All dielectric materials deform under the stress of an electric field, it's called "electrostrictive effect". And some dielectrics exhibit an additional piezoelectric effect, such ceramic capacitors, but the dominant mechanism is piezoelectric effect. I want to know if plastic film capacitors have this phenomenon? And has anyone encountered this?
Electronic – About the audible noise of capacitor
capacitorceramicpiezoelectric-effect
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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.
ESL is almost entirely determined by lead inductance, so the size and type of the package determine this value.
ESR is a function of many things, but one of them is dielectric thickness, as thickness increases so does ESR.
Voltage rating is a function of dielectric strength and thickness.
So, if a package size is fixed, say both 0805, you should expect the same or extremely close ESL, while the dielectric strength or thickness would need to have increased to increase the voltage rating, so ESR would be higher.
This isn't a firm rule, remember that ESR varies with frequency.
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Best Answer
When subjected to an electric field, Dielectrics expands in the direction of the field and contract in a direction transverse to the field. The strain varies roughly with the square of the electric field. Most materials have electrostrictive strains on the order of \$1*10^{-7}\$ for Electric field strengths of \$ 1 \frac{MV}{m}\$. In Perovskite oxide materials the strain is \$ 1* 10^{-3}\$ at the same field strengths. Data comes from *(1) which seems to be highly cited.
While \$ 1 \frac{MV}{m}\$ seems to be unreasonable, a simple plate spacing of 1 um and a drive of 1 volt can easily meet this Field. So a normal capacitor with a 100 nm gap and 10 Volts being driven can easily exceed that number by 100 X.
For a capacitor that is say 1 mm in a given dimension a \$ 1 \frac{MV}{m}\$ Field would cause expansion by about 1 angstrom (0.1 nm) and probably more than 100 X that.
How much of that couples into an audible sound is hard to tell, but US patent # 7689390 B2 uses 100's of nm thick BST (\$Ba_{0.7}Sr_{0.3}TiO_3\$) layers to couple electrical and acoustic modes for a capacitor.
PTFE (Teflon) has high electro-restrictive coefficients. *(2)
So it does look possible.
(1) "Cracking in ceramic actuators caused by electrostriction" W Yang, Z Suo - Journal of the Mechanics and Physics of Solids, 1994 - Elsevier"
(2) Schwodiauer, R.; Neugschwandtner, G.; Bauer-Gogonea, S.; Bauer, S.; Heitz, J.; Bauerle, D., "Dielectric and electret properties of novel Teflon PTFE and PTFE-like polymers," Electrets, 1999. ISE 10. Proceedings. 10th International Symposium on , vol., no., pp.313,316, 1999