"Veritasium" (YT) - "The Big Misconception About Electricity" ?

[adx]

So if energy flows not in wires but near the surface, then if we take a fairly large sheet of copper, ground it, drill a hole in the center with the hole diameter being very close to the wire diameter,  and run the wire thru the hole, then we will block the flow of energy (assuming the hole walls are so close to the wire surface but not touching it ) along the wire surface , and therefore when we close the switch the load will not get any energy in DC steady state because no electromagnetic field can break through the shield . I do not think anyone here would believe that at DC no energy will flow into the load in this experiment. Therefore the claim that energy flows not in wires is bogus.

@Bud: The magnetic field at DC will go straight through a copper plate. The electric field AKA voltage difference will crowd into the gap. From this time of night I can't see how that makes you wrong. The bulk of the magnetic field will be in a place with no electric field.

We're both wrong. (Sort of.)

I just did the integration (in Excel) of this situation:

https://database-physics-solutions.com/buy.php?superlink=use_the_poynting_vector_to_determine_the_3948

(..after being unsatisfied with the Wikipedia one at https://en.wikipedia.org/wiki/Poynting_vector - what is W? This is my main problem with mathematics - people assume you "know" or are willing to go searching from zero context, forward through that material or something else unspecified)

It describes a coax carrying DC. Supplies formulas for electric field and magnetic field using radius r, multiplies together for each r to give the Poynting magnitude at that r, integrates for r over the insulator space, then invokes some magicks to prove something that I am not interested in. I summed Poynting magnitudes in Excel with a radius step of 0.1mm (also tried 0.02mm just to be sure).

Case 1 is coax with 2mm centre conductor (a=1) and 10mm ID shield (b=5).

Case 2 is that same cable, but with a slug of copper (or whatever) inserted into the space, such that it leaves a 1mm gap to centre conductor, and is almost touching the shield but connects at one point (say via a very thin ring round the middle, a tack weld, or tiny piece of wire). The idea is so no current passes along the slug therefore leaves the magnetic field alone, but is electrically connected to the shield so electric field compacts into the smaller gap to the centre conductor. The purpose of this is to test "The bulk of the magnetic field will be in a place with no electric field.", by that I mean a space that once contributed to Poynting power is simply deleted without change to one of the multiplicands (the magnetic field). You'd think this might result in a different number for the sum of the products.

I put 12V on it, and 1A through it. Result for case 1 popped out with the expected 12W (actually -12.3042 because I got a and b round backwards, the excess because the sum over r = 1 to 4.9 step 0.1 is skewed to smaller diameters, out of interest if I change the start number to 1.05 so the table runs to 4.95, the result is -11.997). Verifying that my numerical integration technique works.

Result for case 2 is -12.44361527 (-11.9946 with centred steps, not trying to be confusing, it's just the process and demonstrates the "imprecision" I raised earlier) - confirming that the squeeze in electric field is enough to 'exactly' offset the deletion of the area which no longer contributes to Poynting power.

This confirms what bdunham7 was saying about the results being exactly the same however a circuit is twisted or encased. I don't think anyone was expecting Poynting's math not to work, so hardly a surprise it does, and from looking at the equations it's obvious why there is a problem with Bud's and my intuitions: H = I/2/pi/r irrespective of the size of the cable (I've 'proven' a hidden V=IR in something before, even resorting to graphing it, so this isn't quite as silly as it sounds!). The magnetic field used to calculate the Poynting result in a sense doesn't depend on anything other than the current - a proxy for it? Shifting the 'current' to a place where it can be in line with the electric field calculated in its space between the pressure difference, completely orthogonal to everything else. What more unrealisitc way of measuring power could there be? After all there remain physical charges actually moving in a current of fluid within the copper wires, exerting an actual mechanical pressure (including pressure drop in the load).

On the other hand it works for (and is completely consistent with) AC, with its ability for current to seemingly jump out of the wires and into empty space, into a physical manifestation of that magnetic field. And I've got no complaint over another transverse field (electric) being the driver of energy transfer in circuits, whether that be AC, DC, electric, fluid, string, chain, rotating shaft or whatever.

