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Is any current level safe past capacitor breakdown limits?

Discussion in 'General Electronics Chat' started by valknut, Aug 30, 2017.

  1. valknut

    valknut New Member

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    Say a capacitor is pushed several times higher than its rated voltage limits and begins to conduct current. Would even 5 or 10 milliamps damage the device, or is there a certain level of current that is safe to pass through the component? And if the latter, is it mainly power-dependent, heat-dependent, or something else (or maybe a combination thereof), and moreover is there an algebraic equation that succinctly expresses that relationships which can be used to calculate reasonable limits?
     
  2. valknut

    valknut New Member

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    I found this paper describing the failure mode analysis of certain types of ceramic capacitors, and on page 5 it describes avalanche-breakdown damage occurring at 140V with about 1400A passing through the device, so even scaling that down to a much smaller capacitor equates to a range of like tens or even hundreds of amps. At several hundreds of watts, the resulting damage seems pretty unsurprising, which leads me to believe that much smaller currents would most likely be perfectly safe. Of course, whether that conclusion is correct or not remains to be seen (and I'm sure that it also depends on the construction and physical condition of the particular capacitor in question) but from the outset at least, plus the fact that the continuous tests I've been running for the past couple of days don't seem to indicate any obvious damage in the form of detectable resistance, overheating, changes in capacitance, etc, it looks as though that this may indeed be the case.
     
  3. Pommie

    Pommie Well-Known Member Most Helpful Member

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    If you run the cap above it's rated voltage you will puncture the insulation and the cap will discharge through the insulation making a bigger hole. You may be able to do this short term but prepare yourself for lots of failures down the line. Why would you do this anyway?

    Mike.
     
  4. dave

    Dave New Member

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  5. valknut

    valknut New Member

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    That does seem to be the standard line of reasoning, but I'm just not completely convinced.

    For one thing, you have to consider that the movement of an electron across the dielectric doesn't necessarily result in a reconfiguration of the substance, a chemical reaction, or what have you. Obviously, voltage is a major factor (as it determines the amount of energy propelling the electron) but this alone may not be enough to damage the substrate. An ice cube placed atop a high-temperature system confined to one cubic meter but only containing 10 atoms moving about with great speed and energy may not melt due to the fact that there simply aren't enough particles to heat the box to any appreciable level. By the same token, if avalanche-breakdown occurs within a dielectric, individual electrons may well impart a lot of energy upon impact but because of the overall low energy state of the surrounding medium the energy is safely absorbed by the system.

    And yes, I know that the question itself sounds kind of absurd, like why would anyone want to do such a thing? I'm just curious, really. And if it can be safely done then that could possibly mean some interesting applications. As you pointed out though, I could be very wrong about that and this sort of thing just might be a plain and simple misapplication of the capacitor. Accepted, I've got my fingers crossed but I'm not going to hold my breath!
     
    Last edited: Aug 30, 2017
  6. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Generally there is significant energy stored in a capacitor when it's charged to or above it's breakdown voltage and that energy will usually punch a nice hole through the capacitor dielectric, creating a permanent short, soft of a mini spot-weld.
    Certain metalized film capacitors are an exception to this as the energy can burn away the thin metal film around the punch-through, allowing the cap to live another day.

    But no designer worth his salt operates a capacitor near or above its rated voltage, if he wants the circuit to have decent reliability.
     
  7. valknut

    valknut New Member

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    I guess my main doubt lies in the fact that the breakdown could merely equate to the electrons gaining just enough energy to overcome the low-energy state of the dielectric. Take, for instance, the way that certain substances produce light when current passes through them. Below a given threshold, no light is emitted. Above that, the atomic shells are sufficiently energized and photons are ejected. This process does not in and of itself lead to any subsequent damage to the material. However, if too much current flows then the atoms can become so energetic as to disassociate with their neighbors and change state (melt). Perhaps the same is true of the avalanche breakdown effect of capacitors; as long as the combined current and voltage does not exceed a certain maximum, no harm done.

    Naturally, I'm just thinking out loud here and more or less uncertain either way. Are there some hard numbers I can see which precisely describe what sort of currents/voltages have been observed in the cases where catastrophic breakdown has been documented?
     
  8. dr pepper

    dr pepper Well-Known Member

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    Your question has 2 answers at least.
    Nichicon state for electrolytics you can overvoltage them by 10% for a minute 3 or 4 times an hour without damage, one application for this is tube equipment, where the B+ voltage comes up before the tubes start emitting, this can result in a temporary voltage overshoot.
    However if you get 5 to 10 ma dc through a capacitor of most kinds it would be considered leaky and unserviceable, note 5 to 10ma Dc (not ac).
    As cruts says a cap will flash over at a certain overvoltage, often resulting in destruction of the device.

    Edit: A cap only passes Ac, it will block Dc.
     
    Last edited: Aug 30, 2017
  9. crutschow

    crutschow Well-Known Member Most Helpful Member

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    You really are barking up the wrong tree.
    You are trying to equate the current through a semiconductor with the the current through an insulator, and that's two entirely different things.

    In an insulator you cannot get any significant current flow (most good insulators have gigaohms of resistance) without a physical disruption of the atoms creating physical damage, which then creates an opening for continuous current flow.
     
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  10. valknut

    valknut New Member

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    Hmm, so the behavior of different dielectrics seems to be highly variant. Some pass certain levels of current safely in avalanche mode while others do not at all. From what I have gathered, much longer tests (hundreds or even thousands of hours) might be required in any case to determine just what those levels may be (if they even exist). My 50+ hours of testing simply isn't enough. That said, there may be something yet to pursue here, although I suppose that's more of a material physics problem than one of practical electrical engineering (which is what I am supposed to be studying right now!) so I guess I'll just leave it at that for now.

    Anyway, thanks again for all the input and your patience with my ridiculous questions. Cheers everyone!
     
  11. dr pepper

    dr pepper Well-Known Member

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    No problem.
    P.s. avalanche is a term used for semi conductors rather than dielectrics.
     
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