18
CHAPTER
Transformers and Impedance Matching
WE CAN USE A TRANSFORMER TO OBTAIN THE OPTIMUM VOLTAGE FOR A CIRCUIT, DEVICE, OR SYSTEM. Transformers have various uses in electricity and electronics. For example, they can:
• Match the impedances between a circuit and a load
• Match the impedances between two different circuits or devices
• Provide DC isolation between circuits or devices while letting AC pass
• Make balanced and unbalanced circuits, feed systems, and loads compatible
Principle of the Transformer
When we place two wires near and parallel to each other and then drive a fluctuating current through one of them, a fluctuating current appears in the other, even though no direct physical connection exists between them. We call this effect electromagnetic induction. All AC transformers work according to this principle.
Induced Current and Coupling
If one wire carries sine-wave AC of a certain frequency, then the induced current shows up as sine-wave AC of the same frequency in the other wire. As we reduce the spacing between the two wires, keeping them straight and parallel at all times, the induced current increases for a given current in the first wire. If we wind the wires into coils (making certain that the wires are insulated or enameled so that they can’t “short out” between the coil turns or to anything else nearby) and place the coils along a common axis, as shown in Fig. 18-1, we observe more induced current than we do if the same wires run straight and parallel. We say that the coupling improves when we coil the wires up and place them along a common axis. We can improve the coupling (efficiency of induced-current transfer) still more if we wind one coil directly over the other.
18-1 Magnetic lines of flux between two aligned coils, when one coil carries fluctuating or alternating current.
Primary and Secondary
A transformer comprises two coils of insulated or enameled wire, along with the core, or form, on which we wind them. We call the first coil, through which we deliberately drive current, the primary winding. We call the second coil, in which the induced current appears, the secondary winding. Engineers and technicians usually call them simply the primary and the secondary.
When we apply AC to a primary winding, the coil currents are attended by potential differences between the coil ends, constituting the primary voltage and secondary voltage. In a step-down transformer, the primary voltage exceeds the secondary voltage. In a step-up transformer, the secondary voltage exceeds the primary voltage. Let’s abbreviate the primary voltage as Epri and the secondary voltage as Esec. Unless otherwise stated, we specify effective (RMS) voltages when talking about the AC in a transformer.
The windings of a transformer exhibit inductance because they’re coils. The optimum inductance values for the primary and secondary depend on the frequency of operation, and also on the resistive components of the impedances of the circuits to which we connect the windings. As the frequency increases and the resistive component of the impedance remains constant, the optimum inductance decreases. If the resistive component of the impedance increases and the frequency remains constant, the optimum inductance increases.
Turns Ratio
We define the primary-to-secondary turns ratio in a transformer as the ratio of the number of turns in the primary Tpri to the number of turns in the secondary Tsec. We can denote this ratio as Tpri:Tsec or Tpri/Tsec. In a transformer with optimum primary-to-secondary coupling (Fig. 18-2), we always find that
Epri/Esec = Tpri/Tsec
18-2 The primary voltage (Epri) and secondary voltage (Esec) in a transformer depend on the number of turns in the primary winding (Tpri) versus the number of turns in the secondary winding (Tsec).
Stated in words, the primary-to-secondary voltage ratio equals the primary-to-secondary turns ratio.
Suppose that a transformer has a primary-to-secondary turns ratio of exactly 9:1. We apply 117 V RMS AC across the primary terminals. Do we have a step-up transformer or a step-down transformer? How much voltage can we expect to see across the secondary?
Solution
This device is a step-down transformer. Let’s use the above equation and solve for Esec. We start with
Epri/Esec = Tpri/Tsec
We can plug in the values Epri = 117 and Tpri/Tsec = 9.00 to obtain
117/Esec = 9.00
Using a little bit of algebra, we can solve this equation to obtain
Esec = 117 / 9.00 = 13.0 V RMS
Problem 18-2
Consider a transformer with a primary-to-secondary turns ratio of exactly 1:9. The voltage across the primary equals 121.4 V RMS. Do we have a step-up transformer or a step-down transformer? What’s the voltage across the secondary?
Solution
In this case, we have a step-up transformer. As before, we can plug in the numbers and solve for Esec. We start with
Epri/Esec = Tpri/Tsec
Inputting Epri = 121.4 and Tpri/Tsec = 1/9.000, we get
121.4/Esec = 1/9.000
which solves to
Esec = 9.000 × 121.4 = 1093 V RMS
Which Ratio Is Which?