How to test for its position? A thought experiment with a lop-sided conductor pair, perhaps with breaks in the current path as above. Or use microwaves, something similar with fets switching the current paths and a great big roots-blower-ish beamline centrifuge to weigh the data.

Time to call it a day (as in, it's getting late again). I think I'll pick a side, and choose the mechanical explanation, that is power flows in a combination of wires and the voltage difference. Otherwise it becomes a bit of a circular definition, where everything is "energy" but which behaves like particles.

(I can attach the spreadsheet if anyone wants - it ain't pretty.)

[T3sl4co1l]

At high frequencies, lumped-equivalent methods fail, and transmission line methods must be applied.

The converse is not true.  We can apply transmission line theory all we want at LF -- that want is just not very much, because nothing is gained from the higher complexity, while essentially identical results are obtained from the lumped-equivalent case (or indeed the DC case).

Put another way:
DC equivalent: works at DC only.
Lumped equivalent: works from DC up to frequencies where the wavelength is comparable to line lengths.
Transmission line equivalent: works from DC up to frequencies where the wavelength is comparable to line spacing.
Full fields: DC to light.

Just because we might find some of the apparent insights unsatisfying (e.g., a 12V battery is constantly emitting 12V into its terminals' transmission line equivalent, so, say that happens to be Zo = 200ohm, well that's a continuous 12V/200ohm = 60mA outflow; which is immediately reflected back, in phase, for a 60mA inflow, balancing to zero net, and thus zero power loss), doesn't mean it's inapplicable!

Tim

[vad]

How does a transmission line know it is too long or too short?

The transmission line must have gone to the same Uni where they do not teach about dV/dt and dI/dt

[rfeecs]

Does the transmission line really work from DC to daylight?

At low frequencies, down to audio frequencies, the impedance of the line starts to change because the distributed resistance of the line will become higher than the distributed inductive reactance:

http://k9yc.com/TransLines-LowFreq.pdf

This is what caused problems for the telegraphers.

At high frequencies, you can have higher order modes of propagation.  For example if the spacing of the line is one half wavelength there would be some strange resonance behavior where the wave just bounces back and forth between the two conductors and doesn't propagate down the line.  I'm not sure what this would do for this example, but maybe, if you had a very fast edge when you threw the switch,  there could be some "ringing" as the wave bounced back and forth.

[rfeecs]

Another example of the apparent craziness of applying the Poynting vector to DC:

Take two circuits with batteries and resistors.  Put the resistor of one circuit very close to the battery of the other circuit:



It appears that energy is flowing across the gap from the battery to the resistor.  Yet obviously, if you disconnect the battery on the right, it doesn't affect the circuit with the resistor on the left.  The resistor still dissipates the same amount of power.

[T3sl4co1l]

Does the transmission line really work from DC to daylight?

At low frequencies, down to audio frequencies, the impedance of the line starts to change because the distributed resistance of the line will become higher than the distributed inductive reactance:

http://k9yc.com/TransLines-LowFreq.pdf

This is what caused problems for the telegraphers.

My point exactly.  They'd have never figured it out without the transmission line (Telegrapher's) equations.  That the impedance is dependent on frequency, does not invalidate this fact.   (It might work differently as the relative significance of LRCG varies, but it doesn't stop working as a transmission line!)

Honestly, I'm not sure which way you meant that, as an attack or defense of the claim; normally when one poses a "does it really?" the implication is "no", but you might've been going for the reversal after all?  I feel it reads either way...

At high frequencies, you can have higher order modes of propagation.  For example if the spacing of the line is one half wavelength there would be some strange resonance behavior where the wave just bounces back and forth between the two conductors and doesn't propagate down the line.  I'm not sure what this would do for this example, but maybe, if you had a very fast edge when you threw the switch,  there could be some "ringing" as the wave bounced back and forth.

