EEVBlog 1439 - Analyzing Veritasium's electricity video - Page 1
Forum section: electronics / general

author: Sredni
author_rank: Frequent Contributor
timestamp: November 24, 2021, 03:15:57 pm
content: Dave, you should correct your video, the errors are too big.At minute3 somethingops 9:30 you say that at DC the Poynting vector would be pointing towards the battery. This iswrong.At minute 41 something you show a picture of a resistive wire from Feynman that shows the Poynting vector going inward, into the wire. Well, that's a resistor, power is getting in to be dissipated.Even at DC the Poynting vector is going out from the battery, then goes nearly parallel to the copper wires, and then finally points inside the resistor. You misread Feynman (who, by the way, for fig 27-5 explicitly talks about 'resistive wire' which has a resistance and a voltage drop).Simple experiment:9V battery shorted by copper wire: the wire gets hot. Poynting vector pointing out of the battery and into the wire. (the wire is the dissipative element in this context).9V battery with a 100 ohm resistor connected via copper wires: the resistor gets hot, the wire doesn't. Poynting vector directed out of the battery, nearly parallel to the wires, and then plunging into the resistor. As correctly shown by Veritasium (for DC)And no, skin effect has nothing to do with that.EEVblog 1439 - Analysing Veritasium's Electricity VideoDave analyses Veritasium's video "The Big Misconception About Electricity" and how energy flows in the Poynting vector in the electromagnetic field OUTSIDE the wire instead of inside the wire.00:00 - Veritasium's video "The Big Misconception About Electricity"00:32 - Rection to the points in the video01:11 - This is a bit MISLEADING!02:28 - Electron drift03:51 - Engineers use different tools and theorems04:27 - Every electrical engineer knows this05:17 - Everything he says is correct08:24 - What is current?09:30 - He doesn't address this in the video. Poynting vectors at DC11:12 - How the lightbulb works12:41 - At the physics level, it's correct14:11 - My only problem with this is...15:08 - Is it just an academic discussion?16:17 - The undersea cable is just early transmission line theory17:20 - So what is the answer to the question?22:06 - What about skin effect and DC?25:44 - Let's simulate this and answer the question29:18 - Transient analysis33:00 - DC Steady State analysis34:28 - The quantitative values don't matter38:25 - But what about DC steady state?40:24 - What does Richard Feynman think?

author: golden_labels
author_rank: Super Contributor
timestamp: November 24, 2021, 04:11:27 pm
content: One thing Derek was not wrong about: that video will spawn a deluge of comments, with everyone correcting everybody else.Has anyone so far realized that the video ventured deep into the territory of epistemology and metaphysics? A large part of the whole discussion is caused by making assumptions on the meaning of “truth”. To be more specific: confusing models with absolutely true essence, the inherent nature of the matter discussed. Neither of those explanations is more true. At best they may differ in their practical usefulness or how close they are to the most detailed views of the phenomenon. The Poynting vector approach is closer to the more detailed model, but by no means closer to any kind of truth.Both explanations may be misunderstood and lead to wrong interpretations. It’s not hard to imagine that electrons’ movement play completely no role, after watching just that Veritasium video, if one doesn’t ask a question: how comes magnetic field is non-zero? Also the energy transfer lines (yellow) are reduced to direction only and miss magnitude, which — if considered — would draw a less surprising picture.

author: Sredni
author_rank: Frequent Contributor
timestamp: November 24, 2021, 04:36:42 pm
content: Well, the role of surface charge is very often not mentioned in most introductory books. Chabal and Sherwood have started a new trend in that sense. But one thing is trying to analyze the phenomena with different tools (propagation of fields and redistribution of surface charge, use of transmission line model in the transient phase, even antennas if you will) and another is getting basic physics completely wrong.To say that at DC the Poynting vector is directed toward the battery, or even towards the wires when the complete circuit has a resistor that will sink the power, is just plainly, unmistakenly and uncontroversially wrong.It's not a matter of intepretation. It's wrong.(I corrected the minute in the video were Dave says that)

author: Amaruk
author_rank: Contributor
timestamp: November 24, 2021, 05:11:43 pm
content: Quote from: Sredni on November 24, 2021, 03:15:57 pmI am a big fan of Dave and have watch so many of his videos and I am always amazed by the amount of knowledge shared in them. Thank you! This video is a bit different though but it was kind of forced upon Dave by his viewers. The result of this is that this video does not feel as solid as the others I have watched... But who cares, we all learn stuff on here and who has not said things that are not correct when put on the spot anyway? These are difficult theoretical discussions so it just shows that Dave is human after all!

author: Kalvin
author_rank: Super Contributor
timestamp: November 24, 2021, 05:59:38 pm
content: When people are saying "At DC this and that ..." what do they really mean?- Do they mean the static situation when there is an electric field present but no current is flowing (DC = 0)? In this experiment: Battery connected to the circuit, but no closed circuit present, circuit is in steady state. Because there is no charge moving in the circuit, there is no magnetic field present, thus there cannot be any Poynting vector.- Do they mean that there is an electric field present and there is a  current flowing in the circuit (|DC| > 0)? In this experiment: Battery connected to the circuit, switch closed, and constant current flowing into the load [after transitions have disappeared ie. circuit is in steady state again). Now, as there is charge moving in the circuit creating a magnetic field, there is a Poynting vector, too. The Poynting vector will show the direction of the energy transferred. In this experiment, and if the signs are correct, the direction should be from the battery into the load.When the conductor is lossless, the Poynting vector is parallel to the conductor.How about a situation when there is a good conductor, but somewhat lossy nevertheless? If I understood Feynman correctly, the Poynting vector should tilt a little towards the conductor, creating two vector components. The first component is parallel to the conductor, and which is transferring energy towards the load. And the second one, which is into the conductor, representing the losses.Makes any sense?What is the situation when we have a resistive wire connected directly on the battery, ie. the load is this resistive wire? This should not be solved using a traditional lumped model, because the load is now distributed along this resistive wire: How the Poynting vector is represented along the wire in this case?

