I'm a bit confused by RF grounding. Let's take a HF 1/4 wave vertical antenna for example.

When current flows through the centre of the coax to the antenna, my understanding is that the current then needs a return path, which is back through the outer shield of the coax back to ground.

The antenna is what converts the electrical current into RF radiation.

Therefore, what travels back through the outer coax (or should)? Is this the actual electrical current or the RF signal? Feel free to criticize this question if it doesn't make much sense. I'm still trying to grasp some things. Thanks.


3 Answers 3


If the coax is operating without common-mode current (a typically desirable property of a properly designed antenna system), then it's the case that at any point, at any time, the current in the shield is equal in magnitude but opposite in direction to the current in the center conductor.

When someone says "the coax shield is ground" what they mean is the electric potential of the shield is zero. Or in other words, the electric potential difference between the shield and a point infinitely away from any charged particle in the universe is zero. Since the Earth is pretty big, fairly conductive, and mostly not charged, its electric potential is also mostly zero. So zero electric potential also means the electric potential difference between the shield and a copper rod driven in the ground is zero. This is desirable because it means the coax doesn't radiate.

The reason the electric potential of the shield is zero is that when the electric and magnetic fields associated with the voltage and current of the shield and the center conductor are equal in magnitude but opposite in direction, the sum of these fields is zero for all points outside the shield. So, the wave is contained entirely inside the coax.

However, no common-mode current isn't a given: it's the result of properly designing the antenna system. This is why feeding a dipole with coax requires a balun, for example.

  • $\begingroup$ If the coax shield wasn't being used as the "return path" and current was only flowing in the center conductor, so there were no opposite currents, then there would still be zero radiation outside the coax. A balanced line has opposite currents which cancel resulting in zero net radiation. $\endgroup$
    – Andrew
    Aug 12, 2021 at 22:21
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    $\begingroup$ @Andrew Comments are not for incomplete answers. A comment on a question should be discussing how to clarify the question. A comment on an answer should be discussing how to improve the answer. Comments are ephemeral and for the purpose of improving the post they are commenting on, not for containing the information that post should have. ... $\endgroup$ Aug 12, 2021 at 23:07
  • $\begingroup$ ...The idea is to edit all additional facts into the question so that anyone coming to the page later does not need to read any of the comments — the “conversation in the comments” should be for working to create the question and answer in their best form, not for containing them. (Quoted from Kevin Reid AG6YO♦) $\endgroup$ Aug 12, 2021 at 23:08
  • $\begingroup$ @Andrew Common mode is absolutely relevant. The OQ mentions "which is back through the outer shield of the coax back to ground" and "Therefore, what travels back through the outer coax...?". $\endgroup$ Aug 12, 2021 at 23:18
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    $\begingroup$ @Andrew The shield can only "contain the radiation within" when the currents are equal and opposite. It's impossible to have current "only flowing in the center conductor" due to skin effect. Common-mode current is relevant because if it's nonzero, the shield is no longer ground. I'm making a guess that "shield is ground" is involved in the motivation for the question, even if it wasn't directly stated. If you like, please post your own, more accurate answer, and we can see how it's received. $\endgroup$ Aug 12, 2021 at 23:19

RF energy is AC, not DC. Return path has less meaning here.

Don't think of coax as a center conductor and a shield, think of it as a single transmission line with two halves. The current in each half is (or should be) opposite of the current in the other half. Each current forms an equal and opposite magnetic field from the other half and (ideally) the two fields cancel, preventing the coax from radiating.

The antenna could be considered as a transducer, converting electrical energy into light (at an RF frequency). Or you could consider it as an impedance match to free space. In either case, it takes the AC power from the feed line and radiates a portion of that as radio waves (as radiation resistance). What isn't radiated might be converted to heat (as loss resistance) or reflected back down the feed line (both sides, causing SWR).

There are three "grounds" in radio:

  • RF ground
  • Electrical ground
  • Lightning ground

These three can be connected but are functionally different.

RF ground is not really ground -- it's just the other half of a monopole antenna. Not all antennas need one. It's called ground, because frequently the other half of a monopole vertical is the actual ground, or touching the ground.

Electrical ground, sometimes called safety ground, is part of the electrical code for commercial power. All electrical grounds should be connected in part to reduce noise, and in part to prevent stray voltage in places people can touch.

Lightning ground should be positioned to conduct the majority of a lightning strike around sensitive things and into the physical ground. It should have as few sharp bends as possible and minimize resistance; all lightning grounds should connect to a single point so you don't get voltage differentials between points in the ground system. Typically electrical ground should be connected to lightning ground at a single point.

  • $\begingroup$ Thanks for this $\endgroup$ Aug 6, 2021 at 9:48
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    $\begingroup$ Why doesn't return path have meaning? If charge can not be created nor destroyed, then for any current there must be somewhere an opposite current, right? $\endgroup$ Aug 6, 2021 at 21:16
  • $\begingroup$ It isn't really a return path because the radio pushes energy into both sides in the form of an AC signal, inverted between the two sides. Return path makes more sense with DC, where the power goes along the length, through the load, and back. If RF energy was doing this, the phase of the power would not match on the "return" path. $\endgroup$
    – user10489
    Aug 7, 2021 at 3:43
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    $\begingroup$ I think you may be conflating power and charge. $\endgroup$ Aug 7, 2021 at 4:00
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    $\begingroup$ No, electrons are charged particles (always negative, never positive), and charge is measured in coulombs, never in volts. Current is the first derivative of charge, and so any current, DC, AC, or RF, involves a movement of charge. $\endgroup$ Aug 8, 2021 at 2:29

The antenna requires a ground reference such as radial spokes for a vertical antenna or elevated sufficiently to use the coupling to earth or rear reflector to create radiation patterns.

Impedance matching determines how much of the signal is reflected back in the current loop of RF , called RL, return loss or s11 from scattering parameters.

While the coax is tightly coupled by distributed capacitance and inductance. With controlled impedance $$Z_o=\sqrt{L/C}$$ and the shield reducing radiation of the centre conductor, the conduction loss/100m and transfer loss from the common mode impedance of the braid style, and ground resistance, the emissions in the cable can be minimized. The best being rigid coax, then semi-rigid is determined by the transfer impedance ratio over this spectrum, often done in CATV requirements.

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    $\begingroup$ Respectfully, I suggest editing your answer to have less jargon and equations, and more of a simple explanation to the layman (that's me, haha). Also, surely antenna imbalance figures into the answer somehow. I look forward to hearing more from you, since you obviously know a lot about radio and antennas from your EE background. Cheers! $\endgroup$
    – rclocher3
    Aug 5, 2021 at 22:44

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