Instead of dismissing/waving hands like Dave from EEVblog did, you guys made the whole thing meaningful. And it's actually more helpful than not for even experienced electrical engineers.
@@donmab The difference for me is that I can actually understand what Dave is saying. Maybe it's because I'm also Australian, but the point still stands.
@@ScoochieR Honestly, I think that by introducing complexities like resistance and loss, EEVBlog may be more practical, but he is using it in a way that it just clouds the actual working of the system and saying that everyone already knows about this is just not true. That's not the point of the original video which is trying to explain first principles of EM. I do agree that Derek may have done a poor job by starting with non-ideal components and just claiming them to be ideal, whereas you would just choose a PEC in a lecture, but I guess this makes it more relatable to people.
@@ScoochieR I vant to knov vay you call yourself a "Ruski robot" ven you are Australian? The video Dave did on the Varitasium was not too detailed he pretty much agreed with most of what was said by Varitasium in relationship to principles of transmission lines and propagation. Actually it was probably one of two of his videos I was able to sit through right to the end, normally I have to turn him off because he tends to just go overboard and overcomplicate a generally simple subject.
This is a nice explanation of the problem. It remined me that in my first job after college many years ago, I was testing a circuit that was built from TTL and I wanted to look at the output of a 7474 D flipflop. So I fastened a wire twisted pair to the output and ground and ran it several feet to an oscilloscope to look at it. to my surprise, the Flipflop would not stay set when it got appropriate inputs. This is how I remember it, but it was a long time ago. After further investigation, I determined that the Flip Flop was actually setting then resetting in a very short time. It did set properly when I put the scope probe directly on the output with the wire detached, and with wire attached you could see that it was momentarily set. Then I realized that the scope probe was a very high impedance and the line was reflecting the initial edge from the Flipflop setting back to the chip pin. AND unfortunately the D output was internally connected to the opposite side input of the output latch (not buffered), and the reflection reset the flip flop. So that was a lesson for me. Care is needed in higher speed measurements.
As always, Eric is a pleasure to watch and listen to. I love how this video brings together a lot of different aspects of physics and engineering. Great lecture!
I've seen a few reaction videos on the same subject already but yours is the best 😀 It's great that you explained the difference between the two cases, that was extremely important. Thank you both 😊
I saw the Veritasium video and was waiting for folks like Eric to peel the onion back more. He didn't break out a smith chart. I always liked those in school.
It is not correct to treat the Veritasium arrangement as a pair of single impedance transmission lines. You aren't driving the two ends of your "transmission" lines with the voltage source. For the first few nanoseconds, the lower pair looks like a dipole antenna. The bulb line, also looks like a dipole antenna, receiving the signal. A dipole arrangement does not have a single Zo - the step response is actually pretty challenging to figure out. My point is that you cannot assume the current in the transmitting side is equal to the receiving side - there will be coupling, but I contend a fairly small current - not enough to realistically light the bulb. The beef with the original video was not that some signal arrives 1m/c seconds after switch closure, but that there was implied full power transfer. It won't be.
@bobwhite137 They've got it right. It's not an antenna, because when we place two antennas that close together, we create a highly-coupled transformer, not antennas anymore. (Same thing would happen with loop antennas: if we put them right on top of each other, we now have a transformer, not a xmit/rcv antenna pair.) We also get a capacitor where the plates are formed by two close-spaced wires. So which is it, a transformer, or a capacitor? It's both, and Heaviside discovered, when the two concepts are combined, the reactance of the transformer and the capacitor cancel out to form a virtual resistance. If we could use a DVM to measure Veritasium's million-meter ideal wires, we'll see a 700-ohm resistance, rather than 50-ohms of standard coax cable. To act like dipole antennas, the wires need to be separated by about 1/4th the wire length, so the coupling isn't anywhere near 100%.
Yes, this appeals to my intuition too. The arrangement is like antennas or even a transformer. But probably the antennas make the most sense. This was pointed out in another video discussing the original video. All of it is instructive though!
I think that current starts after 1/c (in seconds) and it is slowly building up until the RC circuit practically is charged after 5 times RC. Since we have a switch, we have to study the step performance (transient) of the circuit and treat it as an RC circuit (until it reaches the steady state which is 5 times RC practically). If we assume R=0 then RC is zero then the circuit charges in 0 seconds. Inconsistencies arise when we assume R=0 and C some value. Using the notion of the displacement current, the transmission line, the transformer, the antenna, the cable impedance is a good tool to understand the circuit. But i consider it as an RC circuit since we have a switch and therefore i am analyzing the step performance of the circuit. The problems arise when we assume a practically infinite and distributed capacitance with no resistance...
@@wbeaty 'a highly-coupled transformer' Not even that highly coupled a meter apart, you'd consider any low voltage transformer a failure if you move the coils one meter apart.. The electrons are literally IN the wire. They don't actually touch, so the energy is in colliding the field shells. The field extends from the electron, and is both inside and outside the wire. That you can tap the energy of the moving electrons both from inside the wire, and somewhat outside the wire, doesn't mean the energy source isn't still the energy in the moving electrons, and is still basically stuck to the wire. Make the wire take a 90 degree turn and the outside field takes that same turn, you can't make it just go straight off the wire because it's not really flowing outside the wire. It takes special pains like many coils and a core to efficiently couple the magnetics of the flowing electrons of one wire into another, and efficiently transfer power from one wire to another. It takes other special pains to use the flow in an antenna and peel the energy off the wire into EM radio waves and similar. If it were really just that easy and all the energy was really outside the wire, transformers and antennas wouldn't be their own specialized fields and take pretty careful planning and adjustment to make sure they even work in an OK manner.. Route a wire all over a room, bending it in different paths, and try to get the waves to do anything that isn't strictly confined to following the path of the wire. Won't happen. Just because you can tap the energy and affect it from the outside doesn't mean that energy isn't coming from the parts that are stuck in the wire, it just means they have a field around them.
I love his way of explaining things. There is even a risk that I might start understanding transmission lines even though I have been avoiding them thus far.
What this explains is how so many of us are conditioned to view things in a defined way. The first circuit makes us feel safe and knowledgeable as we are so used to seeing it this way but when confronted by the second circuit we suddenly develop anxiety and stress because we are conditioned not to really understand circuits but memorize them.
The memorization over understanding is part of the game. Understanding that electricity is like sound and can be amplified to any value without loading the source or adding more energy would make investors in power plants freq. out. You don’t need a new power plant, just an unbalanced transformer.
Finally the correct explanation! It infuriated me to no end that soooo many electrical engineering youtubers' explanations were so misleading to the point of being wrong! This person is probably the very first one to correctly explain the precise physics of signal propagation! Thank you! Now my soul can finally rest knowing that the world (or at least your viewers ;-) ) have received the correct, physically-accurate explanation.
He did it as an EE engineer. Taking the coaxial 50ohms cable as an example was a perfect idea, even though it's also quite confusing. (the original question is very high impedance pair, 1m apart) Physics engineers do explain it with their own formulas. That doesn't make them wrong.
Excellent video, thank you so much for posting it. I have seen others with Dr Bogatin and after watching Veritasium's I hoped that in the responses that were likely to be posted we could see his discussion on the subject. Wish granted, what a treat! To top it up 3 months of free access to his lectures, for "our patience". Dr Bogatin, I get your passion and it takes no patience at all on my part to watch your talks and lectures so I cannot thank you enough for this.
