I've been wondering for the longest time about how a monopole antenna (like the one in a typical FM radio) can work without offering a return path for current.

I have a hypothesis, and I'd really appreciate if the experts here could please either validate it, or contravene it and offer the correct explanation instead (I'm particularly interested in a conceptual explanation rather than a mathematical development).

My hypothesis: current in fact has a loop through the earth because there's an "air capacitor" between the circuit and the earth (the metal of a vehicle can serve the same purpose as the earth if the antenna is inside an airplane for example). Please see image below.

enter image description here

Is this how it works? If not, could a conceptual explanation of how the circuit can work ostensibly without a closed loop (a current return path)? From Kirchhoff's laws, I understand that current has to travel in loops.

Especially if the hypothesis is wrong, if a conceptual explanation could please be provided how the circuit can work without a closed loop for current to flow through.

Note on the image: I know that at that frequency the antenna is also a transmission line. I assume that the characteristic impedance of the line is 75 Ω, so since there's a 75 Ω resistor underneath, the entire voltage will get absorbed by the 75 Ω resistor, and the antenna will behave as a nice short circuit (won't absorb / reflect any voltage).


I've added a transmitter circuit. Question: Current doesn't go between the transmitter and receiver?

Rather voltage entering into the transmitter antenna becomes the energy of the EM wave. The EM wave in turns excites voltage at the receiver. But it's energy / voltage transmission, while the currents of both transmitter and receiver - whether conduction or displacement - are independent. Would that be correct?

enter image description here

Edit: let me summarise my understanding of the answer (comments are appreciated).

There are two kinds of current:

A. Conduction current as through any resistor.

B. Displacement current - when current "flows" through empty space, like between the plates of a capacitor.

  1. Current literally flows down the 75 Ω resistor in mode A.

  2. Then, from the bottom of the 75 Ω resistor and back to the top of the antenna, current "flows" in mode B (in fact current through any capacitor "flows" in mode B).

Hence a capacitor is a good model for the gap between the bottom of the 75 Ω resistor and the top of the monopole antenna.

But as has been pointef out it's not as simple as a capacitor, and there are some complexities involved (as Tim said: "not quite the same, as the curvature of those lines, and the exact magnitude and phase, will all depend on a fields solution, for the D-field around the whole element. But to the extent we can ignore those effects, and especially for 1/4λ mode and below, this is good enough").

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    \$\begingroup\$ You should be careful about applying Kirchhoff's laws to an antenna. An antenna is not a lumped circuit (it is not much smaller than the operating wavelength) and therefore we can't count on Kirchhoff's laws applying to the antenna. \$\endgroup\$
    – The Photon
    Commented Jul 31, 2022 at 6:22
  • \$\begingroup\$ Yes, that's the transmission line problem. Thanks Photon. But let's say all impedances are perfectly matched. (And even if not KLs apply at the edges of a transmission line at any given moment). Then the question remains, how can it work without a loop.. \$\endgroup\$
    – ee_student
    Commented Jul 31, 2022 at 6:50
  • \$\begingroup\$ Does an LED (that produces an EM wave in the visible light region) need a loop so that our eyes can see it? \$\endgroup\$
    – Andy aka
    Commented Jul 31, 2022 at 11:41
  • \$\begingroup\$ An antenna is also not a transmission line. You might feed it with a transmission line, but the antenna itself, except certain specific types, doesn't behave like a transmission line. \$\endgroup\$
    – The Photon
    Commented Jul 31, 2022 at 14:54
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    \$\begingroup\$ I don't know if you can make actual calculations this way, but it's a reasonable way to visualize it. \$\endgroup\$ Commented Aug 22, 2022 at 22:35

3 Answers 3


You are correct that the monopole antenna forms a capacitor with the ground. During a cycle, charge builds up within the the antenna, and then reverses itself.

The changing of the charge within the antenna creates a changing electric field. That changing electric field corresponds to a displacement current through space. Normally, what we think of as current is conduction current, which consists of a net movement of charged particles. A displacement current may "displace" bound charges, as in a material dielectric. However, it can also exist when an electric field exists in empty space.

[Mathematically, the displacement current density is given by

$$J_{displacement}=\epsilon\frac{\partial\vec{E}}{\partial t}$$

where \$\epsilon\$ is the permittivity and \$\vec{E}\$ is the electric field.]

This displacement current, just like conduction current, creates a magnetic field. The magnetic field created by a changing electric field, or displacement current, is part of how electromagnetic waves are able to propagate through a vacuum.

