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Thread: What is a cermet capacitors?

  1. #1 What is a cermet capacitors? 
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    In this article they mention so-called cermet capacitors:

    ``Cermet capacitors
    An alternative way to prepare capacitors with an increased electrode surface
    area and a decreased distance between electrodes is the use of
    metal/dielectric composites with a random distribution of the metal phase. At
    the percolation threshold of the metal phase (the volume fraction at which
    the metal phase starts to form an electrically conducting network inside the
    composite),
    pc, the composite undergoes a transition from an isolating to a
    conducting material. Efros and Shklovskii [8], McLachlan [9] and Dubrov
    [10] described the behaviour of such a composite near its percolation
    threshold. The dielectric constant of the composite increases with the volume
    fraction of metal and diverges in the immediate vicinity of the percolation
    threshold. The increase in the dielectric constant of metal/dielectric composites
    near the insulating-to-conducting transition appears to be comparable to
    the increase obtained during a phase transition in the dielectric phase near
    the Curie temperature. It should be noted that in a conventional capacitor
    only the component of the electrode surface area perpendicular to the electric
    field strength contributes in the polarisation of the dielectric medium. The
    main increase in the capacitance for random composites will result from an
    effective decrease in the distance between the electrodes.``

    Does anyone ever see such type of capacitors?They mention that they allow
    to make smaller distance between electrodes.Does it increase probability of
    dielectric breakdown or leakage?Are they less safe?What does random metalic
    phase distribution means?


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    Quote Originally Posted by Stanley514 View Post
    Does anyone ever see such type of capacitors?They mention that they allow
    to make smaller distance between electrodes.Does it increase probability of
    dielectric breakdown or leakage?Are they less safe?What does random metalic
    phase distribution means?
    All excellent questions. As you surmise, the increase in capacitance comes at the cost of reduced breakdown voltage. However, this sort of tradeoff isn't unique to cermets; it's common to all capacitor technologies.

    As to safety, as long as one operates well below the rated breakdown voltage and other operating limits, you should enjoy high reliability. The typical failure mode is for the capacitor to develop a short circuit, often exhibiting a surprisingly low resistance.

    The phrase "random metallic phase distribution" is indeed mysterious, so you are right to be confused. What it means is this: When you mix metal and ceramic, they could form a mixture that is neither metal nor ceramic, or a composite consisting of metal bits dispersed randomly in an otherwise ceramic matrix. It's this latter composition that is meant by the confusing phrase.


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    I googled "random metal-dielectric" & read some about it & I think I know what this Cermet Capacitor was doing: it develop a network of conductive area within the dielectric/insulating material (imagine it creep into the material) & let this conduction area grow very close to the other-half's conduction area, let it interact, and thus create a very large surface area for capacitive interaction. It also claim that: the random metals inside the ceramic/insulator started to percolating, thus create an area of high dielectic constant on one end but curiously also create an area with divergent (lesser) dielectric constant on other end (thus acting like insulator around it?). -Thus the writer believe this could allow increase in capacitance by using the curious properties of "random metal-dielectric"'s percolation to increase surface area and reduces distance (and "dielectric" as a jargon for insulator).
    Last edited by msafwan; March 14th, 2012 at 01:37 PM.
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    [QUOTE]The phrase "random metallic phase distribution" is indeed mysterious, so you are right to be confused. What it means is this: When you mix metal and ceramic, they could form a mixture that is neither metal nor ceramic, or a composite consisting of metal bits dispersed randomly in an otherwise ceramic matrix. It's this latter composition that is meant by the confusing phrase./QUOTE]

