Practical_Antenna_Handbook_0071639586
C h a p t e r 6 : D i p o l e s a n d D o u b l e t s 177 In wires and other conductors, the length calculated here is too long. The physical length of a conductor having a diameter or cross-sectional side much, much smaller than its length is shortened by about 5 percent because of the velocity factor of the wire, any insulation on it, and capacitive effects of the end insulators. A more nearly correct approximation for such a “slender” half-wave dipole is 468 L (feet) = F(MHz) (6.3) Example 6.1 Calculate the approximate physical length for a half-wavelength dipole operating on a frequency of 7.15 MHz, the middle of the U.S. 40-m band. Solution 468 L = F 468 = 7.15 = 65.45 ft or, stated in feet and inches: L = 65 ft 5½ inches In practice, it would be smart to spool off about 70 ft of wire, fold it in half, and cut it at the midpoint. This leaves a foot or more of wire at each end of the two halves for threading through end and center insulators and wrapping back on itself. Despite the precision in the preceding calculation of L, there’s no real good reason for worrying about the exact length of this dipole. Several local factors—nearby objects, height above ground, and the length/diameter ratio of the conductor to name a few— might make it necessary to either add or trim a short length of wire on each side to achieve exact resonance. But to what end? The primary reason to aim for approximately a half-wavelength is to make the antenna wire resonant, or nearly so, at the frequency of operation so that we can minimize the reactive part of the feedpoint impedance that will have to be tuned out by the transmitter, an antenna tuning unit (ATU), or other matching network. Unfortunately, a lot of people accept Eq. (6.3) as a universal truth, a kind of immutable Law of the Universe. Perhaps abetted by books and articles on antennas that fail to reveal the full story, too many people spend too much time worrying about the wrong things when constructing and installing dipoles. Whether the choice of design frequency in the preceding equations is critical or not depends on the percentage bandwidth of the band in question. For instance, for a half-wave dipole cut to the center (14.175 MHz) of the U.S. 20-m band, the antenna might be used at frequencies ranging between plus and minus 1.2 percent of the center frequency, and there will be a generally insignificant change in the feedpoint impedance over the entire band. An 80-m dipole, on the other hand, could conceivably be called upon to span a range of frequencies (in the United States, at least) up to plus and minus nearly 7 percent of the band center (3.750 MHz). For most home-
178 p a r t I I I : h i g h - F r e q u e n c y B u i l d i n g - B l o c k A n t e n n a s brew and commercially available transmitters, the feedpoint characteristics will change too much for proper matching and full power transfer to the antenna over the entire range without inclusion of an adjustable matching network somewhere between the transmitter and the antenna, or without some method of broadbanding the antenna itself. The Dipole Feedpoint Figure 6.2 shows the voltage (V) and current (I) distributions along a half-wave dipole. If center-fed, the feedpoint is at a voltage minimum (node) and a current maximum (loop). At resonance, the impedance of the feedpoint is purely resistive: R ANT = V/I. There are two contributors to R ANT : • Ohmic losses R Ω that generate nothing but heat when signal is applied. These ohmic losses originate in the small but finite resistance of real conductors and connections at wire junctions, and in any shunt leakage paths. In a properly constructed dipole having a length on the order of l/2 and located in the clear, these losses are usually negligible. Because of skin effect, which is a complex function of frequency and conductor parameters, R Ω is not the same as the dc resistance of the wire as measured with a conventional ohmmeter. • The radiation resistance R RAD of the antenna. This is the value of a fictitious lumped-component resistance that would dissipate the same amount of power that actually radiates from the antenna. Example 6.2 Suppose we use as an antenna a large-diameter conductor with negligible ohmic losses at the transmitter frequency. If 1000 W of RF power is applied to the feedpoint, and a current of 3.7 A RMS is measured, what is the radiation resistance? Solution R P (ohms) = (6.4) I RAD 2 1000 = 2 (3.7) = 73 Ω In this example, matching the antenna feedpoint impedance would appear to be simplicity itself because the free-space feedpoint impedance of a simple dipole is about 73 Ω, seemingly a good match to 75-Ω coaxial cable or hardline. Unfortunately, the 73-Ω feedpoint impedance is almost a myth except in free space, far from the earth or any other conducting surface. Figure 6.3 shows a plot of approximate radiation resistance (R RAD ) versus height above ground (as measured in wavelengths); note that the radiation resistance varies from less than 10 Ω to around 100 Ω as a function of height. As we saw in Chap. 5, this
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C h a p t e r 6 : D i p o l e s a n d D o u b l e t s 177<br />
In wires and other conductors, the length calculated here is too long. The physical<br />
length of a conductor having a diameter or cross-sectional side much, much smaller<br />
than its length is shortened by about 5 percent because of the velocity factor of the wire,<br />
any insulation on it, and capacitive effects of the end insulators. A more nearly correct<br />
approximation for such a “slender” half-wave dipole is<br />
468<br />
L (feet) =<br />
F(MHz)<br />
(6.3)<br />
Example 6.1 Calculate the approximate physical length for a half-wavelength dipole<br />
operating on a frequency of 7.15 MHz, the middle of the U.S. 40-m band.<br />
Solution<br />
468<br />
L =<br />
F<br />
468<br />
=<br />
7.15<br />
= 65.45 ft<br />
or, stated in feet and inches:<br />
L = 65 ft 5½ inches<br />
In practice, it would be smart to spool off about 70 ft of wire, fold it in half, and cut<br />
it at the midpoint. This leaves a foot or more of wire at each end of the two halves for<br />
threading through end and center insulators and wrapping back on itself.<br />
<br />
Despite the precision in the preceding calculation of L, there’s no real good reason<br />
for worrying about the exact length of this dipole. Several local factors—nearby objects,<br />
height above ground, and the length/diameter ratio of the conductor to name a few—<br />
might make it necessary to either add or trim a short length of wire on each side to<br />
achieve exact resonance. But to what end? The primary reason to aim for approximately<br />
a half-wavelength is to make the antenna wire resonant, or nearly so, at the frequency of<br />
operation so that we can minimize the reactive part of the feedpoint impedance that<br />
will have to be tuned out by the transmitter, an antenna tuning unit (ATU), or other<br />
matching network. Unfortunately, a lot of people accept Eq. (6.3) as a universal truth, a<br />
kind of immutable Law of the Universe. Perhaps abetted by books and articles on antennas<br />
that fail to reveal the full story, too many people spend too much time worrying<br />
about the wrong things when constructing and installing dipoles.<br />
Whether the choice of design frequency in the preceding equations is critical or<br />
not depends on the percentage bandwidth of the band in question. For instance, for<br />
a half-wave dipole cut to the center (14.175 MHz) of the U.S. 20-m band, the antenna<br />
might be used at frequencies ranging between plus and minus 1.2 percent of the<br />
center frequency, and there will be a generally insignificant change in the feedpoint<br />
impedance over the entire band. An 80-m dipole, on the other hand, could conceivably<br />
be called upon to span a range of frequencies (in the United States, at least) up<br />
to plus and minus nearly 7 percent of the band center (3.750 MHz). For most home-