The best explanation I have ever seen of signals and transmission theory. @Dr Bogatin: Very well done. And thank you to Robert Feranec for inviting him! Unfortunately, the problem presented to Dr. Bogatin, is NOT the problem presented by Veritasium. Though I totally understand why it was presented in the manner it was, nevertheless, the Veritasium problem has a Physical Demonstration to it that people who view the video get to see. That demonstration makes certain things "tangible" to the viewers and also modifies how we perceive it and how it should be perceived. Let's review: 1. The demo shows a (car) battery, an LED bulb, insulated conductors. These conductors are then given the hypothetical "no internal resistance" and they are 300,000km long, and separated by 1 meter. When the switch is thrown in the demo, the LED turns Solid ON. It simply remains On. There is literally NO mention of Blinking on. Clearly the definition of "ON" here is suspect if one were to consider "blinking" on as "ON". The question offered by Veritasium was after the switch closes, when does the LED turn on? (0.5sec, 1s, 2, 1/c (sic), and None of the above). After listening to Dr. Bogatin, (who clearly knows the material), one would believe that current flow across the bulb would be enough to power the LED. However, let's review some of the facts that everyone seems to leave out. #1 These are insulated conductors not capacitor plates sitting on dielectrics. The conductors are also not Bare Wires, they are insulated. Please get that fact straight. The impedance of insulated conductors 1 meter apart will be markedly different than the simulation because the simulations do not include the fact that there is insulation on the wires. #2 The LOAD is a 12 volt LED Bulb. These "bulbs" are internally made of many LEDs in series. Each LED in series has a forward voltage that MUST BE exceeded in order for the LED to turn on. Nowhere in any simulation are you taking into account the forward voltage requirement of the circuit. #3 THERE IS NOT AC SIGNAL. Dr. Bogatin refers to the signal frequently because he is equating the circuit with typical signal. BUT, there is NOT AN AC Signal in this example. There is only DC, and the instantaneous turning on of the Switch. There will be a very brief pulse. That is it. #4. If you take all these elements: Insulated Wires, DC, Not AC, 12 volts, 12 Volt LED, Forward Voltages of the Diodes, No signal. It results in at best... the microscopic blink of an LED, nothing more, for at best a moment in time. I am sorry, but I simply cannot equate this to "The Light Turns ON". Please show a simulation that corrects the above noted issues, and let's see how it plays out.
@@LiveType I am not sure. I guess it depends on whether he values his integrity. Because, the reason why I watched his videos was to get to the deeper truth of things. If this is how the future of his videos will play out, he lost only 1 viewer. It may only register as a blip on the magic light bulb hehehe but for me, I value the integrity of the process. It really bothers me to see this level of misdirection by someone I really thought knew better.
Sounds like Eric was really getting into the details there, would love to hear him discuss more on how Maxwell's equations have been simplified to the circuit elements we use today.
There's an article in IEEE Spectrum in 2014 called The Long Road to Maxwell's Equations. The article tracks the progress of understanding electromagnetics from Maxwell's predecessors to Oliver Heaviside's transformation of the equations to their modern form.
I was building my first PCB circuit for wired data transmission over RS-485 line (no electrical engineering education), and the terminology was very confusing: termination resistors, impedance matching, etc. While I did all the things "by the book" and followed various instructions, this explanation opened my eyes of what is really going on and what is actually important.
Excellent. A piece of test equipment called Time Domain Reflectometer uses this. You mentioned this as I write. If you know the impedence of the transmission line and put a resistor of the same value at the end the TDR will give you the distance of all the impedence changes down the line. Near shorts and opens and the distance of the same down the line.
In the section "Explaining transmission line..." there's an over-complication. The reason why you have instant current is because this is where you inject the power and make a measurement. The transmission line is irrelevant. What you measure takes another path through your measuring lines. (I didn't have patience to watch further than 10:33 so it may come up later, but there is no excuse for confusing the issue at the beginning)
While agree with John that this video is better than anything else I must say that I am rather frustrated by the original video, the EEVBlog video and also this video, and the overwhelming number of comments. Why is everyone ignoring the simple facts given in the original video: the lines are 1m apart, they do NOT form a 2m thick coax cable, the copper inside the lines must be in the order of 1mm thick, the lamp is big and will consume in the order of 1W, the energy source is a "normal" battery 6 or 12 V, not 1000000V or so. I have by now seen so many "physic professors" that apparently due to their mission to correctly explain the physics behind a transmission line completely oversee or deliberately ignore the problem statement. A decent (and therefore attentive) physics professor should also understand the following very simple facts. a) Coulomp's law according to which the electric field strength decays with 1/r^2, implying that also the time derivative of field strength decays with 1/r^2. b) The time derivative of a constant is zero. Now be my guest to calculate how much energy can be transmitted from some short, e. g. 30cm long section of the wire near the switch to a corresponding wire section near the lamp by a single pulse of e. g. 1ns. Even if one doesn't have that minimum amount intuition to understand that in the order of 1W will certainly not be transmitted (just think what consequences that would have on every day life), it is anyway clear that after 1n the light would be off until the field wave has travelled through the cable. I am also astonished why none of the "professors of physic" goes to their lab, take e. g. 40 m cabele (instead of 4*300000km) , pick a pulse generator to generate a (off->on) pulse, a lamb or resistor and a scope and do the experiment. This takes 10min. The result shows that the answers given in all videos is wrong! I better stop here...
I guess you're right. Eric goes too deeply to solve the problem that he forgets that this is a DC voltage supply that provides DC current. Okay for very tiny amount of time when we push the switch due to maybe high speed transition the LED may get energy wave through the space but that's not enough in terms of both power and time to keep LED turned on forever...the permanent thing that keeps it turned on must be the other form of moving energy in copper which is domino effect in which energy moves from on electron to the other to go ahead, and Eric himself in his book mentions this but here he wrongly goes on to explain it like a high speed pulsed signal.
Unfortunately, any photonic coupling between two wires 1 meter apart is going to be negligible. Photons are generated from the surface of the wire and then spread out before they hit the second wire. If the radius of the wire is 1 millimeter, then the photon cloud will have been attenuated 1000-squared = 1,000,000 times when it hits the second wire. So there's no getting around it. You're really going to have to wait a full second before the bulb lights.
But the assumption from the origional videowas any current turns the light on. Technically correct but misleading. Edit i do wish he had actually gone over the "capacitance"of 2 cables 1m apart to show the true current across it mathematically. Not sure the standard equations apply though so might have to derive for custom circumstances.
The two transmission lines extending left and right have a characteristic impedance of around 800 ohms. When the switch is closed those transmission lines look exactly like an 800 ohm resistor each. Therefore, the circuit at t-0 is equal to a battery and two resistors going to a bulb. This means that the bulb will light dimly after the time delay because of the 1 m separation. This is easy to see if you replace the two transmission lines with resistors that are 1 m long. The light will then flicker as the reflections bounce back and forth and eventually equalize to the simple DC circuit of the bulb and battery.
@@Observ45er this cannot be true. If it was your computer would recieve noticable current every time you plugged it in. The power supply circuits recieve power and then nearly instantly induce or via capacitance would destroy all the ic circuits as they recieve the induced 12V power from the rails. The induced current or capacitance here is negligible and if it wasnt computers as work in life would fail immediately.
Exactly. I can not see how he claims there will be a 20mA return. For an example, if I placed several wires in parallel near the second "receiving" wire, then would all of them have 20mA? Surely that would violate some conservation principle. Neither wire would know the circuit specifics in the time the pulse would propagate over, so what makes this specific wire special?
@@MrThedumbbunny it is however correct, that is just how it goes. It is what it is because it is a distributed system with a uniform structure and hence it obeys those laws.
Another awesome video that is not just providing some useful information but is helping develop an intuitive understanding of signal propagation. Great work, Robert!
Breathtakingly clear explanation of how Electricity works there by Eric Bogatin (I'm at 7:57 right now). I won't forget this explanation. Kudos to Robert Feranec as our host! thanks for posting! :D p.s. and of course to Dr. Derek Muller for having started the conversation in the first place! :D
_Robert, I noticed that both you and Eric were sitting in outer space, albeit on opposite sides of the planet._ 🚀 _So I was waiting for you guys to string up two light-seconds of superconducting wiring for the "bench test."_ 🤣
I loved this video!! unfortunately the last part lacked some detail... The circuit created in the original video, as Eric perfectly described has 2 transmission lines and the reflections can make the situation quite complicated. Depending on the Zo of the Tx lines (in the video two simple cables separated 1m by air) can be in the order of 800 ohms (a 1mm diameter cable and 1m separation using the standard equation for a bifilar tx line), if the propagation speed is considered to be C (time delay to reach the end is 1 sec), and just to take a number consider that the light bulb is 100 ohms, the multiple reflections (and transmissions) from one Tx line to the other will create a time constant that can last tens of seconds (with steps every 2 seconds) to get to the steady state value. This is quite simple to simulate in Pspice. The instantaneous current through the light bulb is V/(R+2*Zo) (neglecting the V internal resistance), which naturally will be extremely small because is displacement current between two wires spaced 1m appart.
I've still not seen anyone explain this correctly, as Federico Di Vruno has mentioned, and now even the Master of SI himself has ignored the details that so many engineers are bothered about on this problem!
@@naszadynastia a wire with a shield around it is nothing more than another transmission line. Displacement current will still occur, depending on how the shield is connected to the other parts of the circuit it can have some effect… conduction current (ie stationary regime)is not affected at all by shielding.
