[An antenna must have] current flowing along its length, so that the resulting fields radiate that energy into space. (Receiving antennas are just this process in reverse).

[This] explains why you can't just stick a small tank circuit on a board and expect it to radiate efficiently.


I understand this is true from experience, but I don't understand why. I guess the dimension of the antenna changes the fields it produces somehow, but how does this make energy radiate away more effectively? What's an energy radiating away look like?

I do understand the need to tune the antenna. I'm just wondering how after we've tuned for maximal power transfer to the antenna, we get more of that energy to go to the receiving antenna.

  • 3
    \$\begingroup\$ Note that in a pure inductor, the impedance Z = 2 pi F j is purely complex, and so the current and voltage will be 90 degrees out of phase and no power transfer will occur. \$\endgroup\$
    – Paul
    Mar 1, 2013 at 14:41

7 Answers 7


Indeed it can be a very good antenna. Look no further than the transistor radios and AM band receivers. In those ubiquitous consumer goods the antenna consisted of a piece of very low loss ferrite with a very high permittivity. This was wrapped in many amp*turns of very fine copper wire. The high permittivity gave the antennas an effective cross-sectional area -due to the permittivity- (If I recall correctly) of a square mile or so, thus bringing the antenna's electrical size up to the dimensions of the wavelength that it was receiving.

On a technical bent, you could consider that the antennae interacted with the magnetic field portion of the radiating Poynting vector.

enter image description here

  • \$\begingroup\$ "Poynting vector" maybe me remember an article by Bill Beaty on the subject I read long ago. \$\endgroup\$
    – Phil Frost
    Feb 14, 2013 at 20:03
  • \$\begingroup\$ I'm having a harder time understanding what you mean by "capture cross-section" and why it's relevant. Is this the same thing as effective aperture? Would this antenna also work well as a transmitting antenna? While I understand transmit and receive are symmetrical, an inefficient receive antenna can be compensated easily with higher gain, while this is rather more difficult for a transmitting antenna, given the power levels involved. \$\endgroup\$
    – Phil Frost
    Feb 14, 2013 at 20:09
  • 3
    \$\begingroup\$ @PhilFrost language changed, got rid of "capture cross-section". I think, and I truly mean surmise, I'd like to know for sure. But is suspect this doesn't make a good Tx antenna because of material constraints, i.e saturation effects in the ferrite at high TX powers. Once the ferrite saturates the \$mu_r\$ drops and then the effective area drops. \$\endgroup\$ Feb 14, 2013 at 21:48
  • \$\begingroup\$ That makes sense. So maybe it's not so much that you need current flowing along the antenna, but rather that you need a strong magnetic field, which could be achieved by high current, but also high permeance. \$\endgroup\$
    – Phil Frost
    Feb 14, 2013 at 22:03
  • 1
    \$\begingroup\$ The ferrite core is used for it's permeability (to concentrate magnetic fields), not it's permitivity (which would affect electric fields). Permitivity is also used in antenna construction, as in chip antennas. More info in this great answer: electronics.stackexchange.com/questions/243341/… \$\endgroup\$
    – remcycles
    Jan 26, 2018 at 8:37

The field strength at a distance from the inductor is critically important. If the inductor is well shielded, with zero field in the space nearby, then it won't act like an antenna. Obviously.

So, how can we maximize an inductor's distant field and create a good radio antenna? Well, first we should wonder about the distance involved. The field must be strong at what particular distance from the inductor? The answer: 1/4 wavelength. This is a somewhat 'magic' value which falls out of the physics of traveling EM waves interacting with conductive objects. If the field at 1/4 wavelength from the inductor is insignificant, then the inductor is being electromagnetically shielded for that frequency. But if the field is significant at that distance, then the inductor can perform as an antenna.

Radiation from dipole antenna: MIT E&M Course

YT animation: fields surrounding an antenna

Why 1/4-wavelength? Above is an MPG animation from the intro E&M course at MIT. Examine the animation carefully. AC is applied to the small coil in the center, and blobs of closed circular field-lines are flying off as EM waves. But very close to the coil location, the field pattern isn't flying outwards. Instead is just expanding and collapsing. Close to our coil-antenna, the field resembles that of a simple electromagnet. It expands larger as coil current increases, and collapses inwards when the current decreases. But out at great distance from the coil, the pattern acts very differently, and it just move outwards continuously. Where does the behavior of the field make its change? At 0.25-wavelength distance. At 1/4-wave distance the field lines are "necking down" into a momentary hourglass-shape, then they peel loose and fly outwards as oblong closed circles.

The volume of space within 1/4-wave distance of the coil is called the Nearfield Region, and exhibits the expanding/contracting field patterns of a simple inductor. At greater distance, in the Farfield Region, the fields behave only as traveling EM radiation.

