# Why/how do capacitors resist change in voltage?

Discussion in 'General Electronics Chat' started by shespuzzling, Aug 27, 2009.

1. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
0
In the Capacitors section of All About Circuits (Vol. 1 DC), it says:

"A capacitor's ability to store energy as a function of voltage (potential difference between the two leads) results in a tendency to try to maintain voltage at a constant level. In other words, capacitors tend to resist changes in voltage drop. When voltage across a capacitor is increased or decreased, the capacitor "resists" the change by drawing current from or supplying current to the source of the voltage change, in opposition to the change."

I find this statement to be the opposite of what I thought a capacitor was supposed to do. I thought that if you connected a capacitor to a voltage source, the voltage source would induce a current and a voltage across the capacitor would appear. I don't see how a capacitor could be able to charge and dissipate energy (thereby changing its voltage) and yet still "resist" any change in voltage. I thought a change in voltage was the whole point!

My other question about capacitors is why, if it was connected to a voltage source in a single-loop circuit, the maximum voltage across it is the same as the voltage source. I understand that if it were a resistor, not a capacitor, then there is a maximum voltage, but since a capacitor is designed to accumulate charge on each of it's plates, why does the voltage source dictate how much charge it can accumulate? What if the plates are very large and are capable of storing more charge and thus having a larger voltage drop across it than the voltage source has?

I feel like I'm missing something BIG here! Thanks for your help.

2. ### Ratch New Member

Mar 20, 2007
1,068
3
shespuzzling,

I do too. I think you need to understand the relationship between voltage, charge, capacitance, and energy. Let's start with the basics. First, like charges repel one another. It takes energy to bring them close together. It takes more energy to concentrate like charges even closer together. This energy is stored in an electrostatic field and is potential energy. The total amount of energy accumulated divided by the amount of charge is voltage. So voltage is the energy density of the charge. You can have the same voltage (energy density) with some charge close together or less charge closer together. A capacitor is a charge separation container. When one plate accumulates a charge, the other plate depletes an equal amount of charge. There is a net gain/loss of zero charge, but the separated charge on the plates stores energy in the form of a electrostatic field. So it can be said that the capacitor becomes "charged" with energy. There will be a higher energy density on one plate and a lower energy density on the opposite plate, resulting in a voltage difference between the plates. A smaller capacitor (charge container) concentrates the charge closer together, so for the same charge, a smaller capacitor will show a higher voltage difference between the plates.

Now let's see what happens to a capacitor when a voltage is placed across it. The voltage forces the charge to accumulate at a rate dependent on the resistance in the circuit. This charge will cause a back voltage which is opposite to the charging voltage. So the displacement current becomes less and less until it becomes zero when the back voltage reaches the charging voltage. Is that what they mean when they say a capacitor resists a voltage change? A capacitor cannot have a back voltage higher than the charging voltage no matter how large its plates are. A larger capacitor (from the word capacity) can store more charge at the same voltage than a smaller one. A capacitor does not dissipate energy unless there are imperfections like leakage or dielectric absorption. A capacitor stores and releases energy to/from the circuit thereby raising or lowering its voltage.

Ratch

3. ### count_volta Active Member

Feb 4, 2009
435
24
Capacitors resist changes in voltage because it takes time for their voltage to change. The time depends on the size of the capacitor. A larger capacitor will take longer to discharge/charge than a small one.

The statement that capacitors resist changes in voltage is a relative thing, and is time dependent. For example if you take a resistor and apply a voltage to it using a battery, it will instantly have that voltage across it. When you take the battery away, the resistor will instantly have 0 volts.

Now a capacitor on the other hand takes time to charge, and time to discharge. So for example you connect a capacitor to a 5 volt battery, and ask what is the voltage across the capacitor at that instant? Is it 5 volts like the resistor? The answer is no. The answer is the voltage is slowly being built up. Wait 2 more seconds and it will be 5 volts then.

So really what people should say is, capacitors resist instantaneous changes in voltage!!!!

This is really a fascinating phenomenon, because a capacitor in an AC circuit where the frequency of the AC is high, acts like a piece of wire (short circuit) This is exactly because of its instantaneous resistance to changes in voltage. And it can be used as a filter.

Last edited: Aug 28, 2009
4. ### rjenkins AAC Fanatic!

Nov 6, 2005
1,015
69
These descriptions, though reasonable, are not simple for a newcomer to understand..

A capacitor is an electrical resovoir or accumulator. Think of it a bit like a rechargeable battery, but as it electrostatic rather than chemical the reactions are instantaneous (well, speed of light limits).

A simple analogy is a balloon.
Connect an empty ballon to an airtap, turn the tap on;
Initially, as the balloon is empty, there is no pressure (voltage) in it. As it fills (charges), the pressure increases and the flow (current) slows, until the pressure is equal to the air pipe pressure and everything is stable.

