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.