Grokking Capacitors

I understand what a capacitor fundamentally is, and that it stores charge. But I am having trouble really understanding how it works at a detail level within a DC circuit. Specifically:

Does a capacitor allow DC current to flow through it while charging? Does being 'full' cause the capacitor to then discharge, and at that point is all current coming from the capacitor and not the power source?

Thank you for any patient responses.

Richard
rwp42

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Re: Grokking Capacitors

rwp42 wrote:I understand what a capacitor fundamentally is, and that it stores charge. But I am having trouble really understanding how it works at a detail level within a DC circuit. Specifically:
Does a capacitor allow DC current to flow through it while charging?

No, charge does not actually 'flow thru' the dialectic of the capacitor. Charged ions collect on one plate of the capacitor and this excess of charged ions (positive or negative depending on the source) create an electrostatic field and push the oppositely charged ions on the second plate away which in turn collect on the first plate. What is happening is the excess of charged ions on one plate build up creating an electrostatic field; i.e., voltage.

Does being 'full' cause the capacitor to then discharge, and at that point is all current coming from the capacitor and not the power source?

A capacitor will discharge given the nature of the circuit regardless if it has reached steady state value (whatever that may be - circuit dependent). And you ask a good question about the current path.
If one is assuming bypass capacitors in a Class D amplifier, it is a fact that the majority of current comes from the bypass caps rather than the supply. Not all of it will come from the capacitors, but most of it will. This holds true in most applications I believe since that is the point of the bulk bypass cap - a charge reservoir during circuit transients.

ImaginaryAxis

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Re: Grokking Capacitors

An ideal capacitor acts as an open circuit under DC conditions and as a short circuit under AC conditions.

If it is charged and at a higher voltage than the points it connects, it will discharge down to the voltage across the capacitor. If it is charged at a lower voltage than the points it connects, it will charge up to the voltage across the capacitor (or device limits). Charging and discharging follow exponential decay where RC is the decay constant.

A discharging capacitor can act as a power source. The variation in its own voltage as it discharges allows it to "see AC" and act as a current path, but the resulting circuit can sometimes be analyzed to some extent via DC rules by looking at the momentary state (mainly, keeping in mind the AC vs. DC behavior of the components involved).

L.A.W wrote:
If one is assuming bypass capacitors in a Class D amplifier, it is a fact that the majority of current comes from the bypass caps rather than the supply. Not all of it will come from the capacitors, but most of it will. This holds true in most applications I believe since that is the point of the bulk bypass cap - a charge reservoir during circuit transients.

The trick here is that there are limitations on the instantaneous behavior. An ideal capacitor cannot experience an instantaneous change in its voltage, while an ideal inductor cannot experience instantaneous change in its current. This can be seen by looking at the respective differential equations governing the behavior of these circuit elements. These components are thus often used to provide some rejection of transient variation in voltage and current. In more complex implementations, you get frequency-discriminating filters...
tinsmith

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Re: Grokking Capacitors

Does a capacitor allow DC current to flow through it while charging?

Well, during the time that the capacitor is charging, current is flowing into one terminal and out of the other. Whether that means that current is flowing "through" the capacitor may be a semantic issue. I'd say it is, and Kirchoff's current law would tend to support me, but I realize that some may draw a distinction between the movement of charge carriers and the build up of of an electric field -- if you draw that distinction, then current isn't flowing between the plates.

Once the capacitor reaches a steady state after a DC voltage is applied, no current flows through it. (Theoretically, it only asymtotically approaches a steady state, but for practical purposes, it eventually reaches a steady state).

Does being 'full' cause the capacitor to then discharge, and at that point is all current coming from the capacitor and not the power source?

No. For one thing, a capacitor used well within its ratings never becomes "full" in the sense that it can't hold more charge. It only becomes "full" in the sense that the voltage on its plates reaches equilibrium with the voltage that the rest of the circuit is applying. And when that happens, current stops flowing -- it neither charges nor discharges. What will cause a capacitor to discharge is when the components connected to it try to present a lower voltage than the voltage present on the plates of the capacitor.