author: jan.met
author_rank: Newbie
timestamp: November 24, 2021, 06:42:19 pm
content: Hi,I think answer D for the switch question is wrong. It would be true, if the electric field would build up instantaneouly along the wire, so that current starts flowing immediately through all of the wire. But the change in the field itself is propagating only at the speed of light along the wire, so that the current build up in the wire takes time. Without current at the bulb, the Poynting vector is zero and the bulb can not receive energy.Greetings, Jan

author: Sredni
author_rank: Frequent Contributor
timestamp: November 24, 2021, 06:51:13 pm
content: @Kalvin - "At DC" means in the steady state condition with a current flowing. Now, Veritasium's video is about the transient before DC, but as you correctly stated, in DC with a steady current there are both a magnetic and an electric field.That are directed as shown by Veritasium and that give the direction of the Poyntin vector shown by veritasium.So, near the battery the Poynting vector is directed outwards, then it 'bends' to follow the wires and finally plunges into the load.Your analysis of the importance of the resistivity of the wires is correct.If the wires have zero resistance, the electric field immediately outside their surface is orthogonal. If they have resistance, the electric field will be slanted. The higher the resistance, the higher the voltage drop, the higher the electric field inside the component. This will cause the Poynting vector to point toward the component, in this case the wire. But you need to take into the picture what else is in the circuit. If there is a load resistor, the current will be limited, so the B field will be lower and the bending in the low resistance wires will be only slight (because compared to the load, they drop nearly zero voltage); in the load, the field is strong and even with the same B field going around it, this will give a Poynting vector directed towards the resistor.The resistivity of the wires determine how much the Poynting vector is slanted - how much power is dissipated in the lossy wires.The bigger the conductor, the less the slant.

author: Sredni
author_rank: Frequent Contributor
timestamp: November 24, 2021, 07:08:56 pm
content: Quote from: jan.met on November 24, 2021, 06:42:19 pmNaaa, it's roughly correct.Current in the wires build up gradually, as the surface charge adapts to the new conditions. When you flip the switch, the charge that had accumulated at the open switch terminals starts to recombine and produce a complex distribution of surface charge that gives the electric field inside the conductor.Have a look at my answer on SEhttps://electronics.stackexchange.com/questions/532541/is-the-electric-field-in-a-wire-constant/532550#532550in particular the last picture, where the process of "E field forming inside the wire" is shown.According to this picture (which consider the rest of the circuit so far away as to not influence the charge distribution) we should wait for the perturbation to go halfway to the moon and back to reach the lamp. Surface charge is extremely fast and the perturbation travels nearly at the speed of light.But the surface charge is also influenced by nearby portions of the circuit. As you can see from the different configurations the electric field can assume near the surface of a conductor (and highlighted in the simulation shown in that answer - from a paper reference by Veritasium, as well).In fact, the charge that is redistributing near the switch produces an electric field that travels across the 1 meter space between the wires to the lamp and the nearby portion of wires. This will try to shape the electric field inside those conductors to the final steady state, but it still lacks the contribute from the charge redistribution that is traveling towards the moon and back.But nevertheless, the local surface charge near the switch is affecting (after 1/c seconds) the surface charge near the lamp (and consequently the field inside the wire in that portion).What will be the magnitude and the direction inside the wire in that transient? I am not sure, it might be complicated. A simulation might be required. And you need to take into account that you cannot use the inductance per unit length when the steady state is reached. Probably only the internal inductance counts until the full current is flowing in all of the circuit... I would love to see a simulation.

author: firewalker
author_rank: Super Contributor
timestamp: November 24, 2021, 07:33:42 pm
content: The capacitor explanation is somehow misleading? What if the lamp was also 300000 km away from the battery? The capacitor is still there.Alexander.

author: Kalvin
author_rank: Super Contributor
timestamp: November 24, 2021, 08:14:39 pm
content: In my opinion using a simple transmission line-model is sufficient for explaining Veritasium's experiment. In addition to that, knowing that it would take 1m/c seconds (3.3ns) before the change of electrical/magnetic field propagates through the distance of 1m will provide the answer.Quote from: firewalker on November 24, 2021, 07:33:42 pmUsing a capacitor as an explanation is not a good model, because it will not explain for example why the lamp/load will not get the full power right after the switch is closed in Veritasium's original circuit. A transmission line and its characteristic impedance will provide better and more realistic model and correct answer.Answer to your question: It would take approx. one second or so before the lamp will be lit / current starts flowing through the load. The actual delay is depending on the cable's signal propagation speed.Pls find my simulation below. I run the simulation for 20 seconds in order to visualize the reflections, and how the circuit is gradually converging towards the steady state.Edit: Please note that I have used a 50 ohm transmission line as the model in my simulation. In Veritasium's experiment the characteristic impedance of the transmission line will be different, and the transient current through the load will be different, but the principles and the explanation will remain the same.

author: Kalvin
author_rank: Super Contributor
timestamp: November 24, 2021, 09:02:32 pm
content: For completeness: Here is a simulation for Veritasium's experiment using a 50 ohm transmission line and a 1 Kohm transmission line. It can be seen that when the characteristic impedance of the transmission line is high, it will take much longer time before the system converges to the steady state.

author: Unixon
author_rank: Frequent Contributor
timestamp: November 24, 2021, 09:16:15 pm
content: Dave, how could you put the ground symbol on the wrong side of the switch?This breaks the definition of the ground reference point.

author: Kleinstein
author_rank: Super Contributor
timestamp: November 24, 2021, 10:17:15 pm
content: Quote from: Kalvin on November 24, 2021, 09:02:32 pmIs there something like a 1 KOhms transmission line ? I though there is a kind of natural limit at some 370 ohms.

