This I created as a follow up for the
"Designed to fail" , where I could not add new response, because the last one was mine. But as this post deserve really separate insertion, with all the "new entry" icon (there were a few people looking to learn about the problem), it is quite long, so I have chosen this way...
So finally it seems we have found the prime suspect: It seems to be the "freewheeling" Schottky barrier diode entering a thermal runaway, going (maybe temporary) effectively short circuit, which killed the IC as a result (and the diode itself could have still survived it).
It was not yet proven (= check with an "adequate" diode, if it operates well), but the hypothesis show a weak spot (and consequently quite common design error; I had to learn it the hard way myself as well...) of Schottky diodes:
The diode was the B0540W from DIOTEC. It's absolute maximum rating say 0.5A/40V, so at first glance it should be OK. But it isn't:
This type exhibit quite a high reverse leakage. So high, ~70% of the total diode power dissipation were caused by the leakage alone. Now the leakage have one important ugly habit: I doubles for each ~10degC the junction temperature increases.
Given the small package, so high thermal resistance too the ambient, just the leakage caused the dissipation so high, the diode won't cool down and in fact heats up faster and faster. Well, until the leakage caused heavy current overload of the IC (so an effective short circuit) and the IC failed.
The IC in the full circuit is supposed to be a current source and tolerate even the output short circuit, but it need the coil, the diode and the input capacitor to work there correctly, as the only means of current control is by switching ON and OFF. And for that it need the coil, where the current can not rise significantly before the IC could respond. But with the diode short, the inductor was not in the circuit, so the current was limited only by the IC switch Ron (about an Ohm, so ~12A).
And beside the direct transistor current overload, this extra current caused way higher dI/dt slopes in the critical loop, exposing the IC to severe overvoltage as well.
It is hard to tell, what is worse, if the high current or voltage, but both were caused by an inadequate diode and yield the