17

CHAPTER

Alternating-Current Power and Resonance

WHEN WE WANT TO OPTIMIZE HOW POWER “TRAVELS,” OR CHANGES FORM, WE FACE A CHALLENGE. A phenomenon called resonance can play an important role in efficient power transfer and conversion, especially at high AC frequencies.

Forms of Power

Scientists and engineers define power as the rate at which energy is expended, radiated, or dissipated. This definition applies to mechanical motion, chemical effects, electricity, sound waves, radio waves, heat, infrared (IR), visible light, ultraviolet (UV), X rays, gamma rays, and high-speed subatomic particles.

Units of Power

The standard unit of power is the watt, abbreviated W and equivalent to a joule per second (J/s). Sometimes, engineers specify power in kilowatts (kW or thousands of watts), megawatts (MW or millions of watts), gigawatts (GW or billions of watts), or terawatts (TW or trillions of watts). In numerical terms,

•   1 kW = 1000 W

•   1 MW = 1,000,000 W

•   1 GW = 1,000,000,000 W

•   1 TW = 1,000,000,000,000 W

We can also express power in milliwatts (mW or thousandths of watts), microwatts (μW or millionths of watts), nanowatts (nW or billionths of watts), or picowatts (pW or trillionths of watts). In numerical terms,

•   1 mW = 0.001 W

•   1 μW = 0.000001 W

•   1 nW = 0.000000001 W

•   1 pW = 0.000000000001 W

Volt-Amperes

In DC circuits, and also in AC circuits having no reactance, we can define power as the product of the voltage E across a device and the current I through that device; that is,

P = EI

If we express E in volts and I in amperes, then P comes out in volt-amperes (VA). Volt-amperes translate directly into watts when no reactance exists in a circuit (Fig. 17-1).

17-1   When an AC component contains no reactance, the power P is the product of the voltage E across the component and the current I through the component.

Volt-amperes, also called VA power or apparent power, can take various forms. A resistor converts electrical energy into heat energy at a rate that depends on the value of the resistance and the current through it. A light bulb converts electricity into light and heat. A radio antenna converts high-frequency AC into radio waves. A loudspeaker or headset converts low-frequency AC into sound waves. A microphone converts sound waves into low-frequency AC. Power figures in these forms give us measures of the intensity of the heat, light, radio waves, sound waves, or AC electricity.

Instantaneous Power

Engineers usually think of electrical power in RMS terms. But for VA power, peak values are sometimes used instead. If the AC constitutes a perfect sine wave with no DC component, the peak current (in either direction) equals 1.414 times the RMS current, and the peak voltage (of either polarity) equals 1.414 times the RMS voltage. If the current and the voltage follow along exactly in phase, then the product of their peak values equals twice the product of their RMS values.

In a reactance-free, sine-wave AC circuit, we see instants in time when the VA power equals twice the effective power. At other points in time, the VA power equals zero; at still other moments, the VA power falls somewhere between zero and twice the effective power level (Fig. 17-2). We call this constantly changing power level, measured or expressed at any particular point in time, the instantaneous power. In some situations, such as with an amplitude-modulated (AM) radio signal, the instantaneous power varies in a complicated fashion.

17-2   Peak versus effective power for a sine wave. The left-hand vertical scale shows relative voltage. The right-hand vertical scale shows relative power. The solid curve represents the voltage as a function of time. The light and heavy dashed waves show peak and effective power, respectively, as functions of time.

Reactive or Imaginary Power

In a pure resistance, the rate of energy expenditure per unit of time (or true power) equals the VA power (also known as apparent power). But when reactance exists, the VA power exceeds the power manifested as heat, light, radio waves, or whatever. The apparent power is greater than the true power. We call the difference reactive power or imaginary power because it exists only within the reactive part of the circuit, represented by the imaginary-number part of the complex-number impedance.

Inductors and capacitors store energy and release it a fraction of a cycle later, over and over, for as long as AC flows. This phenomenon, like true power, manifests itself as the rate at which energy changes from one form to another. But rather than existing in a form that we can employ in some practical way, imaginary power can only go in and out of “storage.” We can’t use it for anything. The storage/release cycle repeats along with the actual AC cycle.

True Power Doesn’t Travel

If you connect a radio transmitter to a cable that runs to an antenna, you might say you’re “feeding power” through the cable to the antenna. Even experienced radio-frequency (RF) engineers and technicians sometimes say that. However, true power always involves a change in form, such as from electrical current and voltage into radio waves, or from sound waves into heat. Power does not actually travel from place to place. It simply happens in one spot or another.

In a radio antenna system, some true power dissipates in a wasteful manner, ending up as heat in the transmitter amplifiers and in the transmission (feed) line, as shown in Fig. 17-3. Obviously, a competent RF engineer seeks to minimize the extent of this waste. The useful dissipation of true power occurs when the imaginary power, in the form of electric and magnetic fields, gets to the antenna, where it changes form and emerges from the antenna as electromagnetic (EM) waves.

