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My monitor's backlight died (can still see screen with torch).

Found a blown ceramic disc capacitor. The problem is it's partly exploded over the values. I'm hoping people here might have a better guess.

The monitor is the Philips BDM4065UC. The capacitor is off the power board. Capacitor

And for reference here are pictures of the board. Can't upload them here due to 2MB filesize. The capacitor is C9818 it's in the middle top of the top of the board picture and the middle right of the bottom of the board.

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  • \$\begingroup\$ It's either 220 pF or 2200 pF rated at 1 kV. Look for another similar one on the PCB that you can show if you resized the image in paint etc.. \$\endgroup\$ – Andy aka Apr 2 at 8:01
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That part appears to be a 220pF 1kV capacitor1. These are readily available (e.g. from here2). These don't typically burn up on their own3 though so there's likely an additional fault on the board that caused this cap to fail.

You should also take a look at this question which is very similar to yours.


1 22 is the multiplier, 1 is the number of zeroes after it, K is usually the tolerance (probably ±10%), value is in pF. Thus 221K == 220pF ±10%.

2 The part links go to Mouser (my preferred distributor) but Digikey, Newark, or RS are also reputable suppliers. I'd avoid purchasing any parts from the likes of ebay, Amazon, or Ali Express.

3 Edit: Having taken another look at the photos you provided it looks like that the failed cap is part of an overvoltage protection circuit. I can't read the markings on the diode on the bottom of the board to determine precisely what it is (type, voltage, etc.)3.

Suffice to say it looks to me like a voltage spike came in and (based on the lack of obvious damage elsewhere on the board) the protection circuit did its job and cut power to the board (using the MOSFET Q98014). The spike, however, exceeded the voltage rating of the bypass capacitor and caused it to burn up.

My guess is the rest of the components are fine though Q9801 may also have been damaged. Cleaning both sides of the board around where it failed6 and replacing the capacitor and maybe the MOSFET should get things working again. One thing that you should probably do is choose a replacement capacitor with a higher voltage rating7 such as one of these. If you're not already doing so make sure your equipment is plugged into a power strip with transient voltage surge protection; that'll prevent a lot of damaging spikes from getting through.

3 Based on the questioner's comment, the diode is marked A27 which may be a DAM1MA27 or DAM3MA27 27V zener. That voltage seems odd though given that it's on the mains (high-voltage) side of the circuit. I've probably misidentified it.

4 Per the questioner's comment, Q9801 is a Toshiba K13A65U which has been discontinued by the manufacturer. It's hard to choose a suitable replacement without doing a detailed analysis of the circuit. The manufacturer recommends either the TK11A65W (datasheet, Mouser) or the TK380A65Y (datasheet, Mouser) as possible replacements. Of the two, the TK11A65W would be my choice given its lower on resistance and higher current rating. When choosing a replacement, the main parameters to try to match are probably the pinout, Vdss, Vgss, Id, V(br)dss, Vth, and Rds(on) with Qq being less important. The part found by the questioner (TK155A65Z, datasheet) also looks like a decent option. Honestly, choosing an appropriate replacement is a whole question unto itself5.

5 I'd encourage you to post such a question.

6 90% denatured or isopropyl alcohol should do fine for cleaning.

7 In my experience, monitor manufacturers are notorious for combining the lowest-speced parts possible with poor thermal design.

Top of board Bottom of board
Top of board Bottom of board
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    \$\begingroup\$ @DavidC.Bishop Can't be sure (what with the hole where the number was) but the kerning on the characters makes anything but 221K seem unlikely. \$\endgroup\$ – Alex Hajnal Apr 2 at 9:15
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    \$\begingroup\$ 22 is the multiplier, 1 is the number of zeroes after it, K is usually the tolerance (probably ±10%) value is in pF. Thus 221K == 220pF ±10%. \$\endgroup\$ – Alex Hajnal Apr 2 at 9:23
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    \$\begingroup\$ @DavidC.Bishop The spike would have come in through the power cord from the building mains. \$\endgroup\$ – Alex Hajnal Apr 2 at 9:25
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    \$\begingroup\$ @DavidC.Bishop If the Mouser post and packing charges are a bit high for such a small order, there are other reliable suppliers such as RS Components who might charge less. \$\endgroup\$ – Andrew Morton Apr 2 at 9:41
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    \$\begingroup\$ @DavidC.Bishop Hard to say without doing a detailed analysis of the circuit. See the note in my answer re: Q9801 for my opinion. \$\endgroup\$ – Alex Hajnal Apr 2 at 19:05

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