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.
Best Answer
Tantalum capacitors are completely unnecessary in this application.
The only reason for choosing tantalum might be lifetime, and this can be designed for with aluminium wet electrolytic caps. It is assumed from here on that lifetime has been properly designed for and is not an issue.
Using a tantalum capacitor as the input capacitor invites capacitor death at any time if the input power rail can have voltage spikes on it from any source. A spike more than a small fraction above a tantalum capacitor's rated value risks it's total destruction in a high energy circuit, such as this one is.
The input capacitor is a typical reservoir capacitor, its value is relatively non critical. Tantalum serves no technical purpose here. If ultra low impedance is desired then use of a smaller parallel ceramic is indicated.
The output capacitor is NOT a filter capacitor in any traditional sense. Its principal role is to provide loop stability for the regulator. (An eg 10 ohm resistor could be placed in series with the capacitor without impeding its functionality. No normal filter cap would tolerate this without impaired functionality).
The characteristics of aluminum wet electrolytic capacitors of the correct capacitance and voltage rating are well suited to the output capacitor's role. There is no reason to not use them there. This 7 cent capacitor pricing / general data / datasheet would be an acceptable choice in many applications. (Longer lifetime applications may indicate 1 2000 hour/105C part).
The LM1117 datasheet provides clear guidance on the essential and desirable characteristics of the input and output capacitors. Any capacitor which meets these specifications is suitable. Tantalum is an OK choice but is not the best choice. There are various factors and cost is one. Tantalum offers OK cost per capability at capacitances from about 10 uF up. The output capacitor is "safe" against spikes in most cases. The input capacitor is at risk from "bad behavior" from other parts of the system. Spikes above rated value will produce a (literally) flaming melt down. (Smoke, flame, noise, bad smell and explosion all optional -
I have seen one tantalum cap do all of these in turn :-))
Input capacitor
The input capacitor is not overly critical when the regulator is fed from an already well decoupled system bus. Under the diagram on the front page they note "Required if the regulator is located far from the power supply filter" - to which you could add "or another well decoupled portion of the supply". ie capacitors used for decoupling in general may make another one here redundant. The output capacitor is more crucial.
Output capacitor
Many modern low drop out high performance regulators are unconditionally unstable as supplied. To provide loop stability they require an output capacitor which has both capacitance and ESR in selected ranges. Meeting these conditions is essential for stability under all load conditions.
Output capacitance required for stability: Stability requires the output output load capacitor to be >= 10 uF when the Cadj pin does not have an added capacitor to ground and >= 20 uF when Cadj has an added bypass capacitor. Higher capacitances are also stable. This requirement could be met by an aluminum wet electrolytic cap or a ceramic cap. As wet electrolytics are generally wide tolerance (up to +100%/-50% if not other wise specified) a 47 uF aluminum wet electrolytic would provide adequate capacitance here even when Cadj was bypassed. BUT it may or may not meet the ESR spec.
Output capacitor ESR required for stability:
ESR is a "Goldilocks requirement" :-) - not too much and not too little.
Required ESR is stated as
This is an extremely wide and unusual requirement. Even quite modest ripple currents in this capacitor would induce far larger than acceptable voltage variations. It's clear that they do not expect high ripple currents and that the capacitor's role is primarily related to loop stability than to noise control per se. Note that "old school" regulators such as eg LM340 / LM7805 often specified no output capacitor or perhaps a 0.1 uF. For example the LM340 datasheet here says "**Although no output capacitor is needed for stability, it does help transient response. (If needed, use 0.1 µF, ceramic disc)".
A tantalum capacitor is not required to meet this specification.
A wet aluminum capacitor will meet this spec with ease. Here are some typical new maximum ESRs for new aluminum wet electrolytic capacitors. The first group are capacitors that might be used in practice in this application at the low end of the capacitance range. The 10 uF, 10V is about half the allowed ESr - perhaps a bit close for comfort across lifetime. The second group are what would be used with Cadj bypassed and could be used anyway - ESRs are far away from limits in both directions. The third group are capacitors chosen to approach the lower limit (and they will get higher resistance = better with age). The 100 uF 63V pushes the lower limit - but there would be no need to use a 63V part here, and it will get higher (= better) with age. .
10uF, 10V - 10 ohm
10 uF, 25V - 5.3 ohm
47uF, 10V - 2.2 ohm
47 uF, 16V - 1.6 ohm 47 uF, 25 V, 1.2 ohm
470 uF, 10V - 024ohm
220uF, 25V - 0.23 ohm
100 uF, 63V - 0.3 ohm
They say in the LM1117 datasheet
1.3 Output Capacitor
The output capacitor is critical in maintaining regulator stability, and must meet the required conditions for both minimum amount of capacitance and ESR (Equivalent Series Resistance).
The minimum output capacitance required by the LM1117 is 10µF, if a tantalum capacitor is used. Any increase of the output capacitance will merely improve the loop stability and transient response.
The ESR of the output capacitor should range between 0.3Ω - 22Ω. In the case of the adjustable regulator, when the CADJ is used, a larger output capacitance (22µf tantalum) is required
ESR is crucial
ADDED - notes
SBCasked:
Regulator instability, in my experience, (and as you'd expect) results in the regulator oscillating, with large level and often high frequency signal at the output and a DC voltage measured with a non-RMS meter that appears to be stable DC at an incorrect value.
The following is comment on what you may see in typical circumstances - actual results vary widely but this is a guide.
Look at the output with an oscilloscope and you may see an eg 100 kHz semi sine wave of 100's of mV to some Volts of amplitude on a nominal 5VDC output.
Depending on feedback parameters you might get low frequency oscillation, slow enough to see as variations on a "DC" meter and you might get more like MHz signals.
I'd expect:
(a) very slow changes to be more liable to be high amplitude (as it suggests that the system is chasing its tail in such a way that it is almost in regulation and that corrective feedback is not bringing it rapidly into line, and
(b) MHz level oscillation to be more liable to be lower than usual amplitude as it suggests that slew rate of the gain path is a major factor in response speed. BUT anything can happen.
The intuitive and the logical do not always match.
A regulator is essentially a feedback controlled power amplifier.
If the feedback is negative overall the system is stable and the output is DC.
If the net loop feedback is positive you get oscillation.
The overall feedback is described by a transfer function involving the components involved. You can look at stability from the point of view of eg Nyquist stability criteria or (related) no poles on right half plane and all poles inside unit circle or ... agh!. It's adequate to say that the feedback from output to input does not reinforce oscillation and that a resistance that is too large or too small may lead to an overall reinforcement when considered as part of the overall system.
Simple, useful.
Only slightly more complex - good
Sueful - stack exchange
Useful
Lots of related pictures
They say " ... 0.3 ohm <= ESR <= 22 ohm ..."
If you has an ESR of 10 Ohms say, then every mA of ripple current will cause 10 mV of voltage variation across the capacitor. 10 mA of ripple current = 100 mV of voltage variation and you'd be very unhappy with your regulator. The active regulator can work to reduce this ripple, but it is nice to not have your filter capacitor adding to the problem you want it to fix.