0
\$\begingroup\$

I have on my circuit an integrated circuit (IC) that has been destroyed by some event. After the event, the component's Vcc pin was shorted to GND inside the IC.

After measurement, it appears that there is a surge spike on the Vcc (5V) net. Here is a screenshot of the measure I made on Vcc:Scope sreenshot

It shows that the spike can rise up to ~15V, but I have seen it over 20V. The half-wave last for approximately 1 µs to 1.5 µs.

The datasheet of the destroyed IC gives an absolute maximum value for VCC of max 6V. This is largely exceeded here.

To protect the ICs, I have thought to use a TVS diode such as SMAJ5.0A/CA (5V stand-off voltage). But such a diode have a minimum breakdown voltage of 6.4V, typical 6.7V, which is sill over the absolute maximum ranking for VCC of my ICs.

Will this protect the ICs, despite that the fact that we are still a bit higher that the spec ?

Do you know know an other mean of protection for the ICs ?

\$\endgroup\$
8
  • \$\begingroup\$ Show a circuit. \$\endgroup\$
    – Andy aka
    Mar 9, 2016 at 10:54
  • \$\begingroup\$ There are tons of TVS diodes with appropriate safe tension, why stick to the SMAJ5.0A/CA ? \$\endgroup\$
    – MaximGi
    Mar 9, 2016 at 10:57
  • \$\begingroup\$ @MaximGi: Yes but for a stand-off voltage of 5V, the ones I have found have ~6.4V breakdown voltage. SMAJ5.0A/CA was just an example. If you have a TVS in mind that fit the purpose, that will answer my question ! \$\endgroup\$
    – Antoine
    Mar 9, 2016 at 11:00
  • \$\begingroup\$ @MaximGi: Or should I use a TVS with lower stand-off voltage ? \$\endgroup\$
    – Antoine
    Mar 9, 2016 at 11:01
  • 3
    \$\begingroup\$ Surely you should be finding out where the "some event" comes from? Why is the power supply regulator not helping here? \$\endgroup\$
    – pjc50
    Mar 9, 2016 at 11:27

1 Answer 1

5
\$\begingroup\$

Trying to clip this mess after the fact is the wrong way to start attacking this problem. The real problem is that there is loads of crap on the power supply. Find out why that is happening and deal with it at the source.

Is this caused by ground bounce? If so, figure out why and address it. In that case it will be affecting other things too.

Is it caused by a bad power supply? If so, fix it.

Is it caused by improperly handled inductive kickback? If so, redesign the offending circuit.

Is it caused by an external circuit sharing the power supply? If so, consider adding a separate (robust) power supply for the external circuit.

In any case, you need to pop up a couple of levels and look at the bigger picture. You are not ready yet to just treat the symptom at the point of consumption.

\$\endgroup\$
5
  • \$\begingroup\$ This was done. The mess is due to ground bounce (due to inrush current inside the GND plane). But there is no way to to address it. As said higher in this thread, this appends when a capacitive load is connected (to other supply rail but same ground). But the load is provided by the customer, so we have no way to change that. I have though of separate ground, but then we will go out of the common mode range of the analog front-end. I think we have to live with this mess and clamp it some way. \$\endgroup\$
    – Antoine
    Mar 9, 2016 at 11:55
  • \$\begingroup\$ The event is rare, I don't care about the transient noise it can cause or whatever. I just don't want my ICs to spark. \$\endgroup\$
    – Antoine
    Mar 9, 2016 at 11:58
  • \$\begingroup\$ Sounds like a great case for a "not quite separate ground"; linked to the dangerous ground by some combination of resistor and inductor so it's at the same DC potential but has spike suppression. \$\endgroup\$
    – pjc50
    Mar 9, 2016 at 12:15
  • \$\begingroup\$ Could consider fully isolating your circuitry from that "unpredictable" ground. \$\endgroup\$
    – rdtsc
    Mar 9, 2016 at 12:57
  • \$\begingroup\$ Added anther suggestion to the list of excellent suggestions: "Is it caused by an external circuit sharing the power supply? If so, consider adding a separate (robust) power supply for the external circuit." \$\endgroup\$ Mar 9, 2016 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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