Antenna Farm – Article 6 – May 30, 2022

By Charles KC6UFM

Dipole Deep Dive

Hello ECR Family, and welcome to The Antenna Farm. This is your friendly Antenna Farmer Charles, KC6UFM.

Right up front, I want to say that I was wrong. I was nearly 100% sure that Antenna Farm #4 (SWR) would be the longest one of the series. I was wrong! As I wrote this article, it became clear that there was WAY more information you need to know—and understand—than would fit into the normal length I try for. With that said, you should know that you’re in for a long read!

In this article, we’re going to take a DDD, that is, a Dipole Deep Dive. This article also marks a (planned) change in direction for the Antenna Farm…in earlier articles, we have focused on theory and general things that apply to all antenna systems. In this article, we’re taking our first close up look at a specific antenna.

Yes, I can see you…some of you are rolling your eyes. Some are thinking, “I don’t care about dipoles! I want a 73 element beam!” Others are thinking, “I want an antenna for VHF/UHF and you can’t use dipoles there! They’re for HF!” And many more are thinking, “You can’t work effectively with a dipole!”

Well, the first group needs to remember that beams—as well as about 95% of all other antennas—are based on the simple dipole. If you don’t understand dipoles, you have zero chance to understand most other antenna designs.

The last two groups are, plain and simple, wrong. Any—and I mean that literally—antenna type/design can be used on any—and again I am being literal—frequency and the performance will be identical. In other words, if a given antenna has 6 dBd gain on 70cm, and you scale the antenna to, let’s say, 160m, it will have just exactly 6 dBd gain. The only problem is one of size. Let’s say you design a 10 element Yagi beam for 70cm and then scale it to 160m. The 70cm version will have a boom about 6 feet long. The 160m version will have a boom just under 2 miles long. Just for fun, the boom for the 160m version will be about 18 feet in diameter. I have no idea what the rotor would look like, but I’m guessing that the engine from an aircraft carrier would be too small.


Trust me for now…you NEED to understand dipoles.

There are many advantages to dipoles…they are easy to design, easy to deploy, easy to use, are reliable, and can be built out of almost anything. To top it off, with a good dipole, you can indeed work the world on 5 watts because they are extremely efficient. I use a 6m dipole to work FT8/CW/SSB (over 2000 contacts as far away as South Africa) and a dipole variant on 10m for FT8/CW/SSB to work into Europe on a regular basis. I get similar results on 15m, 40m, and 80m CW.

I know in earlier articles, I promised to keep the math and techno-speak to a minimum, and I am doing that. It’s just that you need some simple arithmetic to understand dipoles. I suppose this mathematical treatment could be cut down to just a line or two, but I think it’s important for you to understand how we get to a simple equation to calculate dipole length. For our purposes here, we’re going to talk mostly about the old reliable ½ wave dipole. Strictly speaking, and to be 100% correct, we’re talking about an antenna with an overall length of an electrical half wave at the operating frequency. The basic formula is:

Eq. 1 L = W / 2

L = Length for ½ wave (in same units as W)
W = Wavelength

Further, the wavelength (W) is found from the equation:

Eq. 2 W = c / F

c = Speed of light in a vacuum
F = Frequency in Hertz

Now, if we combine these two equations so we only have to work with one, we get:

Eq. 3 L = ( c / F ) / 2

Light in a vacuum travels at about 186,282 miles per second or (better suited for our use) about 983,463,360 feet per second. This is another of those coincidences that makes you believe in a supreme being, but hams mostly measure frequency in megahertz (MHz) which are millions of hertz. So, using basic algebra, we can change some units around to get:

Eq. 4 L = ( 983.463 / F ) / 2

F = Frequency in Megahertz

To prove this, let’s use the 6m band…

L = (983.463 / 50 ) / 2
L = 19.66926 / 2
L = 9.83463 feet = 3 meters

And 3 meters is ½ of the wavelength of the bottom of the 6m band.

Ain’t math great?!?

And we can simplify this equation by one more step by just calculating the ½ wave directly by doing the divide by 2 right up front:

Eq. 5 L = 491.7315 / F

In practical terms, we can even round this off to become:

Eq. 6 L = 492 / F

L = Overall length in feet (½ wave)
F = Frequency in MHz

And for God’s sake, I hope Equation 6 at least looks familiar to everyone who has passed the Technician class exam! Oh…if you like punishment, I can provide you will all of the calculus, trigonometry, and vector analysis for this. Me? I’d just nod my head once and say, “OK, looks good to me.”

