0
\$\begingroup\$

The common wisdom is that Zener diodes are suited for only very low power surge protection because of their low current capacity.

But what about where it was used as a sacrificial device for clamping transient spikes? Surely there would be a very brief period (ie in the order of microseconds) where it shorts completely across the pn junction and the internals become molten but remain temporarily physically intact.

Other typical surge components seem to rely on a massively larger transient current response - for example, the leads on MOVs and the traces on the PCB would normally be capable only of a few amps yet they're usually rated in thousands of amps.

So what kind of instantaneous current handling could you expect from a Zener diode?

Edit: In this particular application, the zener is already used as a voltage regulator. TVS diodes don't seem to be used for steady state voltage regulation - are they unsuitable or is because of cost?

\$\endgroup\$
  • 2
    \$\begingroup\$ Have you never heard of a TVS diode? Semiconductors usually fail short anyways...unless you run so much current through them you expect them to melt or explode instead. \$\endgroup\$ – DKNguyen Aug 26 '19 at 13:49
  • \$\begingroup\$ In my previous company we had a zener triggering a triac to crowbar the voltage rail and thus breaking the fuse. This clamped to a lower voltage during the event than just a straight zener. Worked well. \$\endgroup\$ – winny Aug 26 '19 at 13:53
  • \$\begingroup\$ Even if they did explode or melt, wouldn't there be a few nanoseconds where it was shorted? I've added more context to my original question. \$\endgroup\$ – Mikey Top Aug 26 '19 at 14:12
  • \$\begingroup\$ Are Zener diodes guaranteed to fail open in overcurrent situations? \$\endgroup\$ – Scott Seidman Aug 26 '19 at 15:33
2
\$\begingroup\$

the device your looking for is a TVS diode, it is rated to handle spikes of hundreds of amps, Zeners tend to fail shorted if you overload them,

The current ratings are realistic, with only a few tens of mm of copper, it does not take much voltage to get significant currents.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ I've updated my question with more context. \$\endgroup\$ – Mikey Top Aug 26 '19 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.