    It would be interesting to see a picture of to help visualisation.If they just mixture metalic particles with ceramic particles how then avoid creation of isolated metal areas in ceramic matrix?I suppose it could increase energy density only if metal particles will resamble a nervous system in which all nervous cells are adjustent to each other.If metal particles will get isolated how than they get charge?Or how they fabricate it?
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    The phrase "random metallic phase distribution" is indeed mysterious, so you are right to be confused. What it means is this: When you mix metal and ceramic, they could form a mixture that is neither metal nor ceramic, or a composite consisting of metal bits dispersed randomly in an otherwise ceramic matrix. It's this latter composition that is meant by the confusing phrase.
    It would be interesting to see a picture to help visualisation.If they just mixture metalic particles with ceramic particles how then avoid creation of isolated metal areas in ceramic matrix?I suppose it could increase energy density only if metal particles will resamble a nervous system in which all nervous cells are adjustent to each other.If metal particles will get isolated how then they get charge?Or how they fabricate it?
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    Quote Originally Posted by Stanley514 View Post
    It would be interesting to see a picture of to help visualisation.If they just mixture metalic particles with ceramic particles how then avoid creation of isolated metal areas in ceramic matrix?I suppose it could increase energy density only if metal particles will resamble a nervous system in which all nervous cells are adjustent to each other.If metal particles will get isolated how than they get charge?Or how they fabricate it?
    There's actually a very interesting thing called an "artificial dielectric." Imagine a bunch of metallic ball-bearings, all electrically isolated from each other and embedded in an insulator. For purposes of this discussion, assume that the insulating material has a dielectric constant exactly equal to that of vacuum. Remarkably, that "mixture" will act like a dielectric with a dielectric constant greater than that of vacuum. So you don't have to worry about isolated metal areas; that's actually the idea.

    The cermet acts much in the same way.

    As to "how they get charge," remember that the dielectric doesn't "get charge." All you do is move charge from one plate to another. The dielectric stays neutral. What a dielectric actually does is reduce the electric field produced for a given amount of displaced charge. That field reduction is achieved in conventional dielectrics by the movement of electrical dipoles. When you apply a voltage across the capacitor terminals, the dipoles move as you'd expect -- the positive end is attracted to the negative terminal and the negative dipole end orients toward the positive terminal. Note that the polarity of the dipole field is the opposite of the applied field. The summed effect is to reduce the electric field for a given amount of displaced charge. That description is the same as saying "the capacitance goes up."

    But your comment about a nervous system model is descriptive of another type of capacitor technology. A tantalum electrolytic capacitor employs a quasi-fractal geometry to pack an incredible surface area in a small volume (just like your nervous or circulatory system). That enables large capacitance in a small package.
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    But your comment about a nervous system model is descriptive of another type of capacitor technology. A tantalum electrolytic capacitor employs a quasi-fractal geometry to pack an incredible surface area in a small volume (just like your nervous or circulatory system). That enables large capacitance in a small package.
    Exactly.But I didn`t get how cermet capacitor could be like this.You do not have to confuse capacitance and voltage.Increase in dielectric constant usually leads to ability to store energy at higher voltages.But to increase capacitance we need to increase surface area.Capacitance doesn`t directly depends on voltage and you could have larger capacitance at lower voltage.This is exactly what happens in carbon supercaps where extremely large charge is stored at quite small voltage (up to 5 V).More favorable way to increase energy density is to increase capacitance because high voltages could be dangerous.I still don`t see how incorporation of small metall balls into quite good dielectric coild lead to increase in dielectric constant.Rather it should be opposite.In article they mention increase in capacitance,not in voltage.If we are trying to increase surface area then we need to make sure all this surface are are accessible for electrons to move.This is how electrolytic capacitor works.But if area is isolated???
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    Quote Originally Posted by Stanley514 View Post
    I didn`t get how cermet capacitor could be like this.You do not have to confuse capacitance and voltage.Increase in dielectric constant usually leads to ability to store energy at higher voltages.But to increase capacitance we need to increase surface area.
    You pretty much missed the entirety of what I was saying, so let's try another tack.

    Let's limit the discussion to a parallel-plate capacitor to simplify things. There are only three factors that affect capacitance: Area of the plates; distance between the plates; and the dielectric constant. The cermet material gives you a large dielectric constant. Forget about area, forget about distance, forget about nervous system networks.

    The part of my explanation that you didn't understand was the key part: The cermet acts as an artificial dielectric. To understand an artificial dielectric, I explained how a regular dielectric works. It's clear that you don't quite understand that, so we'll have to fix that before you can understand an artificial dielectric.

    So, how does a dielectric material increase capacitance? As I endeavored to convey earlier, a dielectric works by reducing the electric field associated with a given amount of charge imbalance. If you know that C = Q/V, and that voltage is proportional to electric field here, then you can see mathematically that if Q (the charge imbalance) is fixed, and you reduce the associated electric field E, C goes up.