YEEEEEEEESSSSSSSSSSS!!!! i watched veritasiums video and many responses to that video and this is by FAR the best explanation. I have been lucky enough to sit in seminars with Rick Hartley and Dan Beeker who stress this idea that "energy doesn flow in the wires but the space around them". thanks for posting this video.
In terms of this explanation, the "Veritasium" question then becomes "What is the input impedance of two conductors spaced 1m apart (even allowing Derek his 0 ohms of resistance), and what would be the current flow when a single digit voltage is applied to that transmission line, and would those femtoAmps be enough to light the bulb in 3ns as he claimed?" Thanks for this version of the explanation.
Exactly, his example question has actually nothing to do with the point he was trying to make (the poynting vector). I guess he tried to come with a practical example.. and thats the best he could manage, because for DC the whole poynting vector just doesn't matter, so he couldn't find an example where the standard (simplified) model of "power travels in the wire" breaks down. It get interesting in high AC with the skin effect, so why for example silver coating cables can make sense, while it doesn't for DC etc. but thats again slightly different.
@@georgelionon9050 I dunno... I was taught, decades ago, the transmission lines could be considered an almost infinite number of stacked "low pass" RCL ccts... Even giving Derek his 0 ohm conductor, the miniscule Farads of "capacitance" between the conductors wouldn't cause many electrons to move ("displacement current"?) to establish a field that transfers energy... My gut says the load wouldn't have a noticeable reaction until the fields that travel the length of the conductor meet at his bulb. Consider electrons streaming out of the -ve pole of the battery (DC)... As their "virtual" members charge up one plate of the virtual bridging capacitors, their electric field will repel other virtual electrons across the gap of the virtual capacitor... Those will flee, "seeing" one path to the resistance of the bulb's filament, OR an alternate path in the opposite direction at the exact same voltage... Meanwhile, the travelling magnetic field following the leading edge of the pulse is coaxing electrons to flow in the other direction. Would the two effects cancel ? I dunno...
@@rustycherkas8229 Honestly I don't have a good calculation at hand how long the LED would light up and how strongly. One meter is quite far apart. However note the way Derek sneaked in the conditions for his questions, he said "light up" not "starts lighting" and that he says in the beginning his ideal bulb will light for "any energy flow". So under this conditions the answer is 1m/c. Also the answer is d/c for any such bulb in the universe, no wires required. Thats why I say, his example was bad for the effect/argument he wanted to make.
Wow, I think this is the fourth evaluation of the original video I've watched, and this is probably the best. Thanks. One thing: I thought of the long parallel transmission lines as a sort of transformer or even antennas but I guess that's just transmission line theory in a special case....
That was really interesting, when Mr Eric spoke about Maxwell equations and quantum mechanics. I wouldn't even mind if he continued the talk and then later came back to the explanaition of the circuit below.
Great video! Now understand much better why placement on PCB are important. It was nebulous to me at first, but it's much clearer now! Thank you for you time guys ! Love how internet brings us together. It help a lot !
Great video, and I love Bogatin, but I think he’s not quite right at the end. The “ideal-ish” LED will blink briefly as the displacement current activates it from 1m away, and then it will turn back off until the circuit begins to settle ~1s later. After the displacement wavefront passes the LED, what transient condition exists across the LED terminals? There’s no longer any current flow until after ~1s when we start to see the LED turn on permanently as the reflections settle out.
I'm replying just so that I'll be notified if this gets answered. The wave is propagating down the line, but it isn't clear to me that there are new changes at the origin until a reflection gets there. I told myself it was current being conducted back from new displacement at the wavefront, but ... I'm not confident of that, nor am I confident that a Doppler-like effect wouldn't at least decrease it. Of course, I also don't quite get why it comes up in "10 steps", instead of after 1 step, or at least in a zeno's paradox of 1/2 the way each step forever.
@@jimjjewett I think the displacement current will be enough to sustain the LED, and there's many ways to explain it, but this is my layman understanding using lumped elements (please correct me if I'm wrong): As displacement current is produced through each section of wire, charges are depleted in that section, which causes current to flow from the adjacent section. In the top circuit, only one wavefront is produced, which means there's displacement current in only one section of wire at any given time, but conduction current is supplied by the source, which produces a continuous current between the source and the wavefront. In the bottom circuit, there are two wavefronts producing displacement current in the same direction, starting at the same spot. So immediately the displaced charge from one wavefront is picked up by the other wavefront, such that there's always a continuous current between the two wavefronts. EDIT: Might be completely off-base. This explanation seems to imply that you can throw the switch, remove the source, and the LED will stay lit from the displacement current due to the wavefronts alone. Something's definitely missing here lol
Displacement current is related to transmission line. It flows between the capacitors. Led is far from source. So displacement current is flows between transmission line first. After some time it conventional current reaches to led then led will glow.
When we have a look at Eric's graph, it says, 50OHM all the time until it's back ... so, that would mean 20mA flowing all the time when the edge is travelling the way there and back. If we are talking about the bottom circuit (led, battery, switch, resistor placed in the middle), the current would be flowing through the LED when the edges would be travelling. Or?
I have a fast ohm meter; GHz scope and ns pulses from a photodiode illuminated by a laser. Me yelling “half current before reaching full current!” at the end! Very much enjoyed!
I'm going to do everyone a favor and just tell them that the wires are already charged before the circuit is complete. That's how I always treat them, and for most applications, it works just fine.
but when the switch is closed, the whole electron chain will simultaneously move together creating a current, or only electron at the switch move first and the movement propagate to the rest of the wire?
In the final explanation where the lamp and switch are in the centre and he says that the lamp turns on immediately, I suggest that it actually turns on slowly (in relative terms) because until the wave has propagated far enough down the line to where the capacitance is high enough to couple sufficient current across the lamp can't light. Comments welcome.
Could someone please explain about the resistance and work perform and the current used up by the LED, to the return line, shouldn't after the LED the voltage to be less???
In electromagnetic theory you have LOCAL ENERGY CONSERVATION the relationship between energy density and WORK(fields on charges and charges on fields) and FLUX. Poynting vector is needed for calculation of flux. Poynting vector points in direction it does so that surface integral for flux gives correct answer. Poynting vector is part of local energy conservation and that's the point to remember. Local energy conservation applies everywhere of course, current or no current.
The impedance measured at the input end is Zo only if the far end is terminated with a load of Zo Ohm. If you remove the load, you won't have a 50 Ohm to take your 20mA at the switch.
Hello Robert, great idea to cooperate with Dr. Bogatin in this topic! I have one question regarding this issue: For me both cases described are pretty clear, but what would happened if two wires instead of layout from top picture (transmission line like PCB) and bottom ( from veritasium) would be placed as one big loop (signal lanes very far apart from each other and LED far from voltage source)? The coupling between signal and return would be extremely low. How this signal would propagate then? I would bet that principle is the same but due to very high impedance between lines there would be extremely small displacement current and due to very long risetime of this signal we would need more time for voltage to become stable. What is your/Dr. Bogatin's opinion? Thank you for this video :)
Excellent explanation using the capacitor. Changing electric and magnetic fields occur at the speed of EM radiation (ie light). So, everyone has a clock speed in our brains of the speed of light. Amazing!!!
I think the example below is wrong. the transmission line on the left side of the LED is not a return path, the impedance on that line is not 0. So you can't use the 50ohm impedance to calculate the 20mA transient current from the incoming line to that line. There is no such a model in the real world.
A little hard to follow. I prefer to think of the activity at the atomic level. For example, one end of a 6 inch copper wire is connected to the switch and the other end is left unconnected. The other side of the switch is connected to the negative battery terminal. When the switch is closed, excess electrons will begin jumping from atom to atom as they travel down the wire. After about 1 ns, when they reach the end of the wire, the flow will stop and the potential at the end of the wire will be the same as the switch. If a 6" return wire is added and connected to the positive battery terminal, the electrons will continue for another 1 ns back to the battery, except the flow will not stop. Using light with a mirror to return the light will produce a similar round trip delay, except the delay would be a bit less because light travels faster than electrons in a wire.
There are three situations where electrons DO carry energy ... in a vacuum tube, a semiconductor, and a plasma. Plasma filaments exist in space that are hundreds or thousands of light-years long. They consist of electrons and protons (or positive ions), all on the move. Veritasium doesn't get beyond 18th century electricity experiments where plasma was unknown.