17-3   True power and imaginary power in a radio transmitter and antenna system.

If you do much work with wireless transmitting antenna systems, you’ll hear or read expressions, such as “forward power” and “reflected power,” or “power goes from this amplifier to these speakers.” You can talk or write in such terms if you like, but keep in mind that the notion can sometimes lead to mistaken conclusions. For example, you might get the idea that an antenna system works more or less efficiently than it actually does.

Reactance Consumes No Power

A pure inductance or a pure capacitance can’t dissipate any power. A pure reactance can only store energy and then give it back to the circuit a fraction of a cycle later. In real life, the dielectrics or wires in coils and capacitors dissipate some power as heat, but ideal components wouldn’t do that. A capacitor, as we’ve learned, stores energy as an electric field. An inductor stores energy as a magnetic field.

A reactive component causes AC to shift in phase, so that the current does not follow in step with the voltage, as it would in a reactance-free circuit. In a circuit containing inductive reactance, the current lags the voltage by up to 90°, or one-quarter of a cycle. In a circuit with capacitive reactance, the current leads the voltage by up to 90°.

In a purely resistive circuit, the voltage and current precisely follow each other so that they combine in the most efficient possible way (Fig. 17-4A). But in a circuit containing reactance, the voltage and current do not follow exactly along with each other (Fig. 17-4B) because of the phase difference. In that case, the product of the voltage and current (the VA or apparent power) exceeds the actual energy expenditure (the true power).

17-4   At A, the current I and voltage E are in phase in a nonreactive AC circuit. At B, I and E are not in phase when reactance exists.

Power Parameters

In an AC circuit containing nonzero resistance and nonzero reactance, we can summarize the relationship between true power P_T, apparent (VA) power P_VA, and imaginary (reactive) power P_X in terms of the formula

P_VA = (P_T² + P_X²)^1/2

where P_T < P_VA and P_X < P_VA. If no reactance exists, then P_VA = P_T and P_X = 0. Engineers strive to minimize, and if possible eliminate, the reactance in power-transmission systems.

Power Factor

In an AC circuit, we call the ratio of the true power to the VA power, P_T/P_VA, the power factor (PF). In the ideal case in which we have no reactance, P_T = P_VA, so PF = 1. If the circuit contains reactance but lacks resistance or conductance (zero or infinite resistance), then P_T = 0, so PF = 0.

When a load, or a circuit in which we want power to dissipate or change form, contains both resistance and reactance, then PF falls between 0 and 1. We can also express the power factor as a percentage PF_% between 0 and 100. If we know P_T and P_VA, then we can calculate PF as

PF = P_T/P_VA

and PF_% as

PF_% = 100P_T/P_VA

When a load has nonzero, finite resistance and nonzero, finite reactance, then some of the power dissipates as true power, and some of the power gets “rejected” by the load as imaginary power.

We can determine the power factor in an AC circuit that contains reactance and resistance in two ways:

1.  Find the cosine of the phase angle

2.  Calculate the ratio of the resistance to the absolute-value impedance

Cosine of Phase Angle

Recall that in a circuit having reactance and resistance, the current and the voltage do not follow along exactly in phase. The phase angle (ϕ) constitutes the extent, expressed in degrees, to which the current and the voltage differ in phase. In a pure resistance, ϕ = 0°. In a pure reactance, ϕ = +90° (if the net reactance is inductive) or ϕ = −90° (if the net reactance is capacitive). We can calculate the power factor as

PF = cos ϕ

and

PF_% = 100 cos ϕ

Problem 17-1

Suppose that a circuit comprises a pure resistance of 600 ohms with no reactance whatsoever. What’s the power factor?

Solution

Without doing any calculations, you can sense that PF = 1 because P_VA = P_T in a pure resistance. It follows that P_T/P_VA = 1. But you can also look at this situation by noting that the phase angle equals 0° because the current follows in phase with the voltage. Using a calculator, you’ll find that cos 0° = 1. Therefore, PF = 1 = 100%. Figure 17-5 illustrates the RX half-plane vector for this situation. Remember that you should express the phase angle with respect to the R axis.

17-5   Vector diagram showing the phase angle for a purely resistive impedance of 600 + j0. The R and jX scales are relative.

Problem 17-2

Suppose that a circuit contains a pure capacitive reactance of −40 ohms, but no resistance. What’s the power factor?

Solution

Here, the phase angle equals −90°, as shown in the RX half-plane vector diagram of Fig. 17-6. A calculator will tell you that cos −90° = 0. Therefore, PF = 0, and P_T/P_VA = 0 = 0%. None of the power is true; it’s all reactive.

17-6   Vector diagram showing the phase angle for a purely capacitive impedance of 0 − j40. The R and jX scales are relative.

Problem 17-3

Consider a circuit that contains a resistance of 50 ohms and an inductive reactance of 50 ohms, connected in series. What’s the power factor?