Sometimes, when you read the specifications for a transformer, the manufacturer will quote the secondary-to-primary turns ratio rather than the primary-to-secondary turns ratio. You can denote the secondary-to-primary turns ratio as Tsec/Tpri. In a step-down transformer, Tsec/Tpri is less than 1, but in a step-up unit, Tsec/Tpri is greater than 1.
When you hear someone say that a transformer has a certain “turns ratio” (say 10:1), you’d better make sure that you know which ratio they’re talking about. Do they mean Tpri/Tsec or Tsec/Tpri? If you get it wrong, you’ll miscalculate the secondary voltage by a factor equal to the square of the turns ratio! For example, in a transformer in which Tsec/Tpri = 10:1 and to which an input voltage of 25 V RMS AC is being provided, you don’t want to think that you’ll see only 2.5 V RMS AC across the secondary winding, when in fact you’ll have to contend with 250 V RMS AC.
Ferromagnetic Cores
If we place a sample of ferromagnetic material within a pair of coils composing a transformer, the extent of coupling increases beyond what we get with an air core. However, some energy is invariably lost as heat in a ferromagnetic transformer core. Also, ferromagnetic cores limit the maximum frequency at which a transformer will work efficiently.
The schematic symbol for an air-core transformer looks like two inductor symbols placed back-to-back, as shown in Fig. 18-3A. If the transformer has a laminated-iron (layered-iron) core, we add two parallel lines to the schematic symbol (Fig. 18-3B). If the core consists of powdered iron, we break up the two parallel lines (Fig. 18-3C).
18-3 Schematic symbols for transformers. At A, air core. At B, laminated-iron core. At C, powdered-iron core.
In transformers intended for use with 60-Hz utility AC, and also for low audio-frequency (AF) applications, sheets of an alloy called silicon steel, glued together in layers, are often employed as transformer cores. The silicon steel goes by the nickname transformer iron. The layering breaks up the electrical currents that tend to circulate in solid-iron cores. These so-called eddy currents flow in loops, serving no useful purpose, but they heat up the core, thereby wasting energy that we could otherwise obtain from the secondary winding. We can “choke off” eddy currents by breaking up the core into numerous thin, flat layers with insulation between them.
Transformer Loss
A rather esoteric form of loss, called hysteresis loss, occurs in all ferromagnetic transformer cores, but especially in laminated iron. Hysteresis is the tendency for a core material to act “sluggish” in accepting a fluctuating magnetic field. Air cores essentially never exhibit this type of loss. In fact, air has the lowest overall loss of any known transformer core material. Laminated cores exhibit high hysteresis loss above the AF range, so they don’t work well above a few kilohertz.
At frequencies up to several tens of megahertz, powdered iron can serve as an efficient RF transformer core material. It has high magnetic permeability and concentrates the flux considerably. High-permeability cores minimize the number of turns needed in the coils, thereby minimizing the ohmic (resistive) loss that can take place in the wires.
At the highest radio frequencies (more than a couple of hundred megahertz), air represents the best overall choice as a transformer core material because of its low loss and low permeability.
Transformer Geometry
The properties of a transformer depend on the shape of its core, and on the way in which the wires surround the core. In electricity and electronics practice, you’ll encounter several different types of transformer geometry.
E Core
The E core gets its name from the fact that it has the shape of a capital letter E. A bar, placed at the open end of the E, completes the core assembly (Fig. 18-4A). We can wind a primary and secondary on an E core in either of two ways.
18-4 At A, a utility transformer E core, showing both sections. At B, the shell winding method. At C, the core winding method.
The simplest winding method involves winding both the primary coil and the secondary coil around the middle bar of the E, as shown in Fig. 18-4B. We call this scheme the shell method of transformer winding. It provides maximum coupling, but it also results in considerable capacitance between the primary and the secondary. Such inter-winding capacitance can sometimes be tolerated, but often it cannot. Another disadvantage of the shell geometry is the fact that, when we wind coils one on top of the other, the transformer can’t handle very much voltage. High voltages cause arcing (sparking attended by unwanted current) between the windings, which can destroy the insulation on the wires, lead to permanent short circuits, and even set the transformer on fire.
If we don’t want to use the shell method, we can employ the core method of transformer winding. In this scheme, we place one winding at the bottom of the E section, and the other winding at the top as shown in Fig. 18-4C. The coupling occurs by means of magnetic flux in the core. The core method results in lower inter-winding capacitance than we observe in a shell-wound transformer designed for the same voltage-transfer ratio because the windings are located physically farther apart. A core-wound transformer can handle higher voltages than a shell-wound transformer of the same physical size. Sometimes the center part of the E is left out of the core, in which case we have a transformer with an O core or a D core.