Why does everyone focus on this one part of Veritasium's video? Here are my questions. 1. All standard wiring is done with the hot and neutral wired side-by-side. This is required by the National Electrical Code (NEC 300.20(A)) to prevent inductive heating of metal conduit. It works because the magnetic fields produced by the hot and neutral are in opposite directions and cancel each other out when they are side-by-side. If the magnetic fields of the conductors have been canceled then the cross product of the electric and magnetic fields must be zero, so what remains to conduct the energy outside of the wire? 2. If the energy is carried outside of the wires then why does resistance in the wire dissipate energy, why doesn't the energy just keep going as if nothing was in the way? 3. The cross product is the product of 2 linearly independent vectors and is perpendicular to both of them. Where this is demonstrated at 7:34 in the original video it is not clear what vector Veritasium is using to represent the magnetic field since the field seems to form a cylinder around the battery. How he representing the magnetic field as a vector to form the cross product with the electric field? 4. In the 2 conductors (towards the load and back) the magnetic fields are in opposite directions but the electric field is in the same direction for both conductors. How can the cross products of the electric field and the magnetic field of the two conductors point in the same direction, from the source to the load, when their magnetic fields are in opposite directions? (8:00 in video)
The main problem with Veritasium is the discussion of the Poynting vector, which is an artificial construct to describe EM radiation pressure and behaves inconsistently when trying to use it in electric circuits. Feynman talks about it in his Lectures book volume II
Assuming a 1 meter spacing of the two wires, and let’s say 10 gauge wire… you have a characteristic Zo of 798 ohms on each TL leg (looking left and right from the switch) So if we use a 12v battery … looks like we have a near instantaneous (transient) current of about 7.5 mA before we reach steady state. Not knowing the bulb characteristics- we don’t know if that’s enough to illuminate or not…
I truly loooooove this this video!!! I do like varitasium but this explained it sooo much better and complete. I could follow the explaination with ease. It really explained what would happen if this and that! I do have a question, in the video at 35:59 Eric mentioned that the switch would be the source of the signal. I answer my own question below: If you would have your switch on the moon and your led and battery next to each other and you turn on the switch the wave front wouldn't start at the battery but it would start at the switch cause that is where you open the "gate" let's say. So the source of the signal si still the battery BUT the source of the voltage wave is the switch! Can I also say that Eric, you are so incredibly cool! Your knowledge and your way of explaining such complex systems in a very engaging and just lovely way! It get's me fired up! And I would LOOOVE to be in one of your classes one day... Maybe, it's a little far away but if I ever get the chance, I would take it in a hartbeat!!! Mind you I would never be able to finish a entire class but I would LOVE to be one in or two lessons! A;sp for Robert! The things you can learn from this channel! I started watching you to learn to use altium and it's MANY MANY features, and now I am learning a little bit about the quantum electric behaviour on transmition lines?!?!? Like what? This is just insanity! just incredibly! Love every single video that I have watched!!
Great discussion. It seems that Eric Bogatin should be wearing a shirt saying "Warning: cannot be turned off after being asked a question about a electric circuit".
I have the feeling that more and more videos are popping up to discuss the Veritassium video. Unfortunately, most of the other videos make the subject more fuzzy rather than explaining it. The simple truth (as we know today, unless proven otherwise) is that we have to solve Maxwell’s equations to get the E and B fields for the concrete setup. That means that all materials have properties of permeability, resistivity and susceptibility. That includes the wire and air plus insulation of any sort. If your 600000km long wire has more resistance that the path through insulation and air ( looking at E and B fields, not necessarily just DC resistance), then the fields at the long ends (or where the loop closes) are very small and do not contribute much to the energy transfer. Looking at the poynting vector, if you have ideal wires the you have no E-field along the wire and hence the Poynting vector goes straight from source to load. With a real setup, there will be losses and the Poynting vector field can be distracted outside source and load, representing distribution characteristics of the system. There is no need to talk about distinctions between frequency ranges as the capacitive and inductive properties are already frequency dependent. One just needs to represent them properly in a system setup before solving. And that means that there is no such thing as an ideal 300000km long loop without capacitance, resistance and inductance. If there were such a thing, it would not have a voltage drop and hence can be represented with a short across it at the source and load.
Thank you for the interesting video! I wonder what would happen if you increased the distance between the transmission lines. For example if the switch was on the north pole and the LED on the south pole. Wouldn't the capacitive coupling be negligible?
Thank you kramer3d. TDR is what our call supposed to be about :D But then Derek posted the video, so we talked about it. However TDR is sill on the list.
I think the biggest confusion among all explanations on this topic is that a line represents a connection on a schematic but a transmission line in the context of all of these videos. Once you realize that you have to actually replace the transmission line with RLC components, the solution is at hand.
So to summarize. The answer is that the light source will light up immediately, but only if characteristic impedance of the transmission line, power source output resistance and light source are matched. My problem with Veritasium video is, that I don't thing they are.
And for the people who can't make sense of the idea that there's current from an "open circuit". Another way to think of it is that information will not transfer higher than the speed of light. Meaning you can't know whether the led is connected or not without that information transfer from the far distance through the speed of light.
He is right about how two wires close together will actually form, and have a transmisson line impedance, and this is very inportant in high radio frequency transmission lines, but that is getting way beyond the original queston.
Eric quickly mentions that there are two transmission lines in the second diagram, however, if so, that statement should also apply for the first diagram. In any case, for the second diagram he says that two signals emanate from the switch when it is closed. However, the voltage across the ideal cell remains at 2 volts at all times, and the voltage along the conductor from the cell's negative terminal around the circuit and past the LED remains unchanged, so there is not any associated displacement field to the right of that cell, so no displacement current would flow. Does the pulse also travel through the cell raising the voltage on the 'negative" terminal of the cell? With regards to the voltage divider concept, the voltage on the load side of the source impedance (across the switch) is initially two volts, then as soon as it closes the voltage on the load side of the source impedance drops to one volt with respect to the cell's negative terminal, so what happened to the electric field (represented by the extra volt) that was across the switch contacts? The associated field must have dissipated its energy some place. Was it radiated as an E-M wave? Can anyone clarify? It seems that Eric's comments assume an "instantaneous steady state condition" at various points in time as the pulse travels down the transmission line, but remains unclear what happens after the switch is closed, but before the pulse starts its journey down the conductor. Does the pulse travel in both directions?
Instead of dismissing/waving hands like Dave from EEVblog did, you guys made the whole thing meaningful. And it's actually more helpful than not for even experienced electrical engineers.
Dave from EEVblog is stuck with his own opinion which he claims to be fact.
@@donmab The difference for me is that I can actually understand what Dave is saying. Maybe it's because I'm also Australian, but the point still stands.
@@brugj03 isn't that pretty much what one's opinion is supposed to be? Weird thing to say.
@@ScoochieR Honestly, I think that by introducing complexities like resistance and loss, EEVBlog may be more practical, but he is using it in a way that it just clouds the actual working of the system and saying that everyone already knows about this is just not true. That's not the point of the original video which is trying to explain first principles of EM. I do agree that Derek may have done a poor job by starting with non-ideal components and just claiming them to be ideal, whereas you would just choose a PEC in a lecture, but I guess this makes it more relatable to people.
@@ScoochieR I vant to knov vay you call yourself a "Ruski robot" ven you are Australian?
The video Dave did on the Varitasium was not too detailed he pretty much agreed with most of what was said by Varitasium in relationship to principles of transmission lines and propagation. Actually it was probably one of two of his videos I was able to sit through right to the end, normally I have to turn him off because he tends to just go overboard and overcomplicate a generally simple subject.
This is a nice explanation of the problem.
It remined me that in my first job after college many years ago, I was testing a circuit that was built from TTL and I wanted to look at the output of a 7474 D flipflop. So I fastened a wire twisted pair to the output and ground and ran it several feet to an oscilloscope to look at it. to my surprise, the Flipflop would not stay set when it got appropriate inputs. This is how I remember it, but it was a long time ago.
After further investigation, I determined that the Flip Flop was actually setting then resetting in a very short time. It did set properly when I put the scope probe directly on the output with the wire detached, and with wire attached you could see that it was momentarily set.
Then I realized that the scope probe was a very high impedance and the line was reflecting the initial edge from the Flipflop setting back to the chip pin. AND unfortunately the D output was internally connected to the opposite side input of the output latch (not buffered), and the reflection reset the flip flop.
