EEVblog #486 - Does Current Flow Through A Capacitor?

[EEVblog]

Dave proves he has no fear by opening this can of electronic worms by posing the question - "Does Current Flow Through A Capacitor?"
The answer may surprise you, or drive you into a physics induced rage...

Turns out you can measure the displacement current:
http://personal.rhul.ac.uk/uhap/027/PH2420/PH2420_files/displacement.pdf

NOTE: before commenting, please watch and understand that there are TWO types of currents.



Thanks to KedasProbe for pointing out this doco video on Maxwell & the Displacement current!

[EEVblog]

Should I simply have a blanket rule that I won't respond to any comments on on this video? 
A video that's guaranteed to have something technically wrong with it! 

[AlfBaz]

Great!
So we have electrons whizzing around atoms cause they're attracted to the +ve protons. We then introduce an emf which causes the electrons to be simultaneously repelled by the voltage source's -ve charge and attracted to it's +ve charge. These charges are present on the capacitor plates so the electrons start packing one plate and draining out of the other. So here we have this block (a capacitor) with current flowing in and current flowing out and we are calling this flowing through?

I=C(dV/dt) This tells me that voltage has to be changing over time for a current to flow otherwise I would equal 0. What would happen if we had a fixed capacitance and started a voltage at zero and increased it linearly forever? Isn't there a finite number of electrons and eventually you have drained every free electron from the plate and wire? (assuming the dialectric wouldn't break down as the voltage kept increasing to very high levels)

[qoole]

Hiya,

I have no electronics qualifications whatsoever (waiting to be cooked alive). However when I see the 'Current flowing through a capacitor' argument it makes sense to me (although the wording, notably the 'through', can be a bit off putting)

This is how I see it:

You have two conductive plates A and B. They are held apart by a dielectric.
These plates have a certain amount of electrons residing in their 'sea of electrons'.
At a normal resting state where both plates have the same charge thus zero volts dropped across them.
When one connects plate A to the +Ve side of a battery and B to the -Ve side of the same battery, electrons (aka charge) flow from the battery to plate B.
Because electrons are negative charge the increase in electrons on plate B creates a negative charge which pushes away (repels) electrons from plate A. (As we all know like things repel each other).
Therefore, despite there being a dielectric in the way current can be measured flowing into and out of the capacitor.
When all the electrons have been depleted from plate A the capacitor is 'fully charged'. Thus the more electrons at a resting state the bigger the capacity (aka capacitance) of the.... capacitor...

Please feel free to correct me if my theory is wrong (I'm sure you will)

Excellent blog, keep it up!

Alex

[free_electron]

Its not a 'small resistor' ( when taking about leakage). A small,resistance would mean it leaks like hell... Its a biiiiiig resistance...

Anyway, it can be summed up very simple.

We know like charges repel each other.
A charge creates a field.

So, place one electron more on the left plate of the capacitor than on the right plate.
The field emanating from the left plate is now stronger than the one from the right plate. This will push an electron off the plate at the right.

Keep placing electrons left plate  and electrons on the right plate will be pushed off.
Field strength keeps increasing. We perceive this as voltage.
Since electrons flow in and out of the capacitor we say 'current flows through'.
But that is incorrect. (Nitpicking) as the electrons cant cross the dieelectric.

Its that simple. No need for semesters of maxwell

A really interesting question would be : what amount of electrons do i need to place on the left plate so that the right plate is completely void of electrons ? For a given plate size and plate distance , what voltage would that yield ? Anyone care to work that out ?

[komet]

So we have electrons whizzing around atoms

That is the concern of chemistry, not electronics. Electronics is not(*) concerned with electrons; it is about charges. Note that Dave only mentions the word "electron" once or twice in the video, and perhaps he shouldn't have at all.

In practical electronics you do not need to think about electrons, and certainly not about protons, any more than an architect needs to think about metallurgy.

(*) I know that is not the correct usage of "not", but the argument stands.

[uprightsquire]

A really interesting question would be : what amount of electrons do i need to place on the left plate so that the right plate is completely void of electrons ? For a given plate size and plate distance , what voltage would that yield ? Anyone care to work that out ?

Aren't you meant to be the professional electron wrangler?

[AlfBaz]

A really interesting question would be : what amount of electrons do i need to place on the left plate so that the right plate is completely void of electrons ? For a given plate size and plate distance , what voltage would that yield ? Anyone care to work that out ?

Yeah that's the question I was alluding to in my post's last paragraph, even had your name in it.

According to komet however we should either stop taking about this or move it to a chemistry forum 

[madires]

That is the concern of chemistry, not electronics. Electronics is not(*) concerned with electrons; it is about charges. Note that Dave only mentions the word "electron" once or twice in the video, and perhaps he shouldn't have at all.

In practical electronics you do not need to think about electrons, and certainly not about protons, any more than an architect needs to think about metallurgy.

(*) I know that is not the correct usage of "not", but the argument stands.

Please try explaining how semiconductors work without mentioning electrons! 

[komet]

A really interesting question would be : what amount of electrons do i need to place on the left plate so that the right plate is completely void of electrons ? For a given plate size and plate distance , what voltage would that yield ? Anyone care to work that out ?

Yeah that's the question I was alluding to in my post's last paragraph, even had your name in it.

According to komet however we should either stop taking about this or move it to a chemistry forum 

Well, yes, precisely, because that is an entirely chemical question. If you take too many electrons out of the material it will at some point stop being solid, or what did you think a sea of naked iron nuclei would do? Long before that happens the material will react - chemically - to the air or insulator (including the dielectric) around it; the positive plate (devoid of electrons) will steal electrons from the surroundings, which will then also be positive, and eventually steal the excess electrons from the other plate, and the result is breakdown of the capacitor. It's also how electro-erosion is performed.

If the dielectric is a vacuum then the plates will travel towards each other because the vacuum is not much of a substance to hold them apart, and the mountings will not be solid and therefore able to prevent it happening.

In a real world capacitor, almost all the charge is stored by the dielectric being polarized and the plates do not have to alter their electron count very much at all.

[John Coloccia]

Does current flow from the radio station transmitter to the receiver in your car?   

[uprightsquire]

The point is current is not defined as a movement of electrons. That is best left to thick headed children or people with an MFA.

Consider a wet cell, lead acid battery, lemon with a nail and coin, etc. No electrons are passing, internally, between the terminals, yet a current is flowing.

[komet]

That is the concern of chemistry, not electronics. Electronics is not(*) concerned with electrons; it is about charges. Note that Dave only mentions the word "electron" once or twice in the video, and perhaps he shouldn't have at all.

In practical electronics you do not need to think about electrons, and certainly not about protons, any more than an architect needs to think about metallurgy.

(*) I know that is not the correct usage of "not", but the argument stands.

Please try explaining how semiconductors work without mentioning electrons! 

a) "charge carriers"

b) you don't need to understand semiconductor physics in order to use them

c) Shockley (a physicist) was an arse and Dave (an EE) isn't, and who would you rather i) listen to ii) let design your power supply?

[envisionelec]

Should I simply have a blanket rule that I won't respond to any comments on on this video? 
A video that's guaranteed to have something technically wrong with it! 

Well, you did. And, thanks.

[jpb]

A really interesting question would be : what amount of electrons do i need to place on the left plate so that the right plate is completely void of electrons ? For a given plate size and plate distance , what voltage would that yield ? Anyone care to work that out ?

here is a back of the envelope calculation:

The atomic volume of copper is 7.1cc/mole

if we take capacitor plates of 10cmx10cmx1mm then this is 10 cc which at 1 electron per atom is (10/7.1)x6.022E23 electrons

the charge density is then this number times the electronic charge (1.602E-19) divided by the area of 1E-2 square meters,

if we make the very drastic assumption that all the field is within the capacitor (clearly wrong if the copper is completely depleted!) then
we can calculate the field by dividing the charge density by epsilon zero (8.854E-12),

the answer I get is E = 1.53E18 V/m which for a 1mm gap would be around 1.53E15 V !!!

I suspect that the air or dielectric would break down a bit before that point was reached.

If my maths is right, the energy stored would be the equivalent of around 2,480 million atomic bombs (of the 84TJ variety) so it would be
a rather dangerous capacitor to be near.

[madires]

Please try explaining how semiconductors work without mentioning electrons! 

a) "charge carriers"

Nice try :-)

b) you don't need to understand semiconductor physics in order to use them

Maybe true to some extend but it would be like doing math without understanding it. What could go possibly wrong? :-)

The project I'm working on requires a lot of strong knowledge about (semiconductor) physics, and it's just a simple MCU project.

c) Shockley (a physicist) was an arse and Dave (an EE) isn't, and who would you rather i) listen to ii) let design your power supply?

I'd listen to Dave and design the PSU myself :-)

[EEVblog]

I think I'm going to feel like the fat kid from Stand By Me:

[John Coloccia]

The only real problem with this is that it becomes very difficult to explain how an RC filter works if you simply consider a capacitor as a device that allows current to flow as a function of frequency, and you also loose the subtlety that current lags voltage by 90 degrees.

[RJdaMoD]

That's why you use complex numbers for that.

[KD0CAC John]

Does current flow through a capacitor = the universe

Our understanding of the universe = sometime yes & sometimes no , at this time - not completely

Thanks again Dave

[c4757p]

I meant to say leakage. The resistance is not small, it is very large.

This is why I find conductance much more intuitive for explaining basic concepts like this. It tends to play by the same rules as other electrical quantities. A large capacitance permits a large current in response to a voltage change, a large inductance generates a large voltage in response to a current change, and a large resistance permits a.... small... current in response to voltage? No! A large conductance permits a large current in response to voltage! Much better

[Strada916]

I have not seen the video yet however. Current flow through a Capacitor in a DC circuit will not flow. IE the current is blocked or stopped. It will charge but that is about all that will happen.

However in an AC circuit the capacitor behaves like a resistor.   

Xc = 1/(2piFC)

See what happens when you put 0 or a number close to 0  as your frequency. 

[Mki]

Its that simple. No need for semesters of maxwell

Maxwell's equations is the reason why we have this wonderful lumped model system  Maxwell's are nice to go through once, but really - who remembers them? 

Oops h missing...

[Fratink]

A really interesting question would be : what amount of electrons do i need to place on the left plate so that the right plate is completely void of electrons ? For a given plate size and plate distance , what voltage would that yield ? Anyone care to work that out ?

here is a back of the envelope calculation:

The atomic volume of copper is 7.1cc/mole

if we take capacitor plates of 10cmx10cmx1mm then this is 10 cc which at 1 electron per atom is (10/7.1)x6.022E23 electrons

the charge density is then this number times the electronic charge (1.602E-19) divided by the area of 1E-2 square meters,

if we make the very drastic assumption that all the field is within the capacitor (clearly wrong if the copper is completely depleted!) then
we can calculate the field by dividing the charge density by epsilon zero (8.854E-12),

the answer I get is E = 1.53E18 V/m which for a 1mm gap would be around 1.53E15 V !!!

I suspect that the air or dielectric would break down a bit before that point was reached.

If my maths is right, the energy stored would be the equivalent of around 2,480 million atomic bombs (of the 84TJ variety) so it would be
a rather dangerous capacitor to be near.

