Seems weird that an antenna has no return path. Also, what about capacitive coupling. How are these circuits?

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    \$\begingroup\$ no more than a capacitor. where is all of that charge hiding? \$\endgroup\$ – robert bristow-johnson May 24 '16 at 6:29
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    \$\begingroup\$ actually, the transmission line feeding an antenna has two conductors. and the antenna really begins where the two conductors diverge. so is it at this point of diverging conductors that you posit has no return path?? \$\endgroup\$ – robert bristow-johnson May 24 '16 at 6:37
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    \$\begingroup\$ The RF trasnmitter has two separate conductors. coax has shield and core, balanced sytem has two wires. How do you mean has no return path? If you think about 1/4 lambda then the second part of antenna lies in the ground \$\endgroup\$ – Marko Buršič May 24 '16 at 8:04
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    \$\begingroup\$ An antenna doesn't need a return path - just light a torch emits light or a resistor emits heat. \$\endgroup\$ – Andy aka May 24 '16 at 8:08
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    \$\begingroup\$ If you're asking about a simple "whip" antenna - the whip alone is not actually an antenna yet. The antenna is the whip vs ground plane. So the antenna always has 2 poles thus forming a regular circuit. \$\endgroup\$ – Agent_L May 24 '16 at 13:31

Antennas do have return paths. A normal dipole, for example, has two feed lines. Current flows out one while returning on the other.

Other antennas may appear to be single ended, but there is always a return path somewhere in the larger system if you look closely. Usually the other side of the RF voltage output is connected to ground or a ground plane.

Sometimes that "ground" isn't really ground, just some other large conductor. A good example is a walkie-talkie with a whip antenna. The antenna is fed at one end with no obvious second connection. In this case the chassis or ground plane of the circuitry is the other conductor. This other conductor and the antenna resonate together to cause energy to be radiated into space. Put another way, the whole device is really part of the antenna system, and is being fed from inside. In some cases, the device may have been designed assuming you'll be holding it in your hand and you become part of the antenna system.

Show us a antenna that you think has only one connection and we should be able to show you the second implicit or less obvious connection.

  • \$\begingroup\$ I am also a bit confused by this, take for example this diagram of a dipole. What is confusing is that the two wires just terminate and there doesn't seem to be a potential across the wire for charge to flow through it as is depicted in the picture. \$\endgroup\$ – AlanZ2223 May 24 '16 at 14:15
  • \$\begingroup\$ @Alan: That picture clearly shows charges moving in opposite directions in the two feed lines to the dipole. \$\endgroup\$ – Olin Lathrop May 24 '16 at 17:09
  • \$\begingroup\$ @OlinLathrop I have connected an a male BNC cable to my scanner. At the end of this cable, I have attached to the center conductor to a single long stranded wire which I am using as a functional antenna. However, because I haven't connected a 50 Ohm load to the end of the BNC cable, there is effectively an open circuit where the antenna is connected. Is the antenna's return path through the outer conductor (via capacitance and conductance between connectors) or the scanner chassis? \$\endgroup\$ – cr1901 Aug 22 '16 at 0:59

The gif of a dipole provided by AlanZ2223 is the best illustration that I see in the previously provided anwsers.

the operation of a single ended antenna would operate as one half of the dipole.

Basically, you will charge/discharge the antenna and this current flow automatically results in the change of an electromagnetic field as explained by Maxwell's Laws (emission). And similarly, the presence of a changing electromagnetic field automatically results in a current in the antenna (reception). So the electrons (or the holes) go up and down in the antenna end.

In a way that works like one of the plates of a capacitor - you charge and discharge it. However in a capacitor the energy is not (mostly) lost in the emission of a electromagnetic signal, but in the loss of energy in the resistive path that you used to charge/discharge it (you loose 50% of the energy when charging the capacitor and the other 50% when discharging it). The energy is stored by the build up of an electrical field between the fields.

In an antenna, most of the energy will be converted in or from the electromagnetic wave.

It is different from a light bulb or a resistor as in those elements the current will effectively flow through the element and no charge will be built up on one end of the light bulb or the resistor.

The electroncs flowing into the antenna is temporarily provided by your circuit or the energy source which would temporarily build up a positive charge - an operation which should "immediately" be reversed as you have an alternating signal in your antenna.

The second part of your question about capacitive coupling has some similarities regarding the origin of charge on the source, but we would like to know more about why the "other" signal is subject to capacitive coupling.

It is called capacitive coupling because the function is similar to a capacitor. Two conductors behave partly like the plates of a capacitor. The closer they are, the more they have this capacitive behavior. The characteristic of a capacitor is that it will resist against a change of voltage. Hence a positive change of voltage on one end will be reflected in a positive change of voltage on the other end. The hicher the capactive coupling the higher the resulting voltage change. In terms of current flow in two circuits that have no resistive path between them, there is no electron that goes from one circuit to the other. Each circuit will balance out itself.


No an antenna does not, no more than the heat produced in the circuit violate the circuit laws. Both represent a loss of energy in the system. While an antenna does indeed in some cases represent a capacitance or inductive load, these are manifest in the near field and as such there is usually no energy loss associated with that. A properly matched antenna will manifest itself as a resistive load.

The circuital laws are manifestations of more fundamental laws like the conservation of charge (i.e. you are not producing charge) and conservation of energy.

Just to clarify a point here: When I speak of far field, I speak of the regime in which there is energy transport in the form of radiating waves. So you can have information/energy transported to a distance without a return path.

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    \$\begingroup\$ i think the issue regarding "return path" has more to do with conservation of charge, rather than energy. normally we think that we need a return path, otherwise with a single wire with a current flowing in it, charge builds up at the other end. once you get a coulomb of charge accumulated and just sitting there, you might get forces of something like \$10^9\$ Newtons (if there is another such static charge sitting around). that would require a lotta energy to do that. a force like that might force the charge to flow back to equalize somewhere. (might be an unpleasant resolution.) \$\endgroup\$ – robert bristow-johnson May 24 '16 at 6:34
  • \$\begingroup\$ I'm sure why you've added this comment when I already mentioned conservation of charge. \$\endgroup\$ – placeholder May 24 '16 at 13:00

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