Electronic – Axial vs Radial Film Capacitor

This effect is due to the effects of parasitic characteristics of the device. A capacitor has four basic parasitics:

Equivalent Series Resistance - ESR:

A capacitor is really a capacitor in series with the resistances of its leads, the foil in the dielectric, and other small resistances. This means that the capacitor cannot truly discharge instantly, and also that it will heat up when repeatedly charged and discharged. This is an important parameter when designing power systems.

Leakage current:

The dielectric is not ideal, so you can add a resistance in parallel with your capacitor. This is important in backup systems, and the leakage current of an electrolytic can be much greater than the current required to maintain RAM on a microcontroller.

Dielectric Absorption - CDA:

This is usually of less interest than the other parameters, especially for electrolytics, for which leakage current overwhelms the effect. For large ceramics, you can imagine that there is an RC circuit in parallel with the capacitor. When the capacitor is charged for a long period of time, the imagined capacitor acquires a charge. If the capacitor is rapidly discharged for a brief period and subsequently returned to an open circuit, the parasitic capacitor begins to recharge the main capacitor.

Equivalent Series Inductance - ESL:

By now, you shouldn't be too surprised that, if everything has capacitance as well as nonzero and non-infinite resistance, everything also has parasitic inductance. Whether these are significant is a function of frequency, which leads us to the topic of impedance.

We represent impedance by the letter Z. Impedance can be thought of like resistance, just in the frequency domain. In the same way that a resistance resists the flow of DC current, so does an impedance impede the flow of AC current. Just as resistance is V/R, if we integrate into the time domain, impedance is V(t)/ I(t).

You'll either have to do some calculus, or buy the following assertions about the impedance of a component with an applied sinusoidal voltage with a frequency of w:

\$ \begin{align} Z_{resistor} &= R\\ Z_{capacitor} &= \frac{1}{j \omega C} = \frac{1}{sC}\\ Z_{inductor} &= j\omega L = sL \end{align} \$

Yes, \$j\$ is the same as \$i\$ (the imaginary number, \$\sqrt{-1}\$), but in electronics, \$i\$ usually represents current, so we use \$j\$. Also, \$\omega\$ is traditionally the Greek letter omega (which looks like w.) The letter 's' refers to a complex frequency (not sinusoidal).

Yuck, right? But you get the idea - A resistor doesn't change its impedance when you apply an AC signal. A capacitor has reduced impedance with higher frequency, and it's nearly infinite at DC, which we expect. An inductor has increased impedance with higher frequency - think of an RF choke that's designed to remove spikes.

We can calculate the impedance of two components in series by adding the impedances. If we have a capacitor in series with an inductor, we have:

\$ \begin{align} Z &= Z_C + Z_L\\ &= \frac{1}{j\omega C + j\omega L} \end{align} \$

What happens when we increase the frequency? A long time ago, our component was an electrolytic capacitor, so we'll assume that \$C\$ is very much greater than \$L\$. At first glance, we'd imagine that the ratios wouldn't change. But, some trivial (Note: This is a relative term) complex algebra shows a different outcome:

\$ \begin{align*} Z &= \frac{1}{j \omega C} + j \omega L\\ &= \frac{1}{j \omega C} + \frac{j \omega L \times j \omega C}{j \omega C}\\ &= \frac{1 + j \omega L \times j \omega C)}{j \omega C}\\ &= \frac{1 - \omega^2 LC}{j \omega C}\\ &= \frac{-j \times (1 - \omega^2 LC)}{j \omega C}\\ &= \frac{(\omega^2 LC - 1) * j)}{\omega C} \end{align*} \$

Well, that was fun, right? This is the kind of thing you do once, remember the answer, and then don't worry about it. What do we know from the last equation? Consider first the case where \$\omega\$ is small, \$L\$ is small, and \$C\$ is large. We have, approximately,

\$ \begin{align*} \frac{(small * small * large - 1) \times j}{small * large} \end{align*} \$

which is a negative number (assuming \$small * small * large < 1\$, which it is for practical components). This is familiar as \$Z_C = \frac{-j}{\omega C}\$ - It's a capacitor!

How about, second, your case (High-frequency electrolytic) where \$\omega\$ is large, \$L\$ is small, and \$C\$ is large. We have, approximately,

\$ \begin{align*} \frac{(large * small * large - 1) \times j}{small * large} \end{align*} \$

which is a positive number (assuming \$large * small * large > 1\$). This is familiar as \$Z_L = j \omega L\$ - It's an inductor!

What happens if \$\omega^2 LC = 1\$? Then the impedance is zero!?!? Yes! This is called the resonant frequency - It's the point at the bottom of the curve you showed in your question. Why isn't it actually zero? Because of ESR.

TL,DR: Weird stuff happens when you increase the frequency a lot. Always follow the manufacturers' datasheets for decoupling your ICs, and get a good textbook or take a class if you need to do high speed stuff.

My favorite electronics book is "High Speed Digital Design: A Handbook Of Black Magic". I highly recommend this book. It seems expensive, but it is totally worth the money. This book has 12 pages on choosing a bypass cap! The author, Howard Johnson, also teaches some classes with decoupling caps as one of the topics.

Some important things that I've learned over the years, and have been backed up by this book, is that the "standard practices" with decoupling caps are almost always wrong and there is more art than science when it comes to choosing and routing them.

There are lots of calculations that you can do regarding decoupling caps, but much of those are not accurate due to many things. The caps themselves are vary wildly (especially the higher dielectric caps like X7R). The PCB layout changes things greatly (and you'll need to think in 3-D for this one). Temperature and voltage will change the behavior of the caps. A single cap will behave as both a "power supply smoothing cap" and a "AC signal return bypass cap". Etc.

What Johnson did was, after a lot of experimentation, figure out that inductance is the most important factor and it swamps almost every other consideration. So the goal when selecting and placing decoupling caps is to use a lot of physically small caps, with the highest practical value, and route them so the total inductance is as low as possible.

The ideal would be to use lots of 0.1 uF caps in an 0402 package. Place them under the chip on the back side of the PCB. The cap be routed as in the image below. And the vias go directly to the power/ground planes (not to the chip's power pins, as that would usually increase the inductance). If you place the cap under the chip then sometimes you could share the same via without any issues.

The reason why a 0.1 uF cap was chosen is because it is the highest practical in an 0402 package. The reason why 0402 was chosen is because it is the smallest practical size, and you want to use a lot of them to get the effective ESL/ESR down. Of course all bets are off if you have a 2 layer PCB without power and ground planes.

I don't want to belittle the use of the math, that is important, but the complexity of power supply decoupling and AC return paths often makes the math not so practical in the real world. In the real world, a "rule of thumb" really helps. Of the many rules of thumb for this topic, it has only been Howard Johnson that has proven the other rules don't work and provided this better rule. My experimentation and experiences has shown this to be true.