Solution

The phase angle equals 45° (Fig. 17-7). The resistance and the reactance vectors have equal lengths, forming two sides of a right triangle. The complex impedance vector constitutes the hypotenuse (longest side) of the right triangle. To determine the power factor, you can use a calculator to find cos 45° = 0.707, so you know that P_T/P_VA = 0.707 = 70.7%.

17-7   Vector diagram showing the phase angle for a complex impedance of 50 + j50. The R and jX scales are relative.

The Ratio R/Z

We can calculate the power factor in an RX circuit by finding the ratio of the resistance R to the absolute-value impedance Z. Figure 17-7 provides an example. A right triangle is formed by the resistance vector R (the base), the reactance vector jX (the height), and the absolute-value impedance Z (the hypotenuse). The cosine of the phase angle equals the ratio of the base length to the hypotenuse length, or R/Z.

Problem 17-4

Suppose that a series circuit has an absolute-value impedance Z of 100 ohms, with a resistance R of 80 ohms. What’s the power factor?

Solution

You can set up and calculate the ratio

PF = R/Z = 80/100 = 0.8 = 80%

It doesn’t matter whether the net reactance in this circuit happens to be capacitive or inductive.

Problem 17-5

Consider a series circuit with an absolute-value impedance of 50 ohms, purely resistive. What’s the power factor?

Solution

Here, R = Z = 50 ohms. Therefore

PF = R/Z = 50/50 = 1 = 100 %

Problem 17-6

Consider a circuit with a resistance of 50 ohms and a capacitive reactance of −30 ohms. What’s the power factor? Use the cosine method.

Solution

Remember the formula for phase angle in terms of reactance and resistance:

ϕ = Arctan (X/R)

where X represents the reactance and R represents the resistance. Therefore

ϕ = Arctan (−30/50) = Arctan (−0.60) = −31°

The power factor equals the cosine of this angle, so

PF = cos (−31°) = 0.86 = 86%

Problem 17-7

Consider a series circuit with a resistance of 30 ohms and an inductive reactance of 40 ohms. What’s the power factor? Use the R/Z method.

Solution

First, find the absolute-value impedance using the formula for series circuits:

Z = (R² + X²)^1/2

where R represents the resistance and X represents the net reactance. Plugging in the numbers, you get

Z = (30² + 40²)^1/2 = (900 + 1600)^1/2 = 2500^1/2 = 50 ohms

Now you can calculate the power factor as

PF = R/Z = 30/50 = 0.60 = 60%

You can graph this situation as a 30:40:50 right triangle (Fig. 17-8).

17-8   Illustration for Problem 17-7. The vertical and horizontal scale increments differ; this is a common practice in graphs.

How Much of the Power Is True?

The above formulas allow you to figure out, given the resistance, reactance, and VA power, how many watts constitute true, or real power, and how many watts constitute reactive, or imaginary power. Engineers must consider this situation when working with RF equipment because some RF wattmeters display VA power rather than true power. When reactance exists along with the resistance in a circuit or system, such wattmeters give artificially high readings.

Problem 17-8

Suppose that a circuit has 50 ohms of resistance and 30 ohms of inductive reactance in series. A wattmeter shows 100 W, representing the VA power. What’s the true power?

Solution

We must determine the power factor to figure out the answer to this question. First, we calculate the phase angle as

ϕ = Arctan (X/R) = Arctan (30/50) = 31°

The power factor equals the cosine of the phase angle, so

PF = cos 31° = 0.86

The formula for the power factor in terms of true and VA power is

PF = P_T/P_VA

We can rearrange this formula to solve for true power, obtaining

P_T = PF × P_VA

When we plug in PF = 0.86 and P_VA = 100, we get

P_T = 0.86 × 100 = 86 W

Problem 17-9

Suppose that a circuit has a resistance of 1000 ohms in parallel with a capacitance of 1000 pF. We operate the circuit at a frequency of 100 kHz. If a wattmeter designed to read VA power shows 88.0 W, what’s the true power?

Solution

This problem is rather complicated because the components appear in parallel. To begin, let’s make sure that we have the units in agreement so the formulas work right. We can convert the frequency f to megahertz, getting f = 0.100 MHz. We can convert the capacitance to microfarads, getting C = 0.001000 μF. From the previous chapter, we remember the formula for capacitive susceptance, and calculate it for this situation as

The conductance of the resistor, G, equals the reciprocal of the resistance, R, so

G = 1/R = 1/1000 = 0.001000 S

Now let’s use the formulas for calculating resistance and reactance in terms of conductance and susceptance in parallel circuits. First, we find the resistance as

Next, we find the reactance as

again rounded to four significant figures. Now we can calculate the phase angle as

We calculate the power factor as

PF = cos ϕ = cos (−32.142°) = 0.84673

The VA power P_VA is given as 88.0 W. Therefore, we can calculate the true power, rounding off to three significant figures (because that’s the extent of the accuracy of our input data), getting

Power Transmission

Consider a radio broadcast or communications station. The transmitter produces high-frequency AC. We want to get the signal efficiently to an antenna located some distance from the transmitter. This process involves the use of an RF transmission line, also known as a feed line. The most common type is coaxial cable. Alternatively, we can use two-wire line, also called parallel-wire line, in some antenna systems. At ultra-high and microwave frequencies, another kind of transmission line, known as a waveguide, is often employed.

