EEVblog #486 - Does Current Flow Through A Capacitor?

[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.