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We know that electrons move from the negative to positive terminal, and that holes flow in the conventional direction of current - from the positive to negative terminal.

I've always assumed that this means that electric signals travel in the conventional current direction and that it's the travel of holes that gets close to the speed of light. Upon further research I come across mentions of electric signals being the electromagnetic waves traveling through the medium by the excitation of the electrons.

This begs the question. If electric signals are the propagation of EM waves at close to c in which direction do these waves travel? In the conventional direction? In the direction of electron flow? In both directions?

I can easily imagine that the signal is directionless and propagates from the point of contact/source of signaling in all directions like light in a dark room. But I'm not sure.


Clarification

Note: This merely presents an experiment to perhaps clarify the intent of the question above. I'd accept answers that don't address this part of the question. Note that you can treat this as a thought experiment but I think it's possible to physically set up this circuit. The only problem is we'd have to deal with resistance in the real world.

Consider a circuit that has 3km of conductors. We start with a DC source attached to 1km of conductor. For simplicity you can assume that the conductor has zero resistance. We attach an LED at the end of the conductor. Then we complete the circuit with 2km of conductor back to the DC source:

                1km of wire
        ┌─────────────────────────────────[resistor]─(LED)───┐
        │                                                    │
         ╱ switch                                            │
        │                                                    │
[DC voltage source]                                          │
        └────────────────────────────────────────────────────┘
                      2km of wire

Assume the wires are spooled so that physically the light bulb and the switch are next to each other.

If I close the switch would I expect the LED to turn on roughly 3.34 us later (the time it takes light to travel 1km) or 6.68 us later or somewhere in between?

Does it make a difference if the 1km line is attached to the positive or negative terminal?

Does it make a difference if the wires are not spooled but are physically laid out over the 1.5km distance? Does the physical geometry make a difference eg. a circle vs a straight line and back?

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  • 1
    \$\begingroup\$ @tlfong01 I'm talking about "events" themselves. Like when I turn on a switch the room lights up immediately - which does not correspond with the speed of electron flow which is around 1mm per second and there's at least 5 metres of wire between the switch and my ceiling light. \$\endgroup\$
    – slebetman
    Feb 3 at 7:32
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    \$\begingroup\$ @tlfong01 But if the signal flow is different in semoconductors that would be something interesting as well. For example when electrons start to flow through a diode does it start at the P region, the N region or both simultaneously or does it begin where the circuit first encounter a voltage change (usually at a button or switch somewhere) \$\endgroup\$
    – slebetman
    Feb 3 at 7:32
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    \$\begingroup\$ @tlfong01 OK. Let's assume an AA battery connected to 1 light second of perfectly conducting wire (can assume superconducting minus quantum effects if it makes it easier) to a resistor and an LED and the circuit is completed with another 2 light second of wire. The events are my fingers closing the circuit at the battery and the LED lighting up. We know that when the circuit closes the voltage across the resistor+LED will raise. What I think I'm asking is will the LED light up after 1 second or 2 seconds? Does it matter which wire is connected to the battery's positive terminal? \$\endgroup\$
    – slebetman
    Feb 3 at 11:18
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    \$\begingroup\$ @slebetman I believe the answer depends on the exact topology of your earth-sized circuit - you'll get different answers for a big loop versus a piece of magic coaxial cable. I believe the answers are "slightly more than 1 second" and "no". \$\endgroup\$
    – pjc50
    Feb 3 at 12:45
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    \$\begingroup\$ @tlfong01 It can't be instant because of limitations of relativity. At best it would be 1 second. However I fear we're verging into quantum electrodynamics for a proper explanation. I also fear the true answer is it depends on the type of wire/cable we're using to transmit the signal - which means that the general answer for all case including inside of semiconductors and superconductors can't be answered in a simple manner \$\endgroup\$
    – slebetman
    Feb 3 at 14:14
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If electric signals are the propagation of EM waves at close to c in which direction do these waves travel? In the conventional direction? In the direction of electron flow? In both directions?

"Both" directions, as well as an omnidirectional component.