More MIT animations see especially the very last one

The simplest way to guarantee that the field is strong at a distance of 1/4 wavelength is to build an inductor which acts like a dipole electromagnet. But make an electromagnet where its magnetic poles are roughly a half-wavelength apart. Buy yourself a ferrite rod that's 1/2-wave long, then use that rod as your inductor core. Even simpler: just wind your inductor as a hoop-coil with a radius of about 1/4-wave.

Another way to make the field strong at 1/4-wave distance is to use a very small inductor, but crank up the inductor's current to a much higher value. In this case even a very tiny coil could emit plenty of EM radiation. But this brings in practical problems: small coils are inefficient antennas because of wire-heating. If most of your transmitter wattage is going into creating immense current and antenna heat, rather than emitted EM waves, you're going to run down your batteries (or get large bills from the electric company.) If this doesn't matter in your situation, then no 1/4-wavelength tower is needed. A small loop antenna will work fine, and it can be far smaller than 1/2-wave diameter.

As for portable AM radios and their relatively small antenna coils, in that case we use som more "magic" to increase the coil current. If an inductor is employed as part of a parallel LC resonator, then whenever it's driven with a small signal, the current in the resonating LC loop grows to a very high value. It absorbs incoming EM waves and the coil's current grows progressively larger. Its growth is only limited by wire resistance, and if the resistance is low enough, then it is limited only by losses to EM emission. A zero-resistance coil, at resonance, can grow its surrounding fields until the field strength at 1/4-wave distance from the inductor is as large as the field strength of incoming EM waves. Under these conditions the tiny coil behaves "electrically large," behaving like an EM absorber of about 1/2-wave diameter. (Notice that at the low end of the AM band at 550KHz, a half wave diameter is about 900 feet!)

Unlike other receivers, in AM-band portable radios there are two separate tuning capacitors: one for the local oscillator that's part of the superhet receiver system, and another that's connected in parallel to the ferrite-core antenna coil. Note that the LC resonance is only necessary when the loop-antenna is far smaller than 1/4-wavelength in radius. Conventional "electrically large" loop antennas don't need this capacitor; they're already the proper size for their operating wavelength, and an added tuning capacitor would just make things worse.

Here's another take on the whole issue.

A transformer is not a pair of loop antennas!

For example, take an inch-wide air-core transformer running at 60Hz. As we move the secondary coil far away from the primary, the inductive connection between them falls quickly to zero. This happens because the field pattern surrounding the primary coil is identical to that of a dipole magnet ...and the flux intensity of dipoles drops off as 1/r^3. Increase the primary-secondary distance by 1000x, and the flux at the secondary coil is a billion times weaker.

OK, now increase the drive frequency, but use a constant-current signal generator to keep the primary coil's current the same as before. At first nothing odd will happen. Your transformer works the same over a broad range of frequencies. But at some extremely high frequency, suddenly weird new effects appear. The primary coil, a pure inductor, suddenly seems to develop an internal resistor, and energy starts being lost. Yet the coil isn't heating up! Energy is escaping somehow. And suddenly the value of flux being received by the secondary coil starts increasing. Your two coils are no longer a transformer. They've become a pair of radio antennas: loop antennas. You'll even discover that distant capacitors (pairs of separate electrodes,) have now started picking up the field from the primary coil. The strength of the field pattern no longer drops off as 1/r^3, instead it's more like a light source, and falls with distance as 1/r^2. At what frequency did all this occur? Guess! :)


I see that MIT's Dr. Belcher has ported those original mpegs onto Youtube. Here are three views of a basic radio antenna:

And here's what happens when we suddenly separate a positively-charged pith-ball from a negative one.

  • 1
    \$\begingroup\$ This is a great answer. I have learned a lot. \$\endgroup\$ Mar 15, 2013 at 10:38
  • \$\begingroup\$ Best animation I've seen. +1. \$\endgroup\$ Aug 6, 2015 at 15:23
  • \$\begingroup\$ [Possible spoiler] 11.8Ghz? - 3e8m/s / 0.00254m? \$\endgroup\$
    – Frederick
    Aug 28, 2018 at 20:20
  • \$\begingroup\$ @Frederick yes, an enormous quarter-wave transmission tower at 12GHz is ~6mm tall! At mm-wave frequencies, even your enclosures and groundplanes become antennas. (I think a dielectric rod can be a mm-wave antenna. So, glass plates are antennas, plastic carrying handles are antennas ...also, send your 12ghz along optical fibers! \$\endgroup\$
    – wbeaty
    Aug 28, 2018 at 23:28
  • \$\begingroup\$ @Frederick could you elaborate on why "3e8m/s / 0.00254m"? \$\endgroup\$ Mar 23, 2021 at 6:44

When you make a traditional inductor, you are trying to minimize the leakage inductance. In so doing, you try to get as much of the magnetic field to cut through nearby turns of wire. A toroidal inductor is particularly good at keeping its field to itself.

