EEVBlog 1439 - Analyzing Veritasium's electricity video

[Fungus]

If someone else can assume lossless transmission lines, then I can assume a spherical bulb in a vacuum with no parasitic elements.

Why aren't we using LEDs? This is 2021!

[vad]

If someone else can assume lossless transmission lines, then I can assume a spherical bulb in a vacuum with no parasitic elements.

In Russian they call it spherical stallion in vacuum.

[Fungus]

Another question: How long does it take to turn off the light?

Nobody...? 

[Sredni]

Another question: How long does it take to turn off the light?

Nobody...? 

I 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.

[gamalot]

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.

[Sredni]

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?

[David Hess]

Those results are exactly as predicted.  I have done the same experiment, by accident of course.

[gamalot]

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?

The 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.

[Sredni]

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?

[gamalot]

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?

I 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.

[Sredni]

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.

[David Hess]

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.

A 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.

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?

Which 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.

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 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.

What 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.

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.

The 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.

[Sredni]

Which 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.

You'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.

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.

I 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.

Faster test instrumentation could show the propagation delay across rather than down the transmission line. 

Yes, 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.

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.

Good 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.

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.

The 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.

Yes, 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)

[gamalot]

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.

[David Hess]

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.

I 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.

That 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.

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.

Good 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.

It 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.

[Sredni]

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 simulation


link: https://i.postimg.cc/c4LcZ7cV/screenshot-27.png

If 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?).

[hlevinson]

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.

[Fungus]

I think this can all be simplified to something both physicists and engineers will accept.

Yes.

Make the wires go directly to the light bulb instead of half a light year out into space and back. 

[13hm13]

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.

[Obi_Kwiet]

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.

[13hm13]

13hm13, dude, no one cares what you have to say

Your reply is confused, toots, in that it is YOU that indeed  "cares" enough to reply.

[EEVblog]

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.

You can't just do that:

[EEVblog]

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.

I 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.

[EEVblog]

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.

And that in the bold is 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.

[13hm13]

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