    Ok, so that's the math; a higher capacitance is the same as saying that you get more charge per field (or less field per charge; same statement, mathematically speaking). But what does it mean -- that is, exactly how does a dielectric material achieve this result? It does so by generating its own electric field, with a polarity that subtracts from the electric field you would otherwise get from applying a voltage across a pair of plates in a vacuum. An ordinary dielectric (like plastic) is non-conductive, but charges can still move in place. Imagine that a dielectric consists of vast quantities of tiny dowels with a negative charge on one end, and a positive charge on the other. Imagine too that each dowel has a pin stuck through the middle so that the dowel can rotate, but only in place, kind of like a pinwheel.

    Now put that dowel-dielectric between two metal plates and apply a voltage to those plates. The plus-end of the dowels will be attracted to the negative plate, and the minus-end of the dowels will be attracted to the positive plate. Think about that picture: You've got the plus-end of the dowel close to the negative plate; they are oppositely-polarized things, so they'll tend to offset (partially cancel) one another's effects. An equivalent statement is that the electric field has gone down. From the math, we know that the capacitance has gone up. That is how a dielectric works.

    Ok, so if you've understood that, it's only a short hop to understanding an artificial dielectric (first invented at Bell Labs shortly after WWII). We've just seen that the recipe for a dielectric is an insulating material that possesses charges that move some amount "in place" (as opposed to being able to roam freely throughout the material, as in a conductor). You can create that by distributing metallic balls within an insulating material. The balls don't touch, but they have charges that can move within the ball (but not between balls). This description is the same as for a "natural" dielectric. You now can tune the dielectric properties by adjusting parameters such as the size of the balls and the distance between balls. This flexibility allows you to create dielectrics with properties that nature is otherwise reluctant to provide.

    The cermet is a material arranged to give you something very similar. I'm simplifying things here by ignoring factors like the behavior of metal-ceramic solutions, but imagine that dispersed throughout the ceramic are gajillions of metallic-like regions. It's an artificial dielectric. You can get dielectric constants that are tens to thousands of times that of vacuum by playing with the composition of the cermet.

    Finally, you can enhance that effect by changing geometries from parallel plates to things such as the quasi-fractal shapes of nervous or circulatory system networks, and/or by shrinking the separation between plates.

    Hopefully that will clear things up for you.
    Last edited by tk421; March 15th, 2012 at 08:35 PM.
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  10. #9  
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    So, how does a dielectric material increase capacitance? As I endeavored to convey earlier, a dielectric works by reducing the electric field associated with a given amount of charge imbalance.
    Right.But the problem with dielectric capactors is that you reduce electric field only inside dielectric at given voltage.But in order to increase energy density you still have to increase potential voltage between electrodes.This is very point of dielectric capacitor.If you will short dielectric capacitor or discharge it through a load you will get the higher voltage between the plates the more energy dense it is.You can`t avoid increase of voltage when you increase capacitance if you do not increase surface area.You might`ve try to explain how they tried to increase dielectric constant but it has nothing do to with surface area.The article that I posted at very beginning mentions increased surface area.In traditional sence increased surface area allows you to store larger charge by putting smaller stress on dielectric.It doesn`t have to withstand too strong electric field.You may take a perfectly polarizable material with dielectric constant up to 100.000, but if this is rough piece of material it still would have a tiny surface area.
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    Quote Originally Posted by Stanley514 View Post
    Right.But the problem with dielectric capactors is that you reduce electric field only inside dielectric at given voltage.But in order to increase energy density you still have to increase potential voltage between electrodes.
    You seem to know enough about capacitors to get yourself all twisted up. So unlearn what you think you have learned.

    First, stop confusing the discussion by introducing energy density. We are talking about how cermets increase capacitance. Stay focused, or you'll keep confusing yourself. I have said several times now that capacitance is proportional to the ratio of the stored charge to the electric field. Do you disagree? If so, please show mathematically why.

    Ok, you may assume that the equation I gave you is correct, because it is. Then, for a given Q, reducing E increases C. What more do you need? I'd write QED at this point, except it might get confused with Qd/E. But seriously, this completely answers the question of how a dielectric increases capacitance. That knowledge is necessary to understand how a cermet (or buckshot dispersed in plastic) can be a dielectric, too.

    Now, since you asked about energy, that can be expressed as 0.5CV^2. Note that the stored energy depends on both capacitance and voltage, not just voltage. If we fix the voltage, increasing C by increasing the dielectric constant, say, increases the energy. So you can see that it is not at all true that you have to increase the "potential voltage" (that's redundant, by the way; just say "voltage") to increase energy density. I don't know why you are so sure that you have to. I recommend going slowly, writing down the equations and using them to guide you toward a better understanding. Your instincts are not yet reliable enough to trust; don't shoot from the hip just yet.