Yes, there can be waveguide or free propagation modes.  In this case, we can expect something like a facing pair of end-terminated dipoles, so there will be some reflections between them (and any surroundings if applicable); standing waves likely won't be a big deal as the lamp provides termination to one.

It's actually quite reasonable to expect this in the given experiment, as mechanical switches typically have sub-ns edge rates, and the propagation is multiple ns.

Tim

[MIS42N]

dc switch on of a transmission line.

http://wcchew.ece.illinois.edu/chew/ece350/ee350-12.pdf

There is a (not so) subtle difference between the scenario in the pdf and the circuit in the video. In the pdf, the battery is connected to both wires of the transmission line. So current flows in both wires in opposite directions and satisfies the transmission line concept.

In the video, only one terminal of the battery is connected to the 'transmission line' and the other terminal is connected to another 'transmission line'. There is no opposite direction current when the switch is closed. Does not satisfy the criteria for a transmission line.

A better model is a folded dipole antenna, where the input impedance is a function of both transmission lines and dipole antenna.

[bdunham7]

Just because we might find some of the apparent insights unsatisfying (e.g., a 12V battery is constantly emitting 12V into its terminals' transmission line equivalent, so, say that happens to be Zo = 200ohm, well that's a continuous 12V/200ohm = 60mA outflow; which is immediately reflected back, in phase, for a 60mA inflow, balancing to zero net, and thus zero power loss), doesn't mean it's inapplicable!

I'd call that example mathturbation.   Your mathematical model works to produce an apparently correct result under some theoretically ideal conditions, but does not actually reflect 'what is really going on'.  It isn't that it is unsatisfying, it's that the model fails as soon as you introduce any real components.  The full transmission line equations do depend on frequency and thus any transmission line where either the conductance of the dielectric or the resistance of the wires is not zero, the characteristic impedance rises towards infinity (or the leakage conductance of the dielectric if not zero) as frequency goes to zero and then becomes a divide-by-zero error at DC.

As an educational aside, one of my gripes that I've repeated about the original video is how it is confusing rather than 'mind-blowing'.  Someone looking at your example might easily confuse your fictitious continuous reflected current as something similar to the recirculating current in an AC system with a power factor of less than 1, and then if they see that battery connected to a length of actual 300-ohm twin-lead, they would assume that the resistance of the wires would cause a continuous dissipation of energy.  If you send a pulse down that same line and it is reflected back, you do have a loss due to copper resistance.  You don't in the DC case.  Why?  Only with zero-resistance wires can you just 'imagine' as much current as you like without consequences.

There are times when it might be helpful to model a battery + wires in this way, such as when the circuit is subject to perturbations.  I have no problem with that.  I do object to it being touted as a 'better' model or worse, 'the correct' model.  My model is that when connected to the 'transmission line', charges rearrange themselves until the distribution is such that there is a 12V potential between the conductors and a static electric field reflecting that is established and then the whole thing just sits there doing nothing.  When you introduce real components--wire resistance and dielectric conductance, my model is easily adjustable and doesn't break. 

Shrug.  To anyone less familiar with the subject, a practical answer to your version is fairly trivial -- within a cycle, less than the blink of an eye.  Who could care whether it's 100ns or 10µs, right?

Actually 2000 miles would be approximately 1 cycle, which is why I picked the number.  Now imagine we find a shorter route that is 1000 miles and use that instead, but just for redundancy we connect them in parallel.  This was also sort of a trick question--indeed I could care less if my lights take an extra 8 or 16ms to turn on.  But I do care about connecting my power lines out of phase.  So I need to know if the power goes through (or along) the wires or directly through space?  What does Derek's video indicate to a viewer that is much less advanced than you are?

[bpiphany]

Another example of the apparent craziness of applying the Poynting vector to DC:

Take two circuits with batteries and resistors.  Put the resistor of one circuit very close to the battery of the other circuit:



It appears that energy is flowing across the gap from the battery to the resistor.  Yet obviously, if you disconnect the battery on the right, it doesn't affect the circuit with the resistor on the left.  The resistor still dissipates the same amount of power.

