Wednesday, November 30, 2011

Radiation Resistance of a Half-Wave Dipole

Last week I did some cool calculations of transmission-line impedances, and at the end of my post I said I had some tricks for antennas. Today I'm going to show you how to calculate the radiation resistance of a half-wave dipole.

Let's first recall that the impedance of free space is 377 ohms. What does this mean? That's a tough question in general, but one interesting interpretation is the theory that it tells you how much power radiates from a flat sheet of current. Here is how it works. Suppose you have a metal sheet and current flows on the surface. Let the current density be one amp/meter. (You really have to understand why current density is expressed in amps per meter and not amps per meter-squared!) According to this interpretation, you will be radiating power from the surface at a rate of i-squared-r. If your surface is 100 square meters at 1amp/meter, then you are radiating 37700 watts.


It would be nice if it was that simple, but there are all kinds of complications, and in general it doesn't work this way at all. And yet there are situations where you can get away with this method. The calculation I'm about to do is one such situation. I'm not even sure why it works, but it gives a pretty good answer.

Let's take my favorite local radio station, 680 CJOB. You can easily verify that the wavelength of the station is close to 440 meters. For my half-wave dipole, I'm going to take a giant metal sphere 440 meters in circumference (r = 70m) and feed it at the equator.

You can see I've chosen the size of my antenna so that I've hopefully placed a natural current node at both the north and south poles. The idea is that the current, supposing it to be 1A/m at the equator, diminished to zero at the poles in a natural way. Oddly enough, one "natural" way to do this is by letting the current simply be 1A/m everywhere! Since the lines of latitude get shorter as you go north, I will have a total current of 440A at the equator, 220 A at latitude 60 North (and South), and zero at the poles. The current density is everywhere 1A/m, but the total current goes smoothly to zero at the poles.

That makes my power calculation easy. I have  377 W/m^2 everywhere on the surface, and a total surface area of just over 30,000 m^2, for a total power of 12.6 MW. Taking into account the total current and equating power = i-squared-R, I get an effective resistance of 65 ohms.

(EDIT: I just noticed that I left out something very important from this calculation...where did I get my current??? It's like this: we assume the sphere is cut in half at the equator, and we are feeding current in uniformly around the equatorial circumference. Since the circumference is 440 meters, we need exactly 440 Amps to provide our uniform current density of 1 A/m. That "440 Amps" is the number I used in the i-squared-R formula in the last paragraph.)

This is really close to the "correct" value of 73 ohms, but it seems I was more lucky than I had a right to be. In fact, my current distribution is not all that physical, although it made the calcuation easy. You can see there is something wrong with my current distribution because it gives me a uniform power density over the whole surface of the sphere, and everyone knows that an actual dipole radiates in a donut pattern.

Does this help fix up my calculation? Unfortunately, it does just the opposite. Correcting for the donut pattern, my actual power should be close to half of what I initially estimated, which knocks my resistance down to 32 ohms. It's not really that good, but at least it's in the ballpark. Sometimes it's good just to get into the ballpark.

Regardless of the exact value of the resistance, there is one big question that I've avoided mentioning altogether: a normal antenna is a skinny little wire, and I've used a sphere for my approximation. Why in the world is a big fat spherical antenna a good approximation for a wire?

It turns out that it is, but that is probably a story for another day.

1 comment:

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