There's possibly another meaning to the word "full". If a capacitor is overloaded, stressed way past the voltage printed on the side of the package, then it will eventually break down, often resulting in an explosion and messy goo spattering over the circuit board. That's what happens when a real world capacitor gets too "full", but we try to avoid that.
uoip

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Re: Grokking Capacitors

uoip wrote:There's possibly another meaning to the word "full". If a capacitor is overloaded, stressed way past the voltage printed on the side of the package, then it will eventually break down, often resulting in an explosion and messy goo spattering over the circuit board. That's what happens when a real world capacitor gets too "full", but we try to avoid that.

We try to avoid that when building things.

We do it for fun when we're not. :twisted:
tinsmith

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Re: Grokking Capacitors

uoip wrote:Does a capacitor allow DC current to flow through it while charging?

Well, during the time that the capacitor is charging, current is flowing into one terminal and out of the other. Whether that means that current is flowing "through" the capacitor may be a semantic issue. I'd say it is, and Kirchoff's current law would tend to support me, but I realize that some may draw a distinction between the movement of charge carriers and the build up of of an electric field -- if you draw that distinction, then current isn't flowing between the plates.

It is not a semantic issue, it's physics. Charge does not physically flow thru the dialectic - so your second statement is indeed correct. 8)

tinsmith wrote:The trick here is that there are limitations on the instantaneous behavior. An ideal capacitor cannot experience an instantaneous change in its voltage, while an ideal inductor cannot experience instantaneous change in its current. This can be seen by looking at the respective differential equations governing the behavior of these circuit elements. These components are thus often used to provide some rejection of transient variation in voltage and current. In more complex implementations, you get frequency-discriminating filters...

I should have said "switching currents." If you do the power train analysis of a Class D, PWM H-bridge, the majority of the current is coming from the bypass capacitors, LC filter, and then the supply. It is a clever marketing trick. 92% efficiency! Yeah, with proper bypass and an LC filter...sucka.

Some minor nitpicks that will help the OP.

Tinsmith, I am sure you know the behavior of capacitors under AC conditions, but it should be clarified for the OP that capacitors will have a specific reactance that depends on capacitor value and frequency - they are not simply 'short' during AC. Also, the exponential (dis)charge rate assumes a voltage source. If it is a current source then the (dis)charge is linear.

I only point these out because I have seen so many people fail to catch mistakes during circuit analysis when making these assumptions. The (dis)charge profile of a capacitor when driven by a current source is a favorite interview question. 9 out of 10 get it wrong. :|

ImaginaryAxis

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Re: Grokking Capacitors

L.A.W wrote:It is not a semantic issue, it's physics. Charge does not physically flow thru the dialectic - so your second statement is indeed correct.

For issues of physics, I think I'd trust Maxwell. He invented the concept of "displacement current" to cover this issue.

http://en.wikipedia.org/wiki/Displacement_current

Displacement current is necessary to make Kirchoff's current law work in a capacitor
http://en.wikipedia.org/wiki/Kirchhoff' ... ge_density

Anyway, I think we can agree that current enters one lead of the capacitor and leaves the other lead. Charge accumulates inside the capacitor. If your word "current" includes displacement current, then current is flowing all the way through the capacitor. If your notion of "current" only includes motion of charge carriers, then Kirchoff's current law no longer holds, Ampere's law doesn't hold, and current isn't flowing all the way through the capacitor.

Since displacement current produces magnetic field identical to "real" current, a Hall effect current clamp will show charge moving through a capacitor during the time it's being charged or discharged.
uoip

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Re: Grokking Capacitors

We will have to agree to disagree because I can see this thread going on for weeks - to me Kirchhoff's Laws are an abstraction of field theory and use the motion of charge carriers when completing circuit analysis. You do not need Maxwell's equations.
The dialectic is an insulator so I do not see how charge will flow thru it - it can't. Then again I am not coming from a rigorous Physics background;i.e, Maxwell's equations and dealing with curl integrals of fields. So when I say 'it's physics' that is coming out the mouth of an engineer - and I know mathematicians and physicists get uppity with us engineers from time to time. To me this sounds like "Is a BJT current controlled or voltage controlled?" To which I would reply, "which model are you using?"
Feel free to take it off line and PM me to explain. This is how people learn and remember old concepts.