Loss: the Less, the Better!

The overriding challenge in the design and construction of any power transmission system lies in minimizing the loss. Power wastage occurs almost entirely as heat in the transmission line conductors and dielectric, and in objects near the line. Some loss can take the form of unwanted electromagnetic radiation from the line. Loss also occurs in transformers. Power loss in an electrical system is analogous to the loss of usable work produced by friction in a mechanical system.

In an ideal power-transmission system, all of the power constitutes VA power; that is, all of the power occurs as AC in the conductors and an alternating voltage between them. We don’t want power in a transmission line or transformer to exist in the form of true power because that situation translates into heat loss, radiation loss, or both. True power dissipation or radiation should always take place in the load, usually at the opposite end of the transmission line from the source.

Power Measurement in a Transmission Line

In an AC transmission line, we can measure power by placing an AC voltmeter between the conductors and an AC ammeter in series with one of the conductors, as shown in Fig. 17-9. We should place both meters at the same point on the line (even though they don’t look that way in this diagram!). In that case the power P (in watts) equals the product of the RMS voltage E (in volts) and the RMS current I (in amperes). We can use this technique in any transmission line at any AC frequency, from the 60 Hz of a utility system to many gigahertz in some wireless communications systems. However, when we measure power this way, we don’t necessarily get an accurate indication of the true power dissipated by the load at the end of the line.

17-9   Power measurement in a transmission line. Ideally, we should measure the voltage and the current at the same physical point on the line.

Recall that any transmission line has an inherent characteristic impedance. The value of this parameter Z depends on the diameters of the line conductors, the spacing between the conductors, and he type of dielectric material that separates the conductors. If the load constitutes a pure resistance R containing no reactance, and if R = Z_o, then the power indicated by the voltmeter/ammeter scheme will equal the true power dissipated by the load—provided that we place the voltmeter and ammeter directly across the load, at the end of the line opposite the source.

If the load constitutes a pure resistance that differs from the characteristic impedance of the line (that is, R ≠ Z_o), then the voltmeter and ammeter will not give an indication of the true power. Also, if the load contains any reactance along with the resistance, the voltmeter/ammeter method will fail to give us an accurate reading of the true power, no matter what the resistance.

Impedance Mismatch

If we want a power transmission system to perform at its best, then the load impedance must constitute a pure resistance equal to the characteristic impedance of the line. When we don’t have this ideal state of affairs, the system has an impedance mismatch.

Small impedance mismatches can usually—but not always—be tolerated in power transmission systems. In very-high-frequency (VHF), ultra-high-frequency (UHF) and microwave wireless transmitting systems, even a small impedance mismatch between the load (antenna) and the line can cause excessive power losses in the line.

We can usually get rid of an impedance mismatch by installing a matching transformer between the transmission line and the load. We can also correct impedance mismatches in some situations by deliberately placing a reactive component (inductor or capacitor) in series or parallel with the load to cancel out any existing load reactance.

Loss in a Mismatched Line

When we terminate a transmission line in a pure resistance R equal to the characteristic impedance Z_o of the line, then the RMS current I and the RMS voltage E remain constant all along the line, provided that the line has no ohmic loss and no dielectric loss. In such a situation, we have

R = Z_o = E/I

where we express R and Z_o in ohms, E in volts RMS, and I in amperes RMS. Of course, no transmission line is perfectly lossless. In a “real-world” transmission line, the current and voltage gradually decrease as a signal makes its way from the source to the load. Nevertheless, if the load constitutes a pure resistance equal to the characteristic impedance of the line, the current and voltage remain in the same ratio at all points along the line, as shown in Fig. 17-10.

17-10   Along a matched transmission line, the voltage-to-current ratio E/I holds constant everywhere, although the actual values of E and I decrease with increasing distance from the source.

Standing Waves

If a transmission line and its load aren’t perfectly matched, then the current and voltage alternately rise and fall as they move along the line. We call the maxima and minima loops and nodes, respectively. At a maximum-current point (a current loop), the voltage reaches its minimum (a voltage node). Conversely, at a maximum-voltage point (a voltage loop), the current attains its minimum (a current node).

If we graph the current and voltage loops and nodes along a mismatched transmission line as functions of the position on the line, we see wavelike patterns that remain fixed over time. These orderly patterns of current and voltage don’t move in either direction along the line; they simply “stand there.” For this reason, engineers call them standing waves.