Shell-wound and core-wound transformers are commonly used at 60 Hz in electrical and electronic appliances and devices of all kinds. We can also find transformers of this sort in some older AF systems.
Solenoidal Core
A pair of cylindrical coils, wound around a rod-shaped piece of powdered iron, can operate as an RF transformer, usually as a loopstick antenna in portable radio receivers and radio direction-finding (RDF) equipment. We can wind one of the coils directly over the other, or we can separate them, as shown in Fig. 18-5, to reduce the interwinding capacitance between the primary and secondary.
18-5 A solenoidal-core transformer.
In a loopstick antenna, the primary winding intercepts the electromagnetic waves that carry the wireless signals. The secondary winding provides an optimum impedance match to the first amplifier stage, or front end, of the radio receiver or direction finder. We’ll explore the use of transformers for impedance matching later in this chapter.
Toroidal Core
In recent decades, a donut-shaped transformer core called a toroidal core (or toroid) has become common for winding RF transformers. When we want to construct a toroidal transformer, we can wind the primary and secondary directly over each other, or we can wind them over different parts of the core, as shown in Fig. 18-6. As with other transformers, we observe more interwinding capacitance if we place the coils directly over each other than we do when we keep them physically separated.
18-6 A toroidal-core transformer.
In a toroidal inductor or transformer, practically all of the magnetic flux remains within the core material; almost none ventures outside. This property allows circuit designers to place toroids near other components without worrying about unintended mutual inductance. Also, a toroidal coil or transformer can be mounted directly on a metal chassis, and the external metal has no effect on transformer operation.
A toroidal core provides considerably more inductance per turn, for the same kind of ferromagnetic material, than a solenoidal core offers. We’ll often see toroidal coils or transformers that have inductance values up to several hundred millihenrys.
Pot Core
A pot core takes the form of a ferromagnetic shell that completely surrounds a loop-shaped wire coil. The core is manufactured in two halves (Fig. 18-7). You wind the coil inside one of the shell halves, and then bolt the two shell halves together. In the resulting device, all of the magnetic flux remains confined to the core material.
18-7 Exploded view of a pot-core transformer.
Like a toroid, a pot core is self-shielding. Essentially no magnetic coupling takes place between the windings and external components. You can use a pot core to wind a single, high-inductance coil. You can obtain inductance values of more than 1 H with a reasonable number of wire turns. However, you must wind the primary and secondary right next to each other; the geometry of the core prevents significant physical separation of the windings. Therefore, you’ll always get a lot of interwinding capacitance.
Pot cores find diverse applications at AF, and also in the lowest-frequency part of the RF spectrum. You’ll rarely, if ever, find these types of coils in high-frequency RF systems because other geometries can provide the necessary inductance values without the undesirable interwinding capacitance.
Autotransformer
In some situations, you might not need (or even want) DC isolation between the primary and secondary windings of a transformer. In a case of this sort, you can use an autotransformer that consists of a single, tapped winding. Figure 18-8 shows three autotransformer configurations:
18-8 Schematic symbols for autotransformers. At A, air core, step-down. At B, laminated iron core, step-up. At C, powdered iron core, step-up.
• The unit at A has an air core, and operates as a step-down transformer.
• The unit at B has a laminated-iron core, and operates as a step-up transformer.
• The unit at C has a powdered-iron core, and operates as a step-up transformer.
You’ll sometimes find autotransformers in older radio receivers and transmitters. Autotransformers work well in impedance-matching applications. They also work as solenoidal loopstick antennas. Autotransformers are occasionally, but not often, used in AF applications and in 60-Hz utility wiring. In utility circuits, autotransformers can step the voltage down by a large factor, but they can’t efficiently step voltages up by more than a few percent.
Power Transformers
Any transformer used in the 60-Hz utility line, intended to provide a certain RMS AC voltage for the operation of electrical circuits, constitutes a power transformer. Power transformers exist in a vast range of physical sizes, from smaller than a grapefruit to bigger than your living room.
At the Generating Plant
We’ll find the largest transformers at locations where electricity is generated. Not surprisingly, high-energy power plants have bigger transformers that develop higher voltages than low-energy, local power plants have. These transformers handle extreme voltages and currents simultaneously.