Kirchhoff's Current Law holds in general if and only if we include displacement currents in our analysis, and not merely conduction current. When both displacement and conduction currents are considered, current will, in practice, always flow in a loop. The only exception which is allowed mathematically would be by current coming from an infinite distance and traveling off to infinity. But this is not an exception you need to worry about

  • \$\begingroup\$ Thanks so much for your help. 1. There's a displacement current through space corresponding to EM waves. 2. The displacement current through space results in "conduction" current through the antenna and down to the 75 ohm resistor. 3. My concern is what happens after the conduction current passes through the resistor. 4. Does the "conduction" current remain a conduction current as it flows through the air capacitances (a closed loop of conduction current)? \$\endgroup\$
    – ee_student
    Commented Jul 31, 2022 at 1:18
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    \$\begingroup\$ As it flows through the air, as long as the air does not "break down" and form a spark, the current is displacement current. \$\endgroup\$ Commented Jul 31, 2022 at 1:20
  • \$\begingroup\$ let me correct my points: 1. EM waves (displacement current) trigger voltage at the top of the antenna. 2A. the voltage at the top of the antenna triggers conduction current down the 75 ohm resistor 2B. Below the resistor, the current returns to the antenna "source" through the air capacitances as "displacement" current. 3. It's easier for me to think of the EM waves in terms of voltage, and only speak of current resulting from that voltage - first as conduction current down the 75 ohm resistor, and next as displacement current through the air capacitances back to the "voltage source"?. \$\endgroup\$
    – ee_student
    Commented Jul 31, 2022 at 1:23
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    \$\begingroup\$ Something like that. However, in reality, there are two antennas, a transmitting antenna and a receiving antenna and the receiving antenna forms a small branch of the current loop for transmitting circuit. \$\endgroup\$ Commented Jul 31, 2022 at 1:30
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    \$\begingroup\$ Conduction current does not flow between the plates of a capacitor, but displacement current does. \$\endgroup\$ Commented Jul 31, 2022 at 1:53

Note that, to analyze fields, Kirchoff only applies to differential segments ds: a given volume of space has net zero current into and out of it (equal and opposite displacement currents). Give or take a small (differential) amount with respect to time, corresponding to the charge (voltage) at that point in space.

You can trace the path through many ds, and draw field lines back to ground for example near the element, but you will also find other lines that loop back to the antenna (higher resonant modes), or entirely on themselves (propagating waves). And those lines vary in magnitude and direction with time and position, because they aren't instantaneous (as KLs assume), but propagating at light speed.

We can indeed draw an equivalent circuit for the antenna element, for each resonant mode we are interested in -- but discovering those modes themselves, and determining what RLC values to assign to its equivalent, we really need to look at the fields first.

So, to the extent that it's an RLC circuit, indeed, we can use that; but the values need not be physically relevant. It's an equivalent circuit.

It should at least be the case that, for the lowest (1/4 wave) mode of this antenna, the equivalent capacitance is close to the asymptotic low-frequency equivalent, i.e. what you measure if you connect the element to a capacitance meter at say 1kHz, or 1MHz, or really, most anywhere that's much lower than the self-resonant frequency (which for FM BCB is around 100MHz).


As for rx/tx pairs, if they are sufficiently distant, then no current loops are shared between them, and a lumped-equivalent circuit is not very meaningful. There will be some amount of coupling, but its phase and magnitude is effectively arbitrary (dependent on distance, orientation, reflections, etc.), so we cannot assign a given RLC circuit, at least without first picking a very precise frequency and distance and geometry and all that.

If the antennas are very close ("near field"), some of the above mentioned field lines will indeed share, and some degree of mutual inductance and capacitance can be drawn. The amount again depends on geometry, but the topology will be straightforward.

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    \$\begingroup\$ Thanks very much for your answer. I gather that some EM complexities are involved, but in spite of those, I wonder whether it'd be correct to say that current that flows down the 75 ohm resistor ultimately returns back to its "starting point" at the antenna? I also gathered there's some differentiation between conduction and displacement current, but isn't the way the current "flows back" essentially the same as through any air capacitor? \$\endgroup\$
    – ee_student
    Commented Jul 31, 2022 at 1:38
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    \$\begingroup\$ Yes, "displacement current" == "current flow through a capacitor". To the extent we can apply KLs, the return path is there, certainly; if you have a monopole sticking up from a ground plane for example, fed by a coax lead, the shield currents spread out along the ground plane, picking up the displacement currents from the monopole (give or take delays and all that), and the core current continues up the monopole element, gradually decaying as the displacement currents return that current to GND. \$\endgroup\$ Commented Jul 31, 2022 at 1:45
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    \$\begingroup\$ For such a distribution, see here: chemandy.com/technical-articles/antenna/… the current is unequal along the element, but we can draw the displacement currents around it, and thus satisfy our notion of KCL. \$\endgroup\$ Commented Jul 31, 2022 at 1:46
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    \$\begingroup\$ In a capacitor, the rearrangement of surface charges, is what gives rise to the displacement current. Mechanistically speaking. So, at least on the small scale (ignoring speed of light), we have continuity between ohmic current flow (conduction), causing charges to rearrange on one plate, which causes charges to rearrange on the other plate, and the electric field and displacement current are what communicate between the sides. The same happens near the antenna element, the field lines are just longer and the plates are oddly shaped. \$\endgroup\$ Commented Jul 31, 2022 at 2:03
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    \$\begingroup\$ Specifically, we call it "displacement current" where it flows through space. It's a local kind of current flow mechanism, not concerning the entire current loop. Whereas in conductors, it's ohmic or conduction current. At least, I think that feels right..? \$\endgroup\$ Commented Jul 31, 2022 at 2:06

if a conceptual explanation could please be provided how the circuit can work without a closed loop for current to flow through.

