When dealing with capacitances <= 1uF ceramic is your choice. When you need more than 10uF it is economically justified to use electrolytic cap shunted with small ceramic one. But in range 1-10uF it is questionable. What do you suggest?
Electronic – Decoupling capacitors: low ESR electrolytic vs electrolytic + ceramic vs ceramic
bypass-capacitorcapacitordecoupling-capacitor
Related Solutions
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.
If I could get a ceramic capacitor at the capacitance of 10uF and within my voltage requirements, which from my initial searches I can, what problems would I experience if I were to change, if any?
Some circuits (like some linear regulators, for example), require a certain minimum ESR from the output capacitor, which could cause the circuit to oscillate when using a ceramic but not with an electrolytic.
In a precision circuit, a ceramic might not be preferred due to microphonics, but in those cases you probably wouldn't want an electrolytic either.
Otherwise, ceramics are generally preferred. They'll have lower ESR, they're not polarized, they need less voltage de-rating, and so on.
Finally, when searching SMD footprint standards, the common packages seem to be 0402, 0603, and 0805, where they increase in physical size respectively, but also power rating, which suggests to me I should use as large of a package as possible
Usually you choose the smallest package you can get away with because you want to fit as much circuit as you can in the smallest footprint.
Also, for ceramics, the larger sizes (1210 and higher) can have reliability issues because they can be cracked if the board flexes.
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Best Answer
Your question is naive in that it doesn't consider what type of capacitor best suits the target application. Consider this: -
So, a 1uF ceramic may not be your best choice if wanting an accurate timing circuit (for instance) or some audio application where the microphonic problems associated with a ceramic capacitor cannot be tolerated. Also, these days, it's quite normal to find ceramics up at 100 uF and I'm sure in a few years time 1000 uF will be possible.
Choose a capacitor type that suits the application. If one doesn't exist look at alternatives and if those alternatives look problematic, find a different approach.
Pretty picture taken from here.