If something else takes air from the pipe, the balloon will discharge air back into the pipe to equalise the pressure, supplying the other item.

The balloon is working as a smoothing or resovoir capacitor.

As a cap has two terminals, it may be easier to think of a big pipe with a rubber skin in the middle, which streches one way or the other as pressure is applied. The above still applies, but with this model you can see how varying pressure (voltage) can get through without any flow under static conditions.

5. ### steinar96 Active Member

Apr 18, 2009
239
4
You need to examine the equations describing the capacitor before understanding it. The fundamental equation relating it's capacitance to the voltage applied accross it and amount of stored charge is

$
C = \frac{Q}{V}
$

If we re-arrange this equation and solve for Q we get

$
Q = CV
$

You should notice that Q, the charge that has accumulated between the plates is decided by the capacitance and the voltage. This means that if you apply some voltage accross the capacitor it will charge up the amount of charge Q = CV.

If our capacitor has capacitance 1 Fahrad, and you apply 5V accross it the charge that has accumulated equals 5 coulombs.

If you change the voltage to 4V the charge accumulated will decrease by 1 coulomb. This charge has to go somewhere right ?. In reality what is happening is that the capacitor has pumped out 1 coulomb in attempt to maintain the voltage accross it. If the voltage source changed it's voltage instantly from 5 to 4 volts the capacitor would still maintain the voltage across it at 5V for a limited period of time until (it decreases expontentially from 5V to 4V or until it has reached it's new charge count at 4 coloumbs as governed by the capacitor equation.

Another capacitor equation, which describes the current going in or out of it in terms of the change in voltage across it is

$
i = C\frac{dV}{dT}
$

This equation states that the current going in or out depends entirely on the change in voltage across it. This current exists because the capacitor tries to maintain the 5V voltage (when changed to 4V) across it's terminals for as long as it can (while transitioning from accumulated charge of 5 coloumbs to 4 coloumbs). The voltage decreases exponentially between these two values.

But to keeps things short, if a constant voltage is applied accross the capacitor it charges up to some value decided by the voltage and the capacitance.
If the voltage is always changing over time the capacitor attempts to keep it constant. Charging a capacitor to 5V, then instantly changing the voltage to 4V means the capacitor tries to maintain it at 5V for as long as it can.

6. ### BillB3857 Senior Member

Feb 28, 2009
2,400
348
I was taught that caps store energy by "warping" the normal circular orbit of electrons in the molecules of the electrolyte into an elliptical orbit. Energy is released by allowing the orbits to return to circular.

7. ### Ratch New Member

Mar 20, 2007
1,068
3
count_volta,

No, voltage and energy change begins immediately when a different voltage is applied to a capacitor. It is the final equilibrium voltage and energy that takes time to reach.

I hope you are referring to energy when you say "charge". A capacitor does not store a net coulombic charge, it stores energy. A capacitor is a energy storage device, a resistor is not. So when the energy source (voltage) is removed from a resistor, no energy or voltage remains. In the above example, if the resistance of the circuit it high, it can be a very long time for the energy and voltage to lower.

rjenkins,

I hope you mean an accumulator of energy, and not coulombic charge when you state that.

steinar96,

You should say separated between the plates. The net coulomic charge of a cap is the same before and after a voltage is places across it.

The transistion to 4 volts will begin instantaneously, and take some amount of time to reach the 4 volt equilibrium voltage, depending on the resistance of the circuit.

I hope the value you are referring to is energy.

OK, to summarize. A capacitor is a energy storage device. No one can truly understand its operation unless they realize that. It is syntatically incorrect to say that a capacitor stores voltage. Density cannot be stored, and voltage is the energy density of the charge. A capacitor works by separating the charge, which by electrostatic physics, causes a electrostatic field to form where energy can be stored. Capacitance is a measure of the amount of charge that can be separated by a voltage. The higher the capacitance, the more charge can be separated across the plates of the cap by the same voltage. The voltage of a capacitor does if fact begin to change instantaneously when its stored energy changes, but the final value may take some time to reach.

Ratch

8. ### russ_hensel Well-Known Member

Jan 11, 2009
818
47
I think the comments above are helpful, but would also add that the idea of putting a voltage on the cap. has some problems. The major one is that for an ideal voltage source and cap it would result in infinite current. Perhaps it would help to think of all circuits as having a series resistor that limits the current to a finite value, makes time for change > 0 and limits dv/dt across the cap to a finite value. If you cannot understand this after some thought then drop it as not useful for you.

9. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
0
Thanks for all of the responses. Let me see if I am understanding this correctly...