So that was a lesson for me. Care is needed in higher speed measurements.
I finally understand this now after 45 years. It's never too late! The topic he explains is the course that washes out a lot of would-be EEs.
As always, Eric is a pleasure to watch and listen to. I love how this video brings together a lot of different aspects of physics and engineering. Great lecture!
This is the best explanation of this subject I have listened to so far.
I've seen a few reaction videos on the same subject already but yours is the best 😀
It's great that you explained the difference between the two cases, that was extremely important.
Thank you both 😊
I saw the Veritasium video and was waiting for folks like Eric to peel the onion back more.
He didn't break out a smith chart. I always liked those in school.
It is not correct to treat the Veritasium arrangement as a pair of single impedance transmission lines. You aren't driving the two ends of your "transmission" lines with the voltage source. For the first few nanoseconds, the lower pair looks like a dipole antenna. The bulb line, also looks like a dipole antenna, receiving the signal. A dipole arrangement does not have a single Zo - the step response is actually pretty challenging to figure out. My point is that you cannot assume the current in the transmitting side is equal to the receiving side - there will be coupling, but I contend a fairly small current - not enough to realistically light the bulb. The beef with the original video was not that some signal arrives 1m/c seconds after switch closure, but that there was implied full power transfer. It won't be.
@bobwhite137 They've got it right. It's not an antenna, because when we place two antennas that close together, we create a highly-coupled transformer, not antennas anymore. (Same thing would happen with loop antennas: if we put them right on top of each other, we now have a transformer, not a xmit/rcv antenna pair.) We also get a capacitor where the plates are formed by two close-spaced wires.
So which is it, a transformer, or a capacitor?
It's both, and Heaviside discovered, when the two concepts are combined, the reactance of the transformer and the capacitor cancel out to form a virtual resistance. If we could use a DVM to measure Veritasium's million-meter ideal wires, we'll see a 700-ohm resistance, rather than 50-ohms of standard coax cable.
To act like dipole antennas, the wires need to be separated by about 1/4th the wire length, so the coupling isn't anywhere near 100%.
Yes, this appeals to my intuition too. The arrangement is like antennas or even a transformer. But probably the antennas make the most sense. This was pointed out in another video discussing the original video. All of it is instructive though!
@@wbeaty: "One fourthTH"? Oops!
I think that current starts after 1/c (in seconds) and it is slowly building up until the RC circuit practically is charged after 5 times RC. Since we have a switch, we have to study the step performance (transient) of the circuit and treat it as an RC circuit (until it reaches the steady state which is 5 times RC practically). If we assume R=0 then RC is zero then the circuit charges in 0 seconds. Inconsistencies arise when we assume R=0 and C some value. Using the notion of the displacement current, the transmission line, the transformer, the antenna, the cable impedance is a good tool to understand the circuit. But i consider it as an RC circuit since we have a switch and therefore i am analyzing the step performance of the circuit. The problems arise when we assume a practically infinite and distributed capacitance with no resistance...
@@wbeaty 'a highly-coupled transformer' Not even that highly coupled a meter apart, you'd consider any low voltage transformer a failure if you move the coils one meter apart..
The electrons are literally IN the wire. They don't actually touch, so the energy is in colliding the field shells. The field extends from the electron, and is both inside and outside the wire. That you can tap the energy of the moving electrons both from inside the wire, and somewhat outside the wire, doesn't mean the energy source isn't still the energy in the moving electrons, and is still basically stuck to the wire. Make the wire take a 90 degree turn and the outside field takes that same turn, you can't make it just go straight off the wire because it's not really flowing outside the wire.
It takes special pains like many coils and a core to efficiently couple the magnetics of the flowing electrons of one wire into another, and efficiently transfer power from one wire to another. It takes other special pains to use the flow in an antenna and peel the energy off the wire into EM radio waves and similar.
If it were really just that easy and all the energy was really outside the wire, transformers and antennas wouldn't be their own specialized fields and take pretty careful planning and adjustment to make sure they even work in an OK manner..
Route a wire all over a room, bending it in different paths, and try to get the waves to do anything that isn't strictly confined to following the path of the wire. Won't happen. Just because you can tap the energy and affect it from the outside doesn't mean that energy isn't coming from the parts that are stuck in the wire, it just means they have a field around them.
I'm so lucky that I came across this video, much love and appreciation.
By far the best explanation of this chapter in electromagnetism in just one hour! Highly recommended!
Eric Bogatin is my PCB Mentor
I love his way of explaining things. There is even a risk that I might start understanding transmission lines even though I have been avoiding them thus far.
I watched a lot of videos, and finally understood what is happening due to this great man, Eric. Amazing !
One of your best videos yet !
Thank you for giving us a great learning opportunity from someone who literally wrote the book!
What this explains is how so many of us are conditioned to view things in a defined way. The first circuit makes us feel safe and knowledgeable as we are so used to seeing it this way but when confronted by the second circuit we suddenly develop anxiety and stress because we are conditioned not to really understand circuits but memorize them.
The memorization over understanding is part of the game. Understanding that electricity is like sound and can be amplified to any value without loading the source or adding more energy would make investors in power plants freq. out. You don’t need a new power plant, just an unbalanced transformer.
🤔✊🏾
"sometimes a lie tells more about the truth than the truth itself"..........."strategic simplification"........love it!
Finally the correct explanation! It infuriated me to no end that soooo many electrical engineering youtubers' explanations were so misleading to the point of being wrong! This person is probably the very first one to correctly explain the precise physics of signal propagation! Thank you! Now my soul can finally rest knowing that the world (or at least your viewers ;-) ) have received the correct, physically-accurate explanation.
He did it as an EE engineer. Taking the coaxial 50ohms cable as an example was a perfect idea, even though it's also quite confusing. (the original question is very high impedance pair, 1m apart)
Physics engineers do explain it with their own formulas. That doesn't make them wrong.
Excellent video, thank you so much for posting it. I have seen others with Dr Bogatin and after watching Veritasium's I hoped that in the responses that were likely to be posted we could see his discussion on the subject. Wish granted, what a treat!
To top it up 3 months of free access to his lectures, for "our patience". Dr Bogatin, I get your passion and it takes no patience at all on my part to watch your talks and lectures so I cannot thank you enough for this.
The best explanation I have ever seen of signals and transmission theory. @Dr Bogatin: Very well done. And thank you to Robert Feranec for inviting him!
Unfortunately, the problem presented to Dr. Bogatin, is NOT the problem presented by Veritasium. Though I totally understand why it was presented in the manner it was, nevertheless, the Veritasium problem has a Physical Demonstration to it that people who view the video get to see. That demonstration makes certain things "tangible" to the viewers and also modifies how we perceive it and how it should be perceived.
Let's review:
1. The demo shows a (car) battery, an LED bulb, insulated conductors. These conductors are then given the hypothetical "no internal resistance" and they are 300,000km long, and separated by 1 meter. When the switch is thrown in the demo, the LED turns Solid ON. It simply remains On.
There is literally NO mention of Blinking on. Clearly the definition of "ON" here is suspect if one were to consider "blinking" on as "ON".
The question offered by Veritasium was after the switch closes, when does the LED turn on? (0.5sec, 1s, 2, 1/c (sic), and None of the above).
After listening to Dr. Bogatin, (who clearly knows the material), one would believe that current flow across the bulb would be enough to power the LED. However, let's review some of the facts that everyone seems to leave out.
#1 These are insulated conductors not capacitor plates sitting on dielectrics. The conductors are also not Bare Wires, they are insulated. Please get that fact straight.
The impedance of insulated conductors 1 meter apart will be markedly different than the simulation because the simulations do not include the fact that there is insulation on the wires.
#2 The LOAD is a 12 volt LED Bulb. These "bulbs" are internally made of many LEDs in series. Each LED in series has a forward voltage that MUST BE exceeded in order for the LED to turn on. Nowhere in any simulation are you taking into account the forward voltage requirement of the circuit.
#3 THERE IS NOT AC SIGNAL. Dr. Bogatin refers to the signal frequently because he is equating the circuit with typical signal. BUT, there is NOT AN AC Signal in this example. There is only DC, and the instantaneous turning on of the Switch. There will be a very brief pulse. That is it.
#4. If you take all these elements: Insulated Wires, DC, Not AC, 12 volts, 12 Volt LED, Forward Voltages of the Diodes, No signal. It results in at best... the microscopic blink of an LED, nothing more, for at best a moment in time. I am sorry, but I simply cannot equate this to "The Light Turns ON".