I think you could add a few more atomic bombs if you also tried to remove all the rest of the electrons in the material itself  .

Having completed a degree in Engineering Physics, I often don't realize that people don't understand such relatively simple concepts as capacitors (I don't mean to sound like an ass, it was just first year stuff for me).

Here's one that could blow your mind: A magnetic field can actually be thought of as an electric field, caused to relativistic motion of electrons.  This is why two wires with electrons moving in opposite directions attract one another.  The electrons essentially "see" less than one electron on the other wire, due to their velocity, and are thus attracted to one another.  I don't know if this theory is really established, but I can easily see how to do the math to prove it (at least in this simple example).  Wikipedia has something on it I believe: http://en.wikipedia.org/wiki/Relativistic_electromagnetism.

[cthree]

Another fantastic FF Dave!

"Good on ya" as you say down there.

[IanB]

Forget the subject title, how about this:

Does current flow through a resistor?

I mean, suppose my resistor is 5 mm long and the electron drift velocity is (say) 0.1 mm/s, then any electrons flowing in at one terminal are going to take 50 seconds before they appear at the other terminal    At any instant in time the electrons entering the resistor are not the same electrons that are leaving the other end. The resistor is a giant electron storage container...

[c4757p]

Forget the subject title, how about this:

Does current flow through a resistor?

Elaborating on this would make a great sequel to this video IMHO. Movement of electrons vs. movement of charge vs. current, macroscopic vs. microscopic, etc. I think a failure to establish the relevant definitions along those lines was responsible for most of the last one's confusion. Plus, it's just damn interesting sometimes. But Dave doesn't seem to go for physics much, so maybe someone else would have more fun making this kind of video.

[don.r]

Well done, Dave. Put that puppy to bed and move on... next... the transformer... and Ed, Edd and Eddy currents...

[c4757p]

I don't think they're very current.

[don.r]

I don't think they're very current.

LOL! Very good... 

[sagdahl]

Does Current Flow Through A Capacitor? should maybe be refrased to:
Does Current Flow To A Capacitor? (And From?)
Excellent infotainment by the master of joyful electronic videos!
Does anyone know where Jeri Ellisworth gone?

[madires]

I have not seen the video yet however. Current flow through a Capacitor in a DC circuit will not flow. IE the current is blocked or stopped. It will charge but that is about all that will happen.

Let's see, you switch your device on or off, the DC power could have ripple or maybe there's some switching or oscillation. In any case that involves a voltage change you'll have a current through the cap.

[beaker353]

Had you been teaching my AC theory class last semester, there would have been much less confusion amongst my classmates.  College professors could learn a lot by watching your teaching methods...

-EM

[c4757p]

College professors could learn a lot by watching your teaching methods...

The problem with college professors is that when you know 100 times above what you're teaching, you forget that it's not as obvious to everyone else as it looks to you. This is why Maxwell's equations confuse so many people - a physics professor is so steeped in calculus that he can take one glance at the equation and know what it means. A student looks at the equation and at best sees a sequence of steps to be performed to get the result, not the true, physical relationship it implies.

College professors out there, remember this - whatever it is, it's only 1/100 as obvious as you think it is.

[IanB]

It's been touched on above, but when you consider questions like this you need to be concerned with the system parameters. When analyzing any system you may perform a static (or steady state) analysis, or you may perform a transient (or dynamic) analysis.

It is very appealing, intuitively, to consider systems as static and unchanging, as this reduces the complexity of the analysis. (This is why AC circuits and control systems are often studied in the frequency domain rather than the time domain. You can represent them using time-independent parameter values like frequency, phase angle and impedance and not include time in the calculations.)

So in the capacitor question, we can say that in a static system where nothing is changing with time, then current does not flow through an ideal capacitor.

However, in the real world, most systems of interest are dynamic (or they wouldn't do anything). In a dynamic system, current most certainly can flow through a capacitor.

[firewalker]

Alexander.

[chicken]

OT: Love the whiteboard sessions, but please switch to manual exposure when recording. The brightness jumps all over the place.

Doesn't make the actual content of the video worse, just bugging the photographer in me    (probably like me doing it all wrong when messing with electronics  )

[staxquad]

next week's Fundamentals Friday: "does current flow through a transformer?"

in both cases (capacitors and transformers), they are black voodoo boxes that do magical things, and the lay person sees current flow through both

actual current only flows through both by leakage

and in both cases, voltage is manufactured by one isolated side affecting the other isolated side by their electrowhatever properties, and with a load on the effected other side, there's current flowing 

to use them, you don't need Maxwell

to make them you do need Maxwell

edit: premature blah blah
now that I've seen the video, good one!

[c4757p]

Yes, I removed a stupid brainfart comment so as not to confuse anyone. Sorry if you were replying to it 

[free_electron]

Yeah that's the question I was alluding to in my post's last paragraph, even had your name in it.

According to komet however we should either stop taking about this or move it to a chemistry forum 

the problem is we never fully charge capacitors... a 2200uf 16 volt capacitor is not 'full' a t 16 volts... it flahses over beyond 16 volts. assuming there was no limit to its working voltage it would actually turn out to be a much larger capacitance.

youd have to figure out how many atoms there are in both plates , how many of those atoms can accept extra electrons to find out how many electrons you can actually ram in a plate. then you can work out what the voltage would be.

[sagdahl]

and in both cases, voltage is manufactured by one isolated side affecting the other isolated side by their electromechanical properties, and with a load on the effected other side, there's current flowing 

I presume you mean "electromagnetic properties"?

[Dave]

Achievement unlocked: Inspired an EEVblog episode.

[free_electron]

A really interesting question would be : what amount of electrons do i need to place on the left plate so that the right plate is completely void of electrons ? For a given plate size and plate distance , what voltage would that yield ? Anyone care to work that out ?

here is a back of the envelope calculation:

The atomic volume of copper is 7.1cc/mole

if we take capacitor plates of 10cmx10cmx1mm then this is 10 cc which at 1 electron per atom is (10/7.1)x6.022E23 electrons

the charge density is then this number times the electronic charge (1.602E-19) divided by the area of 1E-2 square meters,

if we make the very drastic assumption that all the field is within the capacitor (clearly wrong if the copper is completely depleted!) then
we can calculate the field by dividing the charge density by epsilon zero (8.854E-12),

the answer I get is E = 1.53E18 V/m which for a 1mm gap would be around 1.53E15 V !!!

I suspect that the air or dielectric would break down a bit before that point was reached.

If my maths is right, the energy stored would be the equivalent of around 2,480 million atomic bombs (of the 84TJ variety) so it would be
a rather dangerous capacitor to be near.

now imagine a technology suffiecinetly advanced enoug to pull of 1/100 of this. a small 10cmx10cm capacitor could power an electric car for millenia !

[free_electron]

Forget the subject title, how about this:

Does current flow through a resistor?

I mean, suppose my resistor is 5 mm long and the electron drift velocity is (say) 0.1 mm/s, then any electrons flowing in at one terminal are going to take 50 seconds before they appear at the other terminal    At any instant in time the electrons entering the resistor are not the same electrons that are leaving the other end. The resistor is a giant electron storage container...

that raises another quesion.. why we need to pay for electrical power .. after all every electron coming out of that wall sockets left prong , goes into the wall sockets right prong ! i should be charging the electricity company for the wear and tear these electroncs cause on my equipment ! it's not like i consume electrons , they get each and every one back.

and at a rate of 50hz or 60hz they don;t travel that far into my equipment either !
electricity is waay overpriced.

far better of with batteries. at leas tthere you know what you get for your money. once the electrons have run from one pole to another and potential differential is zero the cell is empty. not this wishy washy ' we'll jiggle the electrons a bit and you need to pay for that !

[XynxNet]

I like your video, Dave. It hits a sweet point between the point of view of a physicist and an EE.

The funy thing is, we physicists label this problem as experimental physics.  Go figure...

[Ketturi]

This was awesome. They really should show Dave's videos in schools, they would make so much more sense! I was once wondering pretty much same topic at one physics course. Needless to say that my practical opinions differed somewhat to more traditional theoretical ones teachers and books had. And like one instance I used "ohm" in my baccalaureate/final exam instead of "?", which of course is insignificant which one to use in real life, but nah, it's fastidious in school.
Fundamentals Friday is third best segment, or second best. Or what the smeg, they all are as good! Keep up Dave!

[c4757p]

"ohm" in my baccalaureate/final exam instead of "?"

What kind of electronics forum doesn't recognize ? 

Seriously, that symbol needs to die. How impractical to use a symbol that not only isn't supported by a number of computing systems, but also doesn't have a practically useful similar-looking symbol that can be used in its stead ( vs. u, for instance). Just use R...

[jpb]

Even worse than current flow in capacitors is the flow of holes in semiconductors. These behave like electrons with negative mass! Even more weird are phonons which knock electrons around and are quanta of crystal vibrations.

[juani_c]

Don't know if someone already mentioned it, but apparently you can measure displacement current:
Measuring Maxwell's Displacement Current Inside a Capacitor


EDIT: link fixed.

[IanB]

What kind of electronics forum doesn't recognize ? 

Seriously, that symbol needs to die. How impractical to use a symbol that not only isn't supported by a number of computing systems, but also doesn't have a practically useful similar-looking symbol that can be used in its stead ( vs. u, for instance). Just use R...

'Tis an amusing question. Now of course "" is a Greek capital letter "O", the first letter of "ohm". I suspect the reason we don't use "O" for ohms is because of the confusion with zero. So some bright spark thought, "Hey, let's use a Greek 'O' instead!"

[firewalker]

It is Ohm. Thats why in Greek it is . If it was Om it would be O. About 4000 years ago, until the more recent past, Qmega had a different pronunciation from Omicron.

Alexander.

[John Coloccia]

Even worse than current flow in capacitors is the flow of holes in semiconductors. These behave like electrons with negative mass! Even more weird are phonons which knock electrons around and are quanta of crystal vibrations.

It's just convenient bookkeeping, though.  Charge comes in quanta of electrons, so lack of charge has to come in quantities of lack of electrons...holes.  It's like talking about negative money.  We all know that simply means "debt", but it's more convenient to handle it in the same units as regular money because then you can do straightforward math to get a straightforward answer.

It's convenient to consider the changing electric field in a capacitor as "displacement current", because it puts it into the same units as regular current, and the resulting field is something that needs to be considered because it can do work....but that's just book keeping too.  It's just a gimmic to get everything using the same units so that you can figure out where all of the energy goes.  He should have called it something like "equivalent current" to make clear that it's not a real current, namely charge flux through a surface.

I guess it doesn't matter too much.  It's useful to consider current flowing through a capacitor but thinking about it like that sets up an intuition that breaks down when you look too closely.  Better to consider the real process and consciously choose when to simplify it.  In my opinion...just my opinion.

Then again, I always do things "backwards".  Whenever I tutor for programming, I don't start with "Hello World".  I start with nuts and bolts explanations of gates, processors and things like that.  Then I work my way up through memory and stored programs.  By the time we get through that, assembly language seems rather tame.  Then we hit higher level language and it's like, "Hey, this is no big deal...what's all the fuss?".