    This is very point of dielectric capacitor.If you will short dielectric capacitor or discharge it through a load you will get the higher voltage between the plates the more energy dense it is.
    Not necessarily. Again, you're shooting from the hip, and missing your mark.

    You can`t avoid increase of voltage when you increase capacitance if you do not increase surface area.
    Again, Q = CV, so V = Q/C. I can reduce V by increasing C, so your statement can't be true in general.

    You might`ve try to explain how they tried to increase dielectric constant but it has nothing do to with surface area.
    If you read what I wrote, I specifically enumerated the three factors that affect capacitance: Area, spacing and dielectric constant. For an ideal parallel-plate capacitor (one whose plate spacing is tiny compared to any linear dimension of the plates), the formula for capacitance is epsilon*A/d, where epsilon is the dielectric constant, A is the plate area, and d is the plate-to-plate separation. In SI units, epsilon is about 8.85pF/m in vacuum, and will be larger by a factor known as the relative dielectric constant.

    The article that I posted at very beginning mentions increased surface area.In traditional sence increased surface area allows you to store larger charge by putting smaller stress on dielectric.It doesn`t have to withstand too strong electric field.You may take a perfectly polarizable material with dielectric constant up to 100.000, but if this is rough piece of material it still would have a tiny surface area.
    You are again unnecessarily tangling a bunch of things together. Stick to simple and you won't confuse yourself. You can increase capacitance by any combination of bigger area, higher dielectric constant, and reduced plate spacing. Each has different consequences with respect to breakdown voltage, volume, etc. That's why capacitors come in such a large variety of shapes, sizes and dielectric composition. One size doesn't fit all.
    Last edited by tk421; March 15th, 2012 at 11:49 PM.
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    Again, Q = CV, so V = Q/C. I can reduce V by increasing C, so your statement can't be true in general.
    This one point is confusing.Lets to tight analogy to particular number of electrons that should be moved
    from one plate to another.Lets say we take dielectric of qubic inch size and E=5 and move 1 mln. of electrons from one
    plate to another.We obtain voltage=1V.Then we take better dielectric of the same size with E=100.And again move one
    million of electrons from one plate to another.Is energy density going to increase?I don`t think so because charge remained
    the same.Than we increase number of electrons up to 20 mln.Because we have better dielectric now the electric field inside
    of dielectric is possibly going to be not larger than in the first case.(Though dielectric is going to experience higher stress and
    getting more dangerous).But strength of electric field inside of dielectric is not only one important thing.Voltage is pushing force
    with which charges are attracted to each other.When we say short circuit capacitor with a metal wire then charge ignores
    dielectric.Therefore how could it be important only, what electric field exist inside of dielectric?I guess we still going to have
    higher voltage as higher puching force of charge if we have larger charge (when we discharge it through load or short it)?
    You can increase capacitance by any combination of bigger area, higher dielectric constant, and reduced plate spacing. Each has different consequences with respect to breakdown voltage, volume, etc. That's why capacitors come in such a large variety of shapes, sizes and dielectric composition. One size doesn't fit all.
    I agree to believe,but in the article they mention namely increased surface area not just capacitance.You was not able to explain how they practically succeded to increase surface area (not just capacitance in any possible of the ways.)
    Last edited by Stanley514; March 16th, 2012 at 08:58 AM.
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    Quote Originally Posted by Stanley514 View Post
    The article that I posted at very beginning mentions increased surface area.
    Without seeing more, I assume the reason they mention increased area is simply the fact that a thinner dielectric allows you to pack more area into the same volume. The dielectric itself has no relevance to the area.

    And thanks to tk421 for a really good description. I have used capacitors of various types for years and never really thought about how they work!
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    Quote Originally Posted by Stanley514 View Post
    Again, Q = CV, so V = Q/C. I can reduce V by increasing C, so your statement can't be true in general.
    This one point is confusing.Lets to tight analogy to particular number of electrons that should be moved
    from one plate to another.Lets say we take dielectric of qubic inch size and E=5 and move 1 mln. of electrons from one
    plate to another.We obtain voltage=1V.Then we take better dielectric of the same size with E=100.And again move one
    million of electrons from one plate to another.Is energy density going to increase?I don`t think so because charge remained
    the same.
    Here's how to figure out the answer without having to trust anything I say beyond independently verifiable math. Here, your mistake is thinking that counting electrons is the same as accounting for energy.