Don't the magnetic fields cancel out in this plane? Hence, the Poynting vector flow across that plane being equal to zero?
Edit: Actually, they add up, right? Vectors are 1D too many for my brain... There's of course still Poynting pointing into the resistor in the right part. So in the complete system energy is still transferred there. I see no real problem with the energy flowing into the left part resistor from the right part battery as long as energy conservation holds. There is no net energy flowing through that infinitely large plane.

[vad]

For ages people thought they can fool conservation of energy law and invent over unity device. Some are still trying.

With Veritasium’s viral video, new fun begins. People now think they can outsmart Poynting’s theorem.

[T3sl4co1l]

dc switch on of a transmission line.

http://wcchew.ece.illinois.edu/chew/ece350/ee350-12.pdf

There is a (not so) subtle difference between the scenario in the pdf and the circuit in the video. In the pdf, the battery is connected to both wires of the transmission line. So current flows in both wires in opposite directions and satisfies the transmission line concept.

In the video, only one terminal of the battery is connected to the 'transmission line' and the other terminal is connected to another 'transmission line'. There is no opposite direction current when the switch is closed. Does not satisfy the criteria for a transmission line.

Oh, no equal and opposite reaction, right. I'll just throw out literally every circuit I've ever made.  Their operation must be a total illusion.

I'd call that example mathturbation.   Your mathematical model works to produce an apparently correct result under some theoretically ideal conditions, but does not actually reflect 'what is really going on'.  It isn't that it is unsatisfying, it's that the model fails as soon as you introduce any real components.

I've tried to understand what your position is, and seem to have failed hopelessly.  Perhaps you can provide a fields-based solution / proof of what you are actually getting at?

Basic arithmetic is "mathturbation" now?  Guess I better hide my calculator when guests come over... 

The full transmission line equations do depend on frequency and thus any transmission line where either the conductance of the dielectric or the resistance of the wires is not zero, the characteristic impedance rises towards infinity (or the leakage conductance of the dielectric if not zero) as frequency goes to zero and then becomes a divide-by-zero error at DC.

Call up Playboy, that's hard core algebra at least!...

Fortunately, for most EE purposes, any infinity is equivalent, corresponding to the reciprocal of a number approaching zero.  The fact that that number has zero magnitude is all that matters: if the conductance is zero, then no power flows, and the solution is even simpler than for finite Zo!

As an educational aside, one of my gripes that I've repeated about the original video is how it is confusing rather than 'mind-blowing'.  Someone looking at your example might easily confuse your fictitious continuous reflected current as something similar to the recirculating current in an AC system with a power factor of less than 1, and then if they see that battery connected to a length of actual 300-ohm twin-lead, they would assume that the resistance of the wires would cause a continuous dissipation of energy.  If you send a pulse down that same line and it is reflected back, you do have a loss due to copper resistance.  You don't in the DC case.  Why?  Only with zero-resistance wires can you just 'imagine' as much current as you like without consequences.

Good point -- this means it's not just circumstance, but necessity, that a nonzero-R, G ~= 0 transmission line have Zo --> infty.

So, we can still imagine propagating waves all we want, and because their energy is zero no matter the amplitude, we have no power dissipation!

There are times when it might be helpful to model a battery + wires in this way, such as when the circuit is subject to perturbations.  I have no problem with that.  I do object to it being touted as a 'better' model or worse, 'the correct' model.  My model is that when connected to the 'transmission line', charges rearrange themselves until the distribution is such that there is a 12V potential between the conductors and a static electric field reflecting that is established and then the whole thing just sits there doing nothing.  When you introduce real components--wire resistance and dielectric conductance, my model is easily adjustable and doesn't break.

Your model is strictly DC then?  Good for you.

What's it do within a few seconds of hitting the switch?  If your answer is "undefined", then like... why are you in this thread in the first place, man?

