Why is a transmitter dipole's ends the max/min voltage instead of the middle?

Question

If the AC voltage source is connected to the middle of the dipole, I'm confused on why the edges have the max/min voltage instead of the middle.

In my mind, the middle of the dipole is closer to the voltage source, and there is even some resistance between the middle of the dipole and the end, which would imply a voltage drop, making the dipole's end a lower voltage than the middle.

For a receiver, I understand why the ends have the max/min voltage (based on how the wave hits the dipole). However, I'm totally lost when it comes to a transmitter. Any help would be greatly appreciated.

Answer 1

On a dipole there is a spread of current and voltage, in very simple words: at the end of the dipole, no current is built up b/c it is an open end, so there the voltage rise to a maximum. The impedance at the ends of the dipole is very high compared to the impedance at the feed point. I think, the wikipedia Web site explains this quite well.

Answer 2

Current and voltage are not especially good to present what happens in an antenna. The actual thing that happens in an antenna which is connected to a transmitter is that a radiowave comes along the feeding cable (it's in the space between the wires, in coaxial cable it's between the shield and the center wire). In antenna the fields spread to wider area and if the antenna is properly designed a substantial part continues to the free space around the antenna as wanted. The rest is dissipated in material losses or reflected back towards the transmitter.

The key thing is that there's a propagating wave around the antenna and some reflection happens at the far end from the feeding point. The current which the fields of the wave induce to the metal cannot jump out of metal in usual power levels, so the end of the metal surface is a discontinuity which causes reflection.

From the theory of transmission lines we should remember that when a wave meets a strict "no current possible any more" border the electric fields of the arriving and reflecting waves are in every moment to the same direction at that border, so the sum field of the arriving and reflecting waves is as strong as possible with that power level.

With transmission lines we usually call the RMS value of the sum field "standing wave" It's maximum is at the unconnected end of the line.

On the surface of an antenna less far from the feeding point the fields of the different direction propagating waves partially cancel each other (but do not disturb each other by any means) so the sum field can be weaker.

At the same time when waves propagate along the surface of the antenna they also lose energy to the space as radiated wave. That radiation is attempted to be maximized in antenna designs, in transmission lines it's wanted to be kept zero.

Voltage is a coarse measure for electric field, all vector field structure is omitted, but the proportionality to electric field strength is still there. Thus also voltage between the ends of a dipole antenna can be higher than at the feeding point.

Answer 3

If the AC voltage source is connected to the middle of the dipole, I'm confused on why the edges have the max/min voltage instead of the middle.

Basically, an antenna converts the operating impedance of the electrical feed-circuit (circa 50 ohms) to that of free space. Given that free space has an impedance of 377 ohms (\$120\pi\$), the voltage to current ratio at the extremes of the antenna have to be greater than at the electrical feed-point (circa 50 ohms).