When we want to transmit electrical energy over long distances, we must use high voltages. That’s because, for a given amount of power ultimately dissipated by the loads, the current goes down as the voltage goes up. Lower current translates into reduced ohmic loss in the transmission line. Recall the formula for power in nonreactive circuits in terms of the current and the voltage:
P = EI
where P represents the power (in watts), E represents the voltage (in volts), and I represents the current (in amperes). If we can make the voltage 10 times larger, for a given power level, then the current goes down to only 
as much. The ohmic losses in the wires vary in proportion to the square of the current. To understand why, remember that
P = I2 R
where P represents the power (in watts), I represents the current (in amperes), and R represents the resistance (in ohms). Engineers can’t do very much about the wire resistance or the power consumed by the loads in a large electrical grid, but the engineers can adjust the voltage, and thereby control the current.
Suppose that we increase the voltage in a power transmission line by a factor of 10, while the load at the end of the line draws constant power. The increase in voltage reduces the current to of its previous value. As a result, we cut the ohmic loss to ()2, or 
, of its previous amount. We, therefore, enjoy a big improvement in the efficiency of the transmission line (at least in terms of the ohmic loss in the wires).
Now we know why regional power plants have massive transformers capable of generating hundreds of thousands—or even millions—of volts! Up to a certain limit, we get better results for our money if we use high RMS voltage than we do if we use heavy-gauge wire for long-distance utility transmission lines.
Along the Line
Extreme voltages work well for high-tension power transmission, but a 200-kV RMS AC electrical outlet would not interest the average consumer! The wiring in a high-tension system requires considerable precautions to prevent arcing (sparking) and short circuits. Personnel must remain at least several meters away from the wires to avoid electrocution.
In a utility grid, medium-voltage power lines branch out from the major lines, and step-down transformers are used at the branch points. These lines fan out to lower-voltage lines, and step-down transformers are employed at these points, too. Each transformer must have windings heavy enough to withstand the product P = EI, the amount of VA power delivered to all the subscribers served by that transformer, at periods of peak demand.
Sometimes, such as during a heat wave, the demand for electricity rises above the normal peak level, “loading down” the circuit to the point that the voltage drops several percent. Then we have a brownout. If consumption rises further still, a dangerous current load appears at one or more intermediate power transformers. Circuit breakers in the transformers protect them from destruction by opening the circuit. Then we experience a blackout.
In most American homes, transformers step the voltage down to approximately 234 V RMS or 117 V RMS. Usually, 234-V RMS electricity appears in the form of three sine waves, called phases, each separated by 120°, and each appearing at one of the three slots in the outlet (Fig. 18-9A). We’ll commonly see this system employed with heavy appliances, such as ovens, air conditioners, and washing machines. A 117-V RMS outlet, in contrast, supplies only one phase, appearing between two of the three slots in the outlet. The third opening should go directly to a substantial earth ground (Fig. 18-9B). This system is commonly used for basic household appliances, such as lamps, television sets, and computers.
18-9 At A, an outlet for three-phase, 234-V RMS utility AC. At B, a conventional single-phase utility outlet for 117-V RMS utility AC.
In Electronic Equipment
Most consumer electronic systems have physically small power transformers. Most solid-state devices use low DC voltages, ranging from roughly 5 V to 50 V. For operation from 117-V RMS AC utility mains, therefore, such equipment requires step-down transformers in its power supplies.
Solid-state equipment usually (but not always) consumes relatively little power, so the transformers are rarely bulky. In high-powered AF or RF amplifiers, whose transistors can demand more than 1000 watts (1 kW) in some cases, the transformers require heavy-duty secondary windings, capable of delivering RMS currents of 90 A or more.
Older television receivers have cathode-ray tube (CRT) displays that need several hundred volts, derived from a step-up transformer in the power supply. Such transformers don’t have to deliver a lot of current, so they have relatively little bulk or mass. Another type of device that needs high voltage is a vacuum-tube RF amplifier such as the sort used by some amateur radio operators. Such an amplifier may demand 2 kV to 5 kV at around 500 mA.
At Audio Frequencies
Audio-frequency (AF) power transformers resemble those employed for 60-Hz electricity, except that the frequency is somewhat higher (up to 20 kHz), and audio signals exist over a range, or band, of frequencies (20 Hz to 20 kHz) rather than at only one frequency.
Most AF transformers are constructed like miniature utility transformers. They have laminated E cores with primary and secondary windings wound around the cross bars, as shown in Fig. 18-4. Audio transformers can function in either the step-up or step-down mode, and are designed to match impedances rather than to produce specific output voltages.