author: bdunham7
author_rank: Super Contributor
timestamp: November 25, 2021, 01:25:36 am
content: Quote from: Kleinstein on November 24, 2021, 10:17:15 pmThe impedance of free space is 377 ohms or something, but yes you can have characteristic impedances higher than that without any problem except the increasingly improbable dimensions (small wires, large spacing).

author: Kalvin
author_rank: Super Contributor
timestamp: November 25, 2021, 02:32:11 am
content: Quote from: Kleinstein on November 24, 2021, 10:17:15 pmYes, it is possible to build practical 450 ohm or 600 ohm open-wire ladder line transmission lines:https://www.qsl.net/co8tw/openline.htm.Parallel feeders go back to the beginnings of radio. By 1930, the "two-wire untuned feeder system" was a standard ARRL Handbook feature. The Jones Radio Handbook of 1937 provides a table of line losses showing the advantages of open-wire feeders (a 440-Ohm line in the table) over lower impedance twisted-pair feeders (p. 70). The use of 600-Ohm lines was fairly standard, using a spacing of about 6". "To reduce radiation from the feeders to a minimum, the two wires should not be more than 10 to 12 inches apart." (The Radio Amateur's Handbook, 7th Ed., ARRL, 1930, p. 162) Rarely did hams exceed the 6" spacing.I wasn't really paying attention to whether 1 kohm characteristic impedance is practical or not with, because in this context we are talking about theoretical, lossless 300 000 km transmission line anyway.But yeah, I should have probably used 450 ohm or 600 ohm transmission line impedance in my last simulation.

author: David Hess
author_rank: Super Contributor
timestamp: November 25, 2021, 04:06:53 am
content: Dave did not mention it exactly, but while the transmission lines are charging, which powers the bulb immediately, they are the equivalent of a resistance equal to their characteristic impedance. I have occasionally run across circuits which took this into account, like sampling circuits which use a transmission line instead of a capacitor.It seems that Mehdi (Electroboom) is going to do a video about it also.

author: golden_labels
author_rank: Super Contributor
timestamp: November 25, 2021, 10:41:23 am
content: Quote from: firewalker on November 24, 2021, 07:33:42 pmThen the wires in the Battery-C-Lamp-C loop would be 300Mm long. Dave has clearly mentioned it multiple times, that this is a situation with them being 1m long. If that would be 300Mm, then the signal would have to propagate over those 300Mm-long capacitors and it would increase time.My gripe with Dave’s approach is that it’s skipping one important detail. The elements in there are infinitesimals. Which means that while response is immediate,(1)the current and voltage are also infinitesimals. In that case it’s a bit of a stretch to say the load receives power. This is the reason whyincreasingdecreasing time step in a numerical calculation decreases the pulse duration: that pulse has no length.____(1)Subject to propagation time limits.

author: notadave
author_rank: Contributor
timestamp: November 25, 2021, 02:28:57 pm
content: Does not anyone disagree with the statement and picture:"E-field and B-field are in phase." ?I have seen that statement three times on youtube and no one disagrees:3:405:406:30

author: nctnico
author_rank: Super Contributor
timestamp: November 25, 2021, 02:48:23 pm
content: In the end it is highly academic and a lot of the fine details are left out. In an experiment of the proposed size the rotational speed of the earth is also going to play a role depending on the orientation of the wires..

author: bdunham7
author_rank: Super Contributor
timestamp: November 25, 2021, 03:30:59 pm
content: Quote from: nctnico on November 25, 2021, 02:48:23 pmYes, you might get a bit more voltage than you bargained for, enough to vaporize your entire experiment. I'm sure what will happen is a scaled-down version of this will be set up that is claimed to be the equivalent. And at that point it will become obvious that the experiment demonstrates nothing of much interest or value, but that won't matter to the target audience.

author: Sredni
author_rank: Frequent Contributor
timestamp: November 25, 2021, 04:04:30 pm
content: Ben Watson did a very nice simulation of the fields, that shows how the propagation starts from the switch and reaches the lamp first, then propagates along the wire to the left and right end, senses the kind of termination there (open, short, load? you cannot know until your fields get there and back), then bounces back and after a few bounces you reach steady state.Response to Veritasium - Electricity Propagation Time ProblemThere is also Eric Bogadin's take in a recent video by Feranec that is more transmission-line centered. The point is that voltage and current are just approximations and we need to fold back to fields.My personal note: what is most interesting is that the fields that reach the lamp after a few nanosecondsare the same irregardless of the far ends being shorted or opened(it remains to be seen if - when considered in that particular topology the characteristic impedance is the same as when the generator in on the left and the load is on the right... Is that a long and narrow line or a short and wide one?)On second thought, this needs to be clarified: the fields after a few ns would be the same if the far ends are opened simultaneously with the switch closing. If the far end were already open, the charge accumulation would not be exactly the same at both the switch terminals (a different initial condition).

author: vad
author_rank: Frequent Contributor
timestamp: November 25, 2021, 04:29:06 pm
content: Here is another excellent analysis by Eric Bogatin from Teledyne LeCroy:https://youtu.be/Lp_b8gQpxW8

author: Bud
author_rank: Super Contributor
timestamp: November 25, 2021, 05:37:33 pm
content: I remain of the opinion that the problem has nothing to do with " transmission line" approach and has to be explained in classic electromagnetic theory terms. If anyone based his explanation on "parallel wires 1 meter apart" , or " X mm trace on an FR4 PCB, the line impedance and such, if they instead wrap the wires around the Globe in a circular shape rather than running them in parallel, their answer to the challenge may differ. This will right away indicate this approach is wrong.

author: vad
author_rank: Frequent Contributor
timestamp: November 25, 2021, 06:28:03 pm
content: Same phenomenon can be described multiple ways. For example, motion of planets and satellites in Solar System can be described both with Newton’s Law of Universal Gravity and General Relativity (and a future Quantum Gravity Theory that is yet to be discovered). It is well known that both Newton’s Law and GR are wrong (the former does not explain many experiments and observations, the latter epically fails at quantum scales with variances on the scales of 10 to 120th power or so). Yet still Newtonian Law is good enough for JPL and NASA engineers in majority of the cases.Same here. Classical Electrodynamics describes reality better than Ohm’s Law, while being classical approximation of Quantum Electrodynamics. Still, in majority of the cases, electrical engineers are better off with Ohm’s Law. Good luck to anyone wishing to describe a simplest electrical circuit with QED.