Please show a simulation that corrects the above noted issues, and let's see how it plays out.
@@LiveType I am not sure. I guess it depends on whether he values his integrity. Because, the reason why I watched his videos was to get to the deeper truth of things. If this is how the future of his videos will play out, he lost only 1 viewer. It may only register as a blip on the magic light bulb hehehe but for me, I value the integrity of the process. It really bothers me to see this level of misdirection by someone I really thought knew better.
Sounds like Eric was really getting into the details there, would love to hear him discuss more on how Maxwell's equations have been simplified to the circuit elements we use today.
There's an article in IEEE Spectrum in 2014 called The Long Road to Maxwell's Equations. The article tracks the progress of understanding electromagnetics from Maxwell's predecessors to Oliver Heaviside's transformation of the equations to their modern form.
It's frequency that transmits he didn't mention that. So I give Eric thumb 👎
@@fyfoh Missed this, I'll check it out, thanks!
Fantastic job explaining this Eric. This is exactly how I learned these concepts back in college.
As soon as I saw Veritasium Electricity video it reminded me about Eric Bogatin. Thank you Robert for the video!
This is amazing ! Just watched Veritasium's video few days back and wanted to know this very perspective. Thanks Robert !!
Thank you very much Girish
Thanks, best explanation of this idea.
Great talk! Thanks to both of you for taking the time to do this.
This video has changed my level of thinking towards electronics..
What a pleasure to watch this. Thank you Eric.
Thank you so much Prof. Eric Bogatin and Robert for these wonderul talks. Looking forward to more.
I was building my first PCB circuit for wired data transmission over RS-485 line (no electrical engineering education), and the terminology was very confusing: termination resistors, impedance matching, etc. While I did all the things "by the book" and followed various instructions, this explanation opened my eyes of what is really going on and what is actually important.
Excellent. A piece of test equipment called Time Domain Reflectometer uses this. You mentioned this as I write. If you know the impedence of the transmission line and put a resistor of the same value at the end the TDR will give you the distance of all the impedence changes down the line. Near shorts and opens and the distance of the same down the line.
Wow, great video. Using an example makes studying Electromagnetism and Transmission lines much more interesting.
Haha, finally a brilliant explanation from a real physicist. Thank you guys!
Thank you both Eric and Robert!
In the section "Explaining transmission line..." there's an over-complication. The reason why you have instant current is because this is where you inject the power and make a measurement. The transmission line is irrelevant. What you measure takes another path through your measuring lines. (I didn't have patience to watch further than 10:33 so it may come up later, but there is no excuse for confusing the issue at the beginning)
While agree with John that this video is better than anything else I must say that I am rather frustrated by the original video, the EEVBlog video and also this video, and the overwhelming number of comments. Why is everyone ignoring the simple facts given in the original video: the lines are 1m apart, they do NOT form a 2m thick coax cable, the copper inside the lines must be in the order of 1mm thick, the lamp is big and will consume in the order of 1W, the energy source is a "normal" battery 6 or 12 V, not 1000000V or so.
I have by now seen so many "physic professors" that apparently due to their mission to correctly explain the physics behind a transmission line completely oversee or deliberately ignore the problem statement.
A decent (and therefore attentive) physics professor should also understand the following very simple facts. a) Coulomp's law according to which the electric field strength decays with 1/r^2, implying that also the time derivative of field strength decays with 1/r^2. b) The time derivative of a constant is zero.
Now be my guest to calculate how much energy can be transmitted from some short, e. g. 30cm long section of the wire near the switch to a corresponding wire section near the lamp by a single pulse of e. g. 1ns. Even if one doesn't have that minimum amount intuition to understand that in the order of 1W will certainly not be transmitted (just think what consequences that would have on every day life), it is anyway clear that after 1n the light would be off until the field wave has travelled through the cable.
I am also astonished why none of the "professors of physic" goes to their lab, take e. g. 40 m cabele (instead of 4*300000km) , pick a pulse generator to generate a (off->on) pulse, a lamb or resistor and a scope and do the experiment. This takes 10min. The result shows that the answers given in all videos is wrong!
I better stop here...
I guess you're right. Eric goes too deeply to solve the problem that he forgets that this is a DC voltage supply that provides DC current. Okay for very tiny amount of time when we push the switch due to maybe high speed transition the LED may get energy wave through the space but that's not enough in terms of both power and time to keep LED turned on forever...the permanent thing that keeps it turned on must be the other form of moving energy in copper which is domino effect in which energy moves from on electron to the other to go ahead, and Eric himself in his book mentions this but here he wrongly goes on to explain it like a high speed pulsed signal.
The over-simplifications in Veritasium's video did exactly the opposite of the expected results: More confusion.
I learn from you since I took your pcb courses. Great thanks.
Unfortunately, any photonic coupling between two wires 1 meter apart is going to be negligible. Photons are generated from the surface of the wire and then spread out before they hit the second wire. If the radius of the wire is 1 millimeter, then the photon cloud will have been attenuated 1000-squared = 1,000,000 times when it hits the second wire. So there's no getting around it. You're really going to have to wait a full second before the bulb lights.
But the assumption from the origional videowas any current turns the light on. Technically correct but misleading.
Edit i do wish he had actually gone over the "capacitance"of 2 cables 1m apart to show the true current across it mathematically. Not sure the standard equations apply though so might have to derive for custom circumstances.
The two transmission lines extending left and right have a characteristic impedance of around 800 ohms. When the switch is closed those transmission lines look exactly like an 800 ohm resistor each.
Therefore, the circuit at t-0 is equal to a battery and two resistors going to a bulb.
This means that the bulb will light dimly after the time delay because of the 1 m separation. This is easy to see if you replace the two transmission lines with resistors that are 1 m long.
The light will then flicker as the reflections bounce back and forth and eventually equalize to the simple DC circuit of the bulb and battery.
@@Observ45er this cannot be true. If it was your computer would recieve noticable current every time you plugged it in. The power supply circuits recieve power and then nearly instantly induce or via capacitance would destroy all the ic circuits as they recieve the induced 12V power from the rails. The induced current or capacitance here is negligible and if it wasnt computers as work in life would fail immediately.
Exactly. I can not see how he claims there will be a 20mA return. For an example, if I placed several wires in parallel near the second "receiving" wire, then would all of them have 20mA? Surely that would violate some conservation principle. Neither wire would know the circuit specifics in the time the pulse would propagate over, so what makes this specific wire special?
@@MrThedumbbunny it is however correct, that is just how it goes. It is what it is because it is a distributed system with a uniform structure and hence it obeys those laws.
This was a really great example and very well explained by Eric, and I only know the very basic electrics.
Holy crap this guys taskbar is panic inducing 😵💫🤪
Another awesome video that is not just providing some useful information but is helping develop an intuitive understanding of signal propagation. Great work, Robert!
Breathtakingly clear explanation of how Electricity works there by Eric Bogatin (I'm at 7:57 right now). I won't forget this explanation. Kudos to Robert Feranec as our host! thanks for posting! :D p.s. and of course to Dr. Derek Muller for having started the conversation in the first place! :D
Great video! You guys really did this topic justice
I think this is the best explanation about this paradox.
_Robert, I noticed that both you and Eric were sitting in outer space, albeit on opposite sides of the planet._ 🚀
_So I was waiting for you guys to string up two light-seconds of superconducting wiring for the "bench test."_ 🤣
Lol 😂
Thank you very kindly
Great video! I Was waiting for it
This was a really good explanation. I definitely have a deeper understanding about current now.
I loved this video!! unfortunately the last part lacked some detail... The circuit created in the original video, as Eric perfectly described has 2 transmission lines and the reflections can make the situation quite complicated. Depending on the Zo of the Tx lines (in the video two simple cables separated 1m by air) can be in the order of 800 ohms (a 1mm diameter cable and 1m separation using the standard equation for a bifilar tx line), if the propagation speed is considered to be C (time delay to reach the end is 1 sec), and just to take a number consider that the light bulb is 100 ohms, the multiple reflections (and transmissions) from one Tx line to the other will create a time constant that can last tens of seconds (with steps every 2 seconds) to get to the steady state value. This is quite simple to simulate in Pspice. The instantaneous current through the light bulb is V/(R+2*Zo) (neglecting the V internal resistance), which naturally will be extremely small because is displacement current between two wires spaced 1m appart.