The best example might be an Ethernet cable. A bit is represented as a pulse. To transmit this pulse, the driving chip puts one output low and another high briefly. If it left them like that, then a conventional circuit would be formed with current flowing in a loop between them through the termination resistor on the other end. However, it does not, it quickly (let's say one nanosecond) moves the drivers back to equal voltages.

This launches a rising edge on the positive transmit wire and a falling edge in the negative transmit wire. If you have a fast enough oscilloscope clipped to the wire, these will appear as voltages, but really they're a small EM pulse travelling down the wire.

Each pulse also radiates an EM wave outside the conductor. Because the wires are twisted together at a constant distance, the "positive" wave emitted almost completely cancels the "negative" wave emitted. If this wasn't done, you wouldn't be able to use your radio in the same room due to the EM interference.

The cable can be modelled as a chain of tiny inductors along the wire, with a ladder of tiny capacitors between the two conductors. Each inductor stores the energy of the incoming current change in a magnetic field, before re-emitting it forwards along the wire.

The general subject is called a "transmission line". The effective velocity is below the speed of light, but not much (see "velocity factor").

travel of holes that gets close to the speed of light

It's a bit more complicated than that. (I've not bothered to fully understand that myself!) However, I believe this is why NPN transistors (hole mobility in the P region) are preferred to PNP transistors (electron mobility in the N region).

One of these days I'm going to write a canonical article on why electrons are a distraction for learning about electricity. Electricity is fairly straightforward and follows some mathematical laws that can be analysed with basic calculus. Electrons are not little pingpong balls in a tube, they are weird quantum objects that are capable of appearing, disappearing, tunneling through solid objects, and are much harder to model properly.

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  • \$\begingroup\$ pjc holes and electrons move at the same rate.And holes is not a carrier type it is just electrons moving to free states in the valence band of the material. \$\endgroup\$ Feb 3 at 10:17
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    \$\begingroup\$ .. no, it's definitely a different rate, although I misremembered and the holes are slower! courses.cit.cornell.edu/ece315/Lectures/handout3.pdf \$\endgroup\$
    – pjc50
    Feb 3 at 10:19
  • \$\begingroup\$ I would definitely appreciate you creating that article. I've always found the thought of electrons a major distraction in learning because I always confuse myself with the clashing of ideas I've accumulated over the years. Especially in circuit analysis - I tend to over complicate the mechanics because I'm trying to reason my way through with those clashing ideas. \$\endgroup\$
    – K. Millar
    Feb 3 at 10:41
  • \$\begingroup\$ Ooh. Good example! \$\endgroup\$
    – slebetman
    Feb 3 at 11:20
  • \$\begingroup\$ Any chance of goading you to use this question to write the canonical answer at the back of your mind? \$\endgroup\$
    – slebetman
    Feb 3 at 11:22
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signals travel in all directions like waves in a pond, they are a disturbance in the current not necessarily a current themselves.

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  • \$\begingroup\$ Which is exactly why we got the direction of current wrong when we set the convention. We guessed and got it wrong. Now we're stuck with explaining "holes" traveling. Sheesh. \$\endgroup\$ Feb 3 at 19:03
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    \$\begingroup\$ Holes are real in as much as they have positive mass, Negatively charged electrons are unfortunately real too. \$\endgroup\$
    – Jasen
    Feb 4 at 3:43
  • \$\begingroup\$ um no \$\endgroup\$ Feb 4 at 3:50
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The signal in a cable is carried by a radiowave which propagates in the space around the wires. Parallel wires or twisted pair have this capability. Coaxial cable also has it, but the wave is limited to the insulation layer between the middle wire and the shield.

The wave happens as electric and magnetic fields, it's not inside the metal. The electrons in metals only make possible that the wave propagates along the cable. The fields of the wave induce some current to the metal surface and we can also squeeze the strength of the electric field to single number, the voltage which, of course, depends on place and time if there's waves. At low frequencies we can virtually forget the waves and make all calculations with voltages and currents. A 60Hz electric power line must be more than 100 miles long before skipping the existence of a wave starts to cause substantial error.

