The Hairpin Monopole Antenna
by Wilfred Caron
"Hmmm, this is about one of those mysterious magnetic antennas!"

The "hairpin" monopole antenna has been around for a long time. During this time numerous questions have been raised concerning its performance, namely " How can a balanced structure radiate?" and "What is its impedance?" This article will attempt to answer these questions. A hairpin monopole is simply one half of a folded dipole, therefore, a good starting point for our discussion is to briefly describe the electrical properties of the folded dipole antenna.

A folded dipole antenna is a center-fed half wave dipole with another half-wave element close to it and joined at the ends as shown in Figure 1. The spacing between the elements is about 1/64 wavelength and the overall length approximately one-half wavelength. Electrically, a dipole antenna can be expressed as a series resonant circuit consisting of a resistive component and two reactive components. At the resonant frequency, the impedance which results from this combination becomes resistive because at resonance the reactive components are of equal and opposite characteristics and cancel each other. The resultant resistance is 73.16 ohms for a vanishingly thin dipole.


Figure 1


Typically, because of finite physical dimensions, a 1/2-wave dipole antenna will exhibit a resistance of about 68 ohms depending upon its length to diameter ratio.

A folded dipole, in addition to the series resonant circuit, has a parallel resonant circuit. The parallel resonant circuit is the result of the two elements being shorted together at the far ends. The shorted ends hold the voltage at these two points to the same value: therefore the distribution of voltages and currents for the two elements remains the same as it would for a simple dipole. When the two elements of the folded dipole are of the same diameter, the input resistance becomes four times that of a simple dipole. Theoretically, 4 x 73.16 = 293 ohms (less in practice). This is the reason why a 300-ohm twin lead transmission line is used with folded dipoles.

The increase in resistance at the feedpoint point comes about because there is an equal division of the current in the two parallel elements. Since there is a division of current at the feedpoint, there is only one-half the current at the feed point that there would be for a simple dipole. Thus, for the same amount of power fed to the antenna (or received by the antenna) only half the current appears at the feedpoint and the resistance is increased by 4. This is more readily understood by examination of the basic equation:

R = P/I2 ohms (1)

Halving the current (I) but keeping power (P) constant means that, when the 1/2 value of current is squared, its final value in the equation will be only one-forth the value had it not been squared:

R = P/[I/2]2 = P/[I2 /4] 4P/ I2 = 300 ohms (2)

One may wonder, how can a folded dipole, which appears to be a short-circuited balanced transmission line radiate? Balanced lines are not supposed to radiate. The folded dipole is basically an unbalanced (and therefore radiating) transmission line. To see this, let us assume that a transmission line, consisting of two parallel conductors, is fed in the center of one side but not the other side, Se Figure 2(a) and 2(b). This constitutes an unbalanced condition. as we shall soon see. To find the currents and impedance presented to the generator, the unbalanced source is replaced by a pair of equivalent sources, one in the push-push configuration, Figure 2(d), and one in the push-pull configuration as shown in Figure 2(c).


Figure 2


The push-push configuration will first be considered. For simplicity, let the far end impedance be replaced by ZL. These impedances transform along the line so that at the generator they have a value which we shall call Z1, that is:

ZI = ZO [ZL + ZO tan q/ ZO + ZL tan q] (3)

Where:

ZO = Impedance of parallel lines

ZL = Impedance of termination

q = half length of parallel lines in electrical degrees.

Z1 = Impedance seen by generator

For the short-circuited condition where ZL = O, Equation (3) simplifies to:

ZI = ZO tan q (4)

 It follows that the transmission-line current is:

 It = V/2Z1 (5)

In the push-push configuration, Figure 2(d), the voltage sources are in parallel. With this type of arrangement, no current flows through the loads ZL and the antenna current is:

 Ia = V/2Za (6)

When we look at the balanced and unbalanced parts, it becomes clear that the currents are not equal. The current in the left side of the line is:

 Ileft = It + Ia/2 (7)

while that in the right side is:

 Iright = It - Ia/2 (8)

Figure 3 illustrates graphically the sum and the difference currents as a function of frequency for a hairpin monopole antenna.