author: vad
author_rank: Frequent Contributor
timestamp: November 25, 2021, 06:48:23 pm
content: Quote from: Bud on November 25, 2021, 05:37:33 pmIt does not matter if you run wires straight in parallel or wrap around the globe, it does not matter if the length of the wire is 300,000 km or 30 meters. You will see displacement current in the opposite wire in d/c seconds, where d is the shortest distance from the switch to the opposite wire in meters, c is light speed in the medium in m/s.The initial displace current will be as high as voltage divided by intrinsic impedance (e.g. for wires that spaced at significant distance, the impedance will be close to 377 Ohms in air).

author: Kalvin
author_rank: Super Contributor
timestamp: November 25, 2021, 07:51:27 pm
content: Quote from: Bud on November 25, 2021, 05:37:33 pmThis and other EM-field problems can be solved using classic electromagnetic methods, no question about that. Typically solving EM-problems requires numerical methods, unless the problem has a simple geometry for which there exists a closed form solution or for which it is possible to create a simple numerical approximation.Looking at this particular experiment, it was very clear that the setup and circuit topology looks like a transmission line. Because there are already known, well established methods for dealing with transmission lines, why not then use a simple transmission line-model for solving this particular experiment? It is not by accident why transmission lines are being used whenever there is a need to move/guide EM-fields form point A to point B.For some other circuit geometry a simple transmission line-model would not work, and another model would had to be created.

author: Bud
author_rank: Super Contributor
timestamp: November 25, 2021, 09:30:21 pm
content: Then we need to decide if the plumbing model with pipes and shut off valves is all that is needed to undedstand how transistors work. Why go to the trouble of the science.

author: David Hess
author_rank: Super Contributor
timestamp: November 25, 2021, 10:56:21 pm
content: Quote from: Bud on November 25, 2021, 09:30:21 pmAnd how hall effect sensors work.Isn't drift velocity of the charge carrier critical to understanding how they work?

author: EEVblog
author_rank: Unknown
timestamp: November 25, 2021, 11:28:35 pm
content: Quote from: Sredni on November 24, 2021, 03:15:57 pmNo, I am not going to pull the video and nuking all the comments.

author: Sredni
author_rank: Frequent Contributor
timestamp: November 26, 2021, 02:52:30 am
content: Quote from: EEVblog on November 25, 2021, 11:28:35 pmNo need to pull it.If YT no longer allows to add corrections, I think you should amend the description to address those two big blunders:- Poynting vector at DC is no different than at AC (where did you get the idea it would point toward the battery or the highly conductive wires?) - timestamp 09:30- Feynman was expressely showing a resistor in the figure you mention and show from his Lectures, not a highly conductive wire in a circuit with bigger resistors. timestamps 38:25 and 40:24Also, maybe removing the timestamp for "Every electrical engineer knows this?" ;-)I mean, the people you called BS on for their dodgy tech could have a field day with this video if you do not correct it yourself in a clear manner.

author: Kalvin
author_rank: Super Contributor
timestamp: November 26, 2021, 08:55:46 am
content: Quote from: Bud on November 25, 2021, 09:30:21 pmIt is not a question whether we have to select exclusively between a plumbing model and most complex transistor model available. We just need to know when to use and apply a specific transistor model at given situation. There are multiple models for a transistor, each providing increasingly better approximation how the transistor will behave within the limits of the model. A very basic transistor model with pipes and shut off-valves is sufficient for understanding how to control LEDs, actuators etc. in practice, but for AC or RF circuits this simplified model is insufficient and better model shall be used.I bet when analyzing transistor circuits you do not calculate EM-fields in the transistor (the science way), but you are using whatever approximation and transistor model there is suitable for that particular situation (practical way). After all, the art and science of engineering is to understand when to apply particular model and approximation, and understand the limits of these simplified models.

author: AlienRelics
author_rank: Unknown
timestamp: November 26, 2021, 01:34:51 pm
content: Anyone linked to this video yet by Professor Walter Lewin, "Poynting Vector, Oscillating Charges, Radiation Pressure, Comet Tails, Polarization (Linear, Elliptical, and Circular)"?https://youtu.be/6lb040GCs2M

author: AlienRelics
author_rank: Unknown
timestamp: November 26, 2021, 01:55:44 pm
content: Quote from: Bud on November 25, 2021, 09:30:21 pmAll of engineering is selecting and using simplifications that are just good enough for that particular situation. To design a guitar amp, I don't need to understand or use a lick of quantum chromodynamics, but the water analogy just isn't good enough. If I'm wiring up switches and lights in my house, the water analogy is good enough. But if I'm to design a new IC, I'd better understand and use a lot more accurate models!I'm reminded of an early episode of "The Big Bang Theory" where Sheldon assumes that because he is extremely good at theoretical physics, he must therefore find engineering something mechanically simple a doddle. Then he fails spectacularly.I see a LOT of engineers, technicians, and hobbyists who learn the simplifications and rules of thumb, who then think that the most complex of the simplifications they've learned is actually how things work. The simplifications are analogies. Analogies are not the truth. If you see a sign that says "Watch For Falling Rocks", the analogy of "the Earth sucks" is good enough, you don't need to know the (current) theory of gravity that it isn't a force, but due to differential time dilation caused by mass. Einstein's warped space is an analogy, a model. There is no rubber sheet.This is an example of a type the Dunning-Kruger effect, where someone assumes they are very smart about everything because they are very smart about something.

author: Fungus
author_rank: Super Contributor
timestamp: November 26, 2021, 03:18:55 pm
content: Better question: How long before the bulb reaches full brightness?