But there is a problem…remember back when we talked about feed lines in Article 3 and I mentioned that even single wires have a velocity factor? Well, this is where that becomes important. In a nutshell, electromagnetic energy passing through a wire moves slower than it does through a vacuum. When dealing with dipoles, you need to take this into consideration because it impacts the length, sometimes dramatically. This means we have to further modify Eq. 6 to take the velocity factor into account:

Eq. 7 L = ( 492 / F ) * VF

L = Length in Feet (½ wave)
F = Frequency in MHz
VF = Velocity Factor as a Decimal

As the diameter of the wire gets larger, the velocity factor also gets larger. In this case, “larger” is a relative thing…we’re not talking about the actual diameter of the wire, but the ratio of the wire diameter compared to the wavelength. In most cases, the wavelength is going to be VERY much larger than the wire diameter. The actual calculation of the VF of a wire is extremely complex and honestly isn’t worth the effort in most cases below the microwave frequencies.

Another thing to consider is what is called “End Effect,” that, basically, makes the wire appear electrically longer than it’s physical length. While end effect does exist, it’s worth noting that much of what was once called end effect is actually VF.

The bottom line here is that for all practical designs using reasonable, real world wire (or tubing) sizes at frequencies below about 1,000 MHz we can use a value of about 0.95 for the VF in Eq. 7. Applying this, we finally get to:

L = ( 492 / F ) * 0.95

Eq. 8 L = 467 / F

Using our 6m example, we would find:

L = 467 / 50 = 9.34 feet

In practice, Equation 8 will give you a wire length slightly longer than what is really needed, and that’s a good thing. First off, it gives you some wire to make connections to the center and end insulators. It also let’s you trim the wires to get the best match to resonance. An old rule of dipole makers is: It’s Easier To Remove Wire Than To Add Wire. In other words, the number you get from Equation 8 will resonate just a little lower in frequency than you planned for. Just shorten the wires to get “up” to your desired frequency.

OK, we have the serious math and theory out of the way now, so lets look at how a ½ wave dipole works.

A ½ wave dipole is made up of two elements end to end and fed at the junction of the two ends, each element fed by one of the two transmission line wires. It looks something like figure 1. Each of the two elements is simply half of the value of L, that is, a quarter wave.


In free space (like that’s gonna happen) a dipole will have an impedance of about 73 ohms. Height above ground, the type of ground, nearby objects, and other factors will change this impedance. Generally speaking, a dipole mounted at two wavelengths above typical ground and at least two wavelengths away from other objects will show an impedance between 50 and 73 ohms.

A dipole is what is called a “Balanced” antenna. That is to say that the two halves of the antenna that are attached to the feed line conductors are electrically identical. As a general rule, they will usually also be physically identical, but there are exceptions. Again looking back at Article 3, you’ll recall that there are both balanced and unbalanced feed lines. A dipole works best when fed with balanced line. That doesn’t mean that you can’t use unbalanced lines…keep reading! For now, however, we’re going to look at a dipole fed with a balanced line.

You have probably figured out already that the feed line is connected to the antenna at one end and your transmitter at the other. The transmitter (usually called either a “source” or a “generator”) provides voltage and current to the antenna. It just so happens that when you feed a dipole at the center, you are at a voltage “trough” and a voltmeter will read just exactly zero volts. But the source is indeed supplying voltage and current.

Remember that RF is alternating current. In other words, the voltage swings from zero, up to some positive maximum, back down to zero, down to some negative maximum, and finally back up to zero again. This is one cycle. A cycle can be measured in degrees, too. A full cycle, or wave, is 360 degrees, a complete circle. It follows then that ½ wave (the overall electrical length of our dipole) is 180 degrees and ¼ wave (the electrical length of one leg of our dipole) is 90 degrees.

The applied power to a dipole will have the voltage and current 90 degrees out of phase so that when the voltage is at zero, the current will be at maximum. As the value of the voltage moves away from zero, the current will move toward zero. Finally, when the voltage reaches a maximum value, the current will be at a minimum.


Figure 2 shows the distribution of voltage and current on a ½ wave dipole where the solid line marked “V” is the voltage and the dashed line marked “I” is the current. You can use Ohm’s Law to calculate the values of voltage and current at a given wattage, but you will find from Figure 2 that while we feed the dipole at a voltage trough, the ends are at a voltage peak. Again, I’ll ask you to trust me on this, but you do NOT want to touch the ends of an energized dipole. Yes…that’s the voice of experience speaking. You also want to have insulators at the ends because many hundreds of volts could be present there.

Additionally, Figure 2 shows arrows pointing toward the top and bottom of the drawing. This is the direction of radiation from the antenna. In other words, a ½ wave dipole radiates at right angles, or “broadside,” to the antenna element.


Figure 3 shows this radiation pattern in more detail, albeit less detail than is available in an antenna modeling software package. The left side of the figure (a) is the antenna wire mounted horizontally and viewed from above. The right side (b) is the antenna mounted vertically and viewed from the side. Notice how the pattern has maximum radiation broadside to the element with two deep nulls at the ends of the wire.