If you have a capacitor that initially has no voltage across it and then connect it to a voltage source, the capacitor will draw a current from the voltage source. The driving force behind this process is the voltage source, and the free electrons on the 2 plates of the capacitor are simply reacting to the polarity of the voltage source. The free electrons start to move (current flows) and this will go on until there is no longer a difference between the ends of the voltage source and the ends of the capacitor (i.e. it has reached equilibrium). Now, there is a voltage across the capacitor which is equal to the voltage source and you can remove the capacitor form the circuit and this energy is maintained. The actual amount of charge that has accumulated on the plates depends on the capacitance and the voltage across the capacitor. Is this correct?

Comparing a capacitor (which resists instantaneous changes in voltage) to a resistor (which is able to change voltage instantaneously), which physical difference is the key reason why a capacitor can store energy and a resistor cannot? Is it the surface area of the conducting plates of a capacitor? Or, is it the ability of the insulator to create an electrostatic field? Is an insulator similar to a very, very high resistance? If that's the case, then a capacitor could be considered a glorified resistor but with a large surface area at either end (compared to the wire) and a very, very large resistance in between?

Regarding capacitance, is that quantity determined solely on the size, material, and distance between the plates?

10. ### rspuzio Active Member

Jan 19, 2009
77
0

It has to do with what happens to the energy suppled to the device.
In a resistor, the energy turns into heat whilst, in a capacitor, the energy
goes into building up an electric field between the plates. When the energy
turns into heat and flows out the heat sink, it's gone but, when it's tucked
away in an electric field, you can get that energy back by discharging the
capacitor. While we're at it, even though you didn't ask, it might be worth
pointing out that something similar happens in a inductor, except that there
the energy is stored in a magnetic field instead of in an electric field.

The hydraulic analogue mentioned above works here as well. If your water
flows through a rough pipe with a lot of friction, then your energy gets lost
in heat whereat if you store it in a balloon, you can recover it at a later time.
By the way, this analogy isn't just useful for illustration; if you go to the
plumbing store, you will find "water hammer arrestors" like the following:

http://www.accentshopping.com/product.asp?P_ID=154311

which amount to the balloon mentioned above and are used in
exactly the same way as a capacitor --- just as, in an electrical circuit,
you might use a capacitor to smooth out the pulse from turning on a
switch by resisting the change in voltage, so too one uses one of
these gizmos to resist a change in pressure when turning on the
faucet. So, you see, once you learn electricity, you get plumbing
thrown in as a free added bonus

If you find this sort of analogy useful or interesting, there is a
whole book, "Dynamical Analogies" by "Harry Olson" devoted to
the topic:

http://www.pmillett.com/Books/Atwood/Olson 1943 Dynamical Analogies.pdf

The point behind this is that the laws of mechanics are the same,
only the forces involved differ between examples so, once you
understand a concept in one setting, that understanding will
work in other contexts as well. Since it's a lot easier to see what
is going on inside pipes than inside wires, these analogies are
quite useful in building up understanding.

11. ### Ratch New Member

Mar 20, 2007
1,068
3
shespuzzling,

You should say "charge flow". Current does not flow, but charge does. Current already is charge flow. It doesn't flow twice. Current exists during charge movement. If the voltage energy source and the leads of the capacitor are connected, then they have the same voltage at all times. A charge flow will occur until the back voltage of the capacitor equals the voltage source. Then the leads can be disconnected and the capacitor will have the same voltage as the source.

Charge accumulates on one plate and depletes on the other plate. In other words, the charge separates, but the total charge remains the same and does not accumulate. Correct, Q = C*V , where Q is the amount of separated charge.

Nope, I thought I made it clear that a capacitor always changes its voltage instantaneously in accordance with its stored energy. But the rate of energy change and thereby the voltage on the capacitor depends on the resistance in series with the capacitor. A resistor, on the other hand, dissipates its energy instantaeously, no energy storage is present, and the voltage across it follows at the same rate as the source voltage.

Do you mean dielectric? An insulator does not create a electric field. The separation of charges does.

Nope, the insulator between the plates has to have dielectric properties, too. You can read up on what a dielectric is, but it has to do more than allow no leakage between the plates.

That covers it pretty good. A dielectric has to be able to align and change its molecules in accordance with the electrostatic field.

Ratch

12. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
0
Thanks again, everyone, for all the help. I'm learning more...but that always ends up with me having more questions.

I'm still grappling with why.

Why, for instance, does current decay to zero once a capacitor that is connected to a voltage source is fully-charged (i.e. has the same voltage across it as the source voltage)? What force is behind this, or what property of a capacitor is saying, no, I don't want anymore current from that voltage source. Ratch, I do not know what you mean by the term "back voltage."

Why does a capacitor want to try maintain voltage anyway? What is it about a capacitor that enables it to do this? What's its motivation?

