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First of all let me just state that I do not feel confident enough to tell anyone anything about how electric circuits work or anything about the physics behind them because I simply do not know or understand it all.

But I have many times read that there has to be a closed path for current to flow in a circuit, leading to a fact that if there isn't a closed conductive loop nothing can happen.

And I have taken that to be a definitive truth, but I wonder about something(and I might just as well be terribly far of the path of reason here).

If I was to design a circuit board which contains traces through which very high frequency signals(currents) will flow then I have to consider things like signal reflections, I don't know what reflections consists of in purely physical terms(but I have to imagine that a reflected signal is a certain amount of the current(s) that was originally sent through the trace) but apparently if I send a high-frequency signal down a trace(or wire) then under certain conditions the signal can travel down the trace(wire) only to bounce off of something and then travel all the way back to where it first came from. Where it might bounce off of something again and so it can bounce back and forth travelling the length of the trace over and over again getting smaller and smaller until it dies out.

This is just stuff from the top of my head, stuff that I have never acquired a fair understanding of in the first place. But if we restrict the scenario to this very high frequency situation, if a signal or current can be reflected back towards where it came then why would it even have to be relevant whether there are a closed loop or not.

Couldn't a broken loop present paths for such currents to bounce around in?

I know that I am at a relatively very low level of insight into these complex matters but I don't now why that wouldn't be possible. I would be very happy if anyone could enlighten me.

I have one single hypothesis without anything what so ever to support it, but perhaps the very high frequency scenario alters the way that a traces copper is utilized so that it in some respect is a closed loop in it self?

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    \$\begingroup\$ Read here: en.wikipedia.org/wiki/Static_electricity which is about static electricity where the charge is moved somewhere and stays there (for a while). The charge/electrons do not go around in loops. Applications are however limited. It is a very impractical way of processing information. Note that moving a charge onto something requires increasing amounts of energy as the charge which is there already repels the new charge. \$\endgroup\$ – Bimpelrekkie Mar 26 '17 at 17:51
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    \$\begingroup\$ We tend to separate high frequency effects from normal current paths just so we can get a decent model that does not totally confuse everything. The truth is, as frequencies increase there are all kinds of electron movements happening within and without conductors and components, closed circuit or not. Most of it we call noise, some of it, like in an Antenna, we call a signal... go figure. \$\endgroup\$ – Trevor_G Mar 26 '17 at 17:56
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    \$\begingroup\$ So basically, you are right... but you may still be insane ;) \$\endgroup\$ – Trevor_G Mar 26 '17 at 18:01
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    \$\begingroup\$ A circuit must be closed. At high frequencies parts of the circuits may easily work as antennas so you can have effects on 'open circuits'... but you don't even need high frequencies. Any appliance with a transformer sends energy (and sometimes quite a lot of it) without a 'closed loop' from primary to secondary. When electric, magnetic or electromagnetic fields are involved things get more complicated, either at high or at low frequencies. \$\endgroup\$ – Claudio Avi Chami Mar 26 '17 at 18:31
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    \$\begingroup\$ And just to note, if it helps to think of it like this, you can model an open circuit in terms of having an extremely high resistance (nearly infinite) all over the place that closes the circuit, as well as tiny inductances and capacitances. \$\endgroup\$ – Michael Mar 26 '17 at 22:40
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You are completely right.

The "closed loop" rule comes from a simplification that we often use in circuit analysis called the "lumped component model". This model provides a good approximation to actual circuit behavior at DC and low frequencies, where the effects of parasitic inductance, capacitance and the speed of light can be ignored.

However, these factors become significant at high frequencies and can no longer be ignored. Any circuit of nonzero size has inductance and capacitance, and is capable of radiating (or receiving) an electromagnetic wave. This is why radio works at all.

Once you start considering parasitic capacitances, you'll discover that everything is connected to pretty much everything else (moreso to nearby objects), and there are closed loops where you wouldn't normally expect to find them.

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    \$\begingroup\$ As I said in another comment, when fields have their word, circuits get more complicated to analyze. And you don't even need high frequencies. A mains transformer is enough proof of that. \$\endgroup\$ – Claudio Avi Chami Mar 26 '17 at 18:37
  • \$\begingroup\$ @ClaudioAviChami A mains transformer has no current path from primary to secondary and ignoring inter-winding capacitance there's no current flowing between the two. \$\endgroup\$ – Dmitry Grigoryev Mar 27 '17 at 9:04
  • \$\begingroup\$ Dmitry Grigoryev there is not a current flowing from an antenna transmitter and an antenna receiver either. But energy is transported through fields. Magnetic, electric and electromagnetic. \$\endgroup\$ – Claudio Avi Chami Mar 27 '17 at 17:52
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Responding to your title:

Am I insane to question that only with a closed path can electrons move?

