The short answer is: insulator thickness. But that probably will not suffice as an actual answer.
It is helpful to understand how capacitors are made. All capacitors we frequently use in electronics are essentially flat plates with insulation in between them. The plates are some kind of metal and the insulation is often plastic or ceramic material. We usually call capacitors by one of these materials: aluminum electrolytic capacitors are literally aluminum foil with a liquid electrolyte in between. Ceramic multilayer capacitors have lots of layers of conductor and ceramic slurry in between them, and tantalum capacitors have the element Ta as its conductor (plate).
The key to making a capacitor is getting two plates with the largest possible area as close as possible to each other without them conducting to each other. You can imagine that this is pretty hard; mechanically it is very hard to get two surfaces exactly parallel and have only in the order of microns or hundreds of nanometers in between them, which is what you need for the thousands or tens of thousands of microfarads, somtimes even tens to thousands of farads we use in electronics today. So an insulator is used in between two conductive plates, but in order to guarantee that the insulator is mechanically strong enough to survive manufacturing and useful life, it usually needs to be fairly thick - tens to hundreds of microns is typical for for instance MKP capacitors. This increases the distance between the plates to far beyond what is necessary for electrical insulation, only to serve a mechanical purpose.
So, in aluminum electrolytic capacitors (and all electrolytic capacitors in fact), a trick is used to reduce the plate distance without actually putting the plates in very close proximity to each other. What they do:
- One plate is oxidized to form a very thin, well-controlled insulating layer on its surface which can be as little as tens of nanometers thick
- The other plate is left with a conductive surface
- A sponge-like, fairly thick mechanical divider is placed in between the plates, but it is drenched in an electrically conductive liquid called the electrolyte.
This electrolyte effectively moves the plates much closer together, as if only the oxide layer on the first plate is the actual distance between them. This allows for a much, much higher capacitance in electrolytic capacitors than other methods of constructing capacitors.
Alright, with this kind of information under our belt we can see why leakage is more pronounced in electrolytic capacitors. Leakage is caused by four major mechanisms:
- Electrical conductivity through the insulator
- insulation breakages
- Electrolyte deionization
- Electron tunneling
The first one is simple: even the best insulator still conducts a little bit of electricity. The fact is that inside a large-value capacitor, there is a ton of surface area of the oxide layer, and even with a very low conductivity, if you have a lot of surface area the amount of leakage will be significant. This is especially significant for special super high-capacity electrolytics, where the aluminum foil inside is not just a coiled flat piece of foil, but the manufacturer has etched patterns into the foil to dramatically increase the amount of surface area of the aluminum.
The second one is also simple. Electrolytics are made by winding up coil and electrolyte 'sponge' together, and during this winding process mechanical stress and material imperfections cause tiny microfractures in the oxide surface. This allows electrons to directly flow between the plates, i.e. leakage current.
Then there is electrolyte dielectric effects. The electric field between the plates does cause some dielectric effects in the electrolyte, which change with the voltage (strength of electric field) over the plates. This means that usually at higher voltages, there is disproportionally more leakage current but at the same time higher ESR because the electrolyte ions are less free to move and are pulled in one direction by the electric field.
Lastly I should mention electron tunneling over very thin oxide layers. This is not an issue for run-of-the-mill electrolytics, but especially ultracapacitors suffer from this phenomenon a lot. Quantum-mechanically, electrons can be regarded not as a single point of charge, but a probability density of the electron being somewhere (wave-like behaviour of elementary particles). The probability of an electron being on the conductor is very high, and as you move further away from the conductor the probability of the electron being there decreases until it becomes essentially zero a few nanometers away. When an oxide layer is extremely thin, the probability of an electron that sits on the conductor to be on the other side of the oxide layer is nonzero, i.e. there is a probability that the electron 'jumps' over the oxide layer. This is called quantum tunneling and starts playing a role in leakage current when you approach nanometer-scale insulators.
So, to get back to your question: even though it may seem like it, by far the most important reason that electrolytic capacitors seem to leak more is that they simply have more capacitance and, by association, larger surface areas and thinner insulators which both contribute to higher leakage. If you would make any other construction of capacitor with the same specifications, they will exhibit fairly similar leakage characteristics. Yes, there are some unique reasons why electrolytics fare a little bit worse on the leakage side, but this is not something that puts the technology in a completely different order of magnitude of leakage.
One pretty simple reason is that the tolerance on Y5V tends to be much flakier numerically than X5R or X7R. Typically -20% + 80%
A less obvious and possibly more profound reason is that the change in capacitance versus temperature is very poor on Y5V. Typically at low temperatures and high temperatures the capacitance might halve in value.
DC voltage can change the capacitance of most capacitors a bit but Y5V are really prone to it. Typically at full rated voltage the capacitance might be a fraction of what it is when low voltages are present.
Just open this data sheet and look at the first few graphs. See also this: -
Best Answer
Q1) X7R capacitors are Class 2. Class 2 ceramic capacitors are ferroelectric and contain dipoles. The dipoles group together into domains where they point in the same direction. Application of an electric field (AC or DC Voltage) will force the dipoles in the domains to try and align with the field. This movement of the dipoles in the domain cause a release of energy in the form of heat along the walls of each domain. Full polarity reversal of the field (AC Voltage) will generate the most heat because the dipoles have to completely flip their direction each time. Continually applying this condition will heat up the component until it eventually fails. It is very similar to hysteresis losses in a ferromagnetic core.
Q2) The conditions where this will not lead to failure are difficult to define, which is why the manufacturers completely avoid the situation and put the statement you mentioned in their datasheets. It is a function of frequency, voltage, footprint size, and the physical construction of the capacitor. It is particularly bad below 1kHz because this gives the dipoles time to flip with each polarity reversal and build up heat. At higher frequencies, it is more difficult for the dipoles to flip quickly enough and less heat is generated as a result. Again, it is difficult for them to clearly define it in a way that they can guarantee no failures. So, they just won't recommend it.