Losses Caused by Standing Waves

When a transmission line contains standing waves, we observe a corresponding pattern in the extent of the line loss, as follows:

•   At current loops, the loss in the line conductors reaches a maximum

•   At current nodes, the loss in the line conductors reaches a minimum

•   At voltage loops, the loss in the line dielectric reaches a maximum

•   At voltage nodes, the loss in the line dielectric reaches a minimum

It’s tempting to suppose that all of the loss variations ought to average out in a mismatched line, so that the excess loss in some places gets “paid back” in the form of reduced loss in other places. But things don’t work that way. Overall, the losses in a mismatched line always exceed the losses in a perfectly matched line. The extra loss increases as the mismatch gets worse. We call it transmission-line mismatch loss or standing-wave loss; it occurs as heat dissipation, so it constitutes true power. Any true power that heats up a transmission line goes to waste because it never reaches the load. We want the true power to end up in the load—if not all of it, then as much of it as possible.

As the extent of the mismatch in a power-transmission system increases, so does the loss caused by the current and voltage loops in the standing waves. The more loss a line has when perfectly matched, the more loss a given amount of mismatch will cause. Standing-wave loss also increases as the frequency increases, if we hold all other factors constant.

Line Overheating

A severe mismatch between the load and the transmission line can cause another problem besides lost power: physical damage to, or destruction of, the line.

Suppose that a certain transmission line can effectively function with a radio transmitter that generates up to 1 kW of power, assuming that the line and the load are perfectly matched. If a severe mismatch exists and you try to feed 1 kW into the same line, the extra current at the current loops can heat the conductors to the point at which the dielectric material melts and the line shorts out. It’s also possible for the voltage at the voltage loops to cause arcing between the line conductors. This arcing can perforate and/or burn the dielectric, ruining the line.

When we have no choice but to operate an RF transmission line with a significant impedance mismatch, we must refer to derating functions in order to determine how much power the line can safely handle. Manufacturers of prefabricated lines, such as coaxial cable, can (or should) provide this information.

Resonance

We observe resonance in an AC circuit when capacitive and inductive reactance both exist, but they have equal and opposite values so that they cancel each other out. We saw a couple of examples of this phenomenon in Chap. 16. Let’s explore it in more detail.

Series Resonance

Capacitive reactance X_C and inductive reactance X_L can have equal magnitudes although they produce opposite effects. In any circuit containing some inductance and some capacitance, there exists a frequency at which X_L = −X_C. This condition constitutes resonance, and we symbolize the frequency at which it occurs by writing f_o. In a simple LC circuit, we observe only one such frequency; in some circuits involving transmission lines or antennas, we observe many such frequencies. In that case, we call the lowest frequency at which resonance occurs the fundamental resonant frequency, symbolized f_o.

You’ll recognize Fig. 17-11 as a schematic diagram of a series RLC circuit. If we apply a variable-frequency AC signal to the end terminals, we’ll find that at one particular “critical” frequency, X_L = −X_C. This phenomenon will always occur if L and C are both finite and nonzero. The “critical” frequency represents f_o for the circuit. At f_o, the effects of capacitive reactance and inductive reactance cancel out so that the circuit appears as a pure resistance, with a value theoretically equal to R. We call this condition series resonance.

17-11   A series RLC circuit.

If R = 0 (a short circuit), then Fig. 17-11 represents a series LC circuit, and the impedance at resonance theoretically equals 0 + j0. Under these conditions, the circuit offers no opposition to the flow of alternating current at f_o. Of course, “real-world” series LC circuits always contain at least a small amount of resistance so that a little loss occurs in the coil and capacitor; the real-number part of the complex impedance doesn’t exactly equal 0. If the coil and capacitor have high quality and exhibit minimal loss, however, we can usually say that R = 0 “for all intents and purposes.”

Parallel Resonance

Figure 17-12 shows a generic parallel RLC circuit. In this situation, we can think of the resistance R as a conductance G, with G = 1/R. For that reason, some people might say that we should call this arrangement a parallel GLC circuit. We can get away with either term as long as we include the word “parallel”!

17-12   A parallel RLC circuit.

At one particular frequency f_o, the inductive susceptance B_L will exactly cancel the capacitive susceptance B_C, so we have B_L = −B_C. This condition always occurs for some applied AC signal frequency f_o, as long as the circuit contains finite, nonzero inductance and finite, nonzero capacitance. At f_o, the susceptances cancel each other out, leaving theoretically zero susceptance. The admittance through the circuit theoretically equals the conductance G of the resistor. We call this condition parallel resonance.

If the circuit contains no resistor, but only a coil and capacitor, we have a parallel LC circuit, and the admittance at resonance theoretically equals 0 + j0. The circuit will offer a lot of opposition to alternating current at f_o, and the complex impedance will, in an informal sense, equal “infinity” (∞). In a practical “real-world” LC circuit, the coil and capacitor always have a little bit of loss, so the real-number part of the complex impedance isn’t really infinite. However, if we use low-loss components, we can get real-number coefficients of many megohms or even gigohms for parallel LC circuits at resonance, so that we can say that R = ∞ “for all intents and purposes.”