[c4757p]

Then again, I always do things "backwards".  Whenever I tutor for programming, I don't start with "Hello World".  I start with nuts and bolts explanations of gates, processors and things like that.  Then I work my way up through memory and stored programs.  By the time we get through that, assembly language seems rather tame.  Then we hit higher level language and it's like, "Hey, this is no big deal...what's all the fuss?".

OT, I know - but if assembly is confusing and difficult for a programmer, that programmer does not understand computers, which is a bit pathetic. Tedious, perhaps, but not confusing. Kudos to you for starting in the right place, and shame on all these teachers and professors who think they need to start with the flashy stuff to get people interested (if they're not interested now, they'll lose interest again soon). You can't teach passion, why bother teaching something that requires it to someone who doesn't have it?

Not as OT as it sounds, by the way. The same applies to electronics - if you can make an RC filter but can't answer this question for yourself, somebody started you in the wrong place. As wishy-washy as a lot of EEs think lower-level physics is for their profession, I find myself relating things to it with surprising frequency. All you beginners out there - go get a physics textbook.

[IanB]

I guess it doesn't matter too much.  It's useful to consider current flowing through a capacitor but thinking about it like that sets up an intuition that breaks down when you look too closely.  Better to consider the real process and consciously choose when to simplify it.  In my opinion...just my opinion.

We do useful engineering by building abstractions that conceal the complexity of inner layers of detail behind concepts that address what we really care about. Computers are a prime example of this, with logic gates hiding transistors, and processors hiding logic gates, and assembly language hiding 1's and 0's, and compilers hiding assembly language, and so on and so on.

In a similar way, electric current is an abstraction. When we do calculations with it we don't care what the charge carriers are, we just have an abstract quantity measured in "amps" that obeys certain rules and results in good designs if we understand what the rules are. By every normal rule of circuit theory, a current flowing at one terminal of a two terminal capacitor must be balanced by an identical current flowing at the other terminal. If we (or Spice) do not obey that rule, our circuits won't work the way we expect them to.

[dr.diesel]

"Displacement Current Meter"        , good one Dave!

[Matje]

The only real problem with this is that it becomes very difficult to explain how an RC filter works if you simply consider a capacitor as a device that allows current to flow as a function of frequency, and you also loose the subtlety that current lags voltage by 90 degrees.

The differential equation Dave showed (the thing with the 'd's)  captures this perfectly well, for it shows that the instantaneous(!) values really don't depend on the frequency. One must choose the model to fit the application, as Dave said.

Aaahhhh, just thinking about differential equations makes my head hurt...

[John Coloccia]

Hi, Ian.

But you never develop the intuition why adding some other impedance to the mix affects frequency response.  All you can do is plug and chug through an equation to get an answer.  I'm pretty sure SPICE models concern themselves with time constants in order to get the right answer.  I'm not saying that there aren't simple ways to get to an answer without getting into the details, but it's a powerful thing to have that intuition in your back pocket.  That's all I'm saying.  It's really useful if a newbie approaches this by simplifying only after understanding the basic, real process and developing a real gut feeling about what's really going on.

[Matje]

That's why you use complex numbers for that.

Naaaa.

You use complex numbers to find out whether the other guy is a maths or physics person or a proper electronics person.

Because these crazy people use 'i' for the imaginary unit, while electronics people use 'j' as God intended.

Never let a user of 'i' touch your measuring equipment, let alone the soldering iron, for chaos and mayhem might ensue.

Hehehe. ;-)

[firewalker]

What about this capacitor? Does current flow through?

[ChadSeibert]

I guess it doesn't matter too much.  It's useful to consider current flowing through a capacitor but thinking about it like that sets up an intuition that breaks down when you look too closely.  Better to consider the real process and consciously choose when to simplify it.  In my opinion...just my opinion.

We do useful engineering by building abstractions that conceal the complexity of inner layers of detail behind concepts that address what we really care about. Computers are a prime example of this, with logic gates hiding transistors, and processors hiding logic gates, and assembly language hiding 1's and 0's, and compilers hiding assembly language, and so on and so on.

In a similar way, electric current is an abstraction. When we do calculations with it we don't care what the charge carriers are, we just have an abstract quantity measured in "amps" that obeys certain rules and results in good designs if we understand what the rules are. By every normal rule of circuit theory, a current flowing at one terminal of a two terminal capacitor must be balanced by an identical current flowing at the other terminal. If we (or Spice) do not obey that rule, our circuits won't work the way we expect them to.

I totally agree. Without meaningful abstractions in either profession, it's basically impossible to accomplish much. If that means being hopelessly confused some of the time, then so be it. It's the price we pay to work at that level. It doesn't mean that there isn't value in learning the lower level physics required for accurate interpretation of the phenomenon, it just means that depending on what you work on, it may have been better spent learning something else.

[Razor512]

Took apart a capacitor a while back. Had some weird chemical in it (didn't drink any so I am unsure of the flavor...)


Inside of a capacitor by Razor512, on Flickr

[Smithy]

Great video!

A good follow up to this would be the inrush current of a circuit. With maybe USB's inrush current limitations as an example.

[EEVblog]

OT: Love the whiteboard sessions, but please switch to manual exposure when recording. The brightness jumps all over the place.

I do.
The problem is that my camera switches out of manual exposure mode when I switch to playback mode to view a video, I then have to switch it back on and the camera my decide to select a slightly different exposure setting. It's a really annoying bug.

[EEVblog]

I guess it doesn't matter too much.  It's useful to consider current flowing through a capacitor but thinking about it like that sets up an intuition that breaks down when you look too closely.  Better to consider the real process and consciously choose when to simplify it.  In my opinion...just my opinion.

The whole point of the video, which many people still seem to miss (they still seem to think I'm talking about electric/conducted current, when I said half a dozen times I am not), is that I am pointing out there is a different type of current when you look and delve deeply enough into the mathematics of it. So rather than this idea "break down" when you look closer, it actually gets more valid mathematically the deeper down the rabbit hole you go.

[EEVblog]

Achievement unlocked: Inspired an EEVblog episode.

 

[EEVblog]

Don't know if someone already mentioned it, but apparently you can measure displacement current:
Measuring Maxwell's Displacement Current Inside a Capacitor

Sweet.
I was thinking it might be possible with the AIM-TTI current probe if you had a bracket to hold and move it accurately?
Collecting the data = fun
Working the math  to see if you are actually verifying any theory = yuck

[IanB]

Hi, Ian.

But you never develop the intuition why adding some other impedance to the mix affects frequency response.  All you can do is plug and chug through an equation to get an answer.  I'm pretty sure SPICE models concern themselves with time constants in order to get the right answer.  I'm not saying that there aren't simple ways to get to an answer without getting into the details, but it's a powerful thing to have that intuition in your back pocket.  That's all I'm saying.  It's really useful if a newbie approaches this by simplifying only after understanding the basic, real process and developing a real gut feeling about what's really going on.

Actually, instantaneous current balances and Kirchhoff's current law and differential equations really are the deep down low level details of what is really going on. I thought that was what you were advocating people to learn first. But I for one cannot develop much intuition about frequency response at that level.

If you want intuition about impedance and frequency response you have to go up a level and choose a higher level abstraction. That is exactly what AC circuit theory is. It is an abstraction built around frequency, phase angle, reactance, resistance, phase angle and so on. It allows you to develop intuition about how AC circuits work and hides all the gory details of differential equations.

AC theory is full of complex numbers, which is clearly bogus (  ) because real physical quantities are not imaginary and there is certainly no imaginary current or voltage in a real circuit.

[John Coloccia]

I guess it doesn't matter too much.  It's useful to consider current flowing through a capacitor but thinking about it like that sets up an intuition that breaks down when you look too closely.  Better to consider the real process and consciously choose when to simplify it.  In my opinion...just my opinion.

The whole point of the video, which many people still seem to miss (they still seem to think I'm talking about electric/conducted current, when I said half a dozen times I am not), is that I am pointing out there is a different type of current when you look and delve deeply enough into the mathematics of it. So rather than this idea "break down" when you look closer, it actually gets more valid mathematically the deeper down the rabbit hole you go.

I thought your video was good, Dave.

[juani_c]

Don't know if someone already mentioned it, but apparently you can measure displacement current:
Measuring Maxwell's Displacement Current Inside a Capacitor

Sweet.
I was thinking it might be possible with the AIM-TTI current probe if you had a bracket to hold and move it accurately?
Collecting the data = fun
Working the math  to see if you are actually verifying any theory = yuck

the researchers used a SQUID to measure the very low magnetic field, don't know if the AIM-TTI probe is that sensitive

[AlfBaz]

the problem is we never fully charge capacitors... a 2200uf 16 volt capacitor is not 'full' a t 16 volts... it flahses over beyond 16 volts. assuming there was no limit to its working voltage it would actually turn out to be a much larger capacitance.


youd have to figure out how many atoms there are in both plates , how many of those atoms can accept extra electrons to find out how many electrons you can actually ram in a plate. then you can work out what the voltage would be.

I always thought the voltage rating of a cap is simply the withstand voltage of the dielectric. If that's the case It would be interesting to see this "flow" thing in action.

We get a very high voltage capacitor with very low capacitance, I see digikey has some 1pF 6kV caps, or maybe some 100pF 50kV. Then we steadily ramp the voltage from 0V up toward the caps voltage limit.

I would expect to see the current rise from zero and level out at C*(dV/dT). Since the only way to get current to "flow" is to have voltage change with time we continue to raise the voltage at a steady rate. At this stage it would be interesting to see if the current continue to flow at the above mentioned value and what. if anything, happens toward the end.

I can't help but think that it becomes a chicken and the egg thing in that to move enough charge to make this interesting you have to increase the voltages rate of change, which means you'll get to the caps withstand voltage to quickly and if you decrease the rate of change then the current becomes to small. Either way we may not get to that "critical" electron mass 

Might have to get in touch with that photonic induction guy see if he can use his high voltage gear for science

[AlfBaz]

Don't know if someone already mentioned it, but apparently you can measure displacement current:
Measuring Maxwell's Displacement Current Inside a Capacitor

Hey Juani, that link seems to be broken

[c4757p]

It's just an extra letter on the end of the URL.

Measuring Maxwell's Displacement Current Inside a Capacitor

[AlfBaz]

It's just an extra letter on the end of the URL.

Measuring Maxwell's Displacement Current Inside a Capacitor

Ahh... I did look at that but didn't have my glasses on
Thanks

[IanB]

I would expect to see the current rise from zero and level out at C*(dV/dT). Since the only way to get current to "flow" is to have voltage change with time we continue to raise the voltage at a steady rate. At this stage it would be interesting to see if the current continue to flow at the above mentioned value and what. if anything, happens toward the end.

I can't help but think that it becomes a chicken and the egg thing in that to move enough charge to make this interesting you have to increase the voltages rate of change, which means you'll get to the caps withstand voltage to quickly and if you decrease the rate of change then the current becomes too small. Either way we may not get to that "critical" electron mass 

You seem to misunderstand the capacitor equation. It is a linear equation:

I = C dV/dt

This equation is of the form:

y = Cx

It says "y is a constant times x".

