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I am trying to repair a motherboard that had a shorted ceramic capacitor (1206 package) in position B. The voltage across the pads is ~19V. I put in a 10 μF 25V-rated ceramic cap, and the motherboard could be powered on, but the cap quickly became very hot (75°C after a couple of minutes), even when the board was just plugged in, and not powered up. The smaller ceramic cap in position A is in parallel with the cap in position B, and they seem to be decoupling the power supply for the IC in position C. enter image description here Fundamentally, my question is how to puzzle out the capacitance of the shorted cap (or, alternatively, an explanation of why the 10μF cap might be inadequate). But a secondary question is why the need for two ceramics (of different capacitance) in parallel.

One idea is to just get the datasheet for the IC, and see what decoupling cap values are specified. Unfortunately, googling the markings on the IC ("EN=FG 82W") don't lead to any obvious result. Perhaps someone here is knowledgeable enough to identify the chip by its appearance.

Needless to say, I don't have access to the schematic, and the manufacturer wasn't forthcoming with the needed info.

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  • \$\begingroup\$ Though often used, it is not optimal to mix multiple decoupling cap sizes next to each other. Please see this: electronics.stackexchange.com/questions/320363/… \$\endgroup\$
    – DKNguyen
    Commented Dec 17, 2019 at 0:04
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    \$\begingroup\$ 19V is an unusual voltage. Is this a laptop motherboard? Perhaps it is part of the battery charger or battery voltage regulation? \$\endgroup\$ Commented Dec 17, 2019 at 0:11
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    \$\begingroup\$ Unless you have low loss tangent or ESR parts in the 10 mOhm range, you can expect problems \$\endgroup\$
    – D.A.S.
    Commented Dec 17, 2019 at 0:27
  • \$\begingroup\$ Yes, it's a laptop MB. And this seems to be part of the standby circuitry: the power button only has its 3.3V when this IC is powered. \$\endgroup\$
    – tavr
    Commented Dec 17, 2019 at 1:10
  • \$\begingroup\$ have you measured the voltage across the cap? If it is stable, on 1206 package. X5R dielectric take the maximum value capacitor you can get. \$\endgroup\$
    – user19579
    Commented Dec 17, 2019 at 5:08

2 Answers 2

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It's a common conception that smaller capacitors have a lower inductance and are therefore better at knocking out the really high frequency stuff. There are strong opinions both ways, and it comes down to the exact capacitor and how high your high frequencies are. Some reports of them having their own resonance when combined in pairs... I'm slightly sceptical.

Have you considered whether one of the larger bulk capacitors has died? If so, it would result in large currents and voltage fluctuations in the smaller ones and thus heating.

The topology you have there is a switch mode dc-dc converter. It will be operating in the 100kHz to 2MHz region and likely converting a few amps. The caps will be taking quite a chunk of this current.

The chip is probably something like allegro ARG81801. If you read the section on that's datasheet about capacitor selection you'll see you need a fairly large X7R type, they're recommending 13.8uF minimum.

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  • \$\begingroup\$ Thank you, this gets me on the right track. \$\endgroup\$
    – tavr
    Commented Dec 17, 2019 at 1:20
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Assuming your brand new cap is still in good condition, without internal shorts... if it gets hot this is due to ripple current from the nearby DC-DC chip.

Every capacitor has an internal resistance (Equivalent Serial Resistance or ESR) which dissipates power equal to RI^2 when current I flows through it.

If this cap is at the input of a DC-DC converter then it will take a substantial ripple current.

The solution would be to use a cap with lower ESR, which probably means a higher value in X7R or X5R dielectric.

Note a ceramic cap will work fine at 75°C.

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