One common practice is to say that wires longer than 10% of the wavelength must be considered as transmission lines which carry waves. If they were not designed as proper transmission lines the result would be unpredictable. I have found in my experiments that 1% or less of the shortest wavelength is a safer limit for freeform wirings.

The wave starts from any change of electric or magnetic field. One can for ex. switch the cable ends to a voltage source (line transmitter ICs do it) or turn the polarity. The wave propagates along the cable further from the point were the change was initiated. Electrons move locally as the fields force. At the same time the current can well be to different directions at different places of a wire. That doesn't happen with continuous unchanging DC.

ADD due the inserted battery+ switch+ 1km wire + Led+ Resistor+ 2 km wire example.

The geometry of the wires and how they are placed over the ground affect radically how the signal reaches the LED. If you reversed the battery and the LED and tried again no difference would be exist, the polarity doesn't affect the wave. This isn't a transmission line if the wires are not parallel. It's not possible to calculate nor simulate the case without knowing the exact geometry. If the wires are wounded to coils they are inductors which surely has some effect in the circuit. In a lucky other geometry case you emit a measurable radiowave to space because you can have a working antenna.

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  • \$\begingroup\$ Damn. You've hit upon another thing I didn't consider. That the actual transmission medium may make a difference. I was hoping that there would be a simple answer that covers everything from how signals really travel on a PCB to how a blackout event propagate over a single transmission line (ignore transformers) to how signals really work inside semoconductors etc. \$\endgroup\$
    – slebetman
    Feb 3 at 14:16
  • \$\begingroup\$ Resistors and semiconductor components have less than ideal conductive materials. Inside those materials there's also remarkable fields longitudally along the conductive pieces. Waves inside them are much more complicated than around low resistance metal wires. Fortunately the dimensions of the semiconductor chips inside transistors etc... are often well below that 1% of wavelength limit I told. That makes current and voltage still useful. Do not expect the same with microwaves. \$\endgroup\$
    – user287001
    Feb 3 at 14:26
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    \$\begingroup\$ @slebetman actually for transmission lines on a PCB or in coaxial cable, the medium between conductors makes a difference, in particular its dielectric constant. \$\endgroup\$
    – pjc50
    Feb 3 at 21:39
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This answer on how the electric field is established in a DC circuit might help you visualize what happens in the initial moments in your imagined circuit.
When you close the switch, surface charge that had accumulated at the switch's terminal will start to recombine and redistribute in order to establish the uniform electric field \$E = j/\sigma\$ directed along the conductor path.
So, the shape of the circuit and the position of the switch and the load changes the way this redistribution take place, irregardless of the fact that once the steady state has been reached the result is the same.
A wide circular circuit will behave differently from a circuit where the two wires are laid close to each other, and differently from a circuit where the two wires are twisted or even put on inside the other (coaxial cable).

If you want to know what happens you have to solve Maxwell's equations with the correct boundary conditions that depends on the geometry of the circuit and the position of the elements.

You can also try to model the circuit as a transmission line approximating it as a chain of distributed elements, whose capacity and inductance depends on the geometry of the connections and the permeability of the space between them. The capacity per unit length represents the way charge on one side of the wires affects charge on the opposite side, while inductance per unit length represents the interaction between the magnetic flux associated with the forward and backward currents (very roughyl speaking; you can translate this into how much the line resists changes in voltage and current).
Note that this kind of modelling requires that you split your circuit into piecese where the connection between the end points is uniform (uniform spacing of the stripline, uniform twisting of the cable, uniform radii ratio in the coaxial line).