Therefore, as long as dV/dt remains constant, then I remains constant.

This is exactly how the op amp integrator circuit works. For as long as there is a constant current flowing at the input, then the output voltage continues to change at a steady rate. There is no doubt at all that this works exactly as the formula predicts.

[AlfBaz]

I would expect to see the current rise from zero and level out at C*(dV/dT). Since the only way to get current to "flow" is to have voltage change with time we continue to raise the voltage at a steady rate. At this stage it would be interesting to see if the current continue to flow at the above mentioned value and what. if anything, happens toward the end.

I can't help but think that it becomes a chicken and the egg thing in that to move enough charge to make this interesting you have to increase the voltages rate of change, which means you'll get to the caps withstand voltage to quickly and if you decrease the rate of change then the current becomes too small. Either way we may not get to that "critical" electron mass 

You seem to misunderstand the capacitor equation. It is a linear equation:

I = C dV/dt

This equation is of the form:

y = Cx

It says "y is a constant times x".

Therefore, as long as dV/dt remains constant, then I remains constant.

This is exactly how the op amp integrator circuit works. For as long as there is a constant current flowing at the input, then the output voltage continues to change at a steady rate. There is no doubt at all that this works exactly as the formula predicts.

Hi Ian
Not sure if I understand what it is your are trying to point out. The more likely scenario is that I'm not explaining myself properly.

The current is directly proportional to the rate of change in voltage, yes? If so I can't clearly see where in the highlighted text I seem to go against this, except in that I am curious what would happen to his linear relationship as we approach charge saturation/depletion of the capacitor plates, or if even charge/saturation can occur

[IanB]

Hi Ian
Not sure if I understand what it is your are trying to point out. The more likely scenario is that I'm not explaining myself properly.

The current is directly proportional to the rate of change in voltage, yes? If so I can't clearly see where in the highlighted text I seem to go against this, except in that I am curious what would happen to his linear relationship as we approach charge saturation/depletion of the capacitor plates, or if even charge/saturation can occur

You seemed to be suggesting that you would have to increase the rate of change to maintain a constant current. Maybe I misunderstood, or it wasn't clear. But a constant rate of change will continue to produce a constant current indefinitely.

There is no such thing as charge saturation in a capacitor. The voltage can rise until the breakdown voltage of the insulation occurs. Until that time the linear equation is obeyed.

[cthree]

College professors could learn a lot by watching your teaching methods...

The problem with college professors is that when you know 100 times above what you're teaching, you forget that it's not as obvious to everyone else as it looks to you. This is why Maxwell's equations confuse so many people - a physics professor is so steeped in calculus that he can take one glance at the equation and know what it means. A student looks at the equation and at best sees a sequence of steps to be performed to get the result, not the true, physical relationship it implies.

College professors out there, remember this - whatever it is, it's only 1/100 as obvious as you think it is.

I thought they got paid to teach and not just know a lot of stuff they can't communicate effectively. Food for thought.

[cthree]

But does science flow through an engineer?

[AlfBaz]

You seemed to be suggesting that you would have to increase the rate of change to maintain a constant current. Maybe I misunderstood, or it wasn't clear. But a constant rate of change will continue to produce a constant current indefinitely.

Poorly explained by me. What I meant was, increasing the rate of voltage change to increase the current level so as to "deplete/saturate" the plates quicker

There is no such thing as charge saturation in a capacitor. The voltage can rise until the breakdown voltage of the insulation occurs. Until that time the linear equation is obeyed.

I think this is the crux of my confusion. Isn't the "charge" related to the capacitors capacitance. The voltage continues to increase due to the externally applied voltage. If we were to continue to increase the applied voltage to a capacitor past the point were the current is constant (ie charged up to its capacitance) and then suddenly remove that applied voltage, would the voltage across the now disconnected capacitor be the same voltage that was last applied or would it "drop" to a level that is a function of the charge stored in it? (Arghh, seeing I couldn't explain myself before, I'm sure I have done equally bad here )

Oh, and another thing, is it fair to say that I=C*(dv/dt) only holds true once the capacitor is charged or is this some idiosyncratic spice thing

[orin]

You seemed to be suggesting that you would have to increase the rate of change to maintain a constant current. Maybe I misunderstood, or it wasn't clear. But a constant rate of change will continue to produce a constant current indefinitely.

Poorly explained by me. What I meant was, increasing the rate of voltage change to increase the current level so as to "deplete/saturate" the plates quicker

There is no such thing as charge saturation in a capacitor. The voltage can rise until the breakdown voltage of the insulation occurs. Until that time the linear equation is obeyed.

I think this is the crux of my confusion. Isn't the "charge" related to the capacitors capacitance. The voltage continues to increase due to the externally applied voltage. If we were to continue to increase the applied voltage to a capacitor past the point were the current is constant (ie charged up to its capacitance) and then suddenly remove that applied voltage, would the voltage across the now disconnected capacitor be the same voltage that was last applied or would it "drop" to a level that is a function of the charge stored in it? (Arghh, seeing I couldn't explain myself before, I'm sure I have done equally bad here )

Oh, and another thing, is it fair to say that I=C*(dv/dt) only holds true once the capacitor is charged or is this some idiosyncratic spice thing

By definition, C = Q/V so Q = C * V.

Differentiate both sides with respect to time:

dQ/dt = C * dv/dt

What is dQ/dt?

Well, it's charge per unit time, i.e. current so you have I = C * dv/dt.

It holds all the time.  Well, at least until the dielectric breaks down which is at about 3 million volts per meter for air if memory serves correctly.

[IanB]

You seemed to be suggesting that you would have to increase the rate of change to maintain a constant current. Maybe I misunderstood, or it wasn't clear. But a constant rate of change will continue to produce a constant current indefinitely.

Poorly explained by me. What I meant was, increasing the rate of voltage change to increase the current level so as to "deplete/saturate" the plates quicker

But there is no such concept. This is not how capacitors work.

There is no such thing as charge saturation in a capacitor. The voltage can rise until the breakdown voltage of the insulation occurs. Until that time the linear equation is obeyed.

I think this is the crux of my confusion. Isn't the "charge" related to the capacitors capacitance.

No, the charge is related to the voltage established between the plates. The capacitance is a constant, not a capacity. It is not like a volume that can be filled.

The voltage continues to increase due to the externally applied voltage. If we were to continue to increase the applied voltage to a capacitor past the point were the current is constant (ie charged up to its capacitance) and then suddenly remove that applied voltage,

Again, this is a misunderstanding of how capacitors behave. You cannot apply a voltage to a capacitor¹, the capacitor applies the voltage to you. It owns its own voltage and delivers it when measured. You change the voltage of a capacitor by applying current to it. The voltage then changes for as long as you apply current, proportional to the length of time you apply the current (strictly the integral of the current over time, but I simplify).

would the voltage across the now disconnected capacitor be the same voltage that was last applied or would it "drop" to a level that is a function of the charge stored in it? (Arghh, seeing I couldn't explain myself before, I'm sure I have done equally bad here )

As soon as you stop applying current to the capacitor the voltage stops changing and remains forever at whatever value it had reached. (In the real world capacitors leak, but ideal capacitors will hold their voltage forever.)

Oh, and another thing, is it fair to say that I=C*(dv/dt) only holds true once the capacitor is charged or is this some idiosyncratic spice thing

It is not meaningful to say "once the capacitor is charged". The charge on a capacitor is a number that can vary from zero to as much as you like (essentially as much as the capacitor can sustain before voltage breakdown occurs). The equation I = C dV/dt holds always. It is the basic defining equation of a capacitor.

(It is possible you may have trouble with this if you have not studied mathematics or physics to a level where you deal with differential equations. Rates of change can be a difficult concept to grasp until you can relate the mathematical notation to real world behavior. A physical analogy might be mass and acceleration. Newton's second law of motion² says that a body will experience a constant acceleration--a constant rate of change of speed--for as long as a constant force is applied. Think of the force as current and the speed as voltage. The speed goes up and up without limit for as long as the force is applied. There is no "maximum speed" and no "saturation" that occurs, at least until you reach relativistic velocities. The speed can be as much as you like, 100 mph, 1000 mph, 1,000,000 mph. It just goes up without limit for as long as you apply the constant force.)

[1] To understand the statement "you cannot apply voltage to a capacitor", think of how a car changes speed. A similar statement is true: "you cannot apply speed to a car". A car changes speed if it is given a push by the engine, but it always takes time for the new speed to come up on the speedometer. You can measure the speed, but you cannot "set" the speed. If you are driving along at a steady 30 mph and you suddenly press your right foot to the floor, the speed does not immediately change. The instant after you put your foot down the speed is still 30 mph just as it was before. But then the speed gradually ramps up with a curve until it tops out and the car is fully "charged".

Capacitors obey the same rules: you change the voltage on a capacitor by giving it a "push", by feeding current through it. You can't set the voltage directly, but you can move the voltage where you want by applying a current, a "force", to it.

[2] Newton's second law of motion is just like the capacitor formula. The capacitor formula says:

I = C dV/dt

"Current equals capacitance times the rate of change of voltage"

Newton's second law says:

F = m dv/dt

"Force equals mass times the rate of change of velocity"

In this equation the mass is the "capacitance" of the object. More massive objects accelerate more slowly with the same force, and when they are moving they hold more "charge" (momentum) than small objects.

Just as with a capacitor you have the charge equation, Q = CV, so with a moving object you have the momentum equation:

p = mv

where p is the momentum, m is the mass and v is the velocity.

[EEVblog]

the researchers used a SQUID to measure the very low magnetic field, don't know if the AIM-TTI probe is that sensitive

Quite likely, I haven't done any back-of-envelope checks.

[gazza666]

Hi
is it fair to say that current from the battery does not flow through the capacitor
but the batteries influence causes current to flow from the opposite side of the capacitor but its not battery current
so you end up with  a positive and negative charge
I think of a capacitor has having a rubber sheet between it as the current flows it pushes on the rubber sheet and expands it
which forces the current down the line
so I say current from the "battery" does not flow through the capacitor
but current does flow from the other side of the capacitor

[amyk]

Just wanted to say that auto-transcribe surprisingly got most of the "capacitor" references, but it also had this little idiosyncrasy...

[John Coloccia]

I think we can all agree that current doesn't flow through a carrot.

[SeanB]

I think if you ask Photonic he will tell you it does, just needs enough welly " I need MOAR!!!!"

Then again the autocaptioning is farking hilarious at times, it just does not get enough Dave to do a good translation with, not surprising as it mostly only speaks Google english, or US english.

[ftransform]

this is all bullshit. the capacitor will eventually run out of electrons to push through the light bulb, even if you connect ac, if electrons do not flow through the "dielectric".

[c4757p]

this is all bullshit. the capacitor will eventually run out of electrons to push through the light bulb, even if you connect ac, if electrons do not flow through the "dielectric".

Unless I'm completely misunderstanding you, you're dead wrong. A light bulb powered by AC through a series capacitor will certainly run indefinitely, even with an ideal capacitor with zero leakage. It'll keep pushing around the electrons on the other side.

[penfold]

He did of course make the huge simplification that the electrons are particles!

