple Frequency Ratio i f f , ' Fig. 3-16. Attenuation characteristic5 foi a Chehyshe\ filter with 0.1-dB ripple 54 RF Cmcurr DESIGN Table 3-4B. Chebyshev Low-Pass Element Values for 0.01-dB Ripple L6 1 1.000 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 1.101 1.111 1.250 1.429 1.667 2.000 2.500 3.333 5.000 10.000 1 .ooo 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 W W M 0.977 0.880 0.877 0.926 1.019 1.166 1.398 1.797 2.604 5.041 1.547 0.851 0.760 0.545 0.436 0.351 0.279 0.214 0.155 0.100 0.048 1.551 0.913 0.816 0.81 1 0.857 0.943 1.080 1.297 1.669 2,242 4.701 1.559 1.685 1.456 1.235 1.040 0.863 0.699 0.544 0.398 0.259 0.127 1.795 1.796 1.782 1.864 2.038 2.298 2.678 3.261 4.245 6.223 12.171 1.847 1.595 1.362 1.150 0.967 0.803 0.650 0.507 0.372 0.242 0.119 1.867 2.037 2.174 2.379 2.658 3.041 3.584 4.403 5.772 8.514 16.741 1.645 1.841 1.775 1.489 1.266 1.061 0.867 0.682 0.503 0.330 0.162 1.790 2.002 2.089 2.262 2.516 2.872 3.382 4.156 5.454 8.057 15.872 1.866 1.685 1.641 1.499 1.323 1.135 0.942 0.749 0.557 0.368 0.182 1.237 2.027 2.094 2.403 2.735 3.167 3.768 4.667 6.163 9.151 18.105 1.598 1.870 1.722 1.525 1.323 1.124 0.928 0.735 0.546 0.360 0.175 1.765 0.977 1.274 1.607 1.977 2.424 3.009 3.845 5.193 7.826 15.613 0.488 1.631 1.638 1.507 1.332 1.145 0.954 0.761 0.568 0.376 0.187 1.190 2.002 2.202 2.465 2.802 3,250 3.875 4.812 6.370 9.484 18.818 1.563 0.937 1.053 1.504 1.899 2.357 2.948 3.790 5.143 7.785 15.595 0.469 1.595 1.581 1.464 1.307 1.131 0.947 0.758 0.568 0.378 0.188 0.913 1.206 1.538 1,910 2.359 2.948 3.790 5.148 7.802 15.652 1.161 0.456 The transformation is affected through the following formulas : L = the final inductor value, C, = a low-pass prototype element value, L, = a low-pass prototype element value, R = the final load resistor value, f, = the final cutoff frequency. The normalized low-pass prototype source resistor L=- RL, ( 3-13) must also be transformed to its final value by multi- plying it by the final value of the load resistor (Ex- ample 3-5). Thus, the ratio of the two always remains the same. C = - cn (Eq. 3-12) 2d,R and 27rf c C = the final capacitor value, where, FILTER DESIGN 55 Table 3-5A. Chebyshev Low-Pass Prototype Element Values for 0.1-dB Ripple 1.355 1.429 1.667 2.000 2.500 3.333 5.000 10.000 1.000 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 1.355 1.429 1.667 2.000 2.500 3.333 5.000 10.000 2 2 CZI 1.209 0.977 0.733 0.560 0.417 0.293 0.184 0.087 1.391 1.433 1.426 1.451 1.521 1.648 1.853 2.186 2.763 3.942 7.512 1.513 0.992 0.779 0.576 0.440 0.329 0.233 0.148 0.070 1.511 1.638 1.982 2.489 3.054 3.827 5.050 7.426 14.433 0.819 1.594 1.494 1.356 1.193 1.017 0.838 0.660 0.486 0.317 0.155 1.510 2.148 2.348 2.730 3.227 3.961 5.178 7.607 14.887 1.768 1.433 1.622 1.871 2.190 2.603 3.159 3.968 5.279 7.850 15.466 0.716 1.585 1.429 1. 1.185 2. 11 30 13 0.967 2.856 0.760 3.698 0.560 5.030 0.367 7.614 0.180 15.230 1.455 0.673 The process for designing a low-pass filter is a very simple one which involves the following procedure: 1. Define the response you need by specifying the re- quired attenuation characteristics at selected fre- quencies. 2. Normalize the frequencies of interest by dividing them by the cutoff frequency of the filter. This step forces your data to be in the same form as that of the attenuation curves of this chapter, where the 3-dB point on the curve is: EXAMPLE 3-5 Scale the low-pass prototype values of Fig. 3-19 (Exam- ple 3-4) to a cutoff frequency of 50 MHz and a load resis- tance of 250 ohms. Solution Use Equations 3-12 and 3-13 to scale each component as follows: 3.546 2 d 5 0 x 106)(250) c, = = 45 pF 9.127 2 ~ ( 5 0 x 106)(250) c= = 116 pF 7.889 c ” 2 ~ ( 5 0 x 106)(250) = 100pF ( 250) ( 0.295) L z = 2 4 5 0 x 106) = 235nH (250) (0.366) 2Tr(50 x 106) L, = = 291 nH ized value by the final value of the load resistor. The source resistance is scaled by multiplying its normal R a t r i n . ~ , = 0.2( 250) = 50 ohms The final circuit appears in Fig. 3-21. i 45 ill6 pF i f100 pF f 2 5 0 I! i Fig. 3-21. Low-pass filter circuit for Example 3-5. 3. Determine the maximum amount of ripple that you can allow in the passband. Remember, the greater the amount of ripple allowed, the more selective the filter is. Higher values of ripple may allow you to eliminate a few components. 4. Match the normalized attenuation characteristics (Steps 1 and 2 ) with the attenuation curves pro- vided in this chapter. Allow yourself a small “fudge- factor” for good measure. This step reveals the mini- mum number of circuit elements that you can get away with-given a certain filter type. 5. Find the low-pass prototype values in the tables. 6. Scale all elements to the frequency and impedance of the final design. - _ f - l f c Example 3-6 diagrams the process of designing a low- pass filter using the preceding steps. 56 RF Cmcurr DESIGN Table 3-5B. Chebyshev Low-Pass Prototype Element Values for 0.1-dB Ripple 1.000 0,900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 1.355 1.429 1.667 2.000 2.500 3.333 5.000 10.000 1.000 0.900 0.800 0.700 0.600 0.500 0.400 0.300 0.200 0.100 CO 00 M 1.301 1.285 1.300 1.358 1.470 1.654 1.954 2.477 3.546 6.787 1.561 0.942 0.735 0.542 0.414 0.310 0.220 0.139 0.067 1.534 1.262 1.242 1.255 1.310 1.417 1.595 1.885 2.392 3.428 6.570 1.575 1.556 1.433 1.282 1.117 0.917 0.778 0.612 0.451 0.295 0.115 1.807 2.080 2.249 2.600 3.068 3.765 4.927 7.250 14.220 1.884 1.520 1.395 1.245 1.083 0.917 0.753 0.593 0.437 0.286 0.141 1.858 2.241 2.380 2.582 2.868 3.269 3.845 4.720 6.196 9.127 17.957 1.766 1.659 1.454 1.183 0.958 0.749 0.551 0.361 0.178 1.831 2.239 2.361 2.548 2.819 3.205 3.764 4.618 6.054 8.937 17.603 1.921 1.556 1.488 1.382 1.244 1.085 0.913 0.733 0.550 0.366 0.182 1.417 2.247 2.544 3.064 3.712 4.651 6.195 9.261 18.427 1.749 1.680 1.578 1.443 1.283 1.209 0.928 0.742 0.556 0.369 0.184 1.827 1.301 1.488 1.738 2.062 2.484 3.055 3.886 5.237 7.889 15.745 0.651 1.534 1.405 1.185 0.979 0.778 0.580 0.384 0.190 1.394 2.239 2.397 2.624 2.942 3.384 4.015 4.970 6.569 9.770 19.376 1.734 1.277 1.629 2.174 2.794 3.645 4.996 7.618 15.350 0.638 1.520 1.459 1.362 1.233 1.081 0.914 0.738 0.557 0.372 0.186 1.379 1.262 1.447 1.697 2.021 2.444 3.018 3.855 5.217 7.890 15.813 0.631 HIGH-PASS FILTER DESIGN Once you have learned the mechanics of low-pass filter design, high-pass design becomes a snap, You can use all of the attenuation response curves presented, thus far, for the low-pass filters by simply inverting the f/f, axis. For instance, a 5-element, 0.1-dB-ripple Chebyshev low-pass filter will produce an attenuation of about 60 dB at an f / f c of 3 (Fig. 3-16). If you were working instead with a high-pass filter of the same size and type, you could still use Fig. 3-16 to tell you that a t an f / f , of 1/3 (or, f,/f = 3) a 5-element, 0.1- dB-ripple Chebyshev high-pass filter will also produce an attenuation of 60 dB. This is obviously more con- venient than having to refer to more than one set of curves. After finding the response which satisfies all of the requirements, the next step is to simply refer to the tables of low-pass prototype values and copy down the prototype values that are called for. High-pass val- ues for the elements are then obtained directly from the low-pass prototype values as follows (refer to Fig. 3-24): FILTER DEIGN 1.821 1.: 1 i 1 59 Table 3-7A. Chebyshev Low-Pass Prototype Element Values for 1.0-dB Ripple ( A ) Low-pass prototype circuit. 