Why is an infinitely large capacitor a DC open circuit?

Due to \$I=C\frac\$, and \$C\$ is infinitely large, any minor change in \$U\$ would result in infinite current, which is obviously impossible. As a result, the voltage across the capacitor can never change, so it's always zero, and a voltage drop of zero further implies a short circuit, no matter in DC or AC.

• The higher the frequency of the signal, the easier it can pass.
• The larger the capacitor, the easier the signal can pass.

When we say "a large capacitor is a DC open circuit", it actually means "After 5RC (time constant), no DC signal can pass a capacitor, although it's very large."

Clarification:
In fact, 5RC only gets you to 99% of the steady state condition, rather than 100%. However, it's reasonable to simply consider it as 0 in practice, because it's too small to care.

\$\begingroup\$ Your only "wrong" step is to use infinity like it's a number that can be used in calculations. Add your \$\lim_\$ for pedantry. :) \$\endgroup\$

\$\begingroup\$ @pipe Let's consider a simple zero state response circuit then: The voltage across the resistor is exactly the source voltage at the beginning, but after 5RC, it would drop to nearly zero. If C -> inf, 5RC -> inf, and it would take, say, billions of years for the resistor (or any other load) to be zero, that is to say, the larger the capacitor, the longer the load can have (larger) voltage drop. So a super capacitor is a DC short circuit. Am I right now? \$\endgroup\$

\$\begingroup\$ Who said that a large capacitor is DC open circuit? That's wrong. It's the SMALL capacitors which behave as open circuit. An infinitely large capacitor is a zero-ohm conductor, even at DC, while a zero-farad capacitor is an open circuit for DC, even open for AC, even for RF freqs. \$\endgroup\$

\$\begingroup\$ @sun qingyao a zero-farad capacitor is a broken wire . with the broken ends spaced at infinite distance! :) A typical broken wire might have 0.01 picofarads of capacitance, which is far from zero, and might easily conduct a displacement current at GHz frequencies. An ideal open circuit is the same as a zero-value capacitor. And going the other way, an infinite-Henry coil is the same as an open circuit, while a zero-Henry coil is the same as a short. \$\endgroup\$

4 Answers 4

Did you ever pick up a telescope, hold the wrong end to your eye, and find you can't make sense of what you see?

The only thing wrong with using this telescope to see how an infinite capacitance behaves

is that you aren't holding the useful end to your eye.

Try it this way round

What this equation means is that you can shove any finite amount of current, however large, into an infinite capacitor, and its voltage will not change. This is a convenient approximation for the behaviour of 'very large' capacitors, whose voltage 'does not change much' during operation. LTSpice has an infinite capacitor, and I would guess most other spices do as well. These are useful for approximating coupling or decoupling capacitors, when you are not interested in their deviation from ideal.

Now if you rephrase your question as 'I have a component whose terminal voltage cannot change, and I change its terminal voltage . ', well, you get what you get.

You might want to edit your title so that says 'DC short circuit'.

This is one of the many cases that it is important to actually think what is physically going on in a capacitor.

Capacitors do not store charge. They store energy, and they store it in an electric field. A field between two plates. If you force more electrons onto one plate than normally want to be there, you have a net negative charge. This means some of the electrons on the other plate, which before felt zero net charge, are feeling a repulsive force from the negatively charged field generated by the excess of electrons on one plate. So electrons are pushed off that plate, exactly the number needed to make that plate as positive as the other is negative.

The number of electrons in in an ideal capacitor (and real ones for all intents and purposes) is fixed. A charged capacitor has no more or less charge than an uncharged one. For every electron you force onto one plate, an electron will be pushed off the other plate, perfectly balancing the net charge. Or, maybe you're pulling electrons off one plate, which will cause an additional electron to get pulled onto the opposite plate.

As capacitors become charged, they develop a voltage drop across them, and if there is a DC component to the signal on one plate, it will slowly (or quickly, depending on size) but surely charge the voltage drop across the capacitor until it equals the voltage across it, and now you don't have enough EMF to push another electron onto one plate and pop another electron out of the other plate.

This is why larger capacitors can conduct lower frequencies. An AC component alternates direction from the point of view of the capacitor. An ac signal without a DC component can simply use a capacitor as a conductor because current into one plate means an equal amount of current out the other plate, and this can continue on indefinitely as long as the current reverses the direction and balances out the electron displacement.

I like to visualize it as a sheet of rubber stretched over a pipe filled with water, and there is a rod on the other side that you can grab. You can push and pull the rod, and thus membrane, which will push and pull the water, but you can't actually pump any water this way, only move it back and fourth. Pumping is DC.

Lower frequencies require pushing or pulling in one direction for a longer amount of time, more distance, and require more charge to do be able to so. A large membrane 10 feet across, you could stretch several feet forward or backward. One the size of a thimble will move a centimeter. You can also think of it in terms of wavelength - lower frequencies have bigger waves, longer wavelengths, so there simply must be more plate charge available to handle the larger charge displacement, even if its temporary and cancels out overall.

But, back to the original question. If a capacitor had an infinite number of electrons available, because it has infinitely large plates to get that princely sum of infinite farads worth of capacitance, No matter how much charge you push or pull on either plate will ever generate enough of an electric field to produce a voltage across it. Any current will not produce any voltage across the capacitor, which means no energy will be stored, and you can just keep pushing the membrane as far as you want without stretching it. You can just push as many electrons onto one plate and have just as many pop off the other and do this forever.

It will even have a phase difference (the imaginary component) of 0, just like a short.

Except it is not a short.

There is one subtle thing that I've intentionally left out of all this.

An infinite capacitor doesn't have to have zero energy stored. It also doesn't need to have 0V across it. It can have any electric field strength you want, and thus, any voltage. The only thing you can't do is change the that field strength, and thus voltage.

So, it is not correct to call an infinite capacitor a short, because a short will always have 0V across it. An infinite capacitor can have any voltage across it that you wish, it simply cannot be changed.

Which is why an infinite capacitor is usually called by its more familiar name, which is an ideal voltage source.

An infinite capacitor and ideal voltage source are the same theoretical object and concept.

So, to stop rambling on and really answer your question, no, an infinite capacitor is not a short. It's a voltage source, because they behave identically in every way at any voltage, including zero. So the part you got wrong is that an infinite capacitor wasn't partially charged by uh the turtle the universe is riding on the back of before the universe existed, and thus forever has a voltage across it that cannot be changed.

Of course, all of this is silly. There is no implication here, as approximations do not imply reality. And that equation is an approximation, as charge is quantized, and the equation assumes it is continuous. Thus, it does not serve as a source of truth about reality, but its great at making predictions with an error between 0 and ± one election!