Edit: I was interested how is voltage gap maintained across an antenna. It seemed to me that an antenna might be like a wire where if there's a voltage at the top, the same voltage would be at the bottom, and hence no gradient and no current flow.

Several individuals explained to me that antenna is not a simple circuit element and goes beyond basic circuit analysis, and so there is no contradiction in it having different voltage at the top and bottom, which would lead to current flow.

One question that still bothers me:

enter image description here

Let's say that at one moment there's a positive voltage at the top of the antenna (10V relative to bottom, say), and current flows from top to bottom. But at the bottom where the antenna is actually connected, voltage is 0. That seems like the end of the journey, and it's unclear to me how the presence of a gradient in the antenna, also creates a voltage gradient in the circuit that drives the circuit elements?

On the other hand in the pic below it makes sense to me as the circuit is connected at 2 points in the antenna and there's a voltage differential between the 2 points that drives current to the radio circuit:

enter image description here

Old question: In circuit sims and in theory, ground always maintains zero volts.

Thus if the point to which the ground is connected has more than 0V, some current will flow into the ground, while ground still maintains 0V despite incoming current.

But it looks like that in most real cases ground is merely a theoretical construct, and unless the circuit is earthed, the point labeled as "ground" can't actually maintain 0V, if the point to which the "ground" is connected in the circuit is above 0V, and current flows into the "ground".

Is there an element or some way to get "real" grounding, such that when the "ground" is connected to say a 1V point in a circuit, the "ground" will maintain 0V, despite current flowing from the 1V point to the "ground"? Otherwise the point labeled "ground" is just a loose wire that will quickly acquire the 1V of the point it is connected to, and not maintain zero volts.

My motivation is radio antennas where I'd like to have one end of the antenna connected to a permanent zero volts, so that when the top end of the antenna gets 1V, currently will flow from top to ground, while the ground maintains zero volts despite current flow.

enter image description here

  • on the left almost no current flows (just tiny bit to get other end of wire to match voltage of source), while on the right current flows steadily. If this was a real circuit, what can I do to make it behave like the one on the right, and not like one on left? I have antennas in mind, so that's why the simple solution of having 2 sided AC voltage source (i.e. like an AC 'battery') that maintains the voltage gap, like in the pic. below, won't help..

This won't be like an antenna

Thanks for your help!

  • 2
    \$\begingroup\$ Ground is defined to be 0 V. It can't be anything else. Even if its voltage is oscillating by 200 V relative to Earth, that means the earth voltage is changing as far as your analysis is concerned, not that the ground voltage is changing. \$\endgroup\$ – The Photon Oct 22 '18 at 23:51
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    \$\begingroup\$ If there's current in that dirt, "earth" over here may not be the same potential as "earth" over there. And all voltages are relative -- we define ground as being 0V, but it's better to say "so many volts with respect to ground", or "so many volts from point A to point B". \$\endgroup\$ – TimWescott Oct 23 '18 at 0:12
  • \$\begingroup\$ @ThePhoton thanks for answer. I expanded my question with a picture to better illustrate what i meant by my questions. \$\endgroup\$ – Daniel Oct 23 '18 at 0:25
  • \$\begingroup\$ @TimWescott thanks for answer. I expanded my question with a picture to better illustrate what i meant by my questions \$\endgroup\$ – Daniel Oct 23 '18 at 0:26
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    \$\begingroup\$ @Daniel, 1. An antenna is not a lumped circuit element so many of the usual rules of circuit analysis (like KVL and KCL) don't apply when working with one. 2. As a rough model, like Felthry says, you can consider there's a parasitic capacitance between the antenna and ground. 3. If you want to drive the antenna relative to the actual earth ground, you'd want to connect some other part of your circuit to the actual earth ground. \$\endgroup\$ – The Photon Oct 23 '18 at 4:32

New Question:

To answer your newer question, I think you'd be better off studying a little bit about the distributed element model and transmission-line theory first. From the linked Wikipedia article (emphasis mine):

The distributed element model is more accurate but more complex than the lumped element model. The use of infinitesimals will often require the application of calculus whereas circuits analysed by the lumped element model can be solved with linear algebra. The distributed model is consequently only usually applied when accuracy calls for its use. Where this point is dependent on the accuracy required in a specific application, but essentially, it needs to be used in circuits where the wavelengths of the signals have become comparable to the physical dimensions of the components. An often quoted engineering rule of thumb (not to be taken too literally because there are many exceptions) is that parts larger than one tenth of a wavelength will usually need to be analysed as distributed elements.