0 549 r"-r;'T? ( B ) Equioalent high-pass prototype circuit. Fig. 3-24. Low-pass to high-pass filter transformation. 3. Change all shunt branches to series branches, and 4. Change all elements in series with each other into 5. Change all voltage sources into current sources, and Fig. 3-26 shows a ladder network and its dual repre- sentation. Dual networks are convenient, in the case of equal terminations, if you desire to change the topology of the filter without changing the response. It is most often used, as shown in Example 3-7, to eliminate an unnecessary inductor which might have crept into the design through some other transformation process. Inductors are typically more lower-Q devices than capacitors and, therefore, exhibit higher losses. These losses tend to cause insertion loss, in addition to gen- erally degrading the overall performance of the filter. The number of inductors in any network should. therefore, be reduced whenever possible. A little experimentation with dual networks having unequal terminations will reveal that you can quickly get yourself into trouble if you are not careful. This is especially true if the load and source resistance are a design criteria and cannot be changed to suit the needs of your filter. Remember, when the dual of a network with unequal terminations is taken, then, the terminations must, by definition, change value as shown in Fig. 3-26. vice-versa. elements that are in parallel with each other. vice-versa. BANDPASS FILTER DESIGN The low-pass prototype circuits and response curves given in this chapter can also be used in the design of bandpass filters. This is done through a simple 2 3.000 4.000 8.000 3 1.000 0.500 0.333 0.250 0.125 4 3.000 4.000 8.000 m 3c x 0.572 0.365 0.157 1.213 2.216 4.431 6.647 8.862 17.725 1.652 0.653 0.452 0.209 1.350 3.132 4.600 9.658 1.109 1.088 0.817 0.726 0.680 0.612 1.460 4.411 7.083 15.164 2.010 2.216 2.216 2.216 2.216 2.216 1108 0.814 2.535 I) 612 2.848 0428 3.281 1.488 1.106 L transformation process similar to what was done in the high-pass case. The most difficult task awaiting the designer of a bandpass filter, if the design is to be derived from the low-pass prototype, is in specifying the bandpass at- tenuation characteristics in terms of the low-pass re- sponse curves. A method for doing this is shown by the curves in Fig. 3-27. As you can see, when a low- pass design is transformed into a bandpass design, the attenuation bandwidth ratios remain the same. This means that a low-pass filter with a 3-dB cutoff frequency, or a bandwidth of 2 kHz, would transform into a bandpass filter with a 3-dB bandwidth of 2 kHz. If the response of the low-pass network were down 30 dB at a frequency or bandwidth of 4 kHz ( f / f , = 2) , then the response of the bandpass network would be down 30 dB at a bandwidth of 4 kHz. Thus, the nornialized f/f,. axis of the low-pass attenuation curves becomes a ratio of bandwidths rather than fre- quencies, such that: (Eq. 3-14) BW - f BW, f,. where, RW = the bandwidth at the required value of at- BW, = the 3-dB bandwidth of the bandpass filter. tenuation, 60 Table 3-7B. Chebyshev Low-Pass Prototype Element Values for 1.0-dB Ripple RF Cmcurr DFSIGN 0.500 4.414 0.565 0.333 6.622 0.376 0.250 8.829 0.282 0.125 17.657 0.141 01 1.721 1.645 6 3.000 0.679 3.873 4.000 0.481 5.644 8.000 0.227 12.310 W 1.378 2.097 7 1.OOO 2.204 1.131 0.500 4.408 0.566 0.333 6.612 0.377 0.250 8.815 0.283 0.125 17.631 0.141 01 1.741 1.677 4.653 6.205 7.756 13.961 2.061 0.771 0.476 0.198 1.690 3.147 6.293 9.441 12.588 25.175 2.155 n R d R L Cl L2 c3 L4 cti L6 c, 5 1.000 2.207 1.128 3.103 1.128 2.207 1.128 2.207 1.128 2.207 1.128 2.207 1.128 2.207 1.493 1.103 4.711 0.969 2.406 7.351 0.849 2.582 16.740 0.726 2.800 2.074 1.494 1.102 1.194 3.147 1.131 2.204 0.895 3.147 1.131 2.204 0.796 3.147 1.131 2.204 0.747 3.147 1.131 2.204 0.671 3.147 1.131 2.204 1.703 2.079 1.494 1.102 Often a bandpass response is not specified, as in Example 3-8. Instead, the requirements are often given as attenuation values at specified frequencies as shown by the curve in Fig. 3-28. In this case, you must transform the stated requirements into informa- tion that takes the form of Equation 3-14. As an ex- ample, consider Fig. 3-28. How do we convert the data that is given into the bandwidth ratios we need? Before we can answer that, we have to find fa. Use the following method. The frequency response of a bandpass filter exhibits geometric symmetry. That is, it is only symmetric when plotted on a logarithmic scale. The center frequency of a geometrically symmetric filter is given by the formula: fo=a (Eq. 3-15) where f, and f b are any two frequencies (one above and one below the passband) having equal attenua- tion. Therefore, the center frequency of the response curve shown in Fig. 3-28 must be f, = d w ) MHz = 58.1 MHz We can use Equation 3-15 again to find fS. 58.1 = d = ) or, f3 = 27 MHz Now that fs is known, the data of Fig. 3-28 can be put into the form of Equation 3-14. BW40 dB - 125 MHz - 27 MHz BW3dB -- 75 MHz - 45 MHz = 3.27 To find a low-pass prototype curve that will satisfy these requirements, simply refer to any of the pertinent graphs presented in this chapter and find a response which will provide 40 dB of attenuation at an f/f, of 3.27. ( A fourth-order or better Butterworth filter will do quite nicely. ) The actual transformation from the low-pass to the bandpass configuration is accomplished by resonating each low-pass element with an element of the opposite type and of the same value. All shunt elements of the low-pass prototype circuit become parallel-resonant FILTER DESIGN 61 EXAMPLE 3-7 Design an LC high-pass filter with an f, of 60 MHz and a minimum attenuation of 40 dB at 30 MHz. The source and load resistance are equal at 300 ohms. Assume that a 0.5-dB passband ripple is tolerable. Solution First, normalize the attenuation requirements so that the low-pass attenuation curves may be used. f - 30 MHz f , 60MHz = 0.5 Inverting, we get. _ - ; - 2 how, select a normalized low-pass filter that offers at least 40-dB attenuation at a ratio of f,/f = 2. Reference to Fig. 3.17 (attenuation response of 0.5-dB-ripple Chebyshev fil- ters) indicates that a normalized n = 5 Chebyshev will pro- vide the needed attenuation. Table 3-6 contains the ele- ment values for the corresponding network. The normalized low-pass prototype circuit is shown in Fig. 3-25A. Note that the schematic below Table 3-6B was chosen as the low-pass prototype circuit rather than the schematic above the table. The reason for doing this will become obvious after the next step. Keep in mind, however, that the ratio of R ~ / R L is the same as the ratio of R&s, and is unity. Therefore, it does not mattcar which form is used for the prototype circuit. Next, transform the low-pass circuit to a high-pass net- work by replacing each inductor with a capacitor, and vice- versa, using reciprocal element values as shown in Fig. 3-25B. Note here that had we begun with the low-pass pro- totype circuit shown above Table 3-6B, this transformation \\.odd have yielded a filter containing three inductors rather than the two shown in Fig. 3-25B. The object in any of these filter designs is to reduce the number of inductors in the final design. More on this later. The final step in the design process is to scale the net- work in both impedance and frequency using Equations 3-12 and 3-13, The first two calculations are done for you. 1 __ 1.807 2n(60 X 106)(300) c, = = 3.9 uF 1807 2.691 I 807 I 1- I ( A ) Normalized low-pass filter circuit ( B ) High-pass transformation. ( C ) Frequency and impedance-scaled filter circuit. Fig. 3-25. High-pass filter design for Example 3-7. 