I've still not seen anyone explain this correctly, as Federico Di Vruno has mentioned, and now even the Master of SI himself has ignored the details that so many engineers are bothered about on this problem!
@@HelloWorlds__JTS see here for a bit more detail…
And how energy is transferred if wires are closely shielded ?
@@naszadynastia a wire with a shield around it is nothing more than another transmission line. Displacement current will still occur, depending on how the shield is connected to the other parts of the circuit it can have some effect… conduction current (ie stationary regime)is not affected at all by shielding.
Exactly! That was my first answer, the LED will start instantly but very dim. Also Dave on eevblog missed that out.
Thanks Robert and Eric! Great video!
Fantastic video, I really enjoyed this. Thank you for uploading.
Oh wow thank you Eric for your offer I am glad I watched to the end :)
YEEEEEEEESSSSSSSSSSS!!!! i watched veritasiums video and many responses to that video and this is by FAR the best explanation. I have been lucky enough to sit in seminars with Rick Hartley and Dan Beeker who stress this idea that "energy doesn flow in the wires but the space around them". thanks for posting this video.
It's depending on the time (stable vs dynamic state) that we examine the signal I guess?
There aren't a lot of people on earth who understand this subject as well as this guy.
Excellent video about the topic.
Brilliant! Thanks for your work Robert !
In terms of this explanation, the "Veritasium" question then becomes "What is the input impedance of two conductors spaced 1m apart (even allowing Derek his 0 ohms of resistance), and what would be the current flow when a single digit voltage is applied to that transmission line, and would those femtoAmps be enough to light the bulb in 3ns as he claimed?"
Thanks for this version of the explanation.
Exactly, his example question has actually nothing to do with the point he was trying to make (the poynting vector). I guess he tried to come with a practical example.. and thats the best he could manage, because for DC the whole poynting vector just doesn't matter, so he couldn't find an example where the standard (simplified) model of "power travels in the wire" breaks down.
It get interesting in high AC with the skin effect, so why for example silver coating cables can make sense, while it doesn't for DC etc. but thats again slightly different.
@@georgelionon9050 I dunno... I was taught, decades ago, the transmission lines could be considered an almost infinite number of stacked "low pass" RCL ccts... Even giving Derek his 0 ohm conductor, the miniscule Farads of "capacitance" between the conductors wouldn't cause many electrons to move ("displacement current"?) to establish a field that transfers energy... My gut says the load wouldn't have a noticeable reaction until the fields that travel the length of the conductor meet at his bulb.
Consider electrons streaming out of the -ve pole of the battery (DC)... As their "virtual" members charge up one plate of the virtual bridging capacitors, their electric field will repel other virtual electrons across the gap of the virtual capacitor... Those will flee, "seeing" one path to the resistance of the bulb's filament, OR an alternate path in the opposite direction at the exact same voltage... Meanwhile, the travelling magnetic field following the leading edge of the pulse is coaxing electrons to flow in the other direction. Would the two effects cancel ? I dunno...
@@rustycherkas8229 Honestly I don't have a good calculation at hand how long the LED would light up and how strongly. One meter is quite far apart. However note the way Derek sneaked in the conditions for his questions, he said "light up" not "starts lighting" and that he says in the beginning his ideal bulb will light for "any energy flow". So under this conditions the answer is 1m/c. Also the answer is d/c for any such bulb in the universe, no wires required. Thats why I say, his example was bad for the effect/argument he wanted to make.
Wow, I think this is the fourth evaluation of the original video I've watched, and this is probably the best. Thanks. One thing: I thought of the long parallel transmission lines as a sort of transformer or even antennas but I guess that's just transmission line theory in a special case....
That was really interesting, when Mr Eric spoke about Maxwell equations and quantum mechanics. I wouldn't even mind if he continued the talk and then later came back to the explanaition of the circuit below.
Bogatin: a great teacher!
Great video! Now understand much better why placement on PCB are important. It was nebulous to me at first, but it's much clearer now! Thank you for you time guys ! Love how internet brings us together. It help a lot !
Great video, and I love Bogatin, but I think he’s not quite right at the end. The “ideal-ish” LED will blink briefly as the displacement current activates it from 1m away, and then it will turn back off until the circuit begins to settle ~1s later. After the displacement wavefront passes the LED, what transient condition exists across the LED terminals? There’s no longer any current flow until after ~1s when we start to see the LED turn on permanently as the reflections settle out.
I'm replying just so that I'll be notified if this gets answered. The wave is propagating down the line, but it isn't clear to me that there are new changes at the origin until a reflection gets there. I told myself it was current being conducted back from new displacement at the wavefront, but ... I'm not confident of that, nor am I confident that a Doppler-like effect wouldn't at least decrease it. Of course, I also don't quite get why it comes up in "10 steps", instead of after 1 step, or at least in a zeno's paradox of 1/2 the way each step forever.
If we're specifying ideal transmission lines, then there will be no reflections.
@@jimjjewett I think the displacement current will be enough to sustain the LED, and there's many ways to explain it, but this is my layman understanding using lumped elements (please correct me if I'm wrong):
As displacement current is produced through each section of wire, charges are depleted in that section, which causes current to flow from the adjacent section. In the top circuit, only one wavefront is produced, which means there's displacement current in only one section of wire at any given time, but conduction current is supplied by the source, which produces a continuous current between the source and the wavefront.
In the bottom circuit, there are two wavefronts producing displacement current in the same direction, starting at the same spot. So immediately the displaced charge from one wavefront is picked up by the other wavefront, such that there's always a continuous current between the two wavefronts.
EDIT: Might be completely off-base. This explanation seems to imply that you can throw the switch, remove the source, and the LED will stay lit from the displacement current due to the wavefronts alone. Something's definitely missing here lol
Displacement current is related to transmission line. It flows between the capacitors. Led is far from source. So displacement current is flows between transmission line first. After some time it conventional current reaches to led then led will glow.
When we have a look at Eric's graph, it says, 50OHM all the time until it's back ... so, that would mean 20mA flowing all the time when the edge is travelling the way there and back. If we are talking about the bottom circuit (led, battery, switch, resistor placed in the middle), the current would be flowing through the LED when the edges would be travelling. Or?
I love contents like this thanks sir
Great Video, expect more!
This really helped understanding antenna impedance
Thanks for making this! I was really struggling to understand what was going on
Thank you very much Eric and Robert! Loved it
I have a fast ohm meter; GHz scope and ns pulses from a photodiode illuminated by a laser. Me yelling “half current before reaching full current!” at the end! Very much enjoyed!
I'm going to do everyone a favor and just tell them that the wires are already charged before the circuit is complete. That's how I always treat them, and for most applications, it works just fine.
but when the switch is closed, the whole electron chain will simultaneously move together creating a current, or only electron at the switch move first and the movement propagate to the rest of the wire?
In the final explanation where the lamp and switch are in the centre and he says that the lamp turns on immediately, I suggest that it actually turns on slowly (in relative terms) because until the wave has propagated far enough down the line to where the capacitance is high enough to couple sufficient current across the lamp can't light. Comments welcome.
Could someone please explain about the resistance and work perform and the current used up by the LED, to the return line, shouldn't after the LED the voltage to be less???
Wow this cleared everything up and I learned a lot... fantastic, great and informative discussion, Eric is a great teacher, thanks Robert!
cracks are forming in the brick wall of my lack of education in this area.
Thank You Both
In electromagnetic theory you have LOCAL ENERGY CONSERVATION the relationship between energy density and WORK(fields on charges and charges on fields) and FLUX.
Poynting vector is needed for calculation of flux.
Poynting vector points in direction it does so that surface integral for flux gives correct answer.
Poynting vector is part of local energy conservation and that's the point to remember.
Local energy conservation applies everywhere of course, current or no current.
The impedance measured at the input end is Zo only if the far end is terminated with a load of Zo Ohm. If you remove the load, you won't have a 50 Ohm to take your 20mA at the switch.
Hello Robert,
great idea to cooperate with Dr. Bogatin in this topic!
I have one question regarding this issue:
For me both cases described are pretty clear, but what would happened if two wires instead of layout from top picture (transmission line like PCB) and bottom ( from veritasium) would be placed as one big loop (signal lanes very far apart from each other and LED far from voltage source)? The coupling between signal and return would be extremely low. How this signal would propagate then?
I would bet that principle is the same but due to very high impedance between lines there would be extremely small displacement current and due to very long risetime of this signal we would need more time for voltage to become stable.