Actually 2000 miles would be approximately 1 cycle, which is why I picked the number.  Now imagine we find a shorter route that is 1000 miles and use that instead, but just for redundancy we connect them in parallel.  This was also sort of a trick question--indeed I could care less if my lights take an extra 8 or 16ms to turn on.  But I do care about connecting my power lines out of phase.  So I need to know if the power goes through (or along) the wires or directly through space?  What does Derek's video indicate to a viewer that is much less advanced than you are?

What level of advancement are we talking about?  If they are wholly unqualified (i.e. highschool electronics or less), someone made an extremely negligent hiring choice and the material solution is irrelevant; a higher priority practical solution exists.  If they are minimally qualified, then they will simply work with the transmission line equations they were given.  Transmission lines at AC steady state are just trig, plug in the length and parameters and out comes the reactive power, and so you need line reactors and switchgear here, here and here to keep the voltage stable.  Or for that matter, you can't transmit AC over that distance, you use a DC link anyway.  It doesn't even need understanding, it's algorithmic, a solved design process; but if they have an understanding of waves propagating (and potentially reflecting and standing) on transmission lines, that's fine, and likely to lead to better corrections when confusion is found or mistakes made.

Tim

[MIS42N]

dc switch on of a transmission line.

http://wcchew.ece.illinois.edu/chew/ece350/ee350-12.pdf

There is a (not so) subtle difference between the scenario in the pdf and the circuit in the video. In the pdf, the battery is connected to both wires of the transmission line. So current flows in both wires in opposite directions and satisfies the transmission line concept.

In the video, only one terminal of the battery is connected to the 'transmission line' and the other terminal is connected to another 'transmission line'. There is no opposite direction current when the switch is closed. Does not satisfy the criteria for a transmission line.

Oh, no equal and opposite reaction, right. I'll just throw out literally every circuit I've ever made.  Their operation must be a total illusion.

Oh grow up. I did not say equal, I did not say reaction. What I said was for half a second the circuit does not satisfy the criteria for a transmission line. You may take that as it doesn't work, but there are a surprising number of circuits that work even when their designers don't take every factor into account.

Most situations have a two way flow of electrons, the video scenario is unusual because of the folded dipole configuration. If a similar situation were encountered on a PCB using tracks 30mm long the transient situation would exist for 100ps. Most bench scopes wouldn't see it. And it may not impinge on the performance of the circuit anyway. And why would you create a dipole on a PCB unless you wanted one.

You seem to have gone from rational consideration to troll mode. Shouting down a considered opinion resolves nothing.

[snarkysparky]

The circuit has two transmission lines.  Otherwise the bulb would have no current after 1/c   seconds

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[Domagoj T]

A guy did an actual test.

[SiliconWizard]

I still don't know why you'd need 1 km of wire to measure that. Except of course to make clickbait YT videos.

[lapi]

Transmission line equations do not apply when the frequency applied is significantly longer than the line length.

Can you please provide a reference that confirms your statement? I don’t see this restiction in the derivation of transmission line theory. Are you aware of the concept of bounce diagrams?

Transmission line theory is widely used to analyse power systems, where the line length can be a fraction of the wavelength.

Wikipedia " The term applies when the conductors are long enough that the wave nature of the transmission must be taken into account." - DC does not have a "wave nature" although transients do. Which is why I say the transmission line model is insufficient. It handles the transients OK, but that isn't the whole of reality.

The statement that the transmission line behavior requires wavelengths short relative to the length of the line is correct, but the statement that this is a DC problem is not. This is a transient analysis, and steeper the transient, shorter the high-frequency end wavelength. That is why the transmission line analysis fits perfectly this problem, and also explains what is seen in the experiments described in this thread before, and in the recent nice experiment by AplhaPhoenix in

[nixxon]

I still don't know why you'd need 1 km of wire to measure that. Except of course to make clickbait YT videos.

How long a wire do you think is enough to measure the things that are about to be measured? And at what (relative and absolute) distances would you connect the components in the electrical circuit?