274 60 x 106) Lz = = 611 nH The remaining values are: C, = 3.3 pF C, z 4.9 pF L, = 611 nH The final filter circuit is given in Fig. 3-25C. EXAMPLE 3-8 Find the Butterworth low-pass prototype circuit which, \vhen transformed, would satisfy the following bandpass filtrr requirement\: BWa,,, = 2 M H z BWm,in = 6 MHz Solution Note that we arc not concerned with the center frequency of the bandpass response just yet. We are only concerned with the relationship between the above requirements and ~~~ the low-pass response curves. Using Equation 3-14. we have: BW f BW,oan BWv- f , - BWa," 6 MHz - 2 MHz = 3 -- Therefore, turn to the Butterworth response curves shown in Fig. 3-9 and find a prototype value that will provide 40 dB of attenuation at an f / f , = 3. The curves indicate a 5- element Butterworth filter will provide the needed attenu- ation. 64 RF Cmcurr DJBIGN (Eq. 3-19) where, in all cases, R = the final load impedance, R = the 3-dB bandwidth of the final design, f, = the geometric center frequency of the final de- L, = the normalized inductor bandpass element C, = the normalized capacitor bandpass element Example 3-9 furnishes one final example of the pro- sign, values, values. cedure for designing a bandpass filter. SUMMARY OF THE BANDPASS FILTER DESIGN PROCEDURE 1. Transform the bandpass requirements into an equivalent low-pass requirement using Equation 2. Refer to the low-pass attenuation curves provided in order to find a response that meets the requirements of Step 1. 3. Find the corresponding low-pass prototype and write it down. 4. Transform the low-pass network into a bandpass configuration. 5. Scale the bandpass configuration in both impedance and frequency using Equations 3-16 through 3-19. 3-14. BAND-REJECTION FILTER DESIGN Band-rejection filters are very similar in design ap- proach to the bandpass filter of the last section. Only, in this case, we want to reject a certain group of fre- quencies as shown by the curves in Fig. 3-30. The band-reject filter lends itself well to the low- pass prototype design approach using the same proce- dures as were used for the bandpass design. First, define the bandstop requirements in terms of the low- pass attenuation curves. This is done by using the inverse of Equation 3-14. Thus, referring to Fig. 3-30, we have : BWC - f q - fi BW f3 - fz This sets the attenuation characteristic that is needed and allows you to read directly off the low-pass at- tenuation curves by substituting BWJBW for fJf on the normalized frequency axis. Once the number of elements that are required in the low-pass prototype circuit is determined, the low-pass network is trans- formed into a band-reject configuration as follows: Each shunt element in the low-pass prototype circuit is replaced by a shunt series-resonant circuit, and each series-element is replaced by a series paralkhesmnt circuit. --- EXAMPLE 3-9 Design a bandpass filter with the following requirements: f. = 75 MHz BWsdn = 7 MHz BWsede = 35 MHz Passband Ripple = 1 dB R. = 50 ohms RL = 100 ohms Solution Using Equation 3-14: = 5 Substitute this value for f / f , in the low-pass attenuation curves for the 1-dB-ripple Chebyshev response shown in Fig. 3-18. This reveals that a 3-element filter will provide about 50 dB of attenuation at an f/f, = 5, which is more than adequate. The corresponding element values for this filter can be found in Table 3-7 for an R J R L = 0.5 and an n = 3. This yields the low-pass prototype circuit of Fig. 3-32A which is transformed into the bandpass prototype circuit of Fig. 3-32B. Finally, using Equations 3-16 through 3-19, we obtain the final circuit that is shown in Fig. 3-32C. The calculations follow. Using Equations 3-16 and 3-17: 4.431 c - - 274 100) ( 7 x 106) = 1007pF (1OO)(7 X 106) L - - Z?r( 75 x 106)z( 4.431 ) = 4.47 nH Using Equations 3-18 and 3-19: 7 x 106 C = 2n(75 x 106)2(0.817)100 = 2.4 pF ( 100) (0.817) Lz = 2 4 7 x 106) = 1.86pH Similarly, C3 = 504 pF L, = 8.93 nH This is shown in Fig. 3-31. Note that both elements in each of the resonant circuits have the same normalized value, Once the prototype circuit has been transformed into its band-reject configuration, it is then scaled in im- pedance and frequency using the following formulas. For all series-resonant circuits: 6 Fig. 3-31. Low-pass to band-reject transformation. FILTER DESIGN 65 THE EFFECTS OF FINITE Q Thus far in this chapter, we have assumed the in- ductors and capacitors used in the designs to be lossless. Indeed, all of the response curves presented in this chapter are based on that assumption. But we know from our previous study of Chapters 1 and 2 that even though capacitors can be approximated as having infinite Q, inductors cannot, and the effects of the finite-Q inductor must be taken into account in any filter design. The use of finite element Q in a design intended for lossless elements causes the following unwanted effects (refer to Fig. 3-33) : 1. Insertion loss of the filter is increased whereas the final stopband attenuation does not change. The relative attenuation between the two is decreased. 2. At frequencies in the vicinity of cutoff ( f r ) , the response becomes more rounded and usually results in an attenuation greater than the 3 dR that was originally intended. 3. Ripple that was designed into the passband will be reduced. If the element Q is sufficiently low, ripple will be totally eliminated. 4. For band-reject filters, the attenuation in the stop- band becomes finite. This, coupled with an increase in passband insertion loss, decreases the relative attenuation significantly. Regardless of the gloomy predictions outlined above, however, it is possible to design filters, using the approach outlined in this chapter, that very closely resemble the ideal response of each network. The key is to use the highest-Q inductors available for the given task. Table 3-9 outlines the recommended mini- mum element-Q requirements for the filters presented in this chapter. Keep in mind, however, that anytime a low-Q component is used, the actual attenuation response of the network strays from the ideal response to a degree depending upon the element Q. It is, therefore, highly recommended that you make it a habit to use onlv the highest-Q components available. Table 3-9. Filter Element-Q Requirements ( .4 ) Lou;-pa.ss prototype circuit. i B ) Bandpass transformation. % - ( C )Final circuit with frequency and impedance scaled. F i g 3-32. Bandpass filter design for Example 3-9. Ideal Fig. 3-.33. Thr