Audio engineers strive to minimize the system reactance so that the absolute-value impedance, Z, is close to the resistance R for both the input and the output. For that ideal condition to exist, the reactance X must be zero or nearly zero. In the following discussion of impedance-matching transformers (both for AF and RF applications), let’s assume that the impedances always constitute pure resistances of the form Z = R + j0.
Isolation and Impedance Matching
Transformers can provide isolation between electronic circuits. While a transformer can and should provide inductive coupling, it should exhibit relatively little capacitive coupling. We can minimize the amount of capacitive coupling by using cores that minimize the number of wire turns needed in the windings, and by keeping the windings physically separated from each other (rather than overlapping).
Balanced and Unbalanced Loads and Lines
When we connect a device to a balanced load, we can reverse the terminals without significantly affecting the operating behavior. A plain resistor offers a good example of a balanced load. The two-wire antenna input in an old-fashioned analog television receiver provides another example. A balanced transmission line usually has two wires running alongside each other and separated by a constant physical distance, such as old-fashioned TV ribbon, also called twinlead.
An unbalanced load must be connected a certain way; we can’t reverse the terminals. Switching the leads will result in improper circuit operation. In this sense, an unbalanced load resembles a polarized component, such as a battery, a diode, or an electrolytic capacitor. Many wireless antennas constitute unbalanced loads. Usually, unbalanced sources, transmission lines, and loads have one side connected to ground. The coaxial input of a television receiver is unbalanced; the shield (braid) of the cable connects to ground. An unbalanced transmission line usually comprises a coaxial cable of the sort that you see in a cable television system.
Normally, you can’t connect an unbalanced line to a balanced load, or a balanced line to an unbalanced load, and expect to obtain optimum performance. However, a transformer can provide compatibility between these two types of systems. Figure 18-10A illustrates a balanced-to-unbalanced transformer. The balanced (input) side of the transformer has a grounded center tap. Figure 18-10B shows an unbalanced-to-balanced transformer. The balanced (output) side has a grounded center tap.
18-10 At A, a balanced-to-unbalanced transformer. At B, an unbalanced-to-balanced transformer.
The turns ratio of a balanced-to-unbalanced transformer (also called a balun) or an unbalanced-to-balanced transformer (also known as an unbal) can equal 1:1, but that’s not a requirement. If the impedances of the balanced and unbalanced parts of the systems are the same, then a 1:1 turns ratio works well. But if the impedances differ, the turns ratio should match the impedances. Shortly, we’ll see how we can adjust the turns ratio of a transformer to convert a purely resistive impedance into another purely resistive impedance.
Transformer Coupling
Engineers sometimes use transformers between amplifier stages in electronic equipment to obtain a large amplification factor from the system. Part of the challenge in making a radio receiver or transmitter perform well involves getting the amplifiers to operate together in a stable manner. If too much feedback occurs, a series of amplifiers will oscillate, degrading or ruining the performance of the radio. Transformers that minimize the capacitance between the amplifier stages, while still transferring the desired signals, can help to prevent such oscillation.
Impedance-Transfer Ratio
The impedance-transfer ratio of a transformer varies according to the square of the turns ratio, and also according to the square of the voltage-transfer ratio. If the primary (source) and secondary (load) impedances are purely resistive and are denoted Zpri and Zsec, then
Zpri/Zsec = (Tpri/Tsec)2
and
Zpri/Zsec = (Epri/Esec)2
The inverses of these formulas, in which we express the turns ratio or voltage-transfer ratio in terms of the impedance-transfer ratio, are
Tpri/Tsec = (Zpri/Zsec)1/2
and
Epri/Esec = (Zpri/Zsec)1/2
Consider a situation in which we need a transformer to match an input impedance of 50.0 ohms, purely resistive, to an output impedance of 200 ohms, also purely resistive. What’s the required turns ratio Tpri/Tsec?
Solution
The transformer must have a step-up impedance ratio of
Zpri/Zsec = 50.0/200 = 1/4.00
From the above information, we can calculate
Tpri/Tsec = (1/4.00)1/2 = 0.2501/2 = 0.5 = 1/2
Problem 18-4
Suppose that a transformer has a primary-to-secondary turns ratio of 9.00:1. The load, connected to the transformer output, constitutes a pure resistance of 8.00 ohms. What’s the impedance at the primary?
