# Capacitor ESR vs. Impedance

One of the ceramic capacitors that I've found(link to the capacitor) specifies the following characteristics at 10 kHz:

The difference in value is quite significant, almost 100 fold.

Which graph on the picture above better represents real-world use-case?

What is the point of specifying the ESR solely for AC loads?

## 3 answers

Capacitors are lossless with 0 ESR and rise in temperature and become less reliable by 50% for every 10’C rise due to I^2ESR=Pd losses. However the current spectrum must be known with the ESR frequency graph in order to estimate losses.

The ultimate goal is to attenuate ripple voltage by the impedance ratio so impedance is the primary concern for a given current spectrum. Even if the current is a step or impulse, then you can compute or measure the spectrum and do the same. The secondary importance is self heating from ESR losses. Both are important.

Ideally you want to attenuate as much as possible with the least power dissipation factor because dielectrics are also poor thermal conductors and it is better to use a resistive series loss to dampen a response with a heat sink as it is hard to put a heatsink on dielectrics other than SMT on a large copper area of 1W/sqin which is not cost effective for large powers.

By having a power loss budget for each component and a max temperature rise, a designer must also insider the undamped Q of a closed loop and mitigate control instabilities from an extremely low ESR and high Q circuit by using low duty cycles to start up and prevent overshoot with Kd compensation to improve phase margin un a wide range of step loads.

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Which graph on the picture above better represents real-world use-case?

They both represent real-world use cases. The impedance graph informs you how much ESL (effective series inductance) the capacitor has and, at the series resonance point you can evaluate the ESL value with a simple formula: -

Resonance occurs at 1.35 MHz and, if you know the capacitance value, you can derive ESL. Note also that at 1.35 MHz, the ESR value is very similar: -

In other words, at resonance you can also determine the ESR.

The impedance graph also shows what you would expect from your capacitor at frequencies below resonance. At 10 kHz (for instance) the impedance is maybe 2 Ω and, if you calculate the capacitance you get a value of around 8 μF. So, I'm guessing that these graphs are for a 10 μF capacitor. At 1 kHz the impedance is ten times higher as you'd expect from a capacitor. At 100 kHz, the impedance is ten times lower.

But, go beyond 100 kHz a little and you are hitting the effects produced by ESL series resonance.

What is the point of specifying the ESR solely for AC loads?

If you mean what is the point of showing a graph of ESR at various frequencies, then the graph is very informative. ESR represents the effective losses of the capacitor at each frequency (that's dielectric and true series resistance losses) as one resistor called *the effective series resistance*.

So, given that you know the minimum ESR, you can calculate the dielectric losses at other frequencies and get a bigger picture of how the capacitor performs across the spectrum.

I'll try first to give a general introduction to the problem, then I'll switch to your particular questions.

Passive linear components, like capacitors, inductors or resistors are not ideal. For this reason, they are often modeled electrically as a more or less complex circuit involving resistances, capacitances and inductances. For example, a capacitor is often modeled as a capacitor in series with a resistor (the so called ESR) and an inductor (the so called ESL). But even this model fails to represent the electrical behavior of the capacitor at very high frequencies, and must be refined to this end. The same could be said about resistors, inductors etc.

Dealing with complex electrical models would be difficult, but fortunately, there is a much simpler way to understand the behavior of these passive components: the notion of complex impedance (I assume you know it, otherwise, you may want to try this Wikipedia article). So, at a given frequency $f$, you have $$ V = Z I,$$ Where $Z$ is the complex impedance of the component at frequency $f$, $V$ is the complex voltage, and $I$ is the complex current.

As a complex number, $Z$ can be written $$ Z = a + j b, $$ with $a$ and $b$ real. For passive components, $a$ is always positive and represents a resistance. On the other hand, $b$ may be positive or negative: if it is positive, that means that the component exhibits an "inductive" behavior at this frequency; if it is negative, it exhibits a "capacitive" behavior. That's because the complex impedance of an ideal inductor is $$Z = jb = 2 \pi f L \cdot j, $$ (hence $b>0$) and that of an ideal capacitor is $$Z = jb = -{1\over 2 \pi f C} j$$ (hence $b<0$). At this point, I would like to formulate the following remarks:

- As we've just seen, the "good" way to understand passive linear components is through their complex impedance: don't think too much about capacitance, inductance, dielectric loss etc. (unless, of course, you have good reasons to do so). The real point is that at a given frequency, you have something that
**exhibits**an inductive or capacitive behavior, in addition to its resistance. - All what have been said concerns a particular frequency $f$, but as the frequency $f$ varies, the behavior of the components varies as well: at some frequencies, the component may exhibit a capacitive behavior. At another higher frequency, it may exhibit an inductive behavior etc. (more about that below). Mathematically, you have $$Z = Z(f).$$
- At DC (the limit case where $f = 0$), the component is nearly ideal (usually at a very high degree of accuracy): a capacitor is an open circuit, and an inductor is a short. Since this is always the case, this need not be dealt in the datasheet: only the AC behavior is interesting.
- It should also be added that there are dedicated instruments that allow measuring the complex impedance of components as a function of the frequency: the so-called network analyzers, and especially vector network analyzers (VNA). Usually, there is a whole accompanying theory of S_ij parameters, equivalent to the complex impedance theory, but that's another story.

After this introduction, let examine your graphs:

The first graph is a typical capacitor graph: It does not tell you what is the complex impedance $Z = Z(f)$ of the cap, but only its module $|Z(f)|$, which has also the dimension of "Ohm". With some experience, you are supposed to identify the shape of this graph, and to understand that the cap exhibits a capacitive behavior up to a bit more than 1000kHz, where it becomes purely resistive. From this point and up to 1GHz, the component exhibits an inductive behavior. So, half graph is capacitive, and the second half is inductive. But how much is each half capacitive/inductive? That's where the second graph comes into play. Recall that you have $Z = a + jb$, where $a = ESR$ (given in the second graph). The first graph gives you $|Z| = \sqrt{a^2 + b^2}$, hence, with some math, you have $$|b| = \sqrt{|Z|^2 - a^2}.$$ More practically, at 10kHz, you see that the first graph gives you an impedance $|Z| = 1\Omega$, while the second graph gives you $a = 20 m\Omega$ (say). Since $1 \Omega$ is much larger than $20m\Omega$, you see that quantity $b$ dominates widely. In other words, at 10kHz, there is a very small ESR resistance (compared to the capacitive reactance), hence you have something that behaves almost like an ideal capacitor. Near 100kHz, the behavior is still widely capacitive. But the more you approach the resonance frequency, the more the component becomes resistive and the less it behaves like a capacitor. At the resonance frequency (about 1MHz) it is purely resistive. After the resonance, it becomes more and more inductive (and the way to compute the ESL is the same as previously). This continues up to 1 GHz or more. Then, if the graph had been continued, it would be observed that the component becomes again capacitive.

**A final note**: you may ask yourselves what to do if capacitors become inductive above the resonance frequency; Indeed, I've tested many ceramic capacitors with a VNA and have observed that most of them have their resonance point below 50MHz, not so high a frequency!
Well, the truth is that in general, who care? most often, you use a cap to eliminate the DC component in a signal and pass only the AC component. From this point of view, the capacitor does always its work, no matter if it becomes an inductor at high frequencies or not. Since in most applications, you deal only with a given frequency in the signal, the only thing that matters is the absolute value of the impedance $|Z|$.
Of course, if you are designing an application that is supposed to preserve the AC component of a signal in an ultra wide band (like an oscilloscope), that's another story. Fortunately, we usually leave these tasks to great masters and gurus in electronics :-)

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