effect of finite-Q element5 on filter responw. GI c=- 27rRB For all parallel-resonant circuits: I3 C 2dO2RC,, RL, I, = ~ 27rR (Eq. 3-20) ( Eq. 3-21 ) (Eq. 3-22) (Eq. 3-23) where, in all cases, A = the 3-dB bandwidth, R = the final load resistance, f, , = the geometric center frequency, C,, = the normalized capacitor band-reject element I ,,, = the normalized inductor band-reject element value, value. Filter T y p e _____ Bessel Butterworth 0.01-dB Chebyshev 0.1-dB Chebyshev 0.5-dB Chebyshev 1-dB Chebyshe\ Minimum Element Q Required .3 15 23 39 J 7 _-_ _-__ ___ c- I J The insertion loss of the filters presented in this chapter can be calculated in the same manner as was used in Chapter 2. Simply replace each reactive ele- ment with resistor values corresponding to the Q of the element and, then, exercise the voltage division rule from source to load. IMPEDANCE MATCHING Impedance matching is often necessary in the de- sign of rf circuitry to provide the maximum possible transfer of power between a source and its load. Prob- ably the most vivid example of the need of such a transfer of power occurs in the front end of any sensi- tive receiver. Obviously, any unnecessary loss in a circuit that is already carrying extremely small signal levels simply cannot be tolerated. Therefore, in most instances, extreme care is taken during the initial de- sign of such a front end to make sure that each device in the chain is matched to its load. In this chapter, then, we will study several methods of matching a given source to a given load. This will be done both numerically and with the aid of the Smith Chart and, in both cases, exact step-by-step pro- cedures will be presented making any calculations as painless as possible. BACKGROUND There is a well-known theorem which states that, for dc circuits, maximum power will be transferred from a source to its load if the load resistance equals the source resistance. A simple proof of this theorem is given by the calculations and the sketches shown in Fig. 4-1. In the calculation, for convenience, the source is normalized for a resistance of one ohm and a source voltage of one volt. In dealing with ac or time-varying waveforms, however, that same theorem states that the maximum transfer of power, from a source to its load, occurs when the load impedance (ZL) is equal to the com- plex conjugate of the source impedance. Complex conjugate simply refers to a complex impedance hav- ing the same real part with an opposite reactance. Thus, if the source impedance were Z, = R +jX, then its complex conjugate would be Z, = R -jX. If you followed the mathematics associated with Fig. 4-1, then it should be obvious why maximum transfer of power does occur when the load impedance is the complex conjugate of the source. This is shown schematically in Fig. 4-2. The source (Zs) , with a series reactive component of +jX (an inductor), is driving its complex conjugate load impedance con- Rs= l a Proof that Pout MAX occurs when R L = Rs, in the circuit of Fig. 4-1A, is given by the formula: RL ( v 6 ) RS + RL Set V6= 1 and RS= 1, for convenience. Therefore, R L v1=- 1 + R L Then, the power into RL is: VI = " S fFovl - - - ( A ) Circuit. - V12 P1 - - RL - I I ! RL I I I I - RL - 0.1 1.0 10 (1 + R L ) ~ If you plot PI versus R L , as in the preceding equation, the result is shown by the curve of the graph in Fig. 4-1B. RL ( B ) Graph. Fig. 4-1. The power theorem. 66 IMPEDANCE MATCHING 1 EXAMPLE 4 1 Design a circuit to match a 100-ohm source to a 1000- ohm load at 100 MHz. Assume that a dc voltage must also be transferred from the source to the load. bolri tion The need for a dc path between the source and load dic- tates the need for an inductor in the series leg, as shown in Fig. 4-4A. From Equation 4-1, we have: 69 =43 - :3 From Equation 4-2, we get: X, = Q.R. = (3)(100) = 300 ohms (inductive) Then, from Equation 4-3, R x -2 " Q p - lo00 - 3 = 333 ohms (capacitive) Thus, the component values at 100 MHz are: X, I . =- 6J - 300 - 2 r ( 100 x 106) I = 477 nH 1 c=- W X , 1 27r(lOOX 10g)(333) - = 4.8 pF This yields the circuit shown in Fig. 4-10. Notice that what you have done is to design the circuit that was previously given in Fig. 4-5 and, then, analyzed. Fig. 4-10. Final circuit for Example 4-1 tances). It is very rare when such an occurrence ac- tually exists in the real world. Transistor input and output impedances are almost always complex; that is they contain both resistive and reactive components ( R 2jX). Transmission lines, mixers, antennas, and most other sources and loads are no different in that respect. Most will always have some reactive com- ponent which must be dealt with. It is, therefore, necessary to know how to handle these stray reactances and, in some instances, to actually put them to m7ork for you. There are two basic approaches in handling complex impedances : 1. Absorption-To actually absorb any stray reactances into the impedance-matching network itself. This can be done through prudent placement of each matching element such that element capacitors are placed in parallel with stray capacitances, and ele- ment inductors are placed in series with any stray inductances. The stray component values are then subtracted from the culcukated element values leav- ing new element values (C', L'), which are smaller than the calculated element values. 2. Resonance-To resonate any stray reactance with an equal and opposite reactance at the frequency of interest. Once this is done the matching network design can proceed as shown for two pure resis- tances in Example 4-1. Of course, it is possible to use both of the approaches outlined above at the same time. In fact, the majority of impedance-matching designs probably do utilize a little of both. Let's take a look at two simple examples to help clarify matters. Notice that nowhere in Example 4-2 was a conjugate match even mentioned. However, you can rest assured that if you perform the simple analysis outlined in the previous section of this chapter, the impedance looking into the matching network, as seen by the source, will be 100 -jl26 ohms, which is indeed the complex con- jugate of 100 +jl26 ohms. Obviously, if the stray element values are larger than the calculated element values, absorption cannot take place. If, for instance, the stray capacitance of Fig. 4-11 were 20 pF, we could not have added a shunt element capacitor to give us the total needed shunt capacitance of 4.8 pF. In a situation such as this, when absorption is not possible, the concept of resonance coupled with absorption will often do the trick. Examples 4-2 and 4-3 detail some very important concepts in the design of impedance-matching net- works. With a little planning and preparation, the de- sign of simple impedance-matching networks between complex loads becomes a simple number-crunching task using elementary algebra. Any stray reactances present in the source and load can usually be absorbed in the matching network (Example 4-2), or they can 70 RF Cmcurr DESIGN ~ ~ ~~ EXAMPLE 4-2 Use the absorption approach to match the source and load shown in Fig. 4-11 (a t 100 MHz). 