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I'm looking at this Power board of a non-working TV and noticed a ceramic capacitor with a bit of the ceramic blown away. I'm not entirely sure if this is the reason why the TV isn't working; though it is possible. It was soldered between the drain and the S/GND of an IC [STR-Y6456 which is a switching power supply(?)[1]]

From the markings, it's a 390pF 2kV ceramic capacitor (I'm so glad it blew on the other side instead of the marked one).

What would make a ceramic cap blow up? While searching for this topic here, I came across [2], but that's not the case. I know a ceramic capacitor doesn't have polarity and with 2kV limit, so it can't be the voltage [though, that's just my guess] ... so I'm guessing the current has something to do with it? What about heat?

Any clarifications appreciated,

Thanks.

[1] - https://www.semicon.sanken-ele.co.jp/sk_content/str-y6456_ds_en.pdf

[2] - Best options for limiting input ringing with ceramic capacitors

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  • \$\begingroup\$ Welcome. Repair questions are off-topic on this website. We would need to have it in our hands with a lot of test equipment, and that is not possible. Try DIY.SE.com. \$\endgroup\$ – user105652 Jul 9 '20 at 2:05
  • \$\begingroup\$ Thanks for the clarifications, @Sparky256; but, I'm not asking a repair question. Was just curious as to what conditions a ceramic capacitor would blow up. \$\endgroup\$ – ewong Jul 9 '20 at 2:33
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Ceramic capacitors (expecially older types) suffer from micro-cracking. Any mechanical or thermal stress can cause them to crack internally allowing conductive parts to meet, where they should be isolated.

Also a high voltage can cause dielectric breakdown, where the internal insulator fractures or carbonises. This can be caused by a really large static discharge or a lightning strike.

Both of these the initial failure is not the cause of the visible damage. The fault causes the capacitor to become a conductor and the resuting high current from the power supply causes the sudden obvious failure.

The capacitor stops being an insulator and instead becomes a sort of resistor. Power dissipation in that resistor is the actual cause.

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  • \$\begingroup\$ Thanks for the information, Jason! Appreciate it very much. \$\endgroup\$ – ewong Jul 9 '20 at 2:34

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