My adapter broke, so I started my repairs by measuring the output voltage; it was 18V as it's supposed to! So, I figured my laptop charging cicruit was broken, but when I opened the adapter I saw the 400V, 120µF capacitor after the main transformer was broken. I replaced it and my laptop was alive again.

I suppose this has something to do when applying a load, but I cannot understand how the capacitor acts in these cases and give me a good voltage reading even though it is blown. Got any good explanation for this?


Yes, the voltage across a capacitor with a load and without a load is vastly different.

What you are seeing there is the rectified AC signal, which is very very rough, just about making the power supply work. The capacitor filters and smooths that rectified AC signal: enter image description here When the capacitor is faulty, with no load, it can store a little bit of charge and the green line remains fairly smooth. As soon as you apply a load the capacitor is no longer able to hold enough charge, and so the green line more closely follows the blue line. Because of that there then isn't enough power to run the power supply.


In general you can't assume that a failed component will cause a failure of the overall module.

In your case:

  • Your multimeter can only give you one value (a sort-of average), it can't tell you whether there is a fast variation in the voltage, which is likely the case.

  • You say you measured the output voltage, but with what load? Probably not your laptop in charging mode, as you stated that the adapter broke.

  • \$\begingroup\$ I just plugged my multimeter on to the power jack, so no load at all \$\endgroup\$ – chwi Jul 11 '14 at 10:10
  • \$\begingroup\$ So that does not prove much, does it? \$\endgroup\$ – Wouter van Ooijen Jul 11 '14 at 10:30
  • \$\begingroup\$ Well, that was my question wasn't it? I am wondering why it behaves differently when applying a load. I think you are being unreasonable here, but I am glad Majenko actually answered without being rude about it \$\endgroup\$ – chwi Jul 11 '14 at 11:13
  • \$\begingroup\$ My point is that the good voltage reading you got does not mean anything. It does not even mean that the voltage is really good. \$\endgroup\$ – Wouter van Ooijen Jul 11 '14 at 11:52
  • \$\begingroup\$ That is why I asked, and thank you for this information. \$\endgroup\$ – chwi Jul 11 '14 at 12:06

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