author: vad
author_rank: Frequent Contributor
timestamp: November 26, 2021, 05:58:56 pm
content: Quote from: Fungus on November 26, 2021, 03:18:55 pmUsing transmission line model: it can take many seconds until all reflected waves attenuate and the circuit reaches steady state.

author: nctnico
author_rank: Super Contributor
timestamp: November 26, 2021, 06:25:46 pm
content: Quote from: vad on November 26, 2021, 05:58:56 pmWithout any resistivity in the circuit, that state is likely never reached because there is nothing to dampen the oscillations.

author: Fungus
author_rank: Super Contributor
timestamp: November 26, 2021, 07:43:40 pm
content: Quote from: nctnico on November 26, 2021, 06:25:46 pmOK, how long until 98%?

author: thm_w
author_rank: Super Contributor
timestamp: November 26, 2021, 09:56:23 pm
content: Quote from: Fungus on November 26, 2021, 03:18:55 pmWhich is intuitively what the majority of people watching the video would think the question is. IMO.I mean look at the simulation Sredniposted, you'd see that and go "oh some energy reaches the bulb, but its not really well lit until it propagates the length of the wires.

author: David Hess
author_rank: Super Contributor
timestamp: November 26, 2021, 10:00:32 pm
content: Quote from: nctnico on November 26, 2021, 06:25:46 pmIf the transmission lines are shorted at the ends, then the bulb alone provides termination of the transmission lines, so proper termination depends on the bulb resistance. There is no requirement for a transmission line to be properly terminated at both ends; a single termination at either end is sufficient to prevent reflections.With a proper termination by the bulb, and shorts at the ends of the two transmission lines, the bulb sees twice the transmission line impedance in series until the wave propagates to the ends and reflects at which point the bulb sees the full voltage. If the bulb resistance is a mismatch, then there will be reflections.There are some (weird) RF power sensors that work like that with the transmission line terminated into a thermister which is DC biased to operate at a constant temperature, and thereby resistance for proper termination. The required DC bias indicates the RF power level.

author: nctnico
author_rank: Super Contributor
timestamp: November 26, 2021, 10:21:37 pm
content: Quote from: David Hess on November 26, 2021, 10:00:32 pmBut the experiment claims ideal wires (0 resistance) so it also implies ideal battery (0 Ohm), ideal switch and an ideal lamp (infinite resistance).

author: vad
author_rank: Frequent Contributor
timestamp: November 27, 2021, 01:09:21 am
content: … and ideal shorts (0 Ohm) at the end of each transmission line, causing ideal reflection. And when reflected wave travels back, it will pass through the ideal switch and will proceed into the opposite transmission line…

author: David Hess
author_rank: Super Contributor
timestamp: November 27, 2021, 05:44:17 am
content: Quote from: nctnico on November 26, 2021, 10:21:37 pmWhile the transmission line is charging or discharging before the wave is reflected, it appears to have a pure resistance equal to its impedance. It really does look like a resistor as far as the source is concerned.Quote from: vad on November 27, 2021, 01:09:21 amThe bulb has a resistance and if it is the same as the transmission line impedance, it will completely absorb the reflection.

author: vad
author_rank: Frequent Contributor
timestamp: November 27, 2021, 12:23:50 pm
content: Quote from: David Hess on November 27, 2021, 05:44:17 amWhile the bulb has the resistance, impedance between its terminals is not completely resistive and neither it is infinite: there is parallel capacitance between bulb’s terminals, and there is impedance of free space. It will take several round trips until all reflections attenuate below noise floor.

author: AlienRelics
author_rank: Unknown
timestamp: November 27, 2021, 04:34:21 pm
content: An ideal bulb is not infinite or zero resistance.An ideal bulb is an ideal resistor. No parasitic capacitance or inductance, just resistance and no change with temperature.So with all ideal parts, and assuming nothing near the wires to absorb near field radiation, the resistance of the bulb is all that absorbs energy. So it will eventually settle, still an asymptotic delay. I think you could model this as two paralleled transmission lines shorted at the far end, so simplify it to one shorted transmission line of half the characteristic impedance with the battery and bulb at one end.So the wave has to travel to the other end and back again, say 5 times.However, if we arbitrarily make the bulb resistance equal to the characteristic impedance of the paralleled transmission line, then it should immediately (1m/c) reach 1/2 power. And reach full current/power in the time it takes for the step change to propagate to and back from the shorted end.Since it is all perfect parts, a shorted transmission line reflects back in opposite polarity, therefore the reflected step change is in exact peak value but opposite polarity. So as it propagates back to the bulb, it exactly cancels the voltage present.When it reaches the bulb/battery end, it is zero. So full brightness after length of wire/c. So if the wires are 1 light second long, it takes 1 second to reach the end, 1 second for the reflection to return. So the bulb will reach 1/2 brightness in the time set by the spacing of the wires, and 2 seconds later it steps to full brightness.

author: Fungus
author_rank: Super Contributor
timestamp: November 27, 2021, 05:36:36 pm
content: Quote from: AlienRelics on November 27, 2021, 04:34:21 pmThe video should really have pointed that out IMHO.Anything less is a bit dishonest.

author: nctnico
author_rank: Super Contributor
timestamp: November 27, 2021, 07:01:33 pm
content: Quote from: AlienRelics on November 27, 2021, 04:34:21 pmNo, an ideal bulb has infinite resistance by definition! An ideal bulb draws zero current. Anything else is just moving goal posts.The only real conclusion that you can draw is that Veritasium's video is a down to earth introduction to EM field theory but the circuit presented is so much dumbed down that any answer between 1m/c and infinite time is correct. It just depends on how you fill in the blanks.

author: Fungus
author_rank: Super Contributor
timestamp: November 27, 2021, 08:13:48 pm
content: Quote from: nctnico on November 27, 2021, 07:01:33 pmThe big lightbulb corporation will kill you if you try to manufacture them.