OK, put down the mic and listen up…this is important! On the (a) side of Figure 3, notice the perfect circle marked as “Ideal Isotropic Antenna Gain.” One of the attributes of the theoretical isotropic antenna is that it radiates equally in all directions. Notice how the dipole’s pattern is greater in some places than the isotropic and less in other places. If you were to go all around the dipole and measure the angle (vector) and amplitude of the RF, you would find that the dipole and the isotropic antenna are radiating the exact same TOTAL power. The First Law of Thermodynamics states that energy can neither be created nor destroyed, only altered. This leads to the concept:

Antennas do NOT increase your radiated power. Ever. Never.

But, an antenna CAN alter your radiated power by focusing it in desirable directions. This what we call “antenna gain.” We “borrow” some energy from the ends of the dipole and focus it broadside to the element. If the scale were large enough in Figure 3 (or you are looking at an actual model in EzNEC), you would see that the dipole, broadside to the wire, has 2.14 dB gain over the isotropic. That should sound familiar.

This focusing of energy results in what we call “Effective Radiated Power,” or ERP. The ERP can be thought of the amount of energy you would need to radiate from an isotropic antenna to get the same signal as from the desired directions of your antenna.

I can’t stress enough how important the four paragraphs above are. It is imperative that you have a good grasp of those concepts to be able to really understand how antennas work.

Another technical thing we need to look at is the bandwidth of a dipole. Mostly because of the fact that many dipoles are made of small diameter wire, the ability of the antenna to cover wide ranges of frequencies is somewhat limited. Essentially, the larger the diameter of the antenna elements, the wider the bandwidth becomes. But even for wire dipoles, the idea of a dipole being a single-band antenna is largely a myth.


Take a look at Figure 4. This is similar to Figure 2 above with the voltage curve removed. What we see here is the current distribution along the wire of the dipole. Again, you will note that we are feeding the dipole at a current “node,” that is, the point of peak current flow. For now, just know that there are three important factors at play here that can all be derived from Ohm’s Law:

1) In a non-reactive circuit (pure resistance), where there is a current node (highest current) there will be a voltage trough (lowest voltage) and vice versa.
2) Where you have a current node and a voltage trough, the resistance will be low.
3) Where you have a current trough and a voltage node, the resistance will be high.

Again, you can prove this with Ohm’s Law, but for now we’ll just accept these as facts.

As you can probably imagine, as the length of the wire changes, the current distribution begins to distort. This distortion manifests as distorted radiation patterns, a loss of the node/trough relationship between the current and voltage, and other things. Why? Because we are now introducing reactance into the equations. As the element gets shorter than ¼ wave, the load becomes more and more capacitive and as it gets longer, it becomes inductive. But this fact can be our friend…

Take a look now at Figures 5 and 6…


In Figure 5, we have an antenna that is twice as long as a “normal” dipole. We call this the “Second Harmonic.” The drawing again shows the current distribution, and you will note that we are now feeding our antenna at a current trough. Look up above at #3…we are at a high resistance point for the feed. In practice, the feed point could very well be several thousands of ohms, and it WILL be reactive.


In Figure 6, the antenna has again be lengthened to three times it’s “normal” length. This is the “Third Harmonic.” Now notice that the feed point is once again at a current node, and so the resistance will be low. It will also have some inductive reactance, but it will be small.

This all means we can use a dipole designed for some frequency (call it F) on the odd number harmonics, that is to say, 3*F, 5*F, etc.. Just as an example, a dipole cut for 40m (7 MHz) will work pretty well on 15m (21 MHz) but not so good at 20m (14 MHz). With careful considerations, a wire dipole can be used on several bands.

Now, a few words on the feed point impedance of a dipole.

As mentioned above, in free space, a ½ wave dipole will have an impedance of about 73 ohms. This can be best fed with a balanced feed line, but there is one small issue…readily available balanced line has a much higher impedance, with 300 ohms being the closest. You can use this, but you will need a 4:1 matching device (like a transformer) to make it work properly (73 * 4 = 292 ohms…close enough) and that comes with it’s own problems. Frankly, ladder line is a pain in the neck to work with.

In many cases, a typical practical dipole installation will cause the dipole to drop to near 50 ohms impedance, and that is a good match for coax, but that is an unbalanced line while the dipole is a balanced antenna. Also not a good thing, though you will see a lot of hams just do it anyway. The trade off is that your radiation pattern goes nuts and is, to a large degree, unpredictable. If you can live with the distorted pattern, or are forced to live with it, you can go this route.