Currents usually* travel in loops. However, the loops need not be entirely made of conductors (ie, copper). Current is a flow of charge. Therefore, all the following physical phenomena represent current:

  • Electrons flowing in a copper wire
  • Ions (which are charged) moving between the electrodes of a battery (or an electrolytic capacitor)
  • Electrons flying through vacuum (ie, thermionic valves, cathode ray tube)
  • And, last but not least, displacement current

The last one answers the question "how can a current pass through a capacitor's dielectric?". A quick summary is that charges accumulating on one plate of your capacitor will push the charges on the other plate away, and give the illusion that electrons are flowing through the cap's dielectric, while in fact they are not. One plate is filling up with electrons, while the other is getting drained of electrons.

... * Yes of course! You can have currents not travelling in loops: simply shoot an electron beam into deep space, with enough speed to escape the solar system. Obviously, this is not applicable to everyday electronics design.

Also, it has a drawback: you only have a certain number of electrons to shoot... and the more electrons your "gun" shoots away, the more positively charged it becomes, making sending electrons away progressively harder.

Whereas your usual circuit, which is a loop, recycles the same electrons (if DC) or just wiggles them around (AC), and will run as long as the battery / nucular power plant / solar cell has available energy.

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    \$\begingroup\$ The correct spelling is nuclear. (Or does each atom in your country have a nucule?) \$\endgroup\$ – user253751 Mar 26 '17 at 22:27
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    \$\begingroup\$ @immibis He attended the George Bush school of nukular physics. \$\endgroup\$ – Majenko Mar 26 '17 at 22:40
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    \$\begingroup\$ @immibis, I think it's a reference to this Simpson's scene \$\endgroup\$ – Turion Mar 27 '17 at 8:59
  • \$\begingroup\$ "Electrons flying through vacuum" or even thin air will perfectly fly onwards when there is no loop, once they are accelerated. Also see the big brother of cathode rays: beta radiation... \$\endgroup\$ – rackandboneman Mar 27 '17 at 10:50
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    \$\begingroup\$ @Turion it's in reference to good ole Gorge W. Bush ;) I can never resist this one. \$\endgroup\$ – peufeu Mar 27 '17 at 11:20
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Rule #1. There is no such thing as an open circuit except under DC steady state conditions.

Between every wire, every part and even every atom, there is capacitance, resistance and inductance to some other wire, part and atom. Microscopic as it may be, it is there. Even within the wire or part itself.

However, if the circuit you are testing is in a steady DC state, the capacitance and inductance present no load, only the resistance does, and that is high enough not to matter. For current to flow in that "Circuit" it has to have a path from it's start point to it's end point.

Rule #2. There is no such thing as DC Steady State conditions.

We are swimming around in a sea of electromagnetic waves. As such, a steady state DC circuit is actually impossible to achieve. Further every current in your circuit is producing it's own electromagnetic field that interacts with each other AND with those outside fields. There will always be what we call "noise" in your circuit.

Rule #3 : The faster you modulate a voltage / current the more potential circuit paths you need to worry about

Those little invisible circuits I mentioned in Rule #1 have impedances that change as the frequencies you are trying to pass increase. As such the higher we go the more we have to deal with strange effects like signal loss, reflections, and noise emission to name but a few.

Fortunately:

For the most part we can dismiss most of these effects because, at the frequencies you are using, they produce little disturbance.

A 60Hz AC circuit works basically the same as the circuit diagram indicates if the connections are not lengthy. We can safely make the bold statement that circuit needs to be complete for current to flow because the current that is actually flowing is basically not measurable enough to matter.

However, if you are trying to pass a 100GHz signal round the same circuit, you will find the numbers no longer make any sense.

As for broken loops... See Rule #1

Are you insane to question that?

No, actually quite the converse. It is always good to think deep and ask questions like that. However, the answers may drive you there.

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A concept that might help you is the transmission line concept. The idealized transmission line is one with a characteristic impedance, and a fixed delay. Think of the transmission line as a trace on a circuit board. The delay is caused because when a voltage is applied at one side of the line, there is a delay before it can be detected at the end of the line. Hopefully this makes sense. What the trace really does, is allow an electric field to propagate down the line to the load. The field can only travel at the speed of light, not faster. So there is a period of time when the field has been applied, but the load has not felt it yet. Hmmm.

So, what is the characteristic impedance? Let's refer to it as Z. When a voltage (V) is first applied to the input of a transmission line, the current that flows is strictly a function of Z. It does not matter what is at the other end of the line. Maybe it is an open circuit or a short circuit or an inductor or capacitor. Let's just assume it is an open circuit. In spite of this, the current that flows into the transmission line will be V/Z UNTIL the electric field propagates all the way to the end of the line reflects, and comes back to the source. In a sense, the electric field is interrogating the line and the load, and when it gets to the end, a reflection comes back which brings information about the load back to the source. The reflection coming back from the end of the line may reflect again when it gets to the source, depending on details I will get to later.

So, anyway, you are right to think that current can flow into an "open circuit." Of course, when this happens, or when it is significant, what it means is that you need to improve your model of the circuit to account for these transmission lines or parasitic capacitances or whatever. Transmission line theory provides a way to do this.