2 Match 1 ? ‘ 100 -jlZ6 Fig. 4-11. Complex source and load circuit for Example 4-2. Solution The first step in the design process is to totally ignore the reactances and simply match the 100-ohm real part of the source to the 1OOO-ohm real part of the load (at 100 MHz). Keep in mind that you would like to use a matching net- work that will place element inductances in series with stray inductance and element capacitances in parallel with stray capacitances. Thus, conveniently, the network circuit shown in Fig. 4-4A is again chosen for the design and, again, Example 4-1 is used to provide the details of the procedure. Thus, the calculated values for the network, if we ignore stray reactances, are shown in the circuit of Fig. 4-10. But, since the stray reactances really do exist, the de- sign is not yet finished as we must now somehow absorb the stray reactances into the matching network. This is done as follows. At the load end, we need 4.8 pF of capacitance for the matching network. We already have a stray 2 pF available at the load so why not use it. Thus, if we use a 2.8-pF element capacitor, the total shunt capacitance be- comes 4.8 pF, the design value. Similarly, at the source, the matching network calls for a series 477-nH inductor. We already have a +j126-ohm, or 200-nH, inductor available in the source. Thus, if we use an actual element inductance of 477 nH - 200 nH = 277 nH, then the total series in- ductance will be 477 nH-which is the calculated design value. The final design circuit is shown in Fig. 4-12. be resonated with an equal and opposite reactance, which is then absorbed into the network (Example 4-3). THREE-ELEMENT MATCHING Equation 4-1 reveals a potential disadvantage of the 2-element L networks described in the previous sec- tions. It is a fact that once R, and R,, or the source and load impedance, are determined, the Q of the network is defined. In other words, with the L network, the designer does not have a choice of circuit Q and simply must take what he gets. This is, of course, usually the case because the source and load imped- ance are typically given in any design and, thus, R, and R, cannot be changed. The lack of circuit-Q versatility in a matching net- work can be a hindrance, however, especially if a narrow bandwidth is required. The 3-element network overcomes this disadvantage and can be used for narrow-band high-Q applications. Furthermore, the designer can sebct any practical circuit Q that he wishes as long as it is greater than that Q which is possible with the L-matching network alone. In other words, the circuit Q established with an L-matching network is the minimum circuit Q available in the 3- element matching arrangement. The 3-element network (shown in Fig. 4-17) is called a Pi network because it closely resembles the Greek letter T. Its companion network (shown in Fig. 4-18) is called a T network for equally obvious reasons. The Pi Network The Pi network can best be described as two “back- to-back L networks that are both configured to match the load and the source to an invisible or “virtual” re- sistance located at the junction between the two networks. This is illustrated in Fig. 4-19. The signifi- cance of the negative signs for -Xsl and -Xsz is sym- bolic. They are used merely to indicate that the X, values are the opposite type of reactance from X,, and Xp2, respectively. Thus, if XP1 is a capacitor, X,, must be an inductor, and vice-versa. Similarly, if Xp2 is an inductor, Xs2 must be a capacitor, and vice- versa. They do not indicate negative reactances (ca- pacitors). The design of each section of the Pi network pro- ceeds exactly as was done for the L networks in the previous sections. The virtual resistance ( R ) must be smaller than either Rs or R L because it is connected to the series arm of each L section but, otherwise, it can be any value you wish. Most of the time, however, R is defined by the desired loaded Q of the circuit that you specify at the beginning of the design process. For our purposes, the loaded Q of this network will be de- fined as: (Eq. 4-4) where, RH = the largest terminating impedance of R, or RL, R = the virtual resistance. Although this is not entirely accurate, it is a widely accepted Q-determining formula for this circuit, and is certainly close enough for most practical work. Ex- ample 4-4 illustrates the procedure. Any of the networks in Fig. 4-21 will perform the impedance match between the 100-ohm source and the IMPEDANCE MATCHING 71 EXAMPLE 4-3 Design an impedance matching network that will block the flow of dc from the source to the load in Fig. 4-13. The frequency of operation is 75 MHz. Try the resonant ap- proach. Fig. 4-13. Complex load circuit for Example 4-3 Solution The need to block the flow of dc from the source to the load dictates the use of the matching network of Fig. 4-4C. But, first, let’s get rid of the stray 40-pF capacitor by reso- nating it with a shunt inductor at 75 MHz. 1 L=----- W W S t W n Fig. 4-14. Resonating the stray load capacitance. This leaves us with the circuit shown in Fig. 4-14. Now that we have eliminated the stray capacitance, we can proceed with matching the network between the 50-ohm load and the apparent 600-ohm load. Thus, = 3.32 X. = QA = (3 .32)(50) = 166 ohms X G O , R,l GOO 3.32 -- = 181 ohm:, Therefore, the elenierit values are: C = - 1 W X , 1 - -2n(75 x lOO)(l66) = 12.78pF X,! I, = - w - 181 = 384 nH 2n(75 X 108) - Vatching Network . - 384 nH 1 ‘ b Fig. 4-15. The circuit of Fig. 4-14 after impedance matching. These vahes, then, yield the circuit of Fig. 4-15. But notice that this circuit can be further simplified by simply replac- ing the two shunt inductors with a single inductor. ‘l‘here- fore, L,La L n e w =- h + L, - (384) ( 112.6) - 384+ 112.6 = 87nH The final circuit design appears in Fig. 4-16 Matching Network _ ~ ~ ~ . _ . ~ _ ~ _ _ . , ~ ~ . . ~ . _ . ~ ~ ~ 1 Fig. 4-16. Final design circuit for Example 4-3. Fig. 4-17. The three-element Pi network. Fig. 4-18. The three-element T network. 74 RF CIRCUIT DESIGN where, Rsmaller = the smallest terminating resistance, Rlarger = the largest terminating resistance, R1, Rz, . . . R, = virtual resistors. p+m - - - - - This The is shown desig in procedure Fig. 4-26. for these wideband matching - networks is precisely the same as was given for the previous examples. To design for a specific low Q, simply solve Equation 4-7 for R to find the virtual Fig. 4-22. The T network shown as two back-to-back L networks. EXAMPLE 4-5 Using Fig. 4-22 as a reference, design four different net- works to match a 10-ohm source to a %-ohm load. Each network is to have a loaded Q of 10. Solution Using Equation 4-5, we can find the virtual resistance we need for the match. R = R a m a ~ ~ ( Q2 + 1) = lo( 101) = 1010 ohms From Equation 4-2: X a = QR. = 10( 10) = 100 ohms From Equation 4-3: R x,, =Ti = 101 ohms Now, for the L network on the load end, the Q is defined by the virtual resistor and the load resistor. Thus, = 4.4 Therefore, R x -- ” - Qz 1010 4.4 -- = 230 ohms = (4 .4) (50) = 220 ohms x . 