[IanB]

Unless I'm completely misunderstanding you, you're dead wrong. A light bulb powered by AC through a series capacitor will certainly run indefinitely, even with an ideal capacitor with zero leakage. It'll keep pushing around the electrons on the other side.

I think you missed the smiley in that post. ftransform was surely speaking in jest.

[c4757p]

Sorry if I came across as overly blunt, then, it did soar over my head.

[ddavidebor]

Oooh, i've a good question.

Why we can't let an ac 3kw line pass throught a capacitor?

I've seen it only for signal

[IanB]

Why we can't let an ac 3kw line pass throught a capacitor?

We can, given a big enough capacitor. But I don't know of a common reason to do this, which is why you don't see it anywhere.

[ejeffrey]

Even worse than current flow in capacitors is the flow of holes in semiconductors. These behave like electrons with negative mass! Even more weird are phonons which knock electrons around and are quanta of crystal vibrations.

It's just convenient bookkeeping, though.  Charge comes in quanta of electrons, so lack of charge has to come in quantities of lack of electrons...holes.  It's like talking about negative money.  We all know that simply means "debt", but it's more convenient to handle it in the same units as regular money because then you can do straightforward math to get a straightforward answer.

It is more subtle than that.  Holes really behave like positive charge carriers (or negative mass) due to the effect of the crystal lattice.  You can see this (for example) by measuring the hall effect which allows you to measure the carrier charge directly through the effect of applied magnetic fields.

What is more, the thing you think of as 'electrons' (negative carriers) in a conductor are not really electrons either.  They are also quasi-particles that result from the collective behavior of a sea of (physical) electrons interacting with a lattice.  That is why the effective mass of negative carriers also varies from the bare electron mass.

Finally, quantum field theory teaches us that even what we think of as physical electrons like those in a vacuum tube, far away from any lattice potential are *also* quasi-particles whose effective mass is a result of interactions with the quantum vacuum.

I guess it doesn't matter too much.  It's useful to consider current flowing through a capacitor but thinking about it like that sets up an intuition that breaks down when you look too closely.  Better to consider the real process and consciously choose when to simplify it.  In my opinion...just my opinion.

No, it becomes more correct the deeper you look.  As dave said, there are two types of current, but the useful, physically meaningful quantity is the total current including the displacement current.  It is useful precisely because it obeys a conservation law.

[ddavidebor]

Why we can't let an ac 3kw line pass throught a capacitor?

We can, given a big enough capacitor. But I don't know of a common reason to do this, which is why you don't see it anywhere.

How big should it is?

[KedasProbe]

Well I guess Yes* is equally as good as No*
With * meaning: but don't base your reasoning on that.


edit:

[mtdoc]

Nice video Dave.   This is not an easy concept to teach.

I used to teach electrophysiology and current flow "through" a capacitor was always a difficult concept.  Nerve and muscle cells are really just miniature RC circuits with cell membranes having both capacitance and resistance.   Another type of current you didn't talk about is ionic current which is what flows through channels in cell membranes - nerve and muscle cells being the prime examples.

The problem with college professors is that when you know 100 times above what you're teaching, you forget that it's not as obvious to everyone else as it looks to you. ......

True but the problem is that the knowledge base of the students in most classrooms has a wide distribution (though not likely along a bell shaped curve!).  So if you teach to the bottom half and oversimplify, the top students are bored and frustrated.  If you teach to the top students, you've lost the bottom half.  Not as easy as it looks....

[IanB]

How big should it is?

It depends on the operating frequency, the current in the circuit, and the allowable voltage drop.

Suppose we want our 3 kW load to operate at mains voltage, 230 V, 50 Hz, and we want a small voltage drop, say < 10 V. Then a capacitor of about 5000 µF in series with the load should do that, if I have my calculations correct¹. However, this would have to be a non-polarized capacitor with a voltage rating of 400 V or so. It would not be small or cheap.

[1] Assuming a desired reactance of about 0.5 ohms, then:

X_c = 1 / (2 pi FC)

0.5 = 1 / [(2)(3.14)(50)C]

C = 1 / [(2)(3.14 )(50)(0.5)] = 0.0064 F = 6400 µF

[KarlMonster]

Yeah, yeah, yeah, so caps work the same as they always did, whatever.

I just want Dave's busted Ampere analogue meter shirt! That is far-riggin cool!

[JackOfVA]

How big should it is?

It depends on the operating frequency, the current in the circuit, and the allowable voltage drop.

Suppose we want our 3 kW load to operate at mains voltage, 230 V, 50 Hz, and we want a small voltage drop, say < 10 V. Then a capacitor of about 5000 µF in series with the load should do that, if I have my calculations correct¹. However, this would have to be a non-polarized capacitor with a voltage rating of 400 V or so. It would not be small or cheap.

[1] Assuming a desired reactance of about 0.5 ohms, then:

X_c = 1 / (2 pi FC)

0.5 = 1 / [(2)(3.14)(50)C]

C = 1 / [(2)(3.14 )(50)(0.5)] = 0.0064 F = 6400 µF

Power factor correction capacitors are used in distribution networks where they are installed in series (sometimes shunt, but that's less common I believe). The idea is to cancel or at least reduce the inductive reactance of motors and bring the power factor closer to 1.00.  Remember that losses in power distribution systems are mostly current related, so anything that brings VARs closer to watts saves losses and money.

PF correction capacitors are often installed in a plant distribution load center but sometimes you will see them on pole top mounts in normal residential and commercial power feeders. You will also find them in transformer substations. In the end distribution network, they are in the high voltage side of the network, not the 240V side. Easy to recognize if you look - older style are rectangular cans with insulators, but some are now in cylindrical enclosures.

So, yes, it's not uncommon to find a series capacitor in a several KV power distribution feeder.

You also will find some electric motors use a run capacitor in series with one of the field windings. The purpose of the run capacitor is to shift the phase of winding current so that a rotating magnetic field is developed. Start capacitors are probably more common, and they serve a similar purpose but only are connected when the motor is started and then disconnect via a centrifugal switch when it is up to speed.

[EEVblog]

I just want Dave's busted Ampere analogue meter shirt! That is far-riggin cool!

It's not my design, but my mate Rogers. He took the actual photo in an old power station.
If there is sufficient interest I could ask if he wants to run a Teespring crowd funded campaign?

[wbeaty]

Here's a way to measure displacement current.   Use a really thick dielectric.  But make it narrow.

Get yourself a rod of PZT ferroelectric ceramic insulator.  That's the material used in ceramic capacitors. 10cm long by 1cm wide would be good.   Plate the ends with metal, and hook up some wires.  You've built a capacitor.

Now connect this capacitor to a high-volt signal generator, and send a few KHz at a few mA through the device.

Next, connect a clamp-on ammeter probe around the rod.  Take a measurement.  You'll see a few mA.

Finally, slide the clamp-on probe all around the circuit.  The milliamps will be the same in the wire, in the insulating rod, and in the metal plating.   Well duh, current is constant in all parts of any basic circuit loop.  That includes the dielectric of capacitors.

Some may point out that the insulating rod is full of mobile electrons, and these electrons are wiggling.  Thus, a current exists.   Sure, but look at the original concept:  DOES CURRENT FLOW THROUGH CAPACITORS?   Is this rod thingy not a capacitor?  Of course it is.   Is there zero current?  Nope.  So, yes, current flows through capacitors.  And the rod device shows that we can even use apparent insulators as a kind of AC conductor, see: http://amasci.com/elect/mcoils.html.

OK, what about VACUUM-DIELECTRIC CAPACITORS?      (I bet there are some people who actually believe that currents do flow in ceramic capacitors and electrolytics, but not in vacuum capacitors.)

[c4757p]

I bet there are some people who actually believe

Doesn't matter what comes next, you'll always win that bet.

[wbeaty]

Doesn't matter what comes next, you'll always win that bet.

Heh.  The real question is, are they students, or do they see themselves as students?   If not, then very probably their ignorance is willfully maintained, and they'll never learn anything new.   We should all try to be like Einstein and RP Feynman, both of whom saw themselves as students, and tended to look down on "experts."


Also...



Does any water flow through the tank above?  Trick question!  Yes, there is a current, because whenever we force any water into one pipe, an exactly equal volume will come out of the second pipe.   Also, NO, no water flows through, because part of the current is made out of flexing rubber!

And that gives a clue to this whole mess.  It easily becomes a pedantic fight over mis-heard terminology.   Do charges flow through capacitors?   Not necessarily (but see the PZT rod above.)   Yet that's a very different question than this one:  is there a current in the space between the capacitor plates?  Yes, there is.  The current in the dielectric is always the same as the current in the capacitor leads.  A clamp-on ammeter could even detect it.  But the amperes in the dielectric need not be made out of flowing charged particles.

Pedantic Nitpicking is required!  "The ill and unfit choice of words wonderfully obstructs the understanding." - Francis Bacon

"Many errors, of a truth, consist merely in the application of the wrong names of things." - Spinoza

"Lest you think that I am quibbling over minor points of language, I note that in my experience many of the misconceptions people harbor have their origins in imprecise language... Precise language is needed in science, not to please pedants but to avoid absorbing nonsense that will take years, if ever, to purge from our minds."- Dr. Craig F. Bohren

[John Coloccia]

I wonder why everyone keeps treating "displacement current" as a real current?  It was misnamed by Maxwell because Maxwell was completely wrong about it's source.  It refers to nothing but a changing electric field giving rise to a magnetic field.  He imagined it a wiggling around of the aether, I suppose.

But let's talk about current flowing through the dielectric.  What's the resistance of the dielectric?  What happens to the all the Watts you should be dissipating?  Now even more laws fall apart, including Ohm's law.

And yes, water flows in the tank.  You can sit there and count the number of water molecules that scoot by a given point.  If you try to count the number of charges that scoot by a given point in the dielectric of a capacitor, hopefully you'll count 0 (and you'll actual count the leakage current).  So the tank analogy is a good one for the basic concept, but it breaks down when you look too closely.

Anyhow, I don't want to ruffle any feathers.  It think this is a fun discussion.

[IanB]

How big should it is?

In order to derail the thread even further, let's take a minor detour into the subject of language.

The correct phrasing above is, "How big should it be?"

Given the conditional nature of the question we should use the subjunctive mood (il congiuntivo in Italian, as I just learned).

In English, the present subjunctive of the verb "be" is also "be". As famously expressed in the song, Que Sera Sera:

When I was just a little girl
I asked my mother, what will I be
Will I be pretty, will I be rich?
...

However, you will find, as a student of English, that the subjunctive tense is an endangered species. English is being dumbed down, and "proper English" is heard less and less in England these days.

References:

[1] http://italian.about.com/od/verbs/a/italian-verbs-present-subjunctive-tense.htm
[2] http://en.wikipedia.org/wiki/Subjunctive_mood

[c4757p]

And yes, water flows in the tank.  You can sit there and count the number of water molecules that scoot by a given point.  If you try to count the number of charges that scoot by a given point in the dielectric of a capacitor, hopefully you'll count 0 (and you'll actual count the leakage current).  So the tank analogy is a good one for the basic concept, but it breaks down when you look too closely.

The rubber plate in the tank is the dielectric, and if you count water molecules scooting through the rubber... that ain't rubber.

