For same capacitance C or same inductance L, one wishes to shirink the physical size of capacitors and inductors (by whatever methods – shrinking down the size of wire, etc.). What limits size becoming small? Planck length limitation is obvious, so I am asking other sources of limitations.
Electronic – What limits physical size of capacitors and inductors becoming really small
capacitorinductorwire
Related Solutions
Short Answer:
Inductor: at t=0
is like an open circuit
at 't=infinite' is like an closed circuit (act as a
conductor)
Capacitor: at t=0
is like a closed circuit (short circuit)
at 't=infinite' is like open circuit (no current through the
capacitor)
Long Answer:
A capacitors charge is given by \$Vt=V(1-e^{(-t/RC)})\$ where V is the applied voltage to the circuit, R is the series resistance and C is the parallel capacitance.
At the exact instant power is applied, the capacitor has 0v of stored voltage and so consumes a theoretically infinite current limited by the series resistance. (A short circuit) As time continues and the charge accumulates, the capacitors voltage rises and it's current consumption drops until the capacitor voltage and the applied voltage are equal and no current flows into the capacitor (open circuit). This effect may not be immediately recognizable with smaller capacitors.
A nice page with graphs and some math explaining this is http://webphysics.davidson.edu/physlet_resources/bu_semester2/c11_rc.html
For an inductor, the opposite is true, at the moment of power-on, when voltage is first applied, it has a very high resistance to the changed voltage and carries little current (open circuit), as time continues, it will have a low resistance to the steady voltage and carry lots of current (short circuit).
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.
Related Topic
- Electronic – Characterization of bypass capacitors
- Electronic – Laplace Relation to Capacitors and Inductors
- Electronic – Modeling passive elements at high frequencies
- Electronic – energy density comparison between inductors and capacitors
- Electronic – Why are the IV curves of capacitors and inductors ellipses
- Electronic – Is the answer correct to this non-ideal capacitor question
Best Answer
Physics (among other things)!
If you wish to shrink a capacitor in physical size, while keeping the capacitance the same, some other property has to go up, as in every capacitance the actual dimension does matter for the size of the capacitance. In this case you would need to increase the relative permittivity of the dielectric material. As this is a material constant you cannot increase it arbitrarily.
Another noteworthy effect is that you will have a reduced breakdown voltage, so reducing the size will also lead to a reduced voltage handling capability of the capacitor, which is probably unwanted.
On the side of inductors you have similar effects. Shrinking the wire will lead to increased ohmic resistance of the inductor which is generally unwanted (less optimal inductor). You can see that already if you look up an inductor with same inductance but different sizes.
Small wires will also have a problem with the current they are supposed to carry as there is a limit on the current density a material can handle.
Another major factor is the magnetic saturation of the core, which will also limit how small you can make the core for a given inductance.
There are of course practical things like production and handling to keep in mind and you can probably find some more things which will cause trouble. These came to my mind first. Also note that I saw this from an engineering perspective, if you would choose a very minuscule value for C or L, their sizes could become very very small I guess. If you are looking for limits in those regions, I guess Physics SE would be a better place to ask.