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  • \$\begingroup\$ OK. But I think we can ignore all of the calculations when just need to think about the direction the disturbance in the electric field (a.k.a, the signal) propagates. Your answer seem to imply they propagate in all directions simultaneously. So electric signals don't travel from positive to negative or vice versa but in all directions simultaneously - including in some cases the air or even vacuum (which may be old news to you but is an enlightening way to think about antennas for me). \$\endgroup\$
    – slebetman
    Feb 3 at 18:37
  • \$\begingroup\$ I really wish I can accept multiple answers \$\endgroup\$
    – slebetman
    Feb 3 at 18:38
  • \$\begingroup\$ @slebetman well, if you look at the answer I linked, when you close the switch the perturbation in the charge distribution (EDITED) go both ways. Current and voltage are simplifications, you should not rely on them too much. What really counts is... the charge: where is it, how does it move (velocity and acceleration) \$\endgroup\$ Feb 3 at 18:41
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According to Maxwell's laws, there are 2 types of electrical signals:

1 - Conduction current signals

The energy of the signal is proportional to the overall kinetic energy of all electrons moving in a host material like Copper or Silicon.

2 - Displacement current signals

The energy of the signal is proportional to the overall electrical and magnetic energy (Poyinting's vector) of the radio wave carrying the signal.

Example 1 In track of copper, carrying a slow time-varying current, we have mainly conduction current signals.

Example 2 In track of copper, carrying a fast time-varying current, we have both conduction current signals and displacement current signals. In this case, the electric and magnetic fields travels in the track, in the PCB and in the surrounding air.

Example 3 In the air, we can only have electric and magnetic fields traveling at "the speed of light in the air" which is almost equal to "the speed of light in vacuum" called c.

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  • \$\begingroup\$ Hi +1 nice answer ,"In track of copper, carrying a fast time-varying current, we have both conduction current signals and displacement current signals" , does both signals disturb each other or they are completely independent ? Ex- lets assume if initially in wire only conduction signal was present and then I introduce displacement Current signal by using radio waves , so does value of conduction signal changes or not? \$\endgroup\$
    – user215805
    Apr 7 at 9:12
  • \$\begingroup\$ Displacement currents in copper are small compared to conduction currents. \$\endgroup\$ Apr 7 at 9:19
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Just thought I'd chuck in that the average speed of the electrons flowing through a copper wire due to the electric field is only about 1 mile per hour - but they effectively push each other along.

Signals don't propagate at the speed of light in a vacuum (i.e. c = 300,000,000 m/s) of course. If you're thinking about optical fibres then the light particles (photons) travel at about 2/3 rds of that speed (due to the refractive index of the glass) - but it's actually a bit more complicated than that as there's also interactions between the photons in each pulse of light. Similar effects with electrons travelling along transmission lines.

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You asked:

In which direction do electric signals flow?

Which I will answer with another question: what's an "electric signal"?

Is it an individual electron? The net motion of electrons? Maybe not electrons but some other charged particle?

Or is it a wave in the electromagnetic field?

Or sometimes "signal" means "the electric potential difference (voltage) between these two points". If you're wondering how bright an LED is, you probably mean signal in this sense. But a signal of this kind can't "flow", since it's by definition between two points.

Before you can have a rigorous answer, you must ask a rigorous question. "Electric signal" isn't a thing with any meaning unless you give it one.

You may find How does the current know how much to flow, before having seen the resistor? to be insightful.

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  • \$\begingroup\$ Wouldn't the information between the switch and the LED need to travel the 1km or 2km of wire before the LED lights up? Isn't that "flow" since two physical locations separated by 1km would by definition be two points? You can of course mathematically model the wire as taking zero time to propagate the fact that the switch is closed to the LED but that's not how relativity works. Nothing is instantaneous because we're limited by the speed of light \$\endgroup\$
    – slebetman
    Feb 3 at 17:58
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My understanding is that signals along a transmission line do not propagate at the speed of light but at some percentage of 'c' depending upon the properties of the line. In the past when using a time domain reflectometer you had to factor in the 'Velocity of Propagation' (VOP) if you wanted an accurate reading of the distance to say a short or open.

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  • \$\begingroup\$ Yes. It usually depends on the medium. Obviously the wavefront that escapes to free space travel at exactly c in a vacuum (wires are antennas after all and notwithstanding that some of the answers imply that the RF part of an electric signal always exist anyway). The speed of light is more of a limit seeing that some answers require a mental model that all electrons in the wire move simultaneously which is impossible given that event propagation is limited to c \$\endgroup\$
    – slebetman
    Feb 17 at 12:57

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