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Power = current times voltage. However current requires a complete circuit to 'flow'. How can a 1,000 Watt antenna work if the output of the transmitter is connected to a single emitting wire or element without a return path from the 'far end' of the antenna back to the transmitter? I've read answers about fluctuations in the voltage in the wire creating electromagnetic fields, which are radio waves. However, if the transmitter is just raising and lowering the voltage without anywhere for the electrons to go, how can there be a current? If the transmitter tries to 'push' current (via electrons) into the antenna wire, where does it go if there is not a complete circuit? Unless at the 'far end' of the antenna the electron is converted into an electromagnetic wave (which does not happen) there would be no current through the antenna. If the current is zero then the power output will also be zero despite what the voltage is doing. There is a ground connecting the 'near' end of the antenna to the transmitter, but there is no complete circuit from the 'far end' of the antenna to the ground (other than an electromagnetic field - which is not an electric current). Yet radio operators and stations talk about how many Watts they are transmitting. If I were to touch the antenna, or connect the antenna to ground, then there would be current (and I might be fried). However, an antenna connected at just one end can't provide a complete circuit - so there shouldn't be any power output.

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    $\begingroup$ An antenna is not a lumped circuit element. $\endgroup$ – The Photon Jan 31 '16 at 23:56
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    $\begingroup$ "However current requires a complete circuit to 'flow'." Is only true over sufficiently long time scales. The antenna works with inputs that are fast compared to the time scale over which that rule applies to it. You can't treat it as a circuit in the sense that you do "circuits" in a introductory course. (The time scales thing even trips experts working in domains they're not used to, BTW. There are special rules at very high frequencies that other EE's stumble on all the time.) $\endgroup$ – dmckee --- ex-moderator kitten Feb 1 '16 at 0:12
  • $\begingroup$ It's the same thing that happens in a capacitor: a so called "displacement current" flows: en.wikipedia.org/wiki/…. If you don't like open antennas, you are welcome to use magnetic ones, which are "closed". In most instances open antennas are preferable, though. $\endgroup$ – CuriousOne Feb 1 '16 at 0:50
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When you have a capacitor, current flows even though the "circuit" is not complete. This is because it's possible for electrons to bunch up - temporarily - in a conductor and generate a corresponding electric field.

That is what happens in an antenna. An antenna is really a combination of an inductor (a straight wire) and a capacitor (when you put a net charge on it, its potential will rise compared to the surroundings). It is this combination that allows an antenna to work - and to work best at a specific frequency, its resonant frequency.

Now when you drive a current into an antenna at resonance, you will find that the current is in phase with the voltage - that is when the circuit is called matched, and when it is possible to send power into the antenna. When you did not properly match your antenna (which is very easy to do!), then power will be reflected - this is a big problem with transmitters and it can actually fry the output circuitry. For this reason, transmitters have matching circuits between the power amplifier and the antenna, and a "VSWR meter" (Voltage Standing Wave Ratio meter) to measure the presence of reflection (which results in a partial standing wave).

There is a thing called the "radiation resistance" of an antenna that describes how the antenna behaves as a resistor at resonance - and how much power you can send into it for a given voltage. This follows directly from the usual power law:

$$P = \frac{V^2}{R}$$

Where $V$ is the voltage (RMS) and $R$ is the radiation resistance, $P$ is the power absorbed by the antenna (and emitted in the form of E/M radiation).

The key concept here is that the impedance of the antenna will be purely real at resonance. At that point, the capacitance and inductance of the antenna (intrinsic, or added) ensures resonance, and then significant currents can flow in phase with the voltage - which allows power to flow from the antenna into the radiated signal.

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I think that this is a very good question because it makes one think beyond a "standard" explanation.

When you study electromagnetic induction you learn about the magnetic flux change through a closed loop, which produces an induced EMF. However, the loop does not have to be a conducting loop. If it was an ideal dynamo with no resistance, friction etc., no mechanical energy would have to be put into the dynamo to keep it generating the induced EMF.

You can have an induced EMF without an induced current. For example, a dynamo can have a voltage across its terminals and yet no current flows and so Lenz's law is inoperative. There is no opposition to the production of the EMF. However, if the circuit is completed, then a current flows and the speed of rotation of the dynamo decreases and mechanical energy has to be delivered to the dynamo.

When a capacitor is charged by a battery via a resistor, is there a complete conducting circuit? The answer is “No” because the material between the plates of a capacitor is an insulator. To deal with this lack of complete conducting circuit @CuriousOne‘s fudge factor called displacement current was introduced.

So what about an antenna? Well, you can say it has capacitance because it can store charge-energy in the form of an electric field, and it also has inductance because it can store energy in the form of a magnetic field.

The following animated GIF is of an antenna called a dipole. It does not show transmission of an EM wave but rather it shows reception of an EM wave (the magnetic part has been omitted for clarity).

P.S. If the animation does not work in this comment, then please go to Antenna (Radio).

enter image description here

For a transmitter, you just replace the resistor $R$ with an alternating voltage source.

This type of antenna has a length which is related to the wavelength of the EM radiation and so a standing wave is produced – put another way – the system is at resonance.

Note how charges are stored at the ends of the antenna, so much so that at the ends of the antenna, which are voltage nodes, you can get a very nasty electric shock if you touch the end of a high-powered transmitting antenna. At the ends there are current nodes, which is because, as you quite rightly pointed out, there is no complete conducting circuit.

The centre of the dipole is a voltage node, but a current antinode. Current through a wire produces a magnetic field; magnetic field stores energy, so it is acting like an inductor.

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  • $\begingroup$ That is a really helpful picture. $\endgroup$ – Floris Feb 1 '16 at 12:09
  • $\begingroup$ It is one of those animations which are mesmerising as there is so much going on. $\endgroup$ – Farcher Feb 1 '16 at 12:33
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    $\begingroup$ " To deal with this lack of complete conducting circuit @CuriousOne ‘s fudge factor called displacement current was introduced." There is nothing "fudge factor" about dispalcement current, it is as real or more so than any "other" current; energy propagates between the wires of a transmission line not within them. $\endgroup$ – hyportnex Feb 1 '16 at 13:17
  • $\begingroup$ @hyportnex I wrote it that way because that is the way it is introduced when the initial stab at Ampere's Law is found wanting. I appreciate it was a key component when Maxwell's equations were formulated. $\endgroup$ – Farcher Feb 1 '16 at 16:37

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