[adx]

I don't understand quite what the confusion over transmission line theory applicability is, I say confusedly. Nor what is being said here.

Transmission line models can be used at any length, 1% of wavelength is fine if wanting to determine some small phase shift, I would have assumed without a second thought. A singe stage lumped model won't properly show the wideband response (ignoring things like stop bands is an 'engineering' solution, fine, but sometimes it's just going to be easier to treat everything as a transmission line and be done with it, like some power systems analysis software I vaguely remember). SandyCox is right.

Transient analysis on a linear time invariant system is the same as a frequency domain analysis - same results, apart from numerical error.

Perhaps it's a problem with semantics, people don’t so much seem confused, as explaining things in different ways but words missing the mark sometimes. Or there is significant background confusion because like in my case I left university not fully understanding what I evidently must have learnt to pass enough exams to graduate. They did teach the fundamental concepts well, and gave the maths a good treatment too I assume, but they didn't adequately connect the two together for someone as atrocious at maths as me. Leaving small but significant gaping holes in my understanding. Or believing in contradictory lessons. Or a combination of that all, for people in different proportion for a different mix of preventable deficits.

No, this example (Veritasium's) is not purely transmission line, because the common mode is a dipole antenna, circuit is a great big terminated folded dipole as I said. Not my forte, so I guess, but a guess is often perfectly adequate for EMC where 6dB is no big deal (voltages and currents -50% +100%). There will be radiation loss. I don't understand EM radiation very well, but something has to happen: Imagine if the wires stopped at the far end and never came back to a lamp, what you've got there is a sort of a paradox. You close the switch and what? Do the wires instantly charge up to the battery voltage taking zero current because they are not connected in a circuit? Does the voltage travel down each wire, drawing a current pulse only from the close ends because capacitance falls off at a distance? Does it speed up as it gets further away? What if it's already going the speed of light? A lot of this is near field stuff, but there comes a point when even the most unobservant person is going to say "hang on, how does this work?".

[MIS42N]

I really liked AlphaPhoenix video, the behavior is about what I expected. What I would like to see is a calculation that shows the induced current should be 200uA as found experimentally. By modelling the circuit as a capacitor, 2 parallel plates 250mm apart and 1000mm area (1mm x 1m) per meter of length the calculation yields about 50uA. The wires may not be 1mm diameter but they aren't 4mm so capacitive coupling doesn't explain it all. The only other culprit is inductive coupling and my maths isn't up to that. Clearly there is inductive coupling, as shown when the ends are cut. The reflected wave caused by the collapsing magnetic field gives a big spike before decaying to zero.

[bdunham7]

Transmission line models can be used at any length, 1% of wavelength is fine if wanting to determine some small phase shift, I would have assumed without a second thought.

The full set of equations can be used in that instance, with all of the relevant parameters put in.  But that would often be more work than necessary and so the common, simple reduction of a transmission line to a single, uniform characteristic impedance needs to be qualified as to what signals that model works with.  Then you can ignore a lot of complication by deeming things like the resistance of the copper or conductance of the dielectric to be 'negligible', at least in as a first order approximation.

[adx]

I haven't quite decoded what's happening in the AlphaPhoenix video.

It appears he's gone for a good impedance match with 1k far resistor (light bulb) and about 500R (transmission) lines. I didn't catch what the sending end shunts are, but not that far off from the voltage - maybe 470R each?

In that case I'd expect more of an initial step. There appears to be a lot of frequency dependent? resistance? in the reflection - the line will be quite resistive (skin effect), but this doesn't seem to affect the 'charge' (first microsecond). So maybe there is more going on than at first guess.

There is a lot of common mode reflection visible (after 1us) in the purple and cyan traces bouncing up and down. The lines are not being driven in a balanced manner because of the scope probe ground on the driver. Because the lines are referenced to ground (literally) my instincts say the same desired result could be achieved by having the circuit drive only 1 leg of the setup to a ground reference. It sort of follows that to replicate Veritasium's setup better, then the lines should be vertical (into the sky - he's got a drone up there), using this ground plane. And sky drone.