Figure 3


One-half of a folded dipole antenna can be operated above a ground plane so as to function as a folded monopole type of antenna. This is the basic structure of the hairpin monopole antenna. Under these conditions, for a perfectly conducting ground plane, the input impedance is one half that of a half-wave folded dipole, about 150 ohms.

Figure 4 illustrates the basic folded monopole or hairpin antenna. The noted difference between a basic folded monopole antenna and a hairpin is that the hairpin is considerably shorter. Shortening an antenna reduced the radiation resistance. General Dynamics developed a 10(c) kW hairpin antenna only 14 feet high that tuned the 2 to 30-MHz range in two bands: 2 to 8 MHz and 8 to 30 MHz.


Figure 4


The General Dynamics hairpin antenna uses a short-circuiting switch part ways up the antenna to modify the impedance of the antenna while operating in the high band. The switch serves to reduce the length of the transmission line for the transmission line currents and maintains the desired inductive reactance properties of the antenna. This short-circuiting switch does not affect the radiation currents or radiating pattern of the antenna. The radiation pattern is essentially omnidirectional in the horizontal plane. An inductively reactive antenna can be tuned or matched with variable capacitors as shown in Figure 5.


Figure 5


The high power requirements of the General Dynamics hairpin antenna requires the use of special tuning capacitors. Capacitors with a 1000-pF, 500-ampere, 45 kV ratings are mandatory. Figure 6 presents tuning curves for the shunt capacitor (Cp) and the series capacitor (Cs) for the General Dynamics hairpin monopole antenna.


Figure 6


Tests run on the hairpin indicate that a "tuned up" Standing wave ratio of less than 1.3:1 can be achieved using this matching technique. -30- Wilfred Caron.

Editor's Comments:
The article on the Hairpin Monopole is representative of a magnetic antenna, of the same type as the DDRR. I hope someone will use the information found in this very interesting article and build a hairpin monopole. The following formulas will help anyone build a hairpin:

Height = 28/F (in MHz)

Spacing between conductors = Height/28

The results are in feet and in the case of the spacing between the conductors, in percentages of a foot. For an example, a 160 meter hairpin will be computed this way:

Height = 28/1.8 = 15.56 ft. and the spacing = 15.56/28 = .57 ft. or a fraction over 6 inches between radiators.

The tuning capacitors, CP, the parallel capacitor should be a 2000 pF, capable of standing high voltages. The value of CS, the series capacitor should be 150 pF, also with high voltage capabilities. These values come off of the charts in the article.

It should be apparent from the chart on radiation resistance that large conductors need to be used to reduce the losses from skin effect to as low a value as possible. This is very important, and is a factor that is often overlooked in the construction of this type of antenna. A GOOD ground is extremely important!

It is also interesting to note that a 10 meter version is only one foot tall with a spacing of only .035 ft.(.42 inch) and CP is about 55 pF and CS is about 22 pF. That wouldn't take too much copper pipe, and it would be easy to build and tune.

A version for CB would only be one foot and a half inch tall, and the spacing would be not quite 1/2 inch. By using a 5 watt CB transmitter, the voltage requirements for the tuning capacitors would be reduced. It is certainly somewhere to start.

Again it is important to state that all joints need to be soldered, not bolted together or pop riveted. The resistance of such a connection would far exceed the radiation resistance of the antenna. So solder all joints, and use copper tubing, the larger diameter the better. Automotive tail pipe tubing and other large diameter conductors like down spout tubing have too high skin resistance due to the material that is used in their construction. Copper is the cheapest thing we can get for this type of project.

There is a definite need for more experimentation with this kind of antenna. As hams we can exercise our talents experimenting with this kind of antenna. For the unfortunate soul that is in an antenna restricted area, this might be an answer. -30-


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