How does capacitance depend on AC voltage?

Question

I am aware of the dependence of some ceramic capacitors (e.g. X7R) capacitance on DC voltage. I assumed that if a high enough AC voltage is applied, capacitance changes in its rhythm causing distortion of the sine wave. So far so good.

I came across this datasheet for a C0G capacitor that also contains some general info. On p.11 there is the familiar graph of capacitance vs. DC voltage. Following that is another graph of capacitance vs. AC voltage, here it is:

This seems to be an altogether different phenomenon - the most pronounced drop in capacitance is at very low AC voltages. Also, the change is apparently independent of frequency. I can't find anything on this.

Does anyone know how this works? (Dielectric types, frequency dependence, DC and AC phenomena combined, etc.)

Answer 1

This effect in Class 2 ferroelectric MLCC capacitors is due to hysteresis in polarization, as illustrated in this paper.

At low AC voltage the hysteresis causes an effective loss in capacitance.

This graph from this white paper shows the effect clearly:

At low AC voltages (the green curve) the K is low, it increases for higher voltages (yellow) and then decreases again for higher AC voltages (purple and red). Incidentally, the yellow curve represents typical measurement conditions for determining the capacitor datasheet characteristics (1.0V +/- 0.2V RMS).

As to the physics behind the ferroelectric behavior of barium titanate used in X7R capacitors, this paper describes it as follows:

These dipoles arise due to the fact that in the tetragonal unit cell of BaTiO3 , the Ti4+ cation is surrounded by Six O-2 anions in a slightly deformed octahedral configuration, and can occupy one of two asymmetrical sites, as illustrated in Figure H-3. In either position, the Ti4+ cation is not coincident with the negative charge center of the oxygen anions by a small fraction of an Angstrom, creating an electric dipole. The energy barrier between the two possible Ti atom positions is sufficiently low to permit motion of the atom between sites by the coercion of an electric field, and the material can thus be directionally polarized with ease. The interaction between adjacent unit cells, in fact, is sufficient to create domains of parallel polarity the instant the material assumes its ferroelectric state on cooling through the Curie Point.

Answer 2

This has nothing to do with AC voltage, but rather any kind of voltage bias. There is a similar (but typically even more pronounced) loss of capacitance seen under DC bias.

There is a property of certain dielectrics that is very similar to ferromagnetism (what makes things like iron magnetic) called ferroelectricity.

Just like how ferromagnetic materials will, in the presence of an external magnetic field, have magnetic grains align, resulting in an additional magnetic field produced by the material, a ferroelectric material will have dipoles align producing an electric field if suitably excited by an external electric field.

Essentially, certain types of ceramic act as something analogous to a magnetic core in an inductor, but as an 'electric core' in a capacitor. The dielectric serves to massively increase the energy stored in an electric field, increasing the capacitance just like a magnetic core does the same for inductance.

And just like magnetic cores, these ferroelectric dielectrics can become saturated.

With an inductor, as current increases, the effective inductance begins to fall, as the magnetic core can't store any additional energy/strengthen the magnetic field further.

The same is true with ferroelectric materials. Voltage, rather than current, is the determining factor here, as voltage determines the electric field (where current determines the magnetic field). As the voltage increases, the capacitance falls as the dielectric is increasingly unable to further strengthen the electric field in the capacitor.

A common mistake I see made is the assumption that the voltage rating of ceramic capacitors has anything to do with this. It does not. Picking a ceramic capacitor rated for a higher voltage will generally have almost no impact on the amount of capacitance lost vs. voltage bias (AC or DC). The only thing that really matters is the physical size of the capacitor. Bigger ceramic capacitors will have more dielectric material, and more material means it will lose less capacitance as the electric field through it strengthens. Just like physical size does the same in the magnetic cores of inductors.

C0G capacitors do not have this problem at all, as they do not have ferroelectric dielectrics. But nearly all high capacitance value ceramic capacitors do use ferroelectric dielectrics, and will lose (unfortunately, often a lot) of their capacitance under even a few volts of AC or DC bias. The only way to mitigate this is to use larger (and pricier) ceramic capacitors, or simply more of them. At the end of the day, if you want that capacitance, you need more volume of dielectric material, at least if you are relying on ferroelectric dielectrics to do it.

Answer 3

It seems that X7R capacitors have a "kind" property.
There is a piezoelectric effect with X5R or X7R (BaTiO3), the capacitor acts as a microphone (?).
So, no use for an audio amplifier ...

NB: this is a "non-linear" effect, so a DC offset appears vs AC amplitude applied.