# Effective area of a dipole

Effective area of a dipole is $\lambda^{2}/4*\pi$. Consider 1 GHz em waves. Consider two antenna of length 15cm ie half wave length and 1 cm. Both have same effective area. This is very non-intuitive as the area does not depend on the length of the dipole. What explains this non-intuitive formula? Why should one bother about antenna length , particularly when receiving is concerned?

• You might consider also asking about the physics aspects of this question in Physics SE as well, as long as the question is not identical. You should probably show the source where the $(\pi/4)\lambda^2$ comes from. The effective area can be loosely thought of as an absorption cross-section and is really another way to express the absorption probability of an incident photon. So an answer there and an answer here will be complementary but not duplicates. – uhoh Sep 14 '18 at 10:57
• Absorption cross section is a probability metric which is quite different from effective aperture in antenna theory. The origin of the effective aperture formula is simply the spherical construct of isotropic radiation. – Glenn W9IQ Sep 14 '18 at 11:35

The preferred term is effective aperture. Effective aperture is defined as:

$$A_e=\frac{\lambda^2}{4\pi}G \tag 1$$

where $\lambda$ is the wavelength of operation and G is the linear gain of the antenna.

So you can see from equation 1 that you need to include the gain of the antenna - this was not reflected in your formula. The gain of the antenna is defined as:

$$Gain=Efficiency * Directivity \tag 2$$

A 1/2 wave dipole has a directivity of 1.65. When you shorten the dipole, it's directivity decreases but quickly reaches a terminal value of 1.5 no matter how much more it is shortened.

What has the ultimate impact on the Ae in the case of a small dipole is its efficiency. Efficiency is defined as:

$$Efficiency=\frac{R_r}{R_r+R_l} \tag 3$$

where Rr is the radiation resistance and Rl is the resistive losses.

As the dipole size becomes smaller compared to a 1/2 $\lambda$ dipole, its radiation resistance drops precipitously as a function of the square of its fractional wavelength. For like construction materials and methods, Rl only drops proportionally to the fractional wavelength. Thus the efficiency rapidly drops with the shrinking size of the dipole. Since this reduces the gain of the antenna the Ae of the antenna is also reduced.

For a given wavelength, the Ae of the antenna is directly proportional to its gain. But Ae provides a construct that helps to visualize the collection aperture of the antenna as a plane wave intersects it. The larger the aperture, the greater the total received signal power that is made available at the feed point of the antenna. It is this power that determines the signal strength at the receiver input terminals.

The effective area, also called effective aperture, is the power observed at the antenna feedpoint, divided by the irradiance to which the antenna is exposed.

Irradiance is expressed as power per area, for example "1 microwatt per square meter". Effective aperture is an area. For example if the irradiance is 1μW/m2 and I observe 2μW at the antenna feedpoint, the antenna's effective aperture must be 2 square meters.

This is simply another way of expressing gain. Gain ($G$) and effective aperture $(A_e$) are related by:

$$G = {4 \pi A_e \over \lambda^2}$$

So when you say a quarter wave dipole and a much smaller dipole have the same effective aperture, you are saying they have the same gain.

Furthermore, you are saying they have a gain of:

$$G = {4 \pi (\lambda^2 / 4 \cdot \pi ) \over \lambda^2}$$

which simplifies to

$$G = \pi^2 \approx 9.87$$

This is incorrect. A lossless, infinitesimal dipole has a gain of 1.5. A half-wave dipole has a gain of 1.64.

That aside, it is a valid question to ask why the gain of a half-wave and a short dipole are almost the same. It seems like the short dipole, being physically much smaller, would have a smaller effective aperture and thus a smaller gain. And yet, the gain never goes below 1.5 no matter how small the dipole becomes.

The reason is that this ideal, theoretical infinitesimal dipole is lossless, so all the energy is radiated eventually. As such, there's no way to decrease the gain except to decrease the directivity of the antenna, which isn't possible as long as the geometry fits the definition of a dipole.

In practice, as the dipole becomes shorter its impedance becomes more reactive. To transfer power to or from the antenna then requires a matching network which must handle increasing peak current and voltage. Theoretically, this reactive power can circulate between the antenna and matching network without loss, but real matching networks are not lossless. Thus in practice, efficiency (and thus gain and effective aperture) approaches zero as the dipole approaches zero length.

By reciprocity the same is true when receiving: gain does not decrease below 1.5 regardless of the size of the dipole. Comparing a quarter wave dipole and an extremely small lossless dipole, each receives about the same power (a gain of 1.64 vs 1.5 respectively), but the current and voltage in the case of the small dipole will be extremely large due to the tremendous reactive power.

• The implicit condition of antenna gain is a matched feedpoint impedance. When thinking of Ae from a receive perspective, if the antenna impedance is R-jX and the load impedance is R+jX, then the maximum available power is transferred from the antenna to the load. And since this is lossless in the idealized case, there is no additional re-radiation from the antenna. What am I missing? – Glenn W9IQ Sep 14 '18 at 22:14
• @GlennW9IQ Nothing. I'm not sure what you're getting at. I don't disagree with anything in your answer, I just thought I'd take another approach. And also thought I'd point out the first sentence of the question is incorrect. – Phil Frost - W8II Sep 14 '18 at 22:48
• I wasn't contrasting our two answers but I was trying to catch the logic of your third to the last paragraph from a receive perspective. – Glenn W9IQ Sep 15 '18 at 0:38
• @GlennW9IQ More clear after rewording? – Phil Frost - W8II Sep 15 '18 at 2:12
• BTW - I really like your power received / irradiance definition. – Glenn W9IQ Sep 15 '18 at 21:25