1 ~ Q ~ R L The network is now complete and is shown in Fig. 4-23 without the virtual resistor. The two shunt reactances of Fig. 4-23 can again be com- bined to form a single element by simply substituting a value that is equal to the combined equivalent parallel re- actance of the two. The four possible T-type networks that can be used for matching the 10-ohm source to the 50-ohm load are shown in Fig. 4-24. Fig. 4-23. The calculated reactances of Example 4-5. 0 (D) Fig. 4-24. The transformation of circuits from double-L to T-type networks. IMPEDANCE MATCHING 75 ( A ) R in shunt k g t - ( E ) R in series lea. Fig. 4-25. Two series-connected L networks for lower Q applications. Fig. 4-26. Expanded version of Fig. 4-25 for even wider bandwidths. resistance needed. Or, to design for an optimally wide bandwidth, solve Equation 4-6 for R. Once R is known, the design is straightforward. THE SMITH CHART Probably one of the most useful graphical tools available to the rf circuit designer today is the Smith Chart shown in Fig. 4-27. The chart was originally conceived back in the Thirties by a Bell Laboratories engineer named Phillip Smith, who wanted an easier method of solving the tedious repetitive equations that often appear in rf theory. His solution, appropriately named the Smith Chart, is still widely in use. At first glance, a Smith Chart appears to be quite complex. Indeed, why would anyone of sound mind even care to look at such a chart? The answer is really quite simple; once the Smith Chart and its uses are understood, the rf circuit designer’s job becomes much less tedious and time consuming. Very lengthy complex equations can be solved graphically on the chart in seconds, thus lessening the possibility of errors creep- ing into the calculations. Smith Chart Construction The mathematics behind the construction of a Smith Chart are given here for those that are inter- ested. It is important to note, however, that you do not need to know or understand the mathematics surround- ing the actual construction of a chart as long as you understand what the chart represents and how it can be used to your advantage. Indeed, there are so many uses for the chart that an entire volume has been writ- ten on the subject. In this chapter, we wiIl concentrate mainly on the Smith Chart as an impedance matching tool and other uses will be covered in later chapters. The mathematics follow. The reflection coefficient of a load impedance when given a source impedance can be found by the formula : z. - ZL z, + Z L p=- In normalized form, this equation becomes: where Z, is a complex impedance of the form R +jX. The polar form of the reflection coefficient can also be represented in rectangular coordinates: P = P +is Substituting into Step 2, we have: R +jX - 1 P +is = R +jx + 1 If we solve for the real and imaginary parts of p +jq, we get: R2 - 1 + X2 = ( R + 1)2 + X2 and, 2x ‘= (R + 1)2 + X2 Solve Step 5 for X: Then, substitute Step 6 into Step 5 to obtain: Step 7 is the equation for a family of circles whose centers are at: R p=- R + 1 q = o and whose radii are equal to: 1 R + l P These are the constant resistance circles, some of which are shown in Fig. 4-28A. Similarly, we can eliminate R from Steps 4 and 5 to obtain : 2 ( p - 1)2 + (q - i)2 = (k) ( Step 8 ) RF CIRCUIT DFSIGN NAME TITLE SMITH FORM 8 ' & ~ P R ~ S l ~ KAY E L E C T R I C COMPANY, PINE BROOK, N J.. 0 1966 P R I N T E D I N U S A . DWO. NO. DAE IMPEDANCE OR ADMITTANCE COORDINATES CEtiiER 0 I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4 5 6 1 1 1 ORIGIN A YCG*-Q*IT Fig. 427. The Smith Chart. (Courtesy Analog lnstnrments Co.) IMPEDANCE MATCHING 79 NAME TITLE DWG. NO. WITH CHART FORMB*-BSPR@E6)/ KAY E L E C T R I C COMPANY, PINE BRO0K.N J.. @I966 P R I N T E D IN U.SA. IMPEDANCE OR ADMITTANCE COORDINATES Fig. 4-30. Plotting impedances on the chart. 80 RF Cmcurr DESIGN NAYE TITLE DWO. NO. W T H CHART FORM 82-ESPRBS6~ K A Y ELECTRIC COMPANY, PINE BROOK, N a, 01966 PRINTED IN U.SA. IMPEDANCE OR ADMITTANCE COORDINATES Fig. 4-31. More impedances are plotted on the chart. IMPEDANCE MATCHING 81 In many cases, the actual circles will not be present on the chart and you will have to interpolate between two that are shown. Thus, plotting impedances and, therefore, any manipulation of those impedances must be considered an inexact procedure which is subject to "pilot error." Most of the time, however, the error introduced by subjective judgements on the part of the user, in plotting impedances on the chart, is so small as to be negligible for practical work. Fig. 4-31 shows a few more impedances plotted on the chart. Notice that all of the impedance values plotted in Fig. 4-31 are very small numbers. Indeed, if you try to plot an impedance of Z = 100 + jl50 ohms, you will not be able to do it accurately because the R = 100 and X = 150 ohm circles would be (if they were drawn) on the extreme right edge of the chart-very close to infinity. In order to facilitate the plotting of larger impedances, normalization must be used. That is, each impedance to be plotted is divided by a con- venient number that will place the new normalized impedance near the center of the chart where in- creased accuracy in plotting is obtained. Thus, for the preceding example, where Z = 100 + j150 ohms, it would be convenient to divide Z by 100, which yields the value Z = 1 + j1.5. This is very easily found on the chart. Once a chart is normalized in this manner, all impedances plotted on that chart must be divided by the same number in the normalization process. Other- wise, you will be left with a bunch of impedances with which nothing can be done. Impedance Manipulation on the Chart Fig. 4-32 graphically indicates what happens when a series capacitive reactance of - j l .O ohm is added to an impedance of Z = 0.5 + j0.7 ohm. Mathematically, the result is Z = 0.5 + j0.7 - jl.0 = 0.5 - j0.3 ohm which represents a series RC quantity. Graphically, what we have done is move downward along the R = 0.5-ohm constant resistance circle for a distance of X = - j l .O ohm. This is the plotted impedance point of Z = 0.5 - j0.3 ohm, as shown. In a similar manner, as shown in Fig. 4-33, adding a series inductance to a plotted impedance value simply causes a move upward along a constant resistance circle to the new impedance value. This type of construction is very important in the design of impedance-matching networks using the Smith Chart and must be understood. In general then, the addition of a series capacitor to an impedance moves that impedance downward ( counterclockwise ) along a constant resistance circle for a distance that is equal to the reactance of the capacitor. The addition of any series inductor to a plotted impedance moves that impedance upmard (clockwise) along a constant resistance circle for a distance that is equal to the reactance of the inductor. Conversion of Impedance to Admittance The Smith Chart, although described thus far as a family of impedance coordinates, can easily be used to convert any impedance ( Z ) to an admittance ( Y ) , and vice-versa. In mathematical terms, an admittance is simply the inverse of an impedance, or (Eq. 4-9) Y = - z 1 where, the admittance ( Y ) contains both a real and an imaginary part, similar to the impedance ( Z ) . Thus. Y = G * j B (Eq. 4-10) where, G = the conductance in mhos, R = the susceptance in mhos. The circuit representation is shown in