I have TVS diode in parallel with circuitry for ESD/lightening/transient voltage protection. Say I receive an over-voltage condition which is beyond the specifications of the part, what happens internally? I assume that the part is unable to dissipate the heat generated internally fast enough which causes a break down in the junction doping resulting in an open circuit (like a blown fuse). Is this an accurate assessment of what happens? Could an out of spec transient condition ever cause a permanent short instead?
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There's no heat until there's current. Of course, when device features are very small, it doesn't take very much current; localized heating is proportional to localized current density. When there's enough current density and enough heating, the dopant atoms begin to diffuse through the substrate. This destroys the integrity of the structure of the affected PN junction region, so that you no longer have the desired electrical properties.
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\$\begingroup\$ Does too much heat in a diode reliably result in an open circuit condition or a closed circuit? \$\endgroup\$– Joel BCommented Jun 20, 2011 at 14:30
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1\$\begingroup\$ @Joel B - I wouldn't want to rely on either condition. Even if the initial failure mode is to turn into a conductor, if you put enough current through it, you could probably vaporise it and end up with an open. \$\endgroup\$– JustJeffCommented Jun 20, 2011 at 22:13