Interesting to see people set up these real tests.

[T3sl4co1l]

The initial step is sharp, because it's representative of differential line impedance; it sags slightly due to complex line effects (ground proximity, CM-DM conversion and CM radiation/absorption).  The reflected wave is significantly attenuated and dispersed (due to the same effects).  That is to say, there's a small amount of those effects reflected back to the source and thus visible in the initial step, but it's a fairly small amount; the loss is primarily in the forward wave, and then reflected back (taking double loss/dispersion as well).

I don't think skin effect is significant here: the frequency is much higher than the cutoff of wire that size (i.e., the conductor is well into full skin effect), and the difference in effective magnetic field volume / distance between current paths is extremely tiny compared to the on-centers distance between them.  That is, over the relevant frequencies, the skin depth might differ by 0.1mm or so, negligible compared to the 250mm separation distance.  I mean it's an 8-bit scope (or maybe a few more, but we don't really get more than 9-ish bits of vertical resolution, or more like 7 for any closeups shown in the video I think?), it's literally not a difference that can be measured.  Also, generally speaking, twin-lead has very low losses thanks to its high Zo, so we have general justification for ignoring copper losses here.

It does seem weird to me, that the initial step is rather small?  It should be more like half the final amplitude, if it's well terminated.  (And yeah, it seems reasonably well terminated, given that attenuation and Zo varies with frequency due to the above effects; hence why you see so much slop and bounce when the wave returns, different frequencies are matched better or worse.)

There is a lot of common mode reflection visible (after 1us) in the purple and cyan traces bouncing up and down. The lines are not being driven in a balanced manner because of the scope probe ground on the driver. Because the lines are referenced to ground (literally) my instincts say the same desired result could be achieved by having the circuit drive only 1 leg of the setup to a ground reference.

A caution on "ground" -- he's on a mains circuit, somewhere.  Presumably an extension cord; how long isn't clear.  Even if it's an installed circuit, it's going some distance through buried conduit, so will have a CM equivalent like a low-Zo transmission line (perhaps 30-50 ohms -- consider all the wires in the conduit acting in parallel with respect to the conduit wall, be it metallic conduit or the surrounding earth).  That's actually not too bad of a ground with respect to the system in question (~250 ohms), but far from ideal, and may show up in measurements.  Assuming of course the scope is grounded to its mains circuit (and that is grounded in turn).

And yes, the system should be bisectable by symmetry.  It's not obvious whether his pulse generator is grounded symmetrically or not; if it's battery powered, and probed differentially just like the other side, then that's okay -- there's just the small loading effect of the board and battery itself, a few cm across, who cares.  As he says, the source isn't nearly sharp enough to show such effects (regarding line spacing specifically, but same applies to anything else of comparable dimension).

Besides saving on wire length, a bisected (unbalanced, middle-grounded) setup has the advantage that it's easier to measure the common mode.  Look at the sum and difference of currents on the TL, and there you have it.

Mind, you will need a good ground in that case; a ground rod is probably good enough, but a scale model can also be made entirely metallic, given a sub-ns generator and just some foil, or metal table or whatever.  This removes earth loss from the system, simplifying the response.  It can even be enclosed, reflecting would-be radiation, in which case you get the reflected waves in both CM and DM -- the system reduces to a twinax cable.  And, now we can model the system quite effectively even in say SPICE, as the sum of two transmission line models.

Tim

[adx]

Yes, not skin effect. The 1us rise after the reflected pulse arrives is far too short for some sort of frequency dependent tilt now I look. The reason to mention it was the web calculator I found to calc impedance also gave resistance working out to somewhere around 3k at 1MHz, which seemed high. (Also I meant 1.6us for the first step - was looking at the screenshot's "1.00us/" as an annotation without thinking.)

That slop might also be a common mode (from ground proximity) getting into the effective drive voltage, since the sides are not balanced. Makes less sense now I say it. What you say makes more sense than my first thoughts, I'd usually not put too much thought into them and tinker with the setup for an answer, but I don't have a farm and it's dark outside. It's also odd that only the first reflection appears in the common differential mode.