ImaginaryAxis

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Re: Grokking Capacitors

Yeah, the stuff I posted above is what we in physics call a first-order ideal approximation under typical conditions. It's good enough to tell you generally how the thing should work, without invoking the deeper mysteries involving differential equations, field theory, and quantum electrodynamics.

Current supplies I leave off because, honestly, they don't come up that often for people who are still wondering what a capacitor does. Practically, there are some current-driven devices out there which will come up often enough that you're sometimes locally trying to produce a current within a given range (LEDs, audio transducers, some types of motors...) and these will sometimes have capacitors around. Really, the "caps are SS under AC" simplification is most useful for answering "bias capacitors, WTF do they do?" since it's then clear that it shorts out a AC signal between two DC lines while still keeping the DC voltages more or less isolated.

Incidentally, AC behavior of capacitors under steady-state conditions isn't that much more complicated if you don't mind working with a few complex numbers. Understanding the basics of complex impedance will let you do simple AC steady-state analysis including understanding simple filters.
tinsmith

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Re: Grokking Capacitors

rwp42 wrote:I understand what a capacitor fundamentally is, and that it stores charge. But I am having trouble really understanding how it works at a detail level within a DC circuit. Specifically:

Does a capacitor allow DC current to flow through it while charging? Does being 'full' cause the capacitor to then discharge, and at that point is all current coming from the capacitor and not the power source?

Thank you for any patient responses.

Richard

This discussion's gotten a bit technical. The simple answer is that a capacitor doesn't really operate in a DC circuit. At least not in the ideal DC circuit. Consider this circuit:

series2.png (4 KiB) Viewed 3036 times

What happens when you close the switch in this circuit? Well, almost nothing. The bulb will see a tiny current pulse from the switch closing (think of it as the AC portion of the DC current you're sending) that gets transmitted through the capacitor, and then the positive side of the capacitor will charge to 9V and the bulb will stay unlit.

On the other hand the capacitor in parallel with the light bulb is slightly more interesting:

parallel2.png (4.28 KiB) Viewed 3036 times

Here, when you close the switch the light bulb might not light up immediately. It'll slowly get brighter until it's at it's full brightness (how long that takes is dependent on the resistance of the bulb & battery and the capacitance of the capacitor and isn't important right now - it's a bunch of mathematical gobbledygook that you don't need to understand yet) as the capacitor charges from 0V to 9V. Then, when you turn the switch off it'll likewise slowly get dimmer as the capacitor continues to supply current to keep the bulb lit until it discharges from 9V back to 0V.

As for the use capacitors have in DC circuits, well, not very many. What they're best at is regulating the power supply. When you get DC power out of anything except a battery, it's coming from AC mains at one point or another and getting rectified to DC. That process is imperfect and there's often quite a bit of ripple. Also DC-to-DC converters likewise generate quite a lot of ripple as well, and ripple is bad for digital circuits. To combat that, designers throw capacitors at the problem, using them in the parallel configuration to smooth out the ripple voltages so the supply voltage appears quite constant.

That's why you see a gagillion capacitors in your motherboard when you open up your computer: Every place there's noise or noise sensitive circuitry they like to throw a capacitor hooked into ground to smooth that noise out. The capacitor acts like a tank for charge, collecting the noise, sinking current when the noise takes the voltage higher than average, and sourcing current when the noise take the voltage lower than average.

Now, with respect to your questions, specifically:

rwp42 wrote:Does a capacitor allow DC current to flow through it while charging?

Kinda, but not really. A capacitor transmits only AC current. Turning DC current on, however, is a time-dependent action, and it transmits the AC part of that act. Think of turning on the DC like a square-wave AC. The capacitor transmits the square-wave, but since you're only doing half a wavelength's worth of that square wave, it doesn't transmit that impulse for very long and then it reverts to being an open circuit.

rwp42 wrote:Does being 'full' cause the capacitor to then discharge, and at that point is all current coming from the capacitor and not the power source?