A special case of a transmission line is when the load at the end is exactly equal to the characteristic impedance of the line. This could be the case if a PCB trace has a resistor connected to it at the end, and the other end of the resistor goes to GND. When this happens, if the resistor value is the same as Z, there is actually no reflection. So, the current that flows into the line is simply I = V/Z. Since no reflection comes back, the current continues to be V/Z. Now let us consider reflections.

When the end of the line is not terminated in Z, there will be some reflection. That reflection behaves exactly the same way as the original electric field traveling down the line, except that it is coming back toward the source. If the source is terminated with a resistor of value Z, then the reflection will be completely absorbed at the source. In other words, if the source impedance is Z, the reflection from the load will be fully absorbed, just the same way that if the load is Z, there will be no reflection back toward the source.

But if neither the load nor the source are terminated in Z, then the reflection will theoretically continue forever, bouncing back and forth. Of course in the real world, the reflection will die out due to some kind of energy loss. If nothing else, the non-zero resistance of the copper wire will cause losses.

I hope you are able to get something out of this. Transmission line effects can be hard to assimilate at first, especially if you don't have other background information. So I tried to just explain it in a somewhat intuitive way that I hope will help you.

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  • \$\begingroup\$ Upvote upvote upvote!!! Closed loops are a lie-to-children. \$\endgroup\$ – rackandboneman Mar 27 '17 at 10:52
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    \$\begingroup\$ A transmission line is NOT a trace on the board. It is TWO traces on a board, or one trace and its GND return. Transmission lines on PCB are closed loops as closed as a DC source, a resistor and a LED. \$\endgroup\$ – Claudio Avi Chami Mar 27 '17 at 18:55
  • \$\begingroup\$ @ClaudioAviChami, I didn't mention the GND plane. Taking it for granted. But I am well aware. If you review the OP's question, this is exactly the kind of thing being asked about, a trace terminated in an open circuit. \$\endgroup\$ – mkeith Mar 28 '17 at 2:56
  • \$\begingroup\$ Once you are talking signals with GHz harmonics on a large PCB, PCB trace pairs can very well be true "long" transmission lines..... \$\endgroup\$ – rackandboneman Mar 29 '17 at 8:51
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An antenna is an "open circuit" if you look at it closely. When talking about alternating current, particularly radio frequency AC, conductors are not idealized components but interact with their surroundings. If you are talking about reflections, you are talking about properties of the conductor that aren't commensurate with the properties of straightforward connections in a circuit diagram.

There are actual circuits built using just etch-a-sketch kind of conductor arrangements on a PCB. Many microwave circuits and filters don't contain more than an arrangement of conductors that, in connection with the free space in between, actually corresponds to a complex composition of inductivities and capacities.

When viewed at much lower frequencies including DC, the whole microwave circuit may be just one or two conductors, just like an antenna viewed at much lower frequencies than its operating frequencies is just an open connection.

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Where do these hidden/parasitic paths matter?

Consider coupling from YOU to the concrete under the floor: 1cm spacing, area of 0.1meter by 0.3 meter, dielectric constant---use that of air (1.000002 or close).

What is the capacitance from you to the floor? $$Capacitance = Eo * Er * Area/distance$$ or [9e-12Farad/meter * 1] * [0.1 * 0.3] / 0.01 = 9e-12 * 0.03/0.01 Capacitance = 9e-12 * 3 = 36 picoFarad.

So? Now touch a neon-sign transformer, 50,000 volts at 60Hz (377 radian/second). The dV/dT = 50,000 (assumed peak) * d(sin(60Hz)/dT) = 50,000 * 377 ~~~ 20Million volts per second.

What is the current through you? I = C * dV/dt = 36 e-12 * 20e+6 = 700 microAmps.

You want to avoid that. Even if there is no obviously closed circuit.

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Strictly speaking, electrons drift in the opposite direction to that of the current flow. In order for the current to flow (and energy to move), you need a potential difference (voltage) across the start and finish points. Note that electrons also move within atoms, in orbital shells, but nobody really knows how; perhaps they go around in circles.

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It's simply not true, though like many rules it is a good and useful approximation when applied to the appropriate circumstances (DC circuits, low-frequency AC mains circuits, where we are primarily interested in the transfer of electrical power).

Electrons are always moving, except at absolute zero (which you can't reach). Turn up the gain on any amplifier high enough, and even with its input carefully screened from any external influence, a hiss (audio) or other random signal will become apparent. This is the electrons jostling around in the input circuitry under the influence of its ambient temperature.

Charge storage on capacitors is absolutely fundamental to modern solid-state electronics. Logic states are trapped packets of electrons. In a flash memory device, a high voltage drives electrons through a normally insulating barrier onto what is effectively the plate of a capacitor and the gate of a field-effect transistor. When the high voltage is removed, the electrons stay put for years (or longer), and their presence or absence can be determined by whether the transistor conducts. Indeed, it's common to measure a quantity of electrons (which determines the voltage on the gate and hence the transistor's output level) and quantize it to one of eight levels, thus storing three bits as one of eight quantities of electrons within a single transistor.

The circuit eventually gets closed, when these electrons leak out as a result of thermal noise and quantum "tunneling". As mentioned above, this takes many years to happen unless the cell is re-written by re-applying the high voltage.

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