13. ### Ratch New Member

Mar 20, 2007
1,068
3
shespuzzling,

Because the capacitor has separated all the charge it can at the voltage that is applied to it.

The back voltage, see below.

From the physics of electrostatics, whenever there is a difference of charge energy density, there will be a voltage difference. Remember, voltage is the energy density of the charge. So one plate has an over abundance of electrons, and the other plate has a under abundance of electrons. It took energy from the voltage source to make this happen. The energy taken, divided by the number of electrons separated, is the energy density difference of the separated charges, or back voltage. This back voltage is of the opposite polarity with respect to the source voltage. So as the back voltage accumulates, the source voltage is less able to push current into the capacitor. Finally, when the back voltage is equals the source voltage, the current stops.

Its ability to store energy in a electrostatic field. Anytime a charge is kept separated, energy is stored, and a voltage is maintained.

Ratch

14. ### loosewire AAC Fanatic!

Apr 25, 2008
1,584
435
In a pefect world If you disconect your car battery from all sources,
If you let it set for( lets say one half hour)would you lose or gain
voltage. Because you say that you disconnect resistor you have
zero voltage. A side from the resistor and cap voltage reaction
I hope you see where I'm coming from,that depending on voltage
source It may not drop to absolute zero.

15. ### count_volta Active Member

Feb 4, 2009
435
24
It is in fact the insulator between the plates which allows a capacitor to store energy. A capacitor is just a temporary battery. In a battery the two electrodes are also separated by an insulator.

You see, when you connect the positive side of the battery to one plate, and negative side to the other plate, you create a positive and negative charge on those plates. In other words you add electrons to the negative plate and remove electrons from the positive plate.

How could you do that with a resistor? You can't. A copper wire is also a resistor, but its resistance is tiny. And you certainly cant store charge in a wire.

But on that same topic, every wire and resistor does have SOME capacitance. Its just tiny. Its insignificant. Even the human body has capacitance.

The key to storing charge is separation of charges. You have to insulate the two opposite charges from each other, because if you didnt, they would instantly recombine to make a neutral atom. It is by forcing them to stay separated that you create a voltage.

A battery does this with chemical reactions, a capacitor does it using static electricity, but in both cases there is an insulator separating the charges.

Btw, that balloon analogy posted earlier is perfect for capacitors. A balloon acts almost exactly like a capacitor. Air is charge. Voltage is air pressure. Size of the balloon is capacitance. And a larger balloon takes longer to empty out the air. Same thing.

16. ### davebee Well-Known Member

Oct 22, 2008
539
46
When a capacitor charges, you could think of it as the battery shoving more and more charges into the capacitor plates, with the electric field between the plates shoving back harder and harder as the field builds (this is "back" voltage).

As more and more charges get packed in, the electric field grows stronger and stronger, and pushes back harder and harder, so the flow of charge becomes less and less.

When the strength that the field pushes back with just balances the pushing strength of the battery, then the battery can't push any more charges in. At that point the current has dropped to zero,

It's not that a capacitor "wants" or "tries" to maintain a voltage; most capacitors know nothing about what the circuit designer wants. (there are exceptions - the capacitors in my projects, for example, know that if they don't maintain voltage I'll smash them with a hammer)

No, but really, it's more that the capacitor supplies a large pool of charge available to supply as additional current if the load resistance momentarily drops. The additional current is at the capacitor voltage, so the circuit voltage tends to follow the capacitor voltage.

The increase in current flow does lower the overall voltage, but the voltage lowers less than if the capacitor weren't there.

17. ### rjenkins AAC Fanatic!

Nov 6, 2005
1,015
69
Just a correction to a point above:

A capacitor does not need a material with special 'dielectric properties'; insulation is the only requirement and all insulators will act as dielectrics.

The most basic capacitor is plates in a vacuum - vacuum capacitors are used for high voltages, also high power at high frequencies as vacuum is a rather good insulator and has zero loss.

http://www.lbagroup.com/technology/vacuum.php

The properties of the dielectric (whether vacuum, oil, plastic, ceramic or whatever) can drastically change the value of the capacitor even if the capacitor plate size and spacing are otherwise unchanged.

Each substance has a characteristic 'dielectric constant', with vacuum being the reference with a dielectric constant of 1.

eg. if you go from a vacuum dielectric to filling the gap between the plates with a substance that has a DC of 100 (without changing the plate size or spacing), the capacitance will be increased by a factor of 100.

Some ceramics especially have incredibly high DCs.

18. ### Ratch New Member

Mar 20, 2007
1,068
3
rjenkins,

Correction noted and accepted.

Ratch

19. ### shespuzzling Thread Starter Active Member

Aug 13, 2009
88
0
I think I finally get it! Thanks, you all have been really helpful. Now...on to insulators.