Calculating the Resonant Frequency

We can calculate the resonant frequency f_o, of a series RLC or parallel RLC circuit in terms of the inductance L in henrys and the capacitance C in farads, using the formula

f_o = 1/[2π(LC)^1/2]

Considering π = 3.1416, we can simplify this formula to

f_o = 0.15915/(LC)^1/2

You might remember from your basic algebra or precalculus course that the ½ power of a quantity represents the positive square root of that quantity. The foregoing formula will also work if you want to find f_o in megahertz (MHz) when you know L in microhenrys (μH) and C in microfarads (μF).

The Effects of R and G

Interestingly, the value of R or G does not affect the value of f_o in series or parallel RLC circuits. However, the presence of nonzero resistance in a series resonant circuit, or nonzero conductance in a parallel resonant circuit, makes f_o less well-defined than it would be if the resistance or conductance did not exist. If we short out R in the circuit of Fig. 17-11 or remove R from the circuit of Fig. 17-12, we have LC circuits that exhibit the most well-defined possible resonant responses.

In a series RLC circuit, the resonant frequency response becomes more “broad” as the resistance increases, and more “sharp” as the resistance decreases. In a parallel RLC circuit, the resonant frequency response becomes more “broad” as the conductance increases (R gets smaller), and more “sharp” as the conductance decreases (R gets larger). In theory, the “sharpest” possible responses occur when R = 0 in a series circuit, and when G = 0 (that is, R = ∞) in a parallel circuit.

Problem 17-10

Find the resonant frequency of a series circuit with a 100-μH inductor and a 100-pF capacitor.

Solution

We should convert the capacitance to 0.000100 μF. Then we can find the product LC = 100 × 0.000100 = 0.0100. When we take the square root of this, we get 0.100. Finally, we can divide 0.15915 by 0.100, getting f_o = 1.5915 MHz. We should round this off to 1.59 MHz.

Problem 17-11

Find the resonant frequency of a parallel circuit consisting of a 33-μH coil and a 47-pF capacitor.

Solution

Let’s convert the capacitance to 0.000047 μF. Then we find the product LC = 33 × 0.000047 = 0.001551. Taking the square root of this, we get 0.0393827. Finally, we divide 0.15915 by 0.0393827 and round off, getting f_o = 4.04 MHz.

Problem 17-12

Suppose that we want to design a circuit so that it exhibits f_o = 9.00 MHz. We have a 33.0-pF fixed capacitor available. What size coil will we need to obtain the desired resonant frequency?

Solution

Let’s use the formula for the resonant frequency and plug in the values. Then we can use arithmetic to solve for L. We convert the capacitance to 0.0000330 μF and calculate in steps as follows:

f_o = 0.15915/(LC)^1/2
9.00 = 0.15915/(L × 0.0000330)^1/2
9.00² = 0.15915²/(0.0000330 × L)
81.0 = 0.025329/(0.0000330 × L)
81.0 × 0.0000330 × L = 0.025329
0.002673 × L = 0.025329
L = 0.025329/0.002673
= 9.48 μH

Problem 17-13

Suppose that we want to design an LC circuit with f_o = 455 kHz. We have a 100-μH in our “junk box.” What size capacitor do we need?

Solution

We should convert the frequency to 0.455 MHz. Then the calculation proceeds in the same way as with the preceding problem:

f_o = 0.15915/(LC)^1/2
0.455 = 0.15915/(100 × C)^1/2
0.455² = 0.15915²/(100 × C)
0.207025 = 0.025329/(100 × C)
0.207025 × 100 × C = 0.025329
20.7025 × C = 0.025329
C = 0.025329/20.7025
= 0.00122 μF

Adjusting the Resonant Frequency

In practical circuits, engineers often place variable inductors and/or variable capacitors in series or parallel LC circuits designed to function at resonance, thereby allowing for small errors in the actual resonant frequency (as opposed to the calculated value for f_o). We can design a circuit of this sort, called a tuned circuit, so that it exhibits a frequency slightly higher than f_o, and then install a padding capacitance (C_p) in parallel with the main capacitance C, as shown in Fig. 17-13. A padding capacitor is a small component with an adjustable value ranging from around 1 pF up to several picofarads, or several tens of picofarads. If the engineers want the circuit to offer a wider range of resonant frequencies, a variable capacitor, having a value ranging from a few picofarads up to several hundred picofarads, can serve the purpose.

17-13   Padding capacitors (C_p) allow limited adjustment of the resonant frequency in a series LC circuit (as shown at A), or in a parallel LC circuit (as shown at B).

Resonant Devices

Resonant circuits often consist of coils and capacitors in series or parallel, but other kinds of hardware also exhibit resonance.

Piezoelectric Crystals

Pieces of the mineral quartz, when cut into thin wafers and subjected to voltages, will vibrate at high frequencies. Because of the physical dimensions of such a piezoelectric crystal, these vibrations occur at a precise frequency f_o, and also at whole-number multiples of f_o. We call these multiples—2f_o, 3f_o, 4f_o, and so on—harmonic frequencies or simply harmonics. The frequency f_o constitutes the fundamental frequency or simply the fundamental. The fundamental, f_o, is the lowest frequency at which resonance occurs. Quartz crystals can act like LC circuits in electronic devices. A crystal exhibits an impedance that varies with frequency. The reactance equals zero at f_o and the harmonic frequencies.