I'm not an expert so here's my take on it...

  • A receive antenna is like a fishing net in which a fraction of the radiant power from a remote transmit antenna is captured. The bigger the receive antenna the bigger the net's area.

  • The "gap" between transmit antenna and receive antenna does not require the need to think about displacement current.

  • Once a transmit antenna has launched an EM wave, that energy travels exactly the same way as light from a lamp does. Light and radio are the same thing (both EM waves).

  • If you want to understand displacement currents in the "gap", this has nothing to do with antennas - you should study classical EM theory.

  • As mentioned, the receiving antenna casts a net that captures true power. The area of this net (called the effective aperture) is somewhat related to the antenna dimensions.

  • Both E (volts per metre) and H (amps per metre) are captured and, as we know, volts × amps is power and, two lots of "per metre" is area.

  • So, the antenna captures true power from an input of E × H.

  • For the transmitting antenna, it launches power into free space and, some of that power will "excite" the receive antenna (in far-field).

Question: Current doesn't go between the transmitter and receiver?

  • There is no electrical loop between transmit and receive antennas.

  • The incident EM wave (when usually more than a couple of wavelengths from the transmitter) is travelling through free space with an effective impedance of 377 Ω. Again, go study EM waves and forget about antennas to understand this bit (it's quite easy this bit)

  • And, the receive antenna can be loosely thought of as a tapering transmission line that reduces the incident 377 Ω down to (say) 50 Ω.

voltage entering into the transmitter antenna becomes the energy of the EM wave.

No, both voltage and current form the E field and H field parts of the EM wave hence, both "make" the energy or power.

Hence a capacitor is a good model for the gap between the bottom of the 75 Ω resistor and the top of the monopole antenna.

No, that's too simplistic. But, as part of the learning process to get from no knowledge to full knowledge about how antennas work, you might regard it as a staging post. If you looked at the impedance presented by a short antenna i.e. a short monopole then you are steering in the right direction so, if you were analysing the antenna for say a crystal set, you would find that the antenna tends to be shorter than one-tenth of a wavelength and it exhibits a pretty much constant capacitance for a given operating frequency.

So, here's a picture I made some time ago that is for a 1 MHz receiving short-antenna: -

enter image description here

And, at about one-tenth \$\lambda\$ or shorter, the antenna behaves like a fixed value capacitance of 300 pF. But, the impedance rapidly changes as you make the antenna longer (towards a quarter \$\lambda\$)

And that's about it in this very simple explanation.

  • \$\begingroup\$ Thanks Andy. It helped that you said that the gap between Tx and Rx doesn't require considering current. Tx and Rx have their own currents. Within Rx, there's apparently displacement current, from circuit ground to earth and back to the top of the circuit. It seems, according to another answer, that part of the displacement current is related to bound charges (like in a cap.) and some returns back in a loop by an EM kind of displacement current. But looks like the claim is that what runs down the 75 ohm resistor, does return in a loop through displacement (in Rx receiver, not between Tx & Rx). \$\endgroup\$
    – ee_student
    Commented Jul 31, 2022 at 19:44
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    \$\begingroup\$ @ee_student any cable carrying a signal (radio or otherwise) is carrying it as an EM wave until it hits the load. When the wavelength of the signal is much longer than the length of the cable, we "pretend" that there is a circuit loop but, you can't really pretend that when it comes to antennas. I'm not bothered by you pretending this happens in an antenna but I don't think that will get you very much further in analysing this type of circuit. Of course, the word "circuit" implies a loop hence, the loop mentality is attractive (and deeply embedded) but incorrect for antennas and t-line theory. \$\endgroup\$
    – Andy aka
    Commented Jul 31, 2022 at 20:07
  • \$\begingroup\$ Thanks for information & the effort. I'm just thinking, even in T-line, true - there's no loop between one side of it and the other side (just delayed waves), yet on each side, on the edges, the voltage at the edge of a T-line produces a current that flows into resistors in a loop (well it looks like a loop because they put a ground symbol underneath the signal terminal of a T-line). Once the voltage of the antenna translates into current through a resistor, doesn't it want to flow back? I suppose in a T-line we put a ground underneath (as though T-line has two ports), so w/ antenna it's same? \$\endgroup\$
    – ee_student
    Commented Jul 31, 2022 at 20:58

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