Solution
The impedance-transfer ratio equals the square of the turns ratio. Therefore
Zpri/Zsec = (Tpri/Tsec)2 = (9.00/1)2 = 9.002 = 81.0
We know that the secondary impedance Zsec equals 8.00 ohms, so
Zpri = 81.0 × Zsec = 81.0 × 8.00 = 648 ohms
Radio-Frequency Transformers
Some RF transformers have primary and secondary windings, just like utility transformers. Others employ transmission-line sections. Let’s look at these two types, which, taken together, account for most RF transformers in use today.
Wirewound Types
In the construction of wirewound RF transformers, we can use powdered-iron cores up to quite high frequencies. Toroidal cores work especially well because of their self-shielding characteristic (all of the magnetic flux stays within the core material). The optimum number of turns depends on the frequency, and also on the permeability of the core.
In high-power applications, air-core coils are often preferred. Although air has low permeability, it has negligible hysteresis loss, and will not heat up or fracture as powdered-iron cores sometimes do. However, some of the magnetic flux extends outside of an air-core coil, potentially degrading the performance of the transformer when it must function in close proximity to other components.
A major advantage of coil type transformers, especially when wound on toroidal cores, lies in the fact that we can get them to function efficiently over a wide band of frequencies, such as from 3.5 MHz to 30 MHz. A transformer designed to work well over a sizable frequency range is called a broadband transformer.
Transmission-Line Types
As you recall, any transmission line has a characteristic impedance, denoted as Zo, that depends on the line construction. This property allows us to construct impedance transformers out of coaxial or parallel-wire line for operation at some radio frequencies.
Transmission-line transformers usually consist of quarter-wave sections. From the previous chapter, remember the formula for the length of a quarter-wave section:
Lft = 246v/fo
where Lft represents the length of the section in feet, v represents the velocity factor expressed as a fraction or ratio, and fo represents the operating frequency in megahertz. If we want to specify the length Lm in meters, then
Lm = 75v/fo
Suppose that a quarter-wave section of line, with characteristic impedance Zo, is terminated in a purely resistive impedance Rout. In this situation (Fig. 18-11), the impedance that appears at the input end of the line, Rin, also constitutes a pure resistance, and the following relations hold:
Zo2 = Rin Rout
and
Zo = (Rin Rout)1/2
18-11 A quarter-wave matching section of transmission line. The input impedance equals Rin, the output impedance equals Rout, and the characteristic impedance equals Zo.
We can rearrange the first formula to solve for Rin in terms of Rout, and vice versa:
Rin = Zo2/Rout
and
Rout = Zo2/Rin
These equations hold true at the frequency fo for which the line length measures ¼ wavelength. We can replace the word “wavelength” by the italic lowercase Greek letter lambda (λ), denoting the length of a quarter-wave section as (¼)λ or 0.25λ.
Neglecting line losses, the above relations hold at all odd harmonics of fo, that is, at 3fo, 5fo, 7fo, and so on, at which we have sections measuring 0.75λ, 1.25λ, 1.75λ, and so on. At other frequencies, a length of transmission line fails to function as a simple transformer. Instead, it behaves in a complex manner, the mathematical details of which would take us beyond the scope of this discussion.
Quarter-wave transmission-line transformers work well in antenna systems, especially at the higher frequencies (above a few megahertz) at which their dimensions become practical. A quarter-wave matching section should be constructed from an unbalanced line if the load is unbalanced, and from a balanced line if the load is balanced.
A disadvantage of quarter-wave sections arises from the fact that they work only at specific frequencies, depending on their physical length and on the velocity factor. But this shortcoming is often offset by the ease with which we can construct them, if we intend to use a piece of radio equipment at only one frequency, or at odd-numbered harmonics of that frequency.
Problem 18-5
Suppose an antenna has a purely resistive impedance of 100 ohms. We connect it to a ¼-wave section of 75-ohm coaxial cable. What’s the impedance at the input end of the section?
Solution
Use the formula from above to calculate
Rin = Zo2/Rout = 752/100 = 5625/100 = 56.25 ohms
We can round this result off to 56 ohms.
Problem 18-6
Consider an antenna known to have a purely resistive impedance of 300 ohms. You want to match it to the output of a radio transmitter designed to work into a 50.0-ohm pure resistance. What’s the characteristic impedance needed for a quarter-wave matching section?
Solution
You can calculate it using the formula for that purpose, as follows:
Zo = (Rin Rout)1/2 = (300 × 50.0)1/2 = 15,0001/2 = 122 ohms
Few, if any, commercially manufactured transmission lines have this particular characteristic impedance. Prefabricated lines come in standard Zo values, and a perfect match might not be obtainable. In that case, you can try the closest obtainable Zo. In this case, it would probably be 92 ohms or 150 ohms. If you can’t find anything near the characteristic impedance needed for a quarter-wave matching section, then you’re better off using a coil-type transformer.