I wonder why everyone keeps treating "displacement current" as a real current?  It was misnamed by Maxwell because Maxwell was completely wrong about it's source.  It refers to nothing but a changing electric field giving rise to a magnetic field.  He imagined it a wiggling around of the aether, I suppose.

1) Newton was wrong about many things as well but we still use his equations and terms when they are sufficiently good approximations. I don't need relativity to figure out how far something will go if I throw it.
2) That's exactly what we're saying displacement current is. We've established already that we are drawing a distinction between current and electric current, with the latter one type of the former. To paraphrase the bastard on YouTube who started all of this, you're saying I don't have fruit in my basket because you don't consider my tomatoes to be sufficiently fruity to be called fruit, when for the purposes of basket-measuring convenience, I've defined them as such.

[duskglow]

I've been following this topic for a while and staying out of it, because I don't have a whole lot to add mathematically.  It's been very educational to me, though, in multiple ways.

I've learned that there is a such thing as displacement current.  I learned how capacitors work in a better way than I did previously.  I even learned how inductors work in a better way than I did previously.

And I learned that in every discipline, there's always *someone* who insists on splitting hairs in such a way that newbies will get hopelessly confused if they listen to that person.

Here's the deal, from my perspective.  If you want to take electric current and displacement current, add them together, call it current, and suddenly all sorts of equations work and things light up and buzz and doesn't release the magic smoke, then trying to split hairs about it from an electronics perspective really isn't helpful.  It may be, in some obscure and very deep field of study, technically correct, but it's utterly useless in the context of getting things to beep, buzz, and release the magic smoke.

I know as a Linux guy, there are a few "traps for young players" too.  For example, if someone were to ask me what "tar" does and I were to go into a huge dissertation about tape drives and streaming data and all that...  yes, that's what tar originally did (tape archive) but all a hobbyist really wants to know is, how to I turn this .tar file into a directory I can do stuff with?  "tar xvf" and move on.  Read the man page if you care.  But there is absolutely nothing essential about the history of the tape drive and tape archiving, etc., and all it does is confuse the issue. 

And there is absolutely nothing essential in this case about knowing where the individual electrons are going.  The important thing is "how do I make the circuit with this capacitor in it do what I want?"  And the answer is...  you think of electrons as flowing through it.

Please, for the sake of beginners and newbies and people like me who have been at it for a while and are only just now learning some of the deeper stuff - don't confuse the issue.  It doesn't help.

*gets off my soapbox* 

[IanB]

I think this thread has exhausted the subject of electricity, and definitely should be moving on to the subject of language:

[c4757p]

I know as a Linux guy, there are a few "traps for young players" too.  For example, if someone were to ask me what "tar" does and I were to go into a huge dissertation about tape drives and streaming data and all that...  yes, that's what tar originally did (tape archive) but all a hobbyist really wants to know is, how to I turn this .tar file into a directory I can do stuff with?  "tar xvf" and move on.  Read the man page if you care.  But there is absolutely nothing essential about the history of the tape drive and tape archiving, etc., and all it does is confuse the issue. 

FWIW, that wouldn't answer the question anyway, tar doesn't really do tapes. (OK, it has a few tiny little "helper" functions for them, like tape-swapping, but most of it is medium-independent, taking advantage of the Unix device file concept.) The correct answer is "concatenates files, retaining metadata, and separates these concatenated streams back into individual files".

And using that word - "concatenate" - explains why you can easily append files to an existing archive, but removing a file or extracting one from in the middle of a thousand others takes ages.

Also - screw tar. What a vile pain in the ass. I only use it because it's the only format that I always remember (without a Google search) supports all the Unix file attributes.

You know, this forum would stay on topic about 75% more if I left.

[duskglow]

 

[IanB]

What's Unix? 

[c4757p]

Wait, did I brainfart?

What's Unix? 

Castrated men.

[duskglow]

Wait, did I brainfart?

No, just took the one paragraph I wrote that could have been left out without changing the meaning at all, and gave a really great example of what I was asking people not to do wrt the displacement current argument.    

[c4757p]

I suppose so. My point was that it wasn't really a close analogy - going into a discussion about tape backup when describing tar is more analogous to going into a discussion about pre-electron theories of electricity. It's even more irrelevant than arguing about current types, because the distinction between currents is the underlying behavior, but the niche applications of tar are more historical facts.

And I thought my description involving concatenation was pretty short and not confusing.

But you're probably right.

gave a really great example of what I was asking people not to do

I'm spectacularly good at not doing what people ask me to do. That's why I derail so many threads. I'm not sure if it's belligerence, stupidity, forgetfulness, or a mixture of both.

[ejeffrey]

And there is absolutely nothing essential in this case about knowing where the individual electrons are going.  The important thing is "how do I make the circuit with this capacitor in it do what I want?"  And the answer is...  you think of electrons as flowing through it.

Yes.  My point is that 'current flows through the capacitor' is not just right at a practical convenience, but also at a fundamental level.  The people who try to split hairs are not just unhelpful, but also fundamentally wrong.

[duskglow]

I agree it wasn't the best analogy.  I just pulled something out of my rear in an attempt to tie it to something from *my* field of expertise.  I wasn't trying to do an exact one for one.  I was just trying to make the point that this discussion, while I'm sure very interesting and esoteric and I'd love to learn more about it, just isn't very helpful when it comes to the topic at hand, which is making things beep and buzz.

Sorry if I came off a bit harsh there, I have a killer headache.

[duskglow]

Yes.  My point is that 'current flows through the capacitor' is not just right at a practical convenience, but also at a fundamental level.  The people who try to split hairs are not just unhelpful, but also fundamentally wrong.

But even if they were right, that wouldn't matter, I guess is my point; for the practical purpose of getting things to beep and buzz, it's not helping.

[c4757p]

Sorry if I came off a bit harsh there, I have a killer headache.

Ah, it's fine. I think I came off a bit harsh too - I wasn't feeling eloquent enough to make it sound polite, and I wasn't feeling intelligent enough to just decide not to say it 

The more of my posts you see, the more you'll start to realize that I generally fail to understand that people can't hear the tone of voice I hear in my head... 

[duskglow]

Ah, it's fine. I think I came off a bit harsh too - I wasn't feeling eloquent enough to make it sound polite, and I wasn't feeling intelligent enough to just decide not to say it 

And I just took another ibuprofen, so my crankiness should ebb at some point.

I probably shouldn't have spoken up, I'm not adding a whole lot.  Things like this kind of rub me wrong, though.  I get confused enough.

[c4757p]

You're doing better than I would. When I have a massive headache, the only things I understand are "aspirin", "air conditioning", "shut the hell up" and "piss off".

[IanB]

And there is absolutely nothing essential in this case about knowing where the individual electrons are going.  The important thing is "how do I make the circuit with this capacitor in it do what I want?"  And the answer is...  you think of electrons as flowing through it.

No, no, no, no. You think of current flowing through it. If you think of electrons when doing basic circuit design, then you have sprung a leak in your abstractions and your analysis has failed. Electrons do not exist in basic electronic circuit theory.

[duskglow]

I was going to reply, but then I realized I'd just be continuing the argument, so I used the wrong term.   It didn't take away from the point I was trying to make, though.

[IanB]

I was going to reply, but then I realized I'd just be continuing the argument, so I used the wrong term.   It didn't take away from the point I was trying to make, though.

I agree with the point you were making, but the whole existence of the word electron in the dictionary does major harm to the subject of circuit analysis. If you are not studying physics, electrons do not exist.

[croberts]

It's time to bring Larry Fine of The Three Stooges into this.

Teacher to Larry Fine:
If I give you a dollar and your father gives you a dollar how many dollars will you have?

Larry Fine To Teacher:
One dollar.

Teacher to Larry Fine:
You don't know your arithmetic.

Larry Fine to Teacher:
You don't know my father.

[wbeaty]

And I learned that in every discipline, there's always *someone* who insists on splitting hairs in such a way that newbies will get hopelessly confused if they listen to that person.

Yeah, that's one class of internet troll.   They're trying to appear expert by playing the "oneupmanship game."

On the other hand, I frequently find myself in just that role, because often the conventional jargon needs to be avoided because it leads to beginners' misconceptions.  Or sometimes the beginners obviously learned some grade-school science which is simply wrong, and their unsuspected learning barriers could use some debunking.

In that case I must switch into pedantic nitpicking physics teacher mode, not because I'm trying to be a self-important troll, but because I've seen the harm that the bad language can do to students' concepts.  I can attack it and head it off early:  clarify with rigorously correct and clearly-defined words to shatter the learning barriers even before anyone runs up against them.

For example, terminology:  "electric current" is commonly defined the same as current.   Electric current isn't usually defined to mean electron flow except in this thread.   Perhaps "electron current" was meant?

But what we're really talking about here are currents ...versus "charge flows."     Currents include electron flows, ion flows, proton flows, charged particle beams, and displacement currents.  If something creates a closed loop of  magnetic field, then it's an electric current.    But when discussing these issues, the term "charge flow" is a bit better than the term "electric current" because charge-flow specifically excludes displacement current, while "electric current" doesn't  ...unless we SAY it does, and unless everyone reads the first messages and then adheres to the definitions there.

To cut through any BS and put it as clearly as possible, capacitor dielectrics do support "electric currents," even though there is no need to have any "charge-flows" in those currents.

Or simplicity from a non-theory standpoint: if a Rogowski probe (a clamp-on ammeter) says that there's a current through the center of its ring, then indeed there's a current, even if that current is entirely made of a vacuum with a changing radial pattern of e-fields.  Clamp-on ammeters measure charge-flows in wires and displacement-currents in empty space, and they can't tell the difference.   Displacement current is real.   But it's not charge-flow.

PS

Another issue:  are people curious about what REALLY happens inside capacitors?  Delve deeply?  Confront actual physics, remove the mysteries, yet try to do it without a pile of equations?   If so, that's a different thread than one about circuit design, or about the One True Meaning of some piece of jargon which is defined totally differently by scientists, engineers, and the public.

[wbeaty]

In order to derail the thread even further, let's take a minor detour into the subject of language.

Abbot and Costello attempt to clarify basic power-transmission issues:

WATT IS A VOLT?

[vk6zgo]

The point is current is not defined as a movement of electrons. That is best left to thick headed children or people with an MFA.
You mean as distinct from members of the Flat Earth society who still persist with Conventional Current Flow?

Consider a wet cell, lead acid battery, lemon with a nail and coin, etc. No electrons are passing, internally, between the terminals, yet a current is flowing.

Actually,positive ions which  still have electrons in their outer shell are passing internally,as part of the chemical reaction.

In an ideal electrochemical cell,there is no such flow except when electrons or pretend positive charge carriers are involved with the flow of current in an external circuit.

[NiHaoMike]

Oooh, i've a good question.

Why we can't let an ac 3kw line pass throught a capacitor?

I've seen it only for signal

In switching power supplies, it's pretty common to use series capacitors with half and full bridges.

[ftransform]

Did you know that you can actually break a circuit by feeding a unbalanced AC signal into it? If you send a binary signal into a capacitor, that is +1, -1, -1, -1, -1, +1, you create a electron vacuum in the capacitor, and you have to transmit several more positive pulses before you can unblock it?