I found his circuit diagram at 7:30 (still haven't watched it completely) after noticing it on his screen. All resistors 1k, wire 24 AWG, and no need to go deciphering trace colours by cleverly looking at scope and the setup! I was guessing the mains might be an inverter in one of the cars in one of the wide shots I briefly saw, or as you say long extension cord. In either case on the ground a fair distance. Guessing the supply to the generator is a small SMPS plugged into that power board, so is better isolated than the scope.

If the scope were truly isolated (or ground lifted, depending on where EMC caps go) then the green trace should rise sharply more like the yellow.

The reason I got to thinking about the bisected test is that it wouldn't work for Veritasium's situation (wires to space), but would for this (wires over ground plane). Hence the vertical idea, which should work for both, give or take some dirt? I've seen someone in some YT video (years ago) wiring up some massive chicken wire ground plane for a test like this, and finding it made little difference I think.

[bdunham7]

It does seem weird to me, that the initial step is rather small?  It should be more like half the final amplitude, if it's well terminated.  (And yeah, it seems reasonably well terminated, given that attenuation and Zo varies with frequency due to the above effects; hence why you see so much slop and bounce when the wave returns, different frequencies are matched better or worse.)

For a wire of radius r=0.25mm, separation of 250mm and distance from ground of 1000mm I get:

Z₀ ~ 600R  (simplified model, assumes wavelength shorter than TL)

Capacitance w-w ~4pF/m, w-ground (each) ~6pF/m  (assumes ground is a conductor)

Mutual inductance w-w ~2.8uH/m  w-ground (each) 1.8uH/m

Self-inductance w 1.6uH/m.

DC Resistance of wire ~21R per leg (250m) or ~84R for the entire loop.

That's all making assumptions ranging from stated or obvious to probably not really correct.  The DC steady state result is just 5V/3084R or ~1.6mA.  The initial 1.6uS of response actually seems pretty flat, and if you just use the basic transmission line characteristic impedance of 600R, you might expect 1.2mA somehow, but in practice there seems to be a lot less.  I'm not yet convinced that model works even in an ideal model, but in any case the losses from coupling to ground look significant.  I also don't see the multiple steps predicted in other models and experiments and I would guess that this actual version is sufficiently well damped and lossy that the reflections subside quickly.

If one wants to look at the initial step response, the first few 10s of ns, a 5GSa/s scope would probably do, but you'd need a lot more attention to the geometry and things like the self-capacitance of the components themselves. I didn't calculate the self-capacitance of the wire because it is going to be a few fF/m actually ~4pF for 1 meter, 330pF for the whole 250m, self-capacitance apparently doesn't scale linearly even for a thin wire, but the self capacitance of an oscilloscope and the other components may matter quite a bit.

[adx]

My calcs went awry, I had gone searching for a differential microstrip calculator but found this instead - the page I used in my lazy dash to a result:

https://cecas.clemson.edu/cvel/emc/calculators/TL_Calculator/index.html

The 3k line resistance I found is from somehow leaving the default 300e6 Hz in the calculator. Maybe didn't press enter. (Assuming the 2 in the formula counts both wires ie multiply by 250m per leg, here I count a leg as 2 wires.)

So there may be something in the skin effect effect after all:
At 30Mhz, 449.4R(AC) per leg
At 3MHz, 142.11R(AC) per leg

Exacerbated by it using the simple Z0=sqrt(L/C) formula. This comes down to instead using a lossy transmission line model in a simulator with "all of the relevant parameters put in", making it easy to check if it has this same effect on the leading edge of the first reflection over the first ~1us, while leaving the initial step alone (it does have a very slight lift).

I think my thought draws from memories of SMPS transformer workings, perhaps proximity effect. Those memories getting a bit long in the tooth, so it's less a job for a calculator on the web, and more one for hard work. Or a farm and drone.