What about Reactance?
When no reactance exists in an AC circuit using transformers, things stay simple. But often, especially in RF antenna systems, pure resistance doesn’t occur naturally. We have to “force it” by inserting inductors and/or capacitors to cancel the reactance out. The presence of reactance in a load makes a perfect match impossible with an impedance-matching transformer alone.
Inductive and capacitive reactances effectively oppose each other, and their magnitudes can vary. If a load presents a complex impedance R + jX, then we can cancel the reactance X by introducing an equal and opposite reactance −X in the form of an inductor or capacitor connected in series with the load. This action gives us a pure resistance with a value equal to (R + jX) − jX, or simply R. The reactance-canceling component should always be placed at the point where the load connects to the line.
When we want to conduct wireless communications over a wide band of frequencies, we can place adjustable impedance-matching and reactance-canceling networks between the transmission line and the antenna. Such a circuit is called a transmatch or an antenna tuner. These devices not only match the resistive portions of the transmitter and load impedances, but they cancel reactances in the load. Transmatches are popular among amateur radio operators, who use equipment capable of operation from less than 2 MHz up to the highest known radio frequencies.
Optimum Location
Whenever we use a transformer or ¼-wave section to match the characteristic impedance of an RF feed line to the purely resistive impedance of a well-designed antenna, that transformer or section should be placed between the line and the antenna. We call this location the feed point. If we place the transformer or ¼-wave section anywhere else, it won’t work at its best. In some cases, an improperly located transformer or ¼-wave section, even when tailored to the correct specifications, will make the overall impedance-mismatch situation worse!
Quiz
Refer to the text in this chapter if necessary. A good score is 18 or more correct. Answers are in the back of the book.
1. An autotransformer
(a) can automatically match impedances over a wide range.
(b) can effectively step down an AC voltage or impedance.
(c) has an automatically variable center tap.
(d) consists of a variable length of transmission line.
2. Which of the following core types works best if you need a coil winding inductance of 35 nH?
(a) Air
(b) Powdered-iron solenoid
(c) Toroid
(d) Pot core
3. Which of the following statements concerning air cores as compared with powdered-iron cores is true?
(a) Air concentrates the magnetic lines of flux more than powdered iron, allowing for higher inductance for a given number of turns per winding.
(b) Air-core transformers have greater loss than powdered-iron-core transformers, so air-core transformers work poorly for impedance matching.
(c) Air-core transformers require toroidal windings, limiting the possible configurations, while powdered-iron-core transformers have no such constraints.
(d) Air-core transformers operate best at the highest frequencies, while powdered-iron-core transformers are generally necessary at the lowest frequencies.
4. If a load contains no reactance whatsoever, then in theory we can use a transformer to match it perfectly to a transmission line
(a) having any characteristic impedance within reason.
(b) whose characteristic impedance is the same as the load impedance or lower, but not higher.
(c) whose characteristic impedance is the same as the load impedance or higher, but not lower.
(d) at only one frequency.
5. The primary-winding impedance always exceeds the secondary-winding impedance in
(a) an autotransformer.
(b) a step-up transformer.
(d) a balanced-to-unbalanced transformer.
6. We can minimize eddy currents in a 60-Hz AC utility transformer by
(a) using an air core.
(b) using a laminated-iron core.
(c) using the core winding method.
(d) winding the primary directly over the secondary.
7. If we wind the secondary directly over the primary in an air-core transformer, we should always expect to see
(a) a large impedance-transfer ratio.
(b) excellent performance at high frequencies.
(c) relatively high capacitance between the windings.
(d) significant hysteresis loss.
8. Suppose that a transformer has a primary-to-secondary turns ratio of exactly 4.00:1. We apply 20.0 V RMS AC across the primary terminals. How much AC RMS voltage can we expect to see across the secondary?
(a) 80.0 V RMS
(b) 40.0 V RMS
(c) 10.0 V RMS
(d) 5.00 V RMS
9. Suppose that a transformer has a primary-to-secondary turns ratio of exactly 1:4.00. The voltage at the primary equals 20.0 V RMS. What’s the AC RMS voltage at the secondary?
(a) 80.0 V RMS
(b) 40.0 V RMS
(c) 10.0 V RMS
(d) 5.00 V RMS
10. Suppose that a transformer has a secondary-to-primary turns ratio of exactly 2.00:1. We apply 20.0 V RMS AC across the primary terminals. How much AC RMS voltage can we expect to see across the secondary?
