So, I’ve got a laptop screen that’s giving up on me. 2/3 of the screen runs alright but the 1/3 on the left edge is acting weird. Half of the broken section displays an image but the image smudged and weird, while the other half is just dead. I opened it up to see what’s up and, lo and behold, a wee tiny capacitor is missing (I know it’s a capacitor 'cause I looked of the board marking, C248).

Now I’m wondering, since ordering a single capacitor just for fixing this screen is not worth the effort, can I just… put some solder in there to at least get power to where it needs to go? I know it’s definitely not ideal but, this is an ancient laptop. Putting in the effort to fix it perfectly is not exactly a great value proposition. What I want to know most is, will the screen be damaged if I do this, or what could go wrong if I do this?

I’m pretty new to DIY electronics fixing so sorry if this is a stupid question. Thanks in advance y’all. Cheers!

  • MajorHavoc@programming.devEnglish
    223·
    6 months ago

    The purpose of a capacitor is to build up charge slowly, and release a larger charge suddenly, under very specific conditions. The specific conditions are determined by the exact model of capacitor.

    A direct connection at a point where a capacitor once was won’t do the job. It’ll both provide too much power (compared to zero) and too little (compared to the burst of extra power on release).

    Shorting a spot where a capacitor once was will almost certainly damage other components.

    If you’re sure there was a capacitor there, and can find out which one, then you want to install (solder in) a fresh capacitor of the exact same make and model.

    • SmoothIsFast@lemmy.worldEnglish
      9·
      6 months ago

      Shorting where a capacitor was won’t provide any power, it will most likely cause a short circuit, potentially resulting in sparks and flames.

      • MajorHavoc@programming.devEnglish
        1·
        6 months ago

        I think you’re trying to clarify that shorting where a capacitor goes won’t provide any useful power.

        But electrons will certainly flow and do things.

        In useful terms, there will be no useful power. In safety terms, there will be some dangerous unexpected-by-the-designer-of-the-circuit power.

        • SmoothIsFast@lemmy.worldEnglish
          2·
          6 months ago

          Sure, it won’t provide any power to the circuits that are supposed to get power was the intent of my comment.