Fig. 4-34. No- tice that the susceptance is positive for a capacitor and negative for an inductor, whereas, for reactance, the opposite is true. To find the inverse of a series impedance of the form Z = R + jX mathematically, you would simply use Equation 4-9 and perform the resulting calcula- tion. But, how can you use the Smith Chart to perform the calculation for you without the need for a calcu- lator? The easiest way of describing the use of the chart in performing this function is to first work a prob- lem out mathematically and, then, plot the results on the chart to see how the two functions are related. Take, for example, the series impedance Z = 1 + j l . The inverse of Z is: 1 y = - 1 + j l 1 1.414 /45" - = 0.7071 / -45" = 0.5 - j0.5 mho If we plot the points 1 + j l and 0.5 - j0.5 on the Smith Chart, we can easily see the graphical relationship between the two. This construction is shown in Fig. 4-35. Notice that the two points are located at exactly the same distance ( d ) from the center of the chart but in opposite directions (180") from each other. Indeed, the same relationship holds true for any im- pedance and its inverse. Therefore, without the aid of a calculator, you can find the reciprocal of an im- pedance or an admittance by simply plotting the point on the chart, measuring the distance ( d ) from the center of the chart to that point, and, then, plot- ting the measured result the same distance from the center but in the opposite direction (180") from the original point. This is a very simple construction tech- nique that can be done in seconds. Another approach that we could take to achieve the same result involves the manipulation of the actual chart rather than the performing of a construc- 84 RF Cmcurr DESIGN Y = G + j B Y = G - j B Fig. 4-34. Circuit representation for admittance. tion on the chart. For instance, rather than locating a point 180" away from our original starting point, why not just rotate the chart itself 180" while fixing the starting point in space? The result is the same, and it can be read directly off of the rotated chart without performing a single construction. This is shown in Fig. 4-36 (Smith Chart Form ZY-01-N ) * where the rotated chart is shown in black. Notice that the impedance plotted (solid lines on the red coordinates) is located at Z = 1 + j l ohms, and the reciprocal of that (the admittance) is shown by dotted lines on the black coordinates as Y = 0.5 - j0.5. Keep in mind that be- cause we have rotated the chart 180' to obtain the ad- mittance coordinates, the upper half of the admittance chart represents negative susceptunce (-jB) which is inductive, while the lower half of the admittance chart represents a positive susceptance (+jB) which is cupucitiue. Therefore, nothing has been lost in the ro- tation process. The chart shown in Fig. 436, containing the super- imposed impedance and admittance coordinates, is an extremely useful version of the Smith Chart and is the one that we will use throughout the remainder of the book. But first, let's take a closer look at the admit- tance coordinates alone. Admittance Manipulation on the Chart Just as the impedance coordinates of Figs. 4-32 and 4-33 were used to obtain a visual indication of what occurs when a se&s reactance is added to an im- pedance, the admittance coordinates provide a visual indication of what occurs when a shunt element is added to an udmittance. 'She addition of a shunt ca- pacitor is shown in Fig. 4-37. Here we begin with an admittance of Y = 0.2 - jO.5 mho and add a shunt capacitor with a susceptance (reciprocal of reactance) of fj0.8 mho. Mathematically, we know that parallel susceptances are simply added together to find the equivalent susceptance. When this is done, the result becomes: Y '= 0.2 - j0.5 + j0.8 = 0.2 + j0.3 mho If this point is plotted on the admittance chart, we quickly recognize that all we have done is to move along a constant conductance circle (G) downward (clockwise) a distance of jB = 0.8 mho. In other words, * Smith Chart Form ZY-01-N is a copyright of Analog Instruments Com- pany, P..O. Box 808, New Providence, NJ 07974. It and other Smith Chart accessories are available from the company. the real part of the admittance has not changed, only the imaginary part has. Similarly, as Fig. 4-38 indicates, adding a shunt inductor to an admittance moves the point along a constant conductance circle upward (counterclockwise) a distance (-jB) equal to the value of its susceptance. If we again superimpose the impedance and admit- tance coordinates and combine Figs. 4-32, 4-33, 4-37, and 4-38 for the general case, we obtain the useful chart shown in Fig. 4-39. This chart graphically illus- trates the direction of travel, along the impedance and admittance coordinates, which results when the particular type of component that is indicated is added to an existing impedance or admittance. A simple ex- ample should illustrate the point ( Example 4-6). IMPEDANCE MATCHING ON THE SMITH CHART Because of the ease with which series and shunt components can be added in ladder-type arrangements on the Smith Chart, while easily keeping track of the impedance as seen at the input terminals of the struc- ture, the chart seems to be an excellent candidate for an impedance-matching tool. The idea here is simple. Given a load impedance and given the impedance that the source would like to see, simply plot the load im- pedance and, then, begin adding series and shunt elements on the chart until the desired impedance is achieved-just as was done in Example 4-6. Two-Element Matching Two-element matching networks are mathematically very easy to design using the formulas provided in earlier sections of this chapter. For the purpose of il- lustration, however, let's begin our study of a Smith Chart impedance-matching procedure with the simple network given in Example 4-7. To make life much easier for you as a Smith Chart user, the following equations may be used. For a series-C component: 1 C=- wXN For a series-L component: (Eq. 4-11) L=- XN (Eq. 4-12 w For a shunt-C component: For a shunt-L component: where, w = 2 d , X = the reactance as read from the chart, (Eq. 4-13 (Eq. 4-14) MPEDANCE MATCHING 85 NAME TITLE SMlTHCHART FORM8%BSPR(ICEbl~ KAY ELECTRIC COMPANY, PINE BROOK, N J . 01966 PRINTED IN U.SA IMPEDANCE OR ADMITTANCE COORDINATES ~ ~~ Fig. 4-35. Impedance-admittance conversion on the Smith Chart. 86 RF Cmcurr DFSIGN TITLE CHART FORM ZV-Il-N] ANALOG INSTRUMENTS COMPANY. NEW PROVIDENCE, N.J. 070 NORMALIZED IMPEDANCE AND ADMITTANCE COORDINATES RADIALLY SCALED PARAMETERS Fig. 4-36. Superimposed admittance coordinates. IMPEDANCE MATCHING 89 Fig. 4-39. Summary of component addition on a Smith Chart. 