Cavities

Lengths of metal tubing, cut to specific dimensions, exhibit resonance at very-high, ultra-high, and microwave radio frequencies. They work in much the same way as musical instruments resonate with sound waves. However, the waves take the form of electromagnetic fields rather than acoustic disturbances. Such cavities, also called cavity resonators, have reasonable physical dimensions at frequencies above about 150 MHz. We can get a cavity to work below this frequency, but we’ll find it difficult to construct because of its great length and clumsiness. Like crystals, cavities resonate at a specific fundamental frequency f_o, and also at all of the harmonic frequencies.

Sections of Transmission Line

When we cut a transmission line to any whole-number multiple of ¼ wavelength, it behaves as a resonant circuit. The most common length for a transmission-line resonator is ¼ wavelength, giving us a so-called quarter-wave section.

When we short-circuit a quarter-wave section at one end and apply an AC signal to the other end, the section acts like a parallel-resonant LC circuit, and it has an extremely high (theoretically infinite) resistive impedance at the fundamental resonant frequency f_o. When we leave one end of a quarter-wave section open and apply an AC signal to the other end, the section acts like a series-resonant LC circuit, and it has an extremely low (theoretically zero) resistive impedance at f_o. In effect, a quarter-wave section converts an AC short circuit into an AC open circuit and vice-versa—at a specific frequency f_o.

The length of a quarter-wave section depends on the desired fundamental resonant frequency f_o. It also depends on how fast the electromagnetic energy travels along the line. We can define this speed in terms of a velocity factor, abbreviated v, expressed as a fraction or percentage of the speed of light in free space. Manufacturers provide velocity factors for prefabricated transmission lines, such as coaxial cable or old-fashioned two-wire television “ribbon.”

If the frequency in megahertz equals f_o and the velocity factor of a line equals v (expressed as a fraction), then we can calculate the length L_ft of a quarter-wave section of transmission line in feet using the formula

L_ft = 246v/f_o

If we know the velocity factor as a percentage v_%, then the above formula becomes

L_ft = 2.46v_%/f_o

If we know the velocity factor as a fraction v, then the length L_m of a quarter-wave section in meters is

L_m = 75.0v/f_o

For the velocity factor as a percentage v_%, we have

L_m = 0.750v_%/f_o

Note that we use L to stand for “length,” not “inductance,” in this context!

Antennas

Many types of antennas exhibit resonant properties. The simplest type of resonant antenna, and the only kind that we’ll consider here, is the center-fed, half-wavelength dipole antenna (Fig. 17-14).

17-14   The half-wave, center-fed dipole constitutes a simple and efficient antenna.

We can calculate the approximate length L_ft, in feet, for a dipole antenna at a frequency of f_o using the formula

L_ft = 467/f_o

taking into account the fact that electromagnetic fields travel along a wire at about 95% of the speed of light. A straight, thin wire in free space, therefore, has a velocity factor of approximately v = 0.95. If we specify the approximate length of the half-wave dipole in meters as L_m, then

L_m = 143/f_o

A half-wave dipole has a purely resistive impedance of about 73 ohms at its fundamental frequency f_o. But this type of antenna also resonates at harmonics of f_o. The dipole measures a full wavelength from end to end at 2f_o; it measures 3/2 wavelength from end to end at 3f_o; it measures two full wavelengths from end to end at 4f_o, and so on.

Radiation Resistance

At f_o and all of the odd-numbered harmonics, a dipole antenna behaves like a series resonant RLC circuit with a fairly low resistance. At all even-numbered harmonics, the antenna acts like a parallel resonant RLC circuit with a high resistance.

Does the mention of resistance in a half-wave dipole antenna confuse you? Maybe it should! Figure 17-14 shows no resistor. Where, you ask, does the resistance in a half-wave dipole come from? The answer gets into some rather esoteric electromagnetic-wave theory, and it brings to light an interesting property that all antennas have: radiation resistance. This parameter constitutes a crucial factor in the design and construction of all RF antenna systems.

When we connect a radio transmitter to an antenna and send out a signal, energy radiates into space in the form of radio waves from the antenna. Although no physical resistor exists anywhere in the antenna system, the radiation of radio waves acts just like power dissipation in a pure resistance. In fact, if we replace a half-wave dipole antenna with a 73-ohm non-reactive resistor that can safely dissipate enough power, we’ll discover that a wireless transmitter connected to the opposite end of the line won’t “know” the difference between that resistor and the dipole antenna.

Problem 17-14

How many feet long, to the nearest foot, is a quarter-wave section of transmission line at 7.1 MHz if the velocity factor equals 80%?