The "leakage" part is that which radiates away into space, without being captured by the coil. This is regarded as "loss", as far as the coil is concerned. When you make an antenna, you are trying to maximize this leakage, because you want it to radiate into space.

  • \$\begingroup\$ So, is there something about an air coil inductor that makes it worse than a loop antenna? Or is it exactly a loop antenna of identical efficiency? \$\endgroup\$
    – Phil Frost
    Feb 15, 2013 at 0:33
  • 3
    \$\begingroup\$ you are getting less upvotes because you do not have pretty pictures. :) \$\endgroup\$
    – Kortuk
    Mar 3, 2013 at 15:52

You are very likely wondering about the condition that we use in EMF called Reciprocity.

Most antennas, like one of the most simple and useful, is the Electric Dipole. Because the system is both linear and time-invariant you can show with a great deal of math that receiving with an antenna is the same as transmitting. This is used, having had to analyze a few antennas, because solving the equations for radiation, with the antenna source, and measuring the field in free space is much much easier then attempting the opposite.

Above I noted the condition for linearity, antennas that use a magnetic core often can have non-linear behavior, which is often not an issue as long as you stay in an acceptable range of field strength, but it also means that measuring radiation from the antenna often does not correlate to the receive strength. An improvement in the tuning network is an improvement that you will likely see in both cases, but trusting a field measured for a field transmitted into your cable is very easily not going to match the opposite path.

How does the field actually leaving an antenna look? I am going to use one of the simplest again, the electric dipole.

From wikipedia.en.wikipedia.org/wiki/File:Felder_um_Dipol.jpg From: http://en.wikipedia.org/wiki/File:Felder_um_Dipol.jpg

So, when you have a wave in free space, it is propagating without boundaries. When you have a wave in a cable, it is normally bound between the conductors. The Coax cable being an example of a bounded TEM mode waveguide. An antennas job is to match and couple the wave in the waveguide to the impedance of free space and help it radiate. As you look at an electric dipole you can see that the wave is coupling into this structure which will smoothly couple into space as the wires draw apart. That is, at minimum, a way to think about it.

I have also made a point of saying electric dipole as I have spoken and shown examples. An interesting thing to think about is how a loop antenna works. A Magnetic Dipole will have the same field pattern as the electric dipole you have seen, but switching the Electric field lines with the magnetic and vice versa. The issue is that the curving magnetic field there will not be nearly as large a loop as an electric half dipole, and getting to that point is quite hard.


Note that in a pure inductor of inductance L henries, the impedance Z = 2 pi F L j is purely complex, and from the generalized Ohms law V/I = Z so the current and voltage will be 90 degrees out of phase and no power transfer will occur.

That said, real world coils are not pure inductors but also have capacitance and thus could even be self-resonant at some frequency.

At HF frequencies the ARRL handbook notes that approx 0.5 wavelength of wire wrapped on a fiberglass support, with a "capacitance hat" or wire load at the top creates a usable compromise antenna for situations where a half wavelength dipole or quarter wavelength vertical is too large.

I have built such an antenna for 3.8 Mhz, consisting of about 40m of wire spaced about ~1.5 cm per turn spaced with toothpicks glued into holes drilled on a ~4cm diameter pole about 5-6m long. The capacitance hat was 4 thick (~gauge 8) wires at the top approx 2m long. Final tuning was done with an antenna analyzer and a dozen or so additional tightly wound turns of wire at the bottom to achieve a crossing of X=0. R is generally not 50 ohms so an antenna tuner is required. This setup was usable to make contacts around the eastern and central US and from the eastern US to Europe with only 100 Watts SSB. Generally the other stations had superior antenna... but still this was usable.


What's an energy radiating away look like?

This is for transmitting antennas. AM output looks like this (in blue):

enter image description here

Better your antenna tuning, more the transmitted energy.

Better your antenna tuning, less reflected energy.

Better your antenna tuning, better your SWR.

More energy transmitted into air, more energy received into a tuned circuit!

Edit: As asked in comments.

What is it that makes a good antenna good?

The length of the antenna matched to the wavelength of the signal you are trying to receive or transmit. Feedline should also be matched so that the signals are not reflected and close to 100% signal power passes in either direction (tx or rx) and there is low loss.

  • 2
    \$\begingroup\$ But a tuned circuit isn't necessarily a good antenna. A \$50 \Omega\$ resistor can make a well tuned load, but it's a horrible antenna. I can also do the tuning with reactive components, but that's also not necessarily a good antenna. What is it that makes a good antenna good? \$\endgroup\$
    – Phil Frost
    Feb 15, 2013 at 12:20
  • \$\begingroup\$ @PhilFrost answer added to my post. \$\endgroup\$ Feb 15, 2013 at 16:57

You will be delighted to know that even the feed point impedance of an ideal half wave dipole antenna is actually partially inductive.


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