90 RF Cmcurr DFSIGN B = the susceptance as read from the chart, N = the number used to normalize the original im- pedances that are to be matched. If you use the preceding equations, you will never have to worry about changing susceptances into re- actances before unnormalizing the impedances. The equations take care of both operations. The only thing you have to do is read the value of susceptance (for shunt components) or reactance (for series compo- nents) directly off of the chart, plug this value into the equation used, and wait for your actual component values to pop out. Three-Element Matching In earlier sections of this chapter, you learned that the only real difference between two-element and three-element matching is that with three-element matching, you are able to choose the loaded Q for the network. That was easy enough to do in a mathe- matical-design approach due to the virtual resistance concept. But how can circuit Q be represented on a Smith Chart? As you have seen before, in earlier chapters, the Q of a series-impedance circuit is simply equal to the ratio of its reactance to its resistance. Thus, any point on a Smith Chart has a Q associated with it. Alter- nately, if you were to specify a certain Q,. you could find an infinite number of points on the chart that could satisfy that Q requirement, For example, the following impedances located on a Smith Chart have a Q o f 5 : R + jX = 1 j5 = 0.5 f. j2.5 = 0.2 2 jl = 0.1 k j0.5 = 0.05 +- j0.25 These values are plotted in Fig. 4-45 and form the arcs shown. Thus, any impedance located on these arcs must have a Q of 5. Similar arcs for other values of Q can be drawn with the arc of infinite Q being located along the perimeter of the chart and the Q = 0 arc (actually a straight line) lying along the pure resistance line located at the center of the chart. The design of high-Q three-element matching net- works on a Smith Chart is approached in much the same manner as in the mathematical methods pre- sented earlier in this chapter. Namely, one branch of the network will determine the loaded Q of the cir- cuit, and it is this branch that will set the character- istics of the rest of the circuit. The procedure for designing a three-element im- pedance-matching network for a specified Q is sum- marized as follows: 1. Plot the constant-Q arcs for the specified Q. 2. Plot the load impedance and the complex conjugate of the source impedance. 3. Determine the end of the network that will be used to establish the loaded Q of the design. For T networks, the end with the m 2 k r terminating resistance determines the Q. For Pi networks, the end with the larger terminating resistor sets the Q. R, > RL 4. For T networks: EXAMPLE 46 What is the impedance looking into the network shown in Fig. 4-40? Note that the task has been simplified due to the fact that shunt susceptances are shown rather than shunt reactances. Fig. 4-40. Circuit for Example 4-0. Solution This problem is very easily handled on a Smith Chart and not a single calculation needs to be performed. The solution is shown in Fig. 4-42. It is accomplished as follows. First, break the circuit down into individual branches as shown in Fig. 4-41. Plot the impedance of the series RL branch where Z = 1 + j l ohm. This is point A in Fig. 4-42. Next, following the rules diagrammed in Fig. 4-39, begin adding each component back into the circuit-one at a time. Thus, the following constructions (Fig. 4-42) should be noted: C Fig. 4-41. Circuit is broken down into individual branch elements. Arc AB = shunt L = -jB = 0.3 mho Arc BC = series C = -jX = 1.4 ohms Arc CD = shunt C = +jB = 1.1 mhos Arc DE series L = +jX = 0.9 ohm The impedance at point E (Fig. 4-42) can then be read directly off of the chart as Z = 0.2 + j0.5 ohm. Continued on next page IMPEDANCE MATCHING 91 EXAMPLE 46-Cont. WAY€ TITLE SMITH CHART FORM ZY-01-N I ANALOG INSTRUMENTS COMPANY, NEW PROVIDENCE, N.J. NORMALIZED IMPEDANCE AND ADMITTANCE COORDINATES Fig. 4-42. Smith Chart solution for Example 4-6. 94 RF Cmcurr DESIGN ~~ CHART FORM ZY-01-N NORMALIZED IMPEDANCE AND ADMITTANCE COORDINATES Fig. 4-45. Lines of constant Q. IMPEDANCX MATCHING 95 NAME TITLE SMITH CHART FORM LV-Ql-N [ ANALOG INSTRUMENTS COMPANY,- NORMALIZED IMPEDANCE AND ADMITTANCE COORDINATES Fig. 4-46. Smith Chart solution for Example 4-8. 96 RF Cmcurr DESIGN ______ EXAMPLE 4-8 Design a T network to match a Z = 15 + jl5-ohm source to a 225-ohm load at 30 MHz with a loaded Q of 5. Solution Following the procedures previously outlined, draw the arcs for Q = 5 first and, then, plot the load impedance and the complex conjugate of the source impedance. Obvi- ously, normalization is necessary as the impedances are too large to be located on the chart. Divide by a convenient value (choose N = 7 5 ) for normalization. Therefore: Z.’ = 0.2 - j0.2 ohm ZC = 3 ohms The construction details for the design are shown in Fig. The design statement specifies a T network. Thus, the source termination will determine the network Q because Following the procedure for Rs < RL (Step 4, above), first plot point I, which is the intersection of the Q = 5 curve and the R = constant circuit that passes through Zs’. Then, move from the load impedance to point I with two elements. 4-46. Rs < RL. Element 1 = arc AB = series L = j2.5 ohms Element 2 = arc BI = shunt C = j1.15 mhos Then, move from point I to Z.* along the R = constant circle. Element 3 = arc IC = series L = j0.8 ohm Use Equations 4-11 through 4-14 to find the actual element values. Element 1 = series L: L = (2.5)75 2 ~ ( 3 0 x 106) = 995 nII Element 2 = shunt C: 1.15 = 2m(30 x 106)75 = 81 pF Element 3 = series L: L = (0.8)75 2a(30 x 106) = 318 nH The final network is shown in Fig. 4-47. 15 +jl5 Fig. 4-47. Final circuit for Example 4-8. treasure trove containing an infinite number of possible solutions. To get from point A to point B on a Smith Chart, there is, of course, an optimum solution. How- ever, the optimum solution is not the only solution. The two-element network gets you from point A to point B with the least number of components and the three- element network can provide a specified Q by follow- ing a different route. If you do not care about Q, however, there are 3-, 4-, 5-, lo-, and 20-element (and more) impedance-matching networks that are easily designed on a Smith Chart by simply following the constant-conductance and constant-resistance circles until you eventually arrive at point B, which, in OUT case, is usually the complex conjugate of the source impedance. Fig. 4-48 illustrates this point. In the lower right-hand corner of the chart is point A. In the upper left-hand corner is point B. Three of the infinite number of possible solutions that can be used to get from point A to point B, by adding series and shunt inductances and capacitances, are shown. Solu- tion 1 starts with a series-L configuration and takes 9 elements to get to point B. Solution 2 starts with a shunt-L procedure and takes 8 elements, while Solu- tion 3 starts with a shunt-C arrangement and takes 5 elements. The element reactances and susceptances can be read directly from the chart, and Equations 4-11 through 4-14 can be used to calculate the actual component values within minutes. SUMMARY Impedance matching is not a form of “black magic” but is a step-by-step well-understood process that is used to help transfer maximum power from a source to its load. The impedance-matching networks can be designed either mathematically or graphically with the aid of a Smith Chart. Simpler networks of two and three elements are usually handled best mathemati- cally, while networks of four or more elements are very easily handled using the Smith Chart.