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I have a PCB that has >10 ICs intended to be powered by a 3.3V supply. Unfortunately, 12V was applied to the circuit and the board now exhibits a short circuit. The offending power supply is current limited so that now when it is attached to the damaged PCB, the supply limits itself to < 100mV.

My question is this: In situations like this where the supply limits itself after a short develops, removing the overvoltage condition, is it typical for only a single IC to have failed (i.e., the weakest link) at which point the failed IC protects the other ICs on the board, or should I be looking for multiple device failures?

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    \$\begingroup\$ Either can happen. \$\endgroup\$
    – user16324
    Nov 6, 2017 at 17:26
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    \$\begingroup\$ It could honestly be either to help you fault find you may want to set a power supply to a low voltage but let say 3 or 4 amps flow in the circuit. The faulty component is probably getting hot. \$\endgroup\$ Nov 6, 2017 at 17:27
  • \$\begingroup\$ The most important factor here is a regulator. When the board has one, chances are it fried leaving everything after our ok. If there is none, likely most chips that for 12V applied got fried \$\endgroup\$
    – PlasmaHH
    Nov 6, 2017 at 17:42

2 Answers 2

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If 3.3V was the intended supply voltage, I'm guessing that you didn't have an in-line regulator to many of these ICs. I try not to design circuits without some sort of regulator for exactly this reason. Also, you can design in Zener diodes with an inline fuse, or other forms of protection to try to prevent this.

However, it happened, and for finding damaged components, nothing beats a thermal camera. In this situation especially, you've got components that are trying to pull too much current. They will be hot. You can pull them off one at a time until your current consumption stabilizes at what you would expect.

What kind of thermal camera? I really like my Fluke TiS ($2k from Newark), but you can also get models that are intended to be used for your phone for less than $350.

Shorts from supply rails to ground are more difficult. You can find these typically by pushing 1A through the supply rail, then probing different vias to ground on the board and looking at the voltage differences around the board. This will help you find where on the ground plane the current is entering.

Good luck!

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  • \$\begingroup\$ Good suggestion, if only I could convince my boss to buy a thermal camera. Maybe he'll be OK with a phone version. I hadn't realized they were so readily available at such a low price. \$\endgroup\$
    – kdm14
    Nov 6, 2017 at 18:37
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Mike Barber gave some excellent suggestions. However, you may not have a thermal imaging camera available.

In that case, the decidedly-low tech technique simply uses Isopropyl alcohol. Spread a thin layer of alcohol over the entire board surface. Set your power supply to the rated supply voltage (3.3 Vdc) and set the current limit to 500 mA or so. Connect the power supply to the circuit and observe all of the devices.

Increase the current if needed until you see the alcohol evaporating from one device more than the others. Add more alcohol and ensure that the devices is indeed running warm.

Remove that device that is running hotter than the rest and repeat the test. You will eventually be able to find all of the components that can't handle your 3.3V power supply rail.

This technique worked well for several decades for me, until I finally purchased my first Flir thermal imager. I still use this technique if I am in the field and don't have access to a thermal imager.

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    \$\begingroup\$ Another tool is thermochromic liquid crystal sheets or paint, both of which can be had for under $10. \$\endgroup\$
    – user71659
    Nov 6, 2017 at 18:39
  • \$\begingroup\$ Thanks for the suggestion about alcohol. I had tried using my finger, but a lot of the components have exposed pads on the bottom to help dissipate heat. Maybe the alcohol will be more sensitive. \$\endgroup\$
    – kdm14
    Nov 6, 2017 at 18:39

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