author: Fungus
author_rank: Super Contributor
timestamp: November 27, 2021, 08:27:28 pm
content: Another question: How long does it take to turn off the light?

author: David Hess
author_rank: Super Contributor
timestamp: November 28, 2021, 08:20:01 pm
content: Quote from: vad on November 27, 2021, 12:23:50 pmIf someone else can assume lossless transmission lines, then I can assume aspherical bulb in a vacuumwith no parasitic elements. Besides which, maybe the bulb is constructed as part of the transmission line; there are real world parts constructed this way for exactly this reason including diodes and resistors and switches.

author: thinkfat
author_rank: Unknown
timestamp: November 28, 2021, 08:24:06 pm
content: Quote from: Fungus on November 27, 2021, 08:27:28 pmCalculate the current of the arc across the switch.

author: Fungus
author_rank: Super Contributor
timestamp: November 28, 2021, 08:38:01 pm
content: Quote from: David Hess on November 28, 2021, 08:20:01 pmWhy aren't we using LEDs? This is 2021!

author: vad
author_rank: Frequent Contributor
timestamp: November 28, 2021, 09:14:16 pm
content: Quote from: David Hess on November 28, 2021, 08:20:01 pmIn Russian they call it spherical stallion in vacuum.

author: Fungus
author_rank: Super Contributor
timestamp: November 29, 2021, 02:27:45 pm
content: Quote from: Fungus on November 27, 2021, 08:27:28 pmNobody...?

author: Sredni
author_rank: Frequent Contributor
timestamp: November 29, 2021, 04:58:25 pm
content: Quote from: Fungus on November 29, 2021, 02:27:45 pmI had dismissed this as a joke, but you have a point here.When we open the switch, after the lamp had reached full brightness in steady state, surface charge will start to pile up at the switch's terminals. This new distribution will affect the lamp after 1/c seconds, but I believe the lamp will still glow brightly until the redistribution in surface charge has reached the lamp making the whole trip.The asymmetry here is in our minds: we accept the lamp 'is on' when just a minuscule current flows into it, and we accept it is off when less than a minuscule current flows into it.

author: gamalot
author_rank: Super Contributor
timestamp: December 01, 2021, 03:10:53 pm
content: I found a roll of 2-core parallel wire, and measured the parameters of the wire. The length is about 23.4 meters, the DC resistance is about 0.77 ohms, the differential impedance is about 120 ohms, and the velocity factor is about 0.585.Replace the bulb with a 50 ohm resistor, and replace the battery with a function generator (output impedance 50 ohm, pulse width 1us, amplitude 1Vpp), the circuit and measurement results are shown in the attached pictures.

author: Sredni
author_rank: Frequent Contributor
timestamp: December 01, 2021, 07:30:25 pm
content: Yep,. the first edge is nearly instantaneous. But you get half the voltage because charge recombination on the right is basically as immediate, then you need 2x130 ns for the reflection from the distant edge to reach the load.Could you place the generator and the load in the middle of the cable, and redo the measures with short and opened ends?Also, ho much is being filtered out by scope and probes?

author: David Hess
author_rank: Super Contributor
timestamp: December 01, 2021, 08:04:02 pm
content: Those results are exactly as predicted. I have done the same experiment, by accident of course.

author: gamalot
author_rank: Super Contributor
timestamp: December 01, 2021, 08:18:30 pm
content: Quote from: Sredni on December 01, 2021, 07:30:25 pmThe voltage of the first step is 456mV, which is the function generator's output voltage of 2Vpp divided by its internal resistance, transmission line impedance, and load resistance.2V*(50/(50+120+50)) = 0.454545454 ......I did try to cut this parallel wire from the middle, and then reconnect the 11.7m long wire back to my test circuit. The result was that the width of the first step was shortened to 130ns accordingly. If I connect the other half of the wire to the other side and put the generator and the load in the middle as you said, the width of the first step is still 130ns, but the amplitude will be reduced because 120 ohm impedance is added.I don’t think the oscilloscope or probe will influence the conclusion in this experiment.

author: Sredni
author_rank: Frequent Contributor
timestamp: December 01, 2021, 09:23:56 pm
content: Well, I assume generator and load would be separated by about half a meter, on your desk. The time to cover that distance at 300 million meter per second is 1.67 ns. My 100 MHz chinese scope has a rise time of 3-5 ns. You have a 70 MHz quality scope, but I don't think you will be able to appreciate what really happens at that timescale. Yes, as you show in your screenshot you can see there's some delay, but maybe a slightly better scope and a very high impedance front end (to hide the capacitance of probes and scope input) might be required .What you see there is just 'ordinary' reflection, I think.Someone with a 500 MHz scope could build the high-Z front-end designed by Bob Pease (I can add a link, if need be), and use maybe two entire UTP5e spools on the left and right of the generator + load + scope point to get a better view?

author: gamalot
author_rank: Super Contributor
timestamp: December 01, 2021, 09:58:11 pm
content: Quote from: Sredni on December 01, 2021, 09:23:56 pmI don’t really understand what you mean by 'ordinary' reflection. What I am concerned about is a time of several hundred nanoseconds, and the rise time of my oscilloscope is about 1% of it, so I really don’t Think it will seriously affect the conclusion.

author: Sredni
author_rank: Frequent Contributor
timestamp: December 01, 2021, 10:57:28 pm
content: The effect Derek is talking about happens in less than 3 ns and it is due to the propagation of the electromagnetic field from the surface charge where the switch (or generator) is to the load placed one meter away in line of sight.The several hundred nanosecond steps you see are due to the surface charge on the wires redistributing itself following the wire. That requires the perturbation to travel the whole trip over the lenght of the cables.In your case, since you made one leg extremely short, the trip along the wires on one side is nearly as immediate as the 'flight' in the space between switch and load. This is the very first edge you see the one that is a fraction of a single division.And that edge is being 'smoothed' by the impedance of probes and scope.The giant ridges you see are 'ordinary' reflections along a transmission line. The effect Derek is talking about is hidden in the first steep edge and you cannot see it with this asymmetrical configuration.

