Why does impedance decrease in an MLCC as frequency increases?

Question

I understand that the impedance of any capacitor decreases as frequency reaches the resonance frequency, following the formula Xc = 1 / ( 2 pi f C ), but why does the impedance of an MLCC such a better frequency response than other capacitor types like electrolytic and tantalum?

Additionally I read somewhere that "this equation is not ( in math terms ) "rigorously correct", it is actually a crude "rule of thumb", which has some built-in assumptions that are NEVER identified, anywhere." on one of the answers [here][1] Is this true? And why are the assumption never identified anywhere?

Answer 1

MLCC capacitors generally have a much lower effective series resistance and therefore will have a sharper series resonance compared to Electrolytics/Tantalums: -

Picture from this wiki page.

"this equation is not ( in math terms ) "rigorously correct", it is actually a crude "rule of thumb", which has some built-in assumptions that are NEVER identified, anywhere." on one of the answers here Is this true?

I can't address your 2nd paragraph because it does not read correctly i.e. it is gobbledygook. Try fixing your question.

If you are referring to the answer given by someone called "@Bob S" to this question then take note that Bob S has not received upvotes for his answer, has accumulated a total reputation of 1 and hasn't been seen since April this year when he joined stack exchange. In short, his answer is not to be relied upon.

Answer 2

An MLCC capacitor has smaller ESR than electrolytic or tantalum. Due to physical construction and size, the ESL is also lower than on electrolytics. The MLCC curve still approximately follows that curve, but due to smaller parasitics, it performs better, has higher resonance frequency, and narrower resonance peak, before the inductance makes the impedace to rise again.

So the formula for reactance for a capacitance is correct, but a physical real world capacitor has other properties than just capacitance. It has ESR, ESL, dielectric losses, and even the capacitance varies over temperature and voltage applied, so the simple formula just does not model these.

Answer 3

The equation you give is the theoretical reactance of an ideal capacitor. It comes from the mathematical descriptions of capacitance. I don't see how you get the idea that it is a "rule of thumb."

$$Z_C = \frac{1}{2 \times \pi \times f \times C}$$

Clearly shows that reactance becomes smaller as the frequency increases - and that is true for all capacitors.

Wikipedia has a lot of the math and links to other sources if you'd like to read about it.

What you are discussing, however, is a difference between different types of real capacitors.

A real capacitor has resistance and inductance in addition to the capacitance it was designed to have.

Resistance is there because all conductors have at least a little resistance.

Inductance is there because even a straight wire has at least a little inductance.

Look at the construction of an electrolytic capacitor:

It has two very long conductors (the anode and cathode foils.) They are very thin and very long, which gives them considerable resistance. Since the whole thing is rolled together, the length of the rolled conductors also gives it considerable inductance.

While the capacitance itself follows the equation for reactance you gave, you also have to figure in the resistance and the inductance.

$$ Z_T = Z_C + Z_I +R$$

Along with that, the capacitance and inductance form a resonator and make things more fun.

The difference between an electrolytic capacitor and a ceramic capacitor is that the frequency response of the ceramic capacitor more closely follows the ideal equation - ceramic capacitors have lower inductance and lower resistance than electrolytic capacitors.

Part of the difference is due to the size - typical ceramic capacitors are smaller than typical electrolytic capacitors.

Part of the difference is due to materials, and part of it is due to the physical construction.

As the applied frequency rises, the resistance and inductance play a proportionally larger role. Resistance is (more or less) fixed, and the impedance of the inductance increases. At some point, those two out weigh the impedance of the capacitance. That's where your capacitor loses its effectiveness as a capacitor and behaves like an inductor.