"Being full" doesn't mean anything for a capacitor. The more voltage you put in it, the more charge it accumulates. (Unless you wanna worry about voltage rating. Sure, too far past the voltage rating it blows up.)

If you want to use the plumbing metaphor, you don't want to think of it as a steel-walled tank, with a maximum capacity. You want to think of it as an elastic balloon. The balloon fills with water proportional to the water pressure in the pipe it's connected to. Drop the pressure, the balloon slowly shrinks down as it dumps some of it's water back into the pipe. Raise the pressure and the opposite happens. Turn off the water entirely, and sure: It'll be the only source of water until it discharges then there'll be no water at all.
Last edited by stinkbutt on Sat Dec 25, 2010 10:26 pm, edited 1 time in total.
Red M&M, Blue M&M: They all wind up the same color

stinkbutt

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Re: Grokking Capacitors

stinkbutt wrote:Here, when you close the switch the light bulb might not light up immediately. It'll slowly get brighter until it's at it's full brightness (how long that takes is dependent on the resistance of the bulb and the capacitance of the capacitor and isn't important right now - it's a bunch of mathematical gobbledygook that you don't need to understand yet) as the capacitor charges from 0V to 9V.

This is not correct - or simply bad word choice.

Ideal case:

The bulb and the capacitor are shunted across the voltage source. When the switch is closed at t0 the voltage across the capacitor (and bulb) will change instantaneously and measure 9V and therefore will light up instantly.
Before anyone says "But the voltage across a capacitor cannot change instantaneously!, " they are correct if they are making the assumption that the current is not infinite. So, the voltage across a capacitor can change instantaneously if the current is infinite. Since we are talking about the ideal case, the voltage source is ideal and will supply an infinite current while maintaining the desired voltage across its terminals.

Real world:
There will be a rise time when the battery is switched into circuit - but it will be far from "slow" as you implied. The battery or whatever source is being used has finite source resistance, the switch has a finite resistance - rds(on) assuming a FET, the PCB traces have resistance and inductance or if using a breadboard the copper contacts have capacitance, inductance and wire resistance, the cap has an ESR..etc., etc. In a real circuit the rise time is dominated by the source resistance and switch resistance. On a scope there will be a finite rise time as insignificant as it is. Practically, the bulb lights up instantly.

"Mathematical gobblygook" - - - How dare, sir.

ImaginaryAxis

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Re: Grokking Capacitors

LOL! Hey look: LAW is attempting to correct someone. It must be Thursday. Or some other day of the week. That ends in "y".
Red M&M, Blue M&M: They all wind up the same color

stinkbutt

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Re: Grokking Capacitors

stinkbutt wrote:LOL! Hey look: LAW is attempting to correct someone. It must be Thursday. Or some other day of the week. That ends in "y".

I would have sent you a PM, but it's not an option in your profile sooooo....

This is the second time that you have publicly been a dick when I kindly point out errors. What is your problem? What is this with you comparing the size of your ePeen with mine? Seriously.

ImaginaryAxis

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Re: Grokking Capacitors

My issue with threads like this is they start with a beginner looking for useful knowledge, and then "physicists" or semantists start arguing how many angels fit on the head of a pin, and almost totally disregard the desire of the OP to learn something. Notice the OP has not been back. Look at everything that has been written and think about what a beginner can say about most of this. I just don't think it is helpful to a beginner which the OP obviously is. I think stinkbutt's circuits are nice, but he does drink a little too much coffee sometimes...

zener

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Re: Grokking Capacitors

I heard it is a 1000.

From my perspective some things can only be 'dumbed down' so far because it can lead to misconceptions later on. There are concepts that I still struggle with because of what I was originally taught and then I have to reconcile it with 'how it REALLY works' sometime later. Even moreso when an explanation is lacking and/or wrong; i.e, the Waveshield page discusses Nyquist in reference to a 11kHz low pass filter because the sampling frequency is 22kHz - yikes.

I am not perfect and I am anal retentive, but I try to steer my replies back to the OP.

Electronics aint easy man. 8)

ImaginaryAxis

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