author: David Hess
author_rank: Super Contributor
timestamp: December 02, 2021, 04:32:28 am
content: Quote from: Sredni on December 01, 2021, 09:23:56 pmA higher bandwidth edge and oscilloscope will show more detail, but the only thing which will be revealed is the non-ideal characteristics of the implementation.QuoteWhich probe is that? Pease published a design for a low capacitance probe but it was more like 50 MHz. It was optimized for lowest capacitance and low cost but not high speed.Quote from: Sredni on December 01, 2021, 10:57:28 pmWhat do you think faster test instrumentation would reveal? At high frequencies before the reflection can return, the transmission line operates as a series resistance equal to the characteristic impedance between the source and load.Faster test instrumentation could show the propagation delay across rather than down the transmission line. My highest bandwidth oscilloscope can "see" down to about 1/10th of an inch, which is eerie to experience, but I lack a fast enough pulse generator to take full advantage of that.QuoteThe same edge would be present at the same time if a symmetrical arrangement was used or not, but the equivalent series resistance would be doubled and balanced. Coaxial baluns work like that.Treat the missing transmission line as a zero length transmission line with a reflection time of zero. The wave still has to propagate across its width.

author: Sredni
author_rank: Frequent Contributor
timestamp: December 02, 2021, 06:00:18 am
content: Quote from: David Hess on December 02, 2021, 04:32:28 amYou're right. I keep forgetting it was a slow probe.Anyway, if I'm trying to see an edge (and possibly something happening besides a rising edge but in a timeframe of a few nanoseconds at most) with a rise time of less than 2 ns, I doubt I could see it clearly with standard 10x probes and a scope with an input capacitance of 10-15 pF.QuoteI am not at all sure you can use the model of a transmission line before you have current in the whole line (the experiment with the generator/switch and the load in the middle is different from the experiment with generator/switch and load on the same side). There is no current in the load side before the field crosses the air gap, and there is not all of the current until the charge has propagated/redistributed along the line and back (see the simulation I linked in the first page). It's that situation at the load facing the generator at times in between t= d/c and say t= 5 d/c that we are trying to see.QuoteYes, that's exactly what we need to see.But we need to place generator/switch and load in the middle. In a way this is a very wide but extremely short transmission line. My guts says the ordinary model for very narrow but extremely long transmission line breaks down.QuoteGood point. What was the rise time of that fast pulse Williams built? But even with a slower edge for the pulse, if the instrumentation does not smooth it out, maybe a comparison could be made.QuoteYes, but the reflection with the full current gets there at the same time, so we won't be able to tell if something funny happened. We need a delay before any reflection hits the load.(I need to sleep now, I hope I did not write anything stupid)

author: gamalot
author_rank: Super Contributor
timestamp: December 02, 2021, 07:00:54 am
content: I found my old but a little faster scope, and also found a DIY simple pulse generator, these two together can get a rise time of less than 700ps. So I repeated the previous experiment, and it didn't seem to make any difference except the steeper leading edge.

author: David Hess
author_rank: Super Contributor
timestamp: December 02, 2021, 10:15:37 pm
content: Quote from: Sredni on December 02, 2021, 06:00:18 amThat is exactly what gamalot's experiment shows; the transmission line impedance is resistive while it is charging or discharging, before the end termination can have any effect on the source because of the speed of light delay. His measurement shows a resistive voltage divider, with the transmission line impedance in place of one resistor.Fast transmission line circuits take advantage of this. For instance a sampling capacitor can be replaced with a transmission line, and if the sampling gate time is made equal to the propagation time (both ways) through the transmission line, then the transmission line load looks resistive instead of capacitive which has advantages. If the source you were sampling from was a 50 ohm transmission line, then you make the sampling capacitor out of a 25 ohm transmission line, and then while sampling, there is a perfect match of 25 ohms to 25 ohms.If you were to measure the current into the transmission line before any reflection from the end can return, then it is proportional to the applied voltage while the transmission line is charging, and presents a resistive impedance equal to the ratio between the voltage and current. The source cannot "see" the load, open or short or anything in between, through the transmission line until the speed of light delay is satisfied, so it sees only the transmission line impedance, which is resistive.QuoteIt is about 300 picoseconds, which is comparable to the sources I have. That is good for measuring down to about 30 picoseconds or half an inch since you can measure much smaller delays compared to transition times.

author: Sredni
author_rank: Frequent Contributor
timestamp: December 02, 2021, 11:54:50 pm
content: To be clear: we are trying to visualize the green curve in the yellowish frame in this picture compose from screenshots in Ben Watson's simulationlink:https://i.postimg.cc/c4LcZ7cV/screenshot-27.pngIf I got the sizes right (I skimmed through the video this time and I don't remember if he said the dimensions explicitly), Watson simulated a strip 9cm long and half a cm wide.It takes 15-20 ps for light at 2.98 x 10^10 cm/s to cross the strip.At 820 ps, ie 620 ps after the switch on, the charge has redistributed itself along the wire (can we infer a velocity factor of about 0.5?).The times will be different in the experiment. If 50 cm were separating generator on one side and load+scope on the other, we will have to wait 1.5 - 2 ns for the green line to rise. And then some 130 ns for the charge redistribution to reach the load.What we see on the scope is the big green step (that in this tiny line starts at 820 ps). What I am dubious about is: is the setup capable of revealing that small step that starts 20 ps after the "switch" is closed?In the real world, with the probe cable, and the scope attached we do not have just the load. What happens to that tiny current when we attach the whole equipment?Also, wouldn't it be required to have the scope on the load side, without any connection to the generator?This looks like a delicate measure: we are not operating the transmission line from one of its ends, the effect is small in itself and might be killed by the instrument, the geometry of the setup is important (for example, if you bring the coax for the load and the generator in the same scopes-- they are what? One inch apart? We must make them half a meter, one meter apart). We might also have to consider the propagation along the probes (what does that do to the feeble charge whose effects we are trying to measure?).