[KedasProbe]

Nostalgia, in the time when they properly explained things
I looked for it on the internet and luckily I found it

In my time I recorded all episodes on VHS tapes (I treated it as my holy grail )

edit: I just realized I could have added questions
http://en.wikipedia.org/wiki/VHS

[EEVblog]

I looked for it on the internet and luckily I found it

Wow, awesome find! 

[John Coloccia]

So I'll ask the question again.  Does current flow from the radio station antenna to the receiver in your car?  Is current flowing from quasars to the radio antennas we use to study them?  Displacement current simply describes...or rather completes....the description of radiation in free space.  I've never heard anyone seriously consider this as "current" of any kind, especially since it doesn't follow Ohm's law.  It doesn't really have much in common with conventional current other than the name, so it just turns into a word game.  It's definitely unconventional to consider current flowing from a satellite to the dish on your roof.

That video is great, BTW.

[digsys]

And that was the end of that ...... :-)

[KedasProbe]

Wow, awesome find! 

You may also like all other full episodes:
http://www.learner.org/resources/series42.html

[John Coloccia]

I recall a series on Calculus that was presented in much the same style.  Very clear and very good.  I looked and I wasn't able to find it, but I wonder if these guys did that too.

[amyk]

But let's talk about current flowing through the dielectric.  What's the resistance of the dielectric?  What happens to the all the Watts you should be dissipating?  Now even more laws fall apart, including Ohm's law.

This reminds me of someone who told me that CCFL inverters can be tested using a capacitor as a dummy load, and he said it didn't get hot at all... didn't believe him at the time but now that I think about it, it must be related to this?

[IanB]

This reminds me of someone who told me that CCFL inverters can be tested using a capacitor as a dummy load, and he said it didn't get hot at all... didn't believe him at the time but now that I think about it, it must be related to this?

The lack of heat is because no power is being dissipated in the capacitor.

You figure out the power consumed by a device by summing up the product of voltage and current over every instant in time and looking at the average value of that product over a sufficiently long time period relative to the AC frequency. (Mathematically you could do it over one complete AC cycle.)

If you do this computation for a capacitor you find the current is out of phase with the voltage in such a way that the total power is zero. (The voltage wave and the current wave are 90 degrees apart, with the current leading the voltage.)

[Ericho]

Just thank you Dave.

Another useful tool for me     

[ashplant]

Great video!

I don't actually disagree. If you define current as, "anything that makes an Ammeter twitch.", current does flow through an ideal cap. But I define current as "flow of charge". 

I say, "Charge doesn't flow through an ideal capacitor, though it does flow through ideal capacitor leads."

In the dim mists of time, when I was a kid, I was really impressed when a teacher told me how you can get electromagnetism from relativity. Something about how, in a wire with Voltage across it, charge cancelled, but electric fields don't quite. He said that protons (which on average are stationary) have symmetric fields a' al Coulomb, but the moving electrons have asymmetric electric fields courtesy of Lorentz contraction, Then he did some algebra and got
results you usually get with fields here and fields there and right hand rules. I thought the advantage to his approach was it didn't make my head hurt.He said the advantage was you don't  need "ad hoc" notions like displacement current.

[wbeaty]

So I'll ask the question again.  Does current flow from the radio station antenna to the receiver in your car?

Obviously not.  Waves are involved, so displacement currents don't connect between distant antennas in the way that they between connect capacitor plates.

Instead you could have asked: are there any electric currents (displacement currents) associated with propagating EM waves in empty space?

[duskglow]

Ah, but couldn't the space between the radio station antenna and your car be seen as a very, very low value capacitor?  I mean, the fields would be so small as to be vanishing, and for all intents and purposes are zero, but wouldn't they actually be an extremely small (many zeros to the right of the decimal point) value?

Or past a certain distance does the dielectric become perfect?

[IanB]

So I'll ask the question again.  Does current flow from the radio station antenna to the receiver in your car?

Obviously not.  Waves are involved, so displacement currents don't connect between distant antennas in the way that they between connect capacitor plates.

Instead you could have asked: are there any electric currents (displacement currents) associated with propagating EM waves in empty space?

I think it is fair to say that the induction of current between the windings of a transformer and the induction of voltage between the plates of a capacitor are two dimensions of the same theory that allows electromagnetic waves to propagate through space. For instance, if we take the primary and secondary windings of a transformer and gradually move them further and further apart, at what point does the apparatus cease to be a transformer and start to be a transmitting/receiving pair of antennas?

The answer is that there is no defining distance as such. When the coils are close together they are very "transformery" and when they are far apart they are very "transmittery". Somewhere in between they are a bit of one and a bit of the other.

What is actually propagated through the air in electromagnetic waves is energy, and the energy waves/particles can interact with matter to induce voltages and currents in the things that they touch (e.g. a radio antenna or a solar cell).

[duskglow]

The answer is that there is no defining distance as such. When the coils are close together they are very "transformery" and when they are far apart they are very "transmittery". Somewhere in between they are a bit of one and a bit of the other.

I idly wonder what properties such a hybrid would have.

[IanB]

Ah, but couldn't the space between the radio station antenna and your car be seen as a very, very low value capacitor?  I mean, the fields would be so small as to be vanishing, and for all intents and purposes are zero, but wouldn't they actually be an extremely small (many zeros to the right of the decimal point) value?

Or past a certain distance does the dielectric become perfect?

Remember that "displacement current" is not a physical thing. It is the name of a term in a mathematical model that helps when using the model to understand and explain things.

There is of course a vanishingly small capacitance between two very distant objects, but this capacitance is not the explanation for how radio transmitters work. Once the distance is large enough the capacitance disappears in importance and electromagnetic waves grow in importance.

Consider an analogy with water.

If you take a small tub of water and you use a paddle to move the water at one end, you will see the water immediately move at the other end and overflow the tub. This is "capacitance". Any movement you make at one end of the tub immediately induces a similar movement in the water nearby.

Now try the same experiment in a lake. If you move the paddle in the water on one side of the lake, the water will not move at all on the other side. However, all is not lost. If you move the paddle back and forth in a vigorous manner you may make waves on the surface of the lake, and these waves may travel to the other side where they can be detected. No water has moved across the lake with these waves, only the energy of motion has been carried across the lake. This is like radio transmission.

[John Coloccia]

Exactly, Ian.  Displacement current is a necessary part of radiation propagating without the help of charge carriers scooting along.  That's the only point that I'm trying to make.  It's not a phenomenon that has to do with capacitors.  It's a phenomenon necessary to completely describe capacitors, but only because you need to consider the magnetic field in the situation when you have exactly NO real current.  That's the crux of it.  Without displacement current, you don't have radio.  If current flows through the capacitor, then current also flows from the radio station to your car.

That's the only point I'm trying to make.  I really hope I'm not ruffling feathers.  I don't mean to, but I want to link these concepts in a way that makes real sense and is consistent.  I know people have called me pedantic, but when we're talking about basic components and start invoking Maxwell, I think the point is to get a little pedantic and really have an interesting conversation about different implications each point of view has, and where each is useful...and were each breaks down.  I'm just probing the edges.

[JackOfVA]

The answer is that there is no defining distance as such. When the coils are close together they are very "transformery" and when they are far apart they are very "transmittery". Somewhere in between they are a bit of one and a bit of the other.

I idly wonder what properties such a hybrid would have.

If you study electromagnetic theory and antennas you will find that the coupling between two antennas is often expressed in a power series involving 1/rⁿ where r is the distance between the two antennas in terms of wavelengths. Antenna engineers usually describe the "far field" as the distance beginning where the field strength (in volts/meter or Amperes/meter) is dominated by the 1/r component of the power series (energy in this case is 1/r²) and the "near field" as the distance where this simplification no longer applies and the coupling between the two antennas is a combination of radiated, inductive coupled and capacitive coupled modes. 

You can also find  evanescent wave creation from the antenna to air interface discontinuity.

These relationships are incredibly complex to model with closed form solutions to Maxwell's equations in the general case and the numerical simulation tools developed in recent decades make life much easier for the antenna engineer.

[IanB]

If current flows through the capacitor, then current also flows from the radio station to your car.

Everything is OK until that sentence. The one does not imply the other.

We can say that current flows through a capacitor because as a two terminal device, any current flowing in or out of one of the terminals is exactly balanced by the same current flowing out or in at the other terminal. If the current flows in to one terminal and out of the other, then the current must have flowed "through" the capacitor.

To see why this is not true of radio transmission, consider a transformer, which is a four terminal device. When current flows through the primary winding, current can be induced to flow through the secondary winding. However, the two windings are isolated. The primary winding satisfies a current balance (in = out), and the secondary winding also satisfies a current balance (in = out). Consequently, no current flows between the primary and the secondary.

Radio transmission is more like a transformer than a capacitor. The transmitting antenna satisfies a current balance and the receiving antenna satisfies a current balance. However, no current flows from transmitter to receiver.

[John Coloccia]

Oh, now this is getting interesting.  What's really funny is that it's almost easier to show that displacement current implies a real current between transmitter and receiver than it is to show that there is none   You need to have displacement current to have waves propagating in free space.  It's a bit technical, but before Mawell you had an interesting situation where....damn, how do you say this in plane language?  Maybe you say that a jiggling electric field couldn't affect the magnetic field, and thus there was an inconsistency in the laws.  The displacement current is the term that allows the field itself to give rise to the magnetic field, and thus allows all of the laws to be consistent in free space even if you have no charge whatsoever.

Now we're getting to the nitty gritty.

edit:  and the whole thing about 1/r^2 is really just talking about when you're far enough away that you can consider something to be a point source and everything between you and it is a straight line.  That's my physical intuition about it, anyway.  You're so far away that that everything looks like a very simple, textbook, homogenous field.

[IanB]

It might be easier to go the other way and say that "real current" has no more actual existence than "displacement current".

What we have is a mathematical model of the physical world, and in this model we have terms which we give names like "voltage", "current", "magnetic flux" and so on. It turns out that this model accurately describes behavior we observe in the real world, so we say it is a good model.

But just because the model works, it does not mean that the things the model describes have any actual, real existence. We can't see electric current, we can only know of its effects. Because it has observable effects we assume it is there. This assumption is useful, so we hold onto it.

When it comes down to it, nobody really knows how electromagnetic fields work, or what they are made of. You don't need displacement current to enable electromagnetic waves to exist. Electromagnetic waves existed long before humans ever tried to model them. Displacement current is just an abstract mathematical quantity in a bunch of equations that happen to fit the data.

[JackOfVA]

Oh, now this is getting interesting.  What's really funny is that it's almost easier to show that displacement current implies a real current between transmitter and receiver than it is to show that there is none   You need to have displacement current to have waves propagating in free space.  It's a bit technical, but before Mawell you had an interesting situation where....damn, how do you say this in plane language?  Maybe you say that a jiggling electric field couldn't affect the magnetic field, and thus there was an inconsistency in the laws.  The displacement current is the term that allows the field itself to give rise to the magnetic field, and thus allows all of the laws to be consistent in free space even if you have no charge whatsoever.

