I had a failed Chinese charger. When I opened it, I saw a melted wire of solder that caused a short circuit, burning and damaging the board. The wire was thin and printed upon copper so when melted it touched other solder wires.

This is the component side of the charger with the damage marked with a red circle:

enter image description here

This is the soldering side:

enter image description here

The arrows point to other solder wires and the long line is the path of the melted solder wire which was removed. The solder wires are placed on copper.

  • \$\begingroup\$ It's highly unlikely that the tinned trace was the root cause of the failure. More likely the missing diode D3 failed short (that's what they do, for a multitude of reasons), overheated (look how brown the PCB is around both of its pins) and melted the trace as a consequence. If a short between the two tinned traces was the original cause, the diode would be shorted out of the circuit and would never heat up so much as to brown the board. \$\endgroup\$
    – TooTea
    Feb 10, 2023 at 14:28
  • \$\begingroup\$ @TooTea Another thing: the led lamps sold in my area are also low quality and made from cheap components imported from china and it usually fails when a diode is burnt causing an open circuit not a short circuit, I expected the diode to do the same when failed in the charger or the led diode is different? \$\endgroup\$
    – dev65
    Feb 10, 2023 at 14:50
  • \$\begingroup\$ The heat probably desoldered that diode so it fell off. Perhaps you can still find it in the plastic box you took off. Ultimately, a diode typically fails open, but that doesn't mean it can't first undergo breakdown and let a whole lot of current through before it gets blown to pieces. Or the output capacitor has failed. Or perhaps the output of the charger got shorted, which caused excessive current through the diode in the first place (these cheapo power supplies typically have very little in the way of overcurrent protection). \$\endgroup\$
    – TooTea
    Feb 10, 2023 at 14:50
  • 1
    \$\begingroup\$ ROFL I struggled with how one drawing is 90° to the other. \$\endgroup\$ Feb 10, 2023 at 17:55

2 Answers 2


Such tin on the trace is for making the current capacity larger.

Normal routing traces on cheap PCBs have 1oz thickness copper. With the given width, to increase the current capacity, the designer may need to make the current capacity larger. If the trace copper exposed to the out without soldermask layer, in the wave soldering process, such thick lines are made.

The melting temperature of such solder is more than 200°C. Thus, for normal operation, solder traces are safe enough. If it is melt, there is a certain heat source that was already broken before the solder is melt.

Edit - my personal preference

(as several comments are made for more concerns, I add this "Edit" section.)

  • Why is this (whole PCB) in low quality? - I don't expect high quality when using a cheap product.
    • Design costs to meet many standards are expensive.
    • Testing costs for such standards are also expensive. e.g. consider the cost of CAT II or CE certification testings.
    • PCB fabrication and assembly with better materials are expensive.
    • By doing all these, it becomes not a cheap product anymore.
  • Does it (solder on/along the trace) really help current flowing? - Yes, it certainly makes the path resistance smaller and helps overall safety.
    • (normal) Solder conductivity is 9~13% of copper's.
    • The most common thickness of copper in PCB production is 1oz = 34um.
    • 0.34mm thick solder has same resistance as 1oz copper.
    • PCB fabrication cost gets much higher for thicker copper layer.
    • PCB assembly cost is same, no matter whether it has solder-trace or not.

When I am buying a product or reviewing one, I concern more on the price-quality or price-performance ratio, instead of quality only or price only.

  • 1
    \$\begingroup\$ Such a 'hack' is not something I'd ever consider. But an interesting insight nonetheless. \$\endgroup\$
    – Velvet
    Feb 10, 2023 at 12:59
  • 1
    \$\begingroup\$ I wouldn't call that safe enough, I would call it unacceptable quackery. It's about as safe as those multiple cold joints on the picture. This is obviously some cheap arse product which could have been designed with multiple layers but they didn't. \$\endgroup\$
    – Lundin
    Feb 10, 2023 at 13:51
  • \$\begingroup\$ I didn't mention the above design was good. If it is made with multi-layer PCB and better design, it could be no more a "cheap arse product". If one buys a expensive and good quality charger, it will have all the good points as you say. After buying a cheap one and blaming that it's not well designed like an expensive one?! \$\endgroup\$ Feb 10, 2023 at 13:57
  • \$\begingroup\$ Well there are product safety directives and clearance/creepage standards to adhere to. There are also IPC standards for how to solder and assembly electronics proper. None of that is a concern for the buyer of the product. As for if this particular PCB is actually legal to sell or not, I can't tell. \$\endgroup\$
    – Lundin
    Feb 10, 2023 at 15:20
  • \$\begingroup\$ The problem with this is it doesn't actually work; the conductivity of solder is much lower than that of copper, so it has very close to zero effect. \$\endgroup\$
    – Hearth
    Feb 11, 2023 at 2:29

because the pcb's are pulled on a chain through a bath of solder and your looks like it.


"Wave soldering is a bulk soldering process used for the manufacturing of printed circuit boards. The circuit board is passed over a pan of molten solder in which a pump produces an upwelling of solder that looks like a standing wave."

Ofcourse, the soldermask and how it's applied is a factor too.

  • \$\begingroup\$ Might want to mention this happens only if you have no solder mask. \$\endgroup\$
    – DKNguyen
    Feb 10, 2023 at 14:16
  • \$\begingroup\$ @DKNguyen done. \$\endgroup\$ Feb 10, 2023 at 14:18
  • \$\begingroup\$ "Why is there a resistor before the diode? Because someone soldered it to the PCB". This is obvious not what OP asks. \$\endgroup\$
    – pipe
    Feb 11, 2023 at 0:04

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