A better option is to use a “Balun.” Balun is a contraction of the words BALanced and UNbalanced. This is simply a transformer that lets your transmitter and feed line see an unbalanced load while the antenna is, in reality, a balanced device. Most come in the form a center insulator that the two ¼ wave wires attach to and have an SO-239 UHF coax connector. Many baluns also have a builtin impedance matching system as well so you can match the 73 ohm balanced antenna to the 50 ohm unbalanced line. The only word of caution here is that you need to make sure that the balun is designed for the frequency you are using. Many are available that can do 160m-6m.

There are other ways to build a balun, too. You can use sections of feed line, a tube over the antenna end of the coax, gamma matches, ¼ wave stubs, and more. We’ll cover these devices in future articles.

Lastly, while you can mount a dipole vertically, most are mounted horizontally. Horizontal mounting tends to be easier mechanically. How you mount the dipole determines the “polarization” of your signal. Radio waves are, technically speaking, electromagnetic waves of energy. As the name implies, there are both electric fields and magnetic fields that combine to make the final wave. Electric and magnetic fields are always at right angles to each other, and by convention we determine the polarization of a radio wave by the orientation of the electric field. If you mount your dipole horizontally, that is with the wire parallel to the ground, the electric field will come off the antenna parallel to the ground, and so you will have a horizontally polarized signal. Conversely, if you mount the antenna vertically, with the wire perpendicular to the ground, your signal will be vertically polarized.

Contrary to popular belief, on HF where sky wave (that is, signals that bounce off the ionosphere) is more the rule than not, polarization just doesn’t matter. This is because while being reflected by the several layers of atmosphere, the polarization becomes random. This is why there is no real performance difference between dipoles that are horizontal, sloped, inverted V, or even an L configuration. At HF, it is worth noting that most natural noise (and a good deal of man-made noise) is vertically polarized, so a horizontal antenna will tend to be quieter on receive.

At VHF and above, polarization is VERY important. At these higher frequencies, there is little reflection and ground waves rule the day. Most repeaters use vertically polarized antennas. Most mobile antennas are vertical. If your main interest is local repeater operations, you too should be vertically polarized.

And here’s the rub…being “cross polarized” (using the opposite polarization from the station you are trying to reach) will result in 20 dB or more of path loss. So if you are trying to reach the local repeater with 100 watts into your horizontal antenna and the repeater is vertical, your ERP just went down to about 1 watt. Toss in another 3 dB of loss from feed line and connectors, and you’re trying to work a repeater with 500 milliwatts.

On the other hand, if you are into VHF/UHF weak signal operations, most stations use horizontal polarization to avoid noise. Again, if you try to work these folks on your vertical antenna, you have your work cut out for you.

Yes, I said “lastly” above. That’s because you now have all the information needed to build a dipole from just about anything that will conduct electricity.

Let’s say you have the bad luck to end up in the middle of some natural disaster or another. All of your fancy antennas you bought from HRO are a twisted pile of aluminum in the neighbor’s yard. But you need to communicate to get help and, more importantly, get help to others. That’s what you signed up for as payment for the radio spectrum we hams have. Call it the price of admission.

Let’s say you have a 10m all-mode mobile and a battery to power it, and you need to get that working and on the air. You know that the 10m SSB segment is around 28.400 MHz. Using Equation 8 above, you calculate that your dipole needs to be about 16.4 feet long. You root around in the rubble and find two pieces of wire each about 9 feet long…the kind of wire doesn’t matter too much and you could use a couple of lengths of barbed wire if you have it. Some more rummaging finds you a piece of coax with a PL-259 on one end and something that looks like a GP-15 that was run over by a train on the other end. You cut that mass of metal off and strip back the coax to get to the center conductor and the shield. You hook one of your pieces of wire to each half of the coax and get the whole thing as high as you can. Then you can do some tuning to get the SWR down or, hopefully, use an antenna matching device.

And you’re on the air.

And remember, you, as a licensed Amateur Radio Operator, can, in a life-threatening emergency, use ANY frequency available to call for help. You’re “only” a Technician? (I HATE that saying!) So what? Get on 20m if you need to. Get on the local police frequencies if needed. Just get on the air and carry out your emergency communications!

Is this a perfect antenna? Not even close. Is this an antenna you want to use on a daily basis. Nope. Will it have perfect SWR? Not likely. Will the pattern look like Figure 3? Don’t make me laugh!

But you can communicate.

Don’t have 10m? You can do the same for any frequency you care to pick. A simple dipole will work no matter what.

Before we leave the Antenna Farm today, I want to mention one very handy tool on the Everything RF website. Go to for a nice, easy to use dipole calculator. Also, if you scroll down to the bottom of the page, you will find a link to phone apps for both Android and iPhones. The Everything RF app offers many handy calculators for you to use.

Next time, we’re going to look at a few variations of the dipole that are more conducive to vertical operations in the VHF and higher bands. My personal favorite is the 6 dBd onmidirectional vertical you can build for about $20 or less.

Take Care & 73