Now we're getting to the nitty gritty.

edit:  and the whole thing about 1/r^2 is really just talking about when you're far enough away that you can consider something to be a point source and everything between you and it is a straight line.  That's my physical intuition about it, anyway.  You're so far away that that everything looks like a very simple, textbook, homogenous field.

That's not quite right. In the near field, you have reactive coupling - due to capacitive and magnetic field coupling between the antennas - as well as some electromagnetic field.

The far field assumption is that the reactive coupling components of energy transfer are sufficiently attenuated to be negligible. Yes, this then becomes a homogenous field, at least over some small area, but a key element of the far field simplification is that you don't have to worry about the inductive and capacitive coupling between the two antennas.

 

[ejeffrey]

It might be easier to go the other way and say that "real current" has no more actual existence than "displacement current".

This.  Current is a mathematical concept, and as such we are free to define it however we like.  It turns out that 99% of the time, the only useful thing to call current is the total current (charge motion + displacement current).  Taken individually, those concepts are almost entirely useless for describing any sort of electrical behavior, except in the limit where one of them is zero.  In every equation you have ever seen, the symbol 'I' means the total current.

Radiation doesn't look like current flow because the displacement currents are transverse to the propagation direction.  The the current flux through, for example, a plane separating the emitter and receiver is zero.  But if you have a volume of otherwise empty space with radio waves traveling through it, then yes, the current density is non-zero.  That is the difference between near- and far-field.

Every physicist I know regards displacement current as 'just as real' as charge motion.  Electrical engineers probably don't think about it as much, but they implicitly agree because all of the equations they use to design and describe circuits treat the two as equivalent.

Finally, as a trivial point, the bulk of the current in most capacitors is actually due to microscopic charge motion.  The relative permitivity of most capacitor dielectrics is much greater than 1, so the majority of the current comes from reorienting microscopic dipoles withing the dielectric.  If you use the version of maxwell's equations for dielectrics (the one that includes H and D fields instead of just B and E), that would be a displacement current, but it is due to physical charge motion.

[NiHaoMike]

Did you know that you can actually break a circuit by feeding a unbalanced AC signal into it? If you send a binary signal into a capacitor, that is +1, -1, -1, -1, -1, +1, you create a electron vacuum in the capacitor, and you have to transmit several more positive pulses before you can unblock it?

Better known as DC balancing a signal. That is obviously required for channels that are inherently AC coupled, but even DC coupled channels sometimes have to use it as too much low frequency content can "charge up" the transmission line (remember, it's not ideal) and cause ISI.

[KedasProbe]

You shouldn't care too much on the naming of things, just understand what is happening.
You could say: If the magnetic field from the change of the electric field was discovered before they found the magnetic field around a current, then maybe Maxwell would have named that term (the current in a wire) a "Displacement Electric Field"

[Poe]

Just trying to see if I understand Dave's video and the reality of the situation.  I admit to being a neophyte and not yet reading the entire thread.  My take on it so far..

There are not two types of current.  Current is simply defined as the the flow of electric charge.

The equations used in Dave's video do not imply "through", they imply "into/out-of".

The phrase "Displacement current" appears to only be used as Maxwell's way of explaining why Amperes law doesn't discount the electric/magnetic field link.  Specifically, current isn't the only thing that is needed for a magnetic field to be created.  The change of flux in a capacitor is equally adequate.  This changing flux is "displacement current".  Just because it has the word 'current' does not mean it's a different type of current. 

Current flowing 'into and out-of' is not the same as "through" contrary to what Dave is saying in his videos. It appears he doesn't understand the concept and is using the "it's too complex for me to explain, trust me" technique to brush away the topic.  Not to be rude, this is just my interpretation of the information I have available and my own limited knowledge.  I'll edit this post after reading the posts from other more educated members.

Let me say thanks Dave.  I truly appreciate the videos, forum and the conversations they start.

[duskglow]

I think we were just trolled.

[PeteInTexas]

The FF videos are shaping up to be fun series.  Awesome work!

So, current flows through a capacitor.  Does Ohm's Law apply through a capacitor?

[digsys]

OHH NOOOO ... it's happening again :-)

[EEVblog]

I think we were just trolled.

Possibly!

I'll bite a bit and say that there are several things I wanted to point out with this video:
1) The term "through" is commonly used and accepted in not only the industry but in teaching as well.
2) There is another "type" of current that not many people know about, and as such there is more than one way to look at the problem, not just the flow of individual electrons.
3) The thought of current flowing "through" the capacitor is commonly how you explain circuit behaviour and design things.

Any of those alone is justification for viewing current as flowing "through" a capacitor.
You can debate the physics until the cows come home. The fact is it is right to say and think that current does flows "through" a capacitor.
And there is theory to back up assertion as well.

[EEVblog]

There are not two types of current.

Yes, there are.

  Current is simply defined as the the flow of electric charge.

As a one-sided simplistic definition, yes. Otherwise, no.

The equations used in Dave's video do not imply "through", they imply "into/out-of".

No, they don't. They do imply "through".
Where in those high level equations does it imply individual electron charge buildup on the plates etc?
The high level everyday equations most certainly do imply "through" and that is why is is thought of that way and why through is the most common industry term.

[AlfBaz]

If I pick up static in one room, walk into the kitchen and discharge it by touching the sink, did we just witness current flow from one room to the next?

[madires]

If I pick up static in one room, walk into the kitchen and discharge it by touching the sink, did we just witness current flow from one room to the next?

If that walk is caused by your wife watching her favoured TV show we may call that a displacement current :-)

[Robomeds]

Not entirely on topic but this was a general area which I covered a while back in a class on bond graphs. 
http://en.wikipedia.org/wiki/Bond_graph

One of the more interesting parts of the class what a discussion of how inductors work.  In EE we are taught to think of inductors as the electrical equivalent of a mass in a mechanical system.  Current through an inductor has inertia just as a mass has inertia.  This would mean we have electrical kinetic energy.
However it turns out this is not really true as there is no electrical kinetic energy.  Bond graphs are a way of explaining what is really happening.  Please note that I'm doing an abbreviated description here so I will probably miss a step.  Basically the electrical current is turned into a magnetic field in the inductor.  In that magnetic field we get a change in flux (the magnetic equivalent of capacitance).  It turns out the inductor is to the magnetic domain what a capacitor is to the electrical domain.  Thus an inductor is actually a FLUX CAPACITOR!  My professor seemed to think nothing of stating as much in the course of the lecture but I could help but snicker and think of 1985.

[IanB]

Current through an inductor has inertia just as a mass has inertia.  This would mean we have electrical kinetic energy. However it turns out this is not really true as there is no electrical kinetic energy.

There is stored energy though.

I mentioned earlier in this thread how Newton's second law may be simply expressed (for rectilinear motion) as:

F = m dv/dt

where F is force, m is mass and dv/dt is acceration; all being scalar quantities in the straight line motion case.

For a perfect inductor there is the similar equation:

V = L dI/dt

Here V is a "force", L is an "inertia", and I is a quantity that changes in response to an applied force.

The analogy continues. The stored kinetic energy in a moving body is given by:

E = ½mv²

And the stored magnetic energy in an inductor is:

E = ½LI²

Kinetic energy is to massive bodies as magnetic energy is to inductors. Note that it is not the velocity that holds the kinetic energy but the massive body, and it is not the current that holds the magnetic energy but the inductor.

[Robomeds]

Yes, energy is stored.  It is stored as magnetic potential energy rather than electrical kinetic energy (which doesn't actually exist). 

[merlinb]

The term "through" is commonly used and accepted in not only the industry but in teaching as well.

This is true, if a little unfortunate.
Strictly speaking, electric current is not 'made' of anything, and therefore is not a physical 'thing' that can flow anywhere.

Electric current is the rate of change of charge, or the rate of change of electric flux (flux and charge are the same thing to engineers; physicists use a slightly different definition).

i = dq/dt

Electric current is therefore an entirely abstract, mathematical concept. It is scalar. Nothing more than a number that exists wherever you choose to measure it, whether that is in a wire or between the pates of a capacitor.

Electric current cannot 'flow' from A to B any more than wind speed can be said to 'flow' from east to west. It is more correct to say the current in a component, rather than through it. Saying that current 'flows' 'through' something is just casual language that we get used to, as long as you're not thinking too hard about the reality of the thing.

[John Coloccia]

Well, if we're going to really get into the minutia, the Ampere is defined by the force between two parallel wires, and is not an abstract quantity at all.  It has definite units.  Something like the fine structure constant is a pure number.  It has no units.  Assuming the laws of physics apply everywhere in the Universe, anyone describing the fine structure constant will come up with precisely the same number.

So it's definitely not just a number.  In some sense, it's a way of considering the current density when J is uniform.

[tru]

Electric current is the rate of change of charge, or the rate of change of electric flux (flux and charge are the same thing to engineers; physicists use a slightly different definition).
i = dq/dt

I would have thought that electric current is the flow of charge (movement of charge propagated by a medium), and that is measured by it's flow rate by the unit called ampere?

Electric current cannot 'flow' from A to B any more than wind speed can be said to 'flow' from east to west. It is more correct to say the current in a component, rather than through it. Saying that current 'flows' 'through' something is just casual language that we get used to, as long as you're not thinking too hard about the reality of the thing.

You've cheated there!    Your analogy is more correct as wind current, but that can be said to 'flow' from east to west.

[IanB]

Electric current is the rate of change of charge, or the rate of change of electric flux (flux and charge are the same thing to engineers; physicists use a slightly different definition).
i = dq/dt

I would have thought that electric current is the flow of charge (movement of charge propagated by a medium), and that is measured by it's flow rate by the unit called ampere?

There's a subtle difference in physics or applied mathematics between a flow and a rate of change, even though both have the same dimensions.

In analysis, a rate of change is measured within a control volume (or at a point if expressed as a density), whereas a flow is measured across a control surface (which may enclose the control volume).

Both flows and rates of change occur together in equations when time is involved. It invariably happens that change in accumulation within a control volume occurs when there are flows in or out of that volume (or, for completeness, sources or sinks within the control volume).

[Eight8]

You may also like all other full episodes:
http://www.learner.org/resources/series42.html

If you are in the USA. 

[KedasProbe]

You may also like all other full episodes:
http://www.learner.org/resources/series42.html

If you are in the USA. 

or you can google PBS_The_Mechanical_Universe_and_Beyond.part1

[philaburns]

Found this and thought is summarised the answer quite elegantly :-)

[IanB]

Why did you put this at the end of a dead thread instead of starting a new thread?

[Brumby]

This thread was originally started to discuss Dave's video.

Your post was to present a particular application using a capacitor.  Certainly, capacitors and related math are common between the two - but thereafter the discussions diverge ... dramatically.

By that logic, once someone started a thread about an oscilloscope, then all other discussions about oscilloscopes can be tacked on.  This blends multiple topics and makes it harder to sift out what is of interest to a member and what is not.

You are starting a new discussion - not extending an old one - so it really belongs in its own thread.


The other, far more relevant point, is that by tacking onto the end of an old thread, people aren't going to pay as much attention to it - and you will lose the attention of a lot of people that may just have some very useful contributions to offer!