Practical_Antenna_Handbook_0071639586

198 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
Shortened Coil-Loaded Dipoles
At lower frequencies, the half-wavelength dipole is often too long for installations where
real estate is at a premium. The solution for many operators is to use a coil-loaded shortened
dipole such as one of those shown in Fig. 6.14. In the absence of any loading, a
shortened dipole (i.e., one that is appreciably less than a half-wavelength) exhibits capacitive
reactance at its feedpoint. This reactance can be canceled with an inductance
placed at almost any point along the radiator; in Fig. 6.14A and B, coils are placed at 0
percent (i.e., at the feedpoint) and 50 percent of the element length, respectively.
Figure 6.14C is a table of inductive reactances (in ohms) as a function of the shortened
radiator’s length, expressed as a percentage of a half-wavelength. It is likely that
the maximum allowable percentage will be dictated by your specific installation, but
the general rule is to pick the largest percentage that will fit within the available space.
For overall antenna efficiency, coils in the middle of both sides are preferable to coils
at the feedpoint, for two reasons:
• Coils at the feedpoint have more current going through them, hence greater
ohmic (resistive) losses.
• Coils at the feedpoint are replacing the highest current portions of the antenna
with lumped components that don’t radiate anywhere near as well.
Example 6.4 Suppose you have about 40 ft of backyard available for a 40-m antenna that
normally needs about 65 ft for a half-wavelength. What value of inductor do you need?
Solution Because 39 ft is 60 percent of 65 ft, you could use this value as the design point
for this antenna. From the table, a shortened dipole that is 60 percent of a full l/2 dipole’s
length requires an additional inductive reactance of 950 Ω with the loading coils
at the midpoint of each radiator element to allow the feedline to “see” a purely resistive
feedpoint impedance. Rearrange the standard inductive reactance equation (X L = 6.28
FL) to the form
X
L
L ( µ H) = (6.10)
6.28 F(MHz)
where L = required inductance, in microhenrys
F = frequency, in megahertz (MHz)
X L = inductive reactance obtained from the table in Fig. 6.14C.
If the antenna is cut for 7.150 MHz, the calculation is as follows:
XL
L( µ H) =
6.28 F(MHz)
(950)
=
(6.28)(7.15)
= 20.7µ
H
Keep in mind that the inductance calculated here is approximate; it might have to be
altered by cut-and-try methods.