author: hlevinson
author_rank: Newbie
timestamp: December 04, 2021, 10:46:55 am
content: I think this can all be simplified to something both physicists and engineers will accept.The idea is to set up a clean transmission line circuit.Reduce the circuit to one transmission line, say 50 ohm characteristic impedance.Connect the transmission line to a (fast) pulse generator with 50 ohm output impedance. The 50 ohm output impedance is the load or bulb and the open circuit pulse amplitude the battery voltage in the Veritasium circuit.Then switch on pulses with amplitude set at, say, 5 V into an open circuit.Using a fast high impedance scope, you should measure 2.5 V across the transmission line at the pulse generator output socket. The other 2.5 V will be across the 50 ohm internal resistance.The 2.5 V step at the pulse generator output will occur instantly, ie no significant time delay between the pulse generator's rising edge and the rising edge at its output.And, in this case, half the open circuit output will be across the internal resistance and half across the transmission line.To get back to the Veritasium set up, add a couple of short lengths of wire between the pulse gen. output and transmission line. This will introduce a messy, short delay.Further reflections from the transmission line termination and pulse gen. internal resistance will result in the voltage on the transmission line (and hence voltage across the pulse gen internal reasistance) going up and down with decreasing amplitude until the dc state is reached.

author: Fungus
author_rank: Super Contributor
timestamp: December 04, 2021, 12:11:44 pm
content: Quote from: hlevinson on December 04, 2021, 10:46:55 amYes.Make the wires go directly to the light bulb instead of half a light year out into space and back.

author: 13hm13
author_rank: Frequent Contributor
timestamp: December 04, 2021, 07:47:08 pm
content: I'm sorry but people like DAVE JONES -- whom I usually like and agree with -- unfortunately EMBARRASS the CRAP OUTTA THEMSELVES when they makes comments like ... "this is really basic stuff" or "every first year EE student knows this " (can't recall exactly what Jones said in his vid, but it had sentiments like along these lines).The fact that SOOOOOOOOOOOOOOOOOOOOOOOOOOOO many analysis videos and blog posts resulted as consequence of the Veritasium video illustrates that the "facts" are not so clear. Yes, I am aware that hyena content creators -- like Jones -- are cashing in on the Veritasium popularism.

author: Obi_Kwiet
author_rank: Regular Contributor
timestamp: December 05, 2021, 04:55:05 am
content: 13hm13, dude, no one cares what you have to say if you make your text a weird size. What are you thinking?This is a weird problem. I thought Dave's explanation made sense at first, but while I was trying to explain why an objection to the veritasium video was wrong, I noticed that someone else had pointed out that the connection isn't in the right place for the unit step function to create a transmission line between the top and bottom wires. The capacitor explanation doesn't work either, because the two nodes have no voltage reference between them. A capacitor will certainly pass pass a transient like a unit step, but voltage needs a common reference to have any meaning.The transmission line would actually be created between the two wires sticking out of the battery. But that's not a transmission line, that's a dipole antenna. But I don't have a clear understanding of the field and wave equations inside an antenna, and I'm not sure power is actually going to be transferred in this case. I get the sense that a unit step of electric field propagating down one half of an infinite length antenna isn't going to radiate.Anyway, this guy's video seemed pretty confused at first, but when I reviewed it after thinking through the issues I just explained, it started to make sense. I think this is a trickier problem than it first looks, because the terms that we can usually assume to be zero, and given them non-trivial values. That throws off our analysis.

author: 13hm13
author_rank: Frequent Contributor
timestamp: December 05, 2021, 06:24:19 pm
content: Quote from: Obi_Kwiet on December 05, 2021, 04:55:05 amYour reply isconfused, toots, in that it is YOU that indeed "cares" enough to reply.

author: EEVblog
author_rank: Unknown
timestamp: December 05, 2021, 10:29:34 pm
content: Quote from: hlevinson on December 04, 2021, 10:46:55 amYou can't just do that:

author: EEVblog
author_rank: Unknown
timestamp: December 05, 2021, 10:32:23 pm
content: Quote from: 13hm13 on December 04, 2021, 07:47:08 pmI did it because I was asked by so many people to talk about it. Ordinarily I wouldn't have cared.Every EE student does (or should) learn about electron drift velocity et.al. But like Feynmen said, most engineers come out, and spend their entire careers not really caring about it.

author: EEVblog
author_rank: Unknown
timestamp: December 05, 2021, 10:37:05 pm
content: Quote from: Obi_Kwiet on December 05, 2021, 04:55:05 amAnd that in theboldis the trick. Propagation takes time, no matter whether you use capacitive coupling, transformer theory, transmission line theory, or antenna theory. And anything that takes time to propagate ultimately leads away from the 1s/c answer. If you want you 1s/c answer then you have to work on the immediately surrounding physical properties only.

author: 13hm13
author_rank: Frequent Contributor
timestamp: December 06, 2021, 01:46:45 am
content: The Gedankenexperiment Veritasium vlogged on should be taken word for word. Battery, light bulb, etc. But even if it were some sort of closed circuit (TV) signal -- 1m part -- my bet is on LIGHTSPEED. Not Warpspeed or Hyperdrive

author: Fungus
author_rank: Super Contributor
timestamp: December 06, 2021, 07:21:04 am
content: Quote from: EEVblog on December 05, 2021, 10:37:05 pmLaws of Physics: "You can't transmit information faster than light".

author: Obi_Kwiet
author_rank: Regular Contributor
timestamp: December 07, 2021, 03:15:23 pm
content: So I'm pretty sure I was sleepy and confused myself last time. Everything after the switch is pulled down to 0V relative to the positive terminal, so when the switch closes, charges will spread out into the wire and there will be a significant current, so the transmission line approach does work.