(a) 80.0 V RMS
(b) 40.0 V RMS
(c) 10.0 V RMS
(d) 5.00 V RMS
11. Suppose that a transformer has a secondary-to-primary turns ratio of exactly 1:2.00. The voltage at the primary equals 20.0 V RMS. What’s the AC RMS voltage at the secondary?
(a) 80.0 V RMS
(b) 40.0 V RMS
(c) 10.0 V RMS
(d) 5.00 V RMS
12. We want a transformer to match an input impedance of 300 ohms, purely resistive, to an output impedance of 50.0 ohms, also purely resistive. What’s the required primary-to-secondary turns ratio?
(a) 36.0:1
(b) 6.00:1
(c) 2.45:1
(d) 2.00:1
13. Suppose that a transformer has a primary-to-secondary turns ratio of 4.00:1. The load, connected to the transformer output, constitutes a pure resistance of 50.0 ohms. What’s the impedance at the primary?
(a) 12.8 k
(b) 800 ohms
(c) 400 ohms
(d) 200 ohms
14. A transformer has a primary-to-secondary impedance-transfer ratio of 4.00:1. We apply an AC signal of 200 V RMS to the primary. What’s the RMS AC signal voltage at the secondary?
(a) 800 V RMS
(b) 400 V RMS
(c) 141 V RMS
(d) 100 V RMS
15. An antenna has a purely resistive impedance of 600 ohms. We connect it to a ¼-wave section of 92-ohm coaxial cable. What’s the impedance at the input end of the section?
(a) 14 ohms
(b) 55 k
(c) 6.5 ohms
(d) 346 ohms
16. Suppose that we operate the system described in Question 15 at a frequency of 14 MHz. Our 92-ohm coaxial cable has a velocity factor of 0.75. How much cable will we need to construct a ¼-wave section?
(a) 8.0 m
(b) 7.5 m
(c) 5.3 m
(d) 4.0 m
17. A radio transmitter is designed to operate into a purely resistive impedance of 50 ohms. We have an antenna that exhibits a purely resistive impedance of 800 ohms. If we want to build an impedance-matching transformer to match these two impedances, what should its primary-to-secondary turns ratio be if we connect the transmitter to the primary and the antenna to the secondary?
(a) 1:16
(b) 1:8
(d) 1:2
18. Suppose that, in the situation described by Question 17, we want to use a ¼-wave section of transmission line to match the transmitter impedance to the antenna impedance. We need a line whose characteristic impedance equals
(a) 400 ohms
(b) 200 ohms
(c) 141 ohms
(d) 100 ohms
19. Imagine that the situation in the preceding two questions gets more complicated. The antenna has reactance in addition to the 800-ohm resistive component. That reactance results in a complex antenna impedance of 800 + j35 at a frequency of 14 MHz. Our antenna constitutes an unbalanced system, so it’s designed to work with a coaxial-cable transmission line or an unbalanced transformer secondary. How can we modify the antenna so that it will work properly at 14 MHz with either of the impedance-matching systems described in Questions 17 and 18?
(a) We can connect a capacitor in series with the antenna at the point where the transformer secondary or ¼-wave section output meets the antenna, such that the capacitive reactance equals −35 ohms at 14 MHz.
(b) We can connect an inductor in series with the antenna at the point where the transformer secondary or ¼-wave section output meets the antenna, such that the inductive reactance equals 35 ohms at 14 MHz.
(c) We can connect a capacitor in series with the system at the point where the transmitter output meets the transformer primary or ¼-wave section input, such that the capacitive reactance equals −35 ohms at 14 MHz.
(d) We can connect an inductor in series with the system at the point where the transmitter output meets the transformer primary or ¼-wave section input, such that the inductive reactance equals 35 ohms at 14 MHz.
20. Suppose that we want to use the antenna system described in Questions 17 through 19 over a continuous range of frequencies from 10 MHz to 20 MHz. In that case, what can we do to obtain a perfect impedance match at any frequency in that range?
(a) We can place a variable inductor in series with the antenna at the point where the transformer secondary or ¼-wave section output meets the antenna.
(b) We can place a variable capacitor in series with the antenna at the point where the transformer secondary or ¼-wave section output meets the antenna.
(c) We can place a well-engineered transmatch in series with the antenna at the point where the transformer secondary or ¼-wave section output meets the antenna.
(d) We can’t do anything. We can never expect to get a perfect impedance match over a continuous range of frequencies in a situation of this sort.