Solution

We can use the formula for the length L_ft of a quarter-wave section based on the velocity factor v_% = 80, as follows:

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.  A transmission line operates at its best efficiency when

(a)  the load impedance constitutes a pure resistance equal to the characteristic impedance of the line.

(b)  the load impedance constitutes a pure inductive reactance equal to the characteristic impedance of the line.

(c)  the load impedance constitutes a pure capacitive reactance equal to the characteristic impedance of the line.

(d)  the absolute-value impedance of the load equals the characteristic impedance of the line.

2.  The ninth harmonic of 900 kHz is

(a)  100 kHz.

(b)  300 kHz.

(c)  1.20 MHz.

(d)  8.10 MHz.

3.  A pure resistance dissipates or radiates

(a)  complex power.

(b)  imaginary power.

(c)  true power.

(d)  apparent power.

4.  Suppose that we want to build a ½-wave dipole antenna designed to have a fundamental resonant frequency of 14.3 MHz. How long should we make the antenna, as measured from end to end in meters?

(a)  32.7 m

(b)  10.0 m

(c)  16.4 m

(d)  We need more information to answer this question.

5.  When a transmission line exhibits standing waves, we find a voltage maximum

(a)  wherever we find a current maximum.

(b)  wherever we find a current minimum.

(c)  at the transmitter end of the line.

(d)  at the load end of the line.

6.  Standing waves on a transmission line (as compared with a line operating without any impedance mismatch) increase the loss in the wire conductors at the

(a)  current maxima.

(b)  voltage maxima.

(c)  current minima and voltage maxima.

(d)  current minima and voltage minima.

7.  When we take the cosine of the phase angle in an AC circuit or system that contains both resistance and reactance, we get the

(a)  true power.

(b)  imaginary power.

(c)  apparent power.

(d)  power factor.

8.  Which of the following parameters is an example of true power in an AC circuit or system?

(a)  The AC that appears between the plates of a capacitor

(b)  The AC that passes through a wire inductor

(c)  The AC that dissipates as heat in a transmission line

(d)  The AC that travels along a transmission line

9.  Suppose that the apparent power in a circuit equals 40 W and the true power equals 30 W. What’s the power factor?

(a)  60%

(b)  75%

(c)  80%

(d)  We need more information to calculate it.

10.  Suppose that the true power in a circuit equals 40 W and the imaginary power equals 30 W. What’s the power factor?

(a)  60%

(b)  75%

(c)  80%

(d)  We need more information to calculate it.

11.  Consider a series circuit with a resistance of 24 ohms and an inductive reactance of 10 ohms. What’s the power factor?

(a)  42%

(b)  58%

(c)  92%

(d)  18%

12.  Imagine that you encounter a series circuit with a resistance of 24 ohms and a capacitive reactance of −10 ohms. What’s the power factor?

(a)  42%

(b)  58%

(c)  92%

(d)  18%

13.  Suppose that a circuit has 24 ohms of resistance and 10 ohms of inductive reactance in series. A meter shows 100 W, representing the VA power. What’s the true power?

(a)  18 W

(b)  34 W

(c)  85 W

(d)  92 W

14.  Suppose that the true power equals 100 W in a circuit that consists of a resistance of 60.0 ohms in series with an inductive reactance of 80.0 ohms. What’s the VA power?

(a)  167 W

(b)  129 W

(c)  60.0 W

(d)  36.0 W

15.  Suppose that the true power equals 100 W in a circuit that consists of a resistance of 80.0 ohms in series with an inductive reactance of 60.0 ohms. (The resistance and reactance numbers here are transposed from the values in Question 14.) What’s the VA power?

(a)  64.0 W

(b)  80.0 W

(c)  125 W

(d)  156 W

16.  Suppose that we connect a coil and capacitor in series. The inductance is 36 μH and the capacitance is 0.0010 μF. What’s the resonant frequency?

(a)  36 kHz

(b)  0.84 MHz

(c)  2.4 MHz

(d)  6.0 MHz

17.  What will happen to the resonant frequency of the circuit described in Question 16 if we connect a 100-ohm resistor in series with the existing coil and capacitor?

(a)  It will increase.

(b)  It will stay the same.

(c)  It will decrease.

(d)  We need more information to answer this question.

18.  Suppose that we connect a coil and capacitor in parallel, with L = 75 μH and C = 150 pF. What’s f_o?

(a)  1.5 MHz

(b)  2.2 MHz

(c)  880 kHz

(d)  440 kHz

19.  What will happen to the resonant frequency of the circuit described in Question 18 if we connect a 22-pf capacitor in parallel with the existing coil and capacitor?

(a)  It will increase.

(b)  It will stay the same.

(c)  It will decrease.

(d)  We need more information to answer this question.

20.  We want to cut a ¼-wave section of transmission line for use at 18.1 MHz. The line has a velocity factor of 0.667. How long should we make the section?

(a)  9.05 m

(b)  3.62 m

(c)  3.00 m

(d)  2.76 m