    The energy expended (or acquired) in moving an electron around depends on the voltage through which it is moved. So even if the number of electrons moved is the same in two experiments, one cannot say for sure that the energies are the same. I also need an additional piece of information: What are the voltages in the two experiments? If the voltages are the same, then yes, you can use electron count as a proxy for energy. But if you don't know the voltages, then electron count alone doesn't tell you enough.

    More formally, the energy in a capacitor can be written as 0.5*Q*V (if you were taught that it is 0.5*C*V^2, no problem; it is the same expression -- we've simply substituted CV = Q in the second equation to get the first). So the math supports the words in the paragraph above. The rest is left as an exercise for the reader.

    I agree to believe,but in the article they mention namely increased surface area not just capacitance.You was not able to explain how they practically succeded to increase surface area (not just capacitance in any possible of the ways.)
    You have every right to be confused, because the article is badly written. The first paragraph makes you think that they are going to describe how to increase the area and reduce the distance. After setting up those expectations, they abruptly pull a switch on you and talk at length (and only) about how cermets increase the dielectric constant. Then after all that, the very last sentence suddenly jerks your chain again with a comment about reduced distances (but no more mention of increasing area or dielectric constant). Given that dreadful writing, I've chosen to address what I find to be the more important factor (the increase in dielectric constant).

    But if the electrodes start off as smooth parallel plates, but grow bumpy as a result of the manufacturing process, the effective surface area will indeed go up. At the same time, the bumps will mean that the plates are separated, on average, by a diminished distance. Those two effects of bumpy plates multiply the capacitance boost provided by the dielectric.
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    I think that main way in future capacitor development is ability to create
    capacitor with extremely high surface area similar to already createded
    graphene capacitors Super energy storage: Activated graphene makes superior supercapacitors for energy storage
    But which would be able to charge to a relatively high voltage,for example up to 40 V.
    I`m trying to imaging how we could create giant surface area in in good ceramic dielectric.
    Any ideas?
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    Quote Originally Posted by Stanley514 View Post
    I think that main way in future capacitor development is ability to create
    capacitor with extremely high surface area similar to already createded
    graphene capacitors Super energy storage: Activated graphene makes superior supercapacitors for energy storage
    But which would be able to charge to a relatively high voltage,for example up to 40 V.
    I`m trying to imaging how we could create giant surface area in in good ceramic dielectric.
    Any ideas?
    Tantalum capacitors have exploited highly textured surfaces for decades. Google "tantalum capacitor manufacturing process" and look for a pdf from a company called Avx. It's a very informative article on how this miracle is achieved.

    The path to more-super supercapacitors will likely involve a combination of textured "plates" and better dielectrics. There is no shortage of ideas (and one-off demos), but converting those lab prototypes into reliable, low-cost and high volume products will take a lot more work.
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    There is idea to create quantum digital batteries which is no mere then capacitors.
    Digital quantum batteries inspired by plasma TVs
    I think that unfortunately that thing is going to have very high cost and low reliablity.
    If only one nanocapacitor from billion of parallel capacitors will burn then entire array
    will be out of use!
    Last edited by Stanley514; March 18th, 2012 at 02:49 PM.
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    There is one more interesting effect disovered.
    IEEE Xplore - Grain boundary behaviors of ferroelectric ceramics above Curie point
    They mention very large charge 10 Coulombs per square meter.
    Does it mean high energy density?
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    There's actually a very interesting thing called an "artificial dielectric." Imagine a bunch of metallic ball-bearings, all electrically isolated from each other and embedded in an insulator. For purposes of this discussion, assume that the insulating material has a dielectric constant exactly equal to that of vacuum. Remarkably, that "mixture" will act like a dielectric with a dielectric constant greater than that of vacuum.
    Quite interesting idea,actually.And how much they succeed to increase permittivity?Permittivity of metal should be close to infinity, I guess?So what would be energy density?
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    There's actually a very interesting thing called an "artificial dielectric." Imagine a bunch of metallic ball-bearings, all electrically isolated from each other and embedded in an insulator. For purposes of this discussion, assume that the insulating material has a dielectric constant exactly equal to that of vacuum. Remarkably, that "mixture" will act like a dielectric with a dielectric constant greater than that of vacuum.
    Quite interesting idea,actually.And how much they succeeded to increase permittivity?Permittivity of metal should be close to infinity, I guess?So what would be energy density?
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