If you were to excite your antenna with a low frequency source (or a DC source, let's say), then your intuition is correct: current would flow for a very short period of time until the voltage across the antenna was stabilized. But when exciting it with an AC source of a suitable frequency (i.e. a frequency where the wavelength is, let's say four times as large as the antenna, making the antenna a quarter-wave in length), the voltage will never be the same across the whole antenna, since the source is changing as fast as it takes the wave to reach the top of the antenna. In other words, there is no way for the bottom of the antenna to be fixed at a total of 0V, but it can have a DC potential fixed at 0V (with an AC potential swinging above and below 0V). This is what The Photon was mentioning in his comment talking about driving the antenna relative to the actual earth ground:


simulate this circuit – Schematic created using CircuitLab

Here, the DC source is 0V (i.e. it could be omitted), meaning your antenna's bottom end is fixed at earth potential on average. The AC source will excite a sinusoidal voltage at the bottom of the antenna, swinging from +1V to -1V at 1kHz. We'll assume the antenna is a quarter-wavelength (at 1GHz, that's about 3 inches long). I've also included the typical distributed model for a transmission line. That is how regular wires are modeled once the frequency of excitation becomes fast enough.

In your second schematic, with diode-detector circuit, you say the circuit is connected at two points to the antenna, but that's not true. The antenna is only connected to the node at the top-left of the circuit. The resistors are not part of the antenna.

Old question:

A quick answer: I think a \$\frac{\lambda}{4}\$ (quarter-wave) monopole antenna precisely describes what you're asking about. In the case of a monopole antenna, the end of the monopole facing the ground plane is actually a node of the voltage standing wave. Call that point GND, and that point will "maintain \$0\text{V}\$"; it is the node of a standing wave (i.e. no change in magnitude), and by calling it GND you've defined it to be \$0\text{V}\$, full-stop.

I'm still not sure I entirely understand your question, and schematics don't necessarily convey what you're trying to get across. By definition, ground is \$0\text{V}\$. Any part of your schematic that touches a GND is fixed at \$0\text{V}\$, since a wire in a schematic is defined to be a perfect conductor. For example, in your first picture, exactly zero current flows because there is no loop. GND is just a symbol we use to define where \$0\text{V}\$ is, and nothing more. You can't have a GND in a schematic that has any potential other than \$0\text{V}\$.

I think your confusion stems from the following:

But it looks like that in most real cases ground is merely a theoretical construct...

In all cases, ground is merely a theoretical construct.

  • \$\begingroup\$ @Daniel Thanks for the acceptance! However, in StackExchange fashion, it's typically good etiquette to not accept an answer for about a day or so to encourage others to write more answers. \$\endgroup\$ – Shamtam Oct 23 '18 at 1:38
  • \$\begingroup\$ "monopole antenna" - is a telescopic antenna an example of a monopole antenna? I mean, you said: "exactly zero current flows because there is no loop". and that's the problem i have. I wish to create a voltage gradient from top to bottom of antenna, without connecting it to physical earth. It sounds promising if a monopole antenna knows to maintain such a voltage gradient by itself, and current flows across it. (per your comment, I removed the acceptance for now.) \$\endgroup\$ – Daniel Oct 23 '18 at 1:38
  • \$\begingroup\$ @Daniel Yes, I'd say so (but I'm not particularly well versed in antenna theory). Try looking up "whip antenna." \$\endgroup\$ – Shamtam Oct 23 '18 at 1:41
  • \$\begingroup\$ @Daniel Any antenna will have a "voltage gradient from top to bottom of antenna," regardless of if it's touching the earth or not. \$\endgroup\$ – Shamtam Oct 23 '18 at 1:43
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    \$\begingroup\$ @Daniel Antennas are a special kind of black magic ;). With the right excitation (i.e. correct output impedance, correct frequency, etc.), a time-varying source will cause a standing wave in the antenna, and it will "look" like a load that is connected to ground (on the end that is just floating in air!). That's not exactly what happens, but I think is a decent explanation if you haven't studied any electromagnetics, microwave, or antenna theory. \$\endgroup\$ – Shamtam Oct 23 '18 at 1:51

In most electronics, "Ground" is simply the point in the circuit that we choose to call "Zero Volts", and use as a reference when measuring voltages elsewhere in the circuit.

In AC power distribution and in some radio antenna systems, "Ground" is really "a connection to the Earth".

  • \$\begingroup\$ Thanks for your answer. I expanded my question with a picture to better illustrate what I meant by my question.. \$\endgroup\$ – Daniel Oct 23 '18 at 0:27

I'm not an expert on antennas but I have some experience in pcb design which might help. Having a good reliable ground is important in circuits and crucial in sensitive analog circuits. If you have a good power supply which can maintain it's value in all situations, then it's - node is your true ground. All the current in the circuit should come back here (unless there's an antenna involved which I dont know what happens). When a component needs to be connected to the ground you need to make a path with lowest possible impedance to the minus pin of the supply that's one of the reasons they use an entire layer of a pcb for ground node in the circuit, that way a very low impedance path would be available for every component in the circuit.

As I said before I'm not an expert on antennas, but I suspect the same scenario applies to your antenna. You should connect one side with lowest possible impedance to where all the current is supposed to flow.

I hope it helped.

  • \$\begingroup\$ thanks for answer Sadegh. I expanded my question with a picture to better illustrate what i meant by my question.. \$\endgroup\$ – Daniel Oct 23 '18 at 0:26

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