What is your/Dr. Bogatin's opinion?
Thank you for this video :)
Excellent explanation, Thank you very much.
Amazing video
Amazing video, I learned a lot 😮
Excellent explanation using the capacitor. Changing electric and magnetic fields occur at the speed of EM radiation (ie light).
So, everyone has a clock speed in our brains of the speed of light.
Amazing!!!
I think the example below is wrong. the transmission line on the left side of the LED is not a return path, the impedance on that line is not 0. So you can't use the 50ohm impedance to calculate the 20mA transient current from the incoming line to that line. There is no such a model in the real world.
A little hard to follow. I prefer to think of the activity at the atomic level. For example, one end of a 6 inch copper wire is connected to the switch and the other end is left unconnected. The other side of the switch is connected to the negative battery terminal. When the switch is closed, excess electrons will begin jumping from atom to atom as they travel down the wire. After about 1 ns, when they reach the end of the wire, the flow will stop and the potential at the end of the wire will be the same as the switch. If a 6" return wire is added and connected to the positive battery terminal, the electrons will continue for another 1 ns back to the battery, except the flow will not stop. Using light with a mirror to return the light will produce a similar round trip delay, except the delay would be a bit less because light travels faster than electrons in a wire.
There are three situations where electrons DO carry energy ... in a vacuum tube, a semiconductor, and a plasma. Plasma filaments exist in space that are hundreds or thousands of light-years long. They consist of electrons and protons (or positive ions), all on the move. Veritasium doesn't get beyond 18th century electricity experiments where plasma was unknown.
It's funny that you mention those three: I work with travelling wave tubes, solid-state power amplifiers and have to worry about the ionosphere!
Why does everyone focus on this one part of Veritasium's video? Here are my questions.
1. All standard wiring is done with the hot and neutral wired side-by-side. This is required by the National Electrical Code (NEC 300.20(A)) to prevent inductive heating of metal conduit. It works because the magnetic fields produced by the hot and neutral are in opposite directions and cancel each other out when they are side-by-side. If the magnetic fields of the conductors have been canceled then the cross product of the electric and magnetic fields must be zero, so what remains to conduct the energy outside of the wire?
2. If the energy is carried outside of the wires then why does resistance in the wire dissipate energy, why doesn't the energy just keep going as if nothing was in the way?
3. The cross product is the product of 2 linearly independent vectors and is perpendicular to both of them. Where this is demonstrated at 7:34 in the original video it is not clear what vector Veritasium is using to represent the magnetic field since the field seems to form a cylinder around the battery. How he representing the magnetic field as a vector to form the cross product with the electric field?
4. In the 2 conductors (towards the load and back) the magnetic fields are in opposite directions but the electric field is in the same direction for both conductors. How can the cross products of the electric field and the magnetic field of the two conductors point in the same direction, from the source to the load, when their magnetic fields are in opposite directions? (8:00 in video)
The main problem with Veritasium is the discussion of the Poynting vector, which is an artificial construct to describe EM radiation pressure and behaves inconsistently when trying to use it in electric circuits. Feynman talks about it in his Lectures book volume II
Assuming a 1 meter spacing of the two wires, and let’s say 10 gauge wire… you have a characteristic Zo of 798 ohms on each TL leg (looking left and right from the switch) So if we use a 12v battery … looks like we have a near instantaneous (transient) current of about 7.5 mA before we reach steady state. Not knowing the bulb characteristics- we don’t know if that’s enough to illuminate or not…
A single wire is not a coax transmission line.
I truly loooooove this this video!!! I do like varitasium but this explained it sooo much better and complete. I could follow the explaination with ease. It really explained what would happen if this and that!
I do have a question, in the video at 35:59 Eric mentioned that the switch would be the source of the signal.
I answer my own question below:
If you would have your switch on the moon and your led and battery next to each other and you turn on the switch the wave front wouldn't start at the battery but it would start at the switch cause that is where you open the "gate" let's say. So the source of the signal si still the battery BUT the source of the voltage wave is the switch!
Can I also say that Eric, you are so incredibly cool! Your knowledge and your way of explaining such complex systems in a very engaging and just lovely way! It get's me fired up! And I would LOOOVE to be in one of your classes one day... Maybe, it's a little far away but if I ever get the chance, I would take it in a hartbeat!!! Mind you I would never be able to finish a entire class but I would LOVE to be one in or two lessons!
A;sp for Robert! The things you can learn from this channel! I started watching you to learn to use altium and it's MANY MANY features, and now I am learning a little bit about the quantum electric behaviour on transmition lines?!?!? Like what? This is just insanity! just incredibly! Love every single video that I have watched!!
Great discussion. It seems that Eric Bogatin should be wearing a shirt saying "Warning: cannot be turned off after being asked a question about a electric circuit".
32:09 Thank you for explaining the approximation.
Thanks for the video!
I have the feeling that more and more videos are popping up to discuss the Veritassium video. Unfortunately, most of the other videos make the subject more fuzzy rather than explaining it. The simple truth (as we know today, unless proven otherwise) is that we have to solve Maxwell’s equations to get the E and B fields for the concrete setup. That means that all materials have properties of permeability, resistivity and susceptibility. That includes the wire and air plus insulation of any sort. If your 600000km long wire has more resistance that the path through insulation and air ( looking at E and B fields, not necessarily just DC resistance), then the fields at the long ends (or where the loop closes) are very small and do not contribute much to the energy transfer. Looking at the poynting vector, if you have ideal wires the you have no E-field along the wire and hence the Poynting vector goes straight from source to load. With a real setup, there will be losses and the Poynting vector field can be distracted outside source and load, representing distribution characteristics of the system. There is no need to talk about distinctions between frequency ranges as the capacitive and inductive properties are already frequency dependent. One just needs to represent them properly in a system setup before solving. And that means that there is no such thing as an ideal 300000km long loop without capacitance, resistance and inductance. If there were such a thing, it would not have a voltage drop and hence can be represented with a short across it at the source and load.
Thank you for the interesting video!
I wonder what would happen if you increased the distance between the transmission lines. For example if the switch was on the north pole and the LED on the south pole.
Wouldn't the capacitive coupling be negligible?
Fantastic! Can you do video on TDR and other instruments used to debug transmission lines?
Thank you kramer3d. TDR is what our call supposed to be about :D But then Derek posted the video, so we talked about it. However TDR is sill on the list.
I think the biggest confusion among all explanations on this topic is that a line represents a connection on a schematic but a transmission line in the context of all of these videos. Once you realize that you have to actually replace the transmission line with RLC components, the solution is at hand.
Nice lecture, thank you! Could you continue the discussion on how energy flows if an AC source is used instead of DC
It's the same, but electromagnetic fields change vectors of direction 50Hz.
So to summarize. The answer is that the light source will light up immediately, but only if characteristic impedance of the transmission line, power source output resistance and light source are matched.
My problem with Veritasium video is, that I don't thing they are.
And for the people who can't make sense of the idea that there's current from an "open circuit". Another way to think of it is that information will not transfer higher than the speed of light. Meaning you can't know whether the led is connected or not without that information transfer from the far distance through the speed of light.
He is right about how two wires close together will actually form, and have a transmisson line impedance, and this is very inportant in high radio frequency transmission lines, but that is getting way beyond the original queston.
Eric quickly mentions that there are two transmission lines in the second diagram, however, if so, that statement should also apply for the first diagram. In any case, for the second diagram he says that two signals emanate from the switch when it is closed. However, the voltage across the ideal cell remains at 2 volts at all times, and the voltage along the conductor from the cell's negative terminal around the circuit and past the LED remains unchanged, so there is not any associated displacement field to the right of that cell, so no displacement current would flow. Does the pulse also travel through the cell raising the voltage on the 'negative" terminal of the cell? With regards to the voltage divider concept, the voltage on the load side of the source impedance (across the switch) is initially two volts, then as soon as it closes the voltage on the load side of the source impedance drops to one volt with respect to the cell's negative terminal, so what happened to the electric field (represented by the extra volt) that was across the switch contacts? The associated field must have dissipated its energy some place. Was it radiated as an E-M wave? Can anyone clarify? It seems that Eric's comments assume an "instantaneous steady state condition" at various points in time as the pulse travels down the transmission line, but remains unclear what happens after the switch is closed, but before the pulse starts its journey down the conductor. Does the pulse travel in both directions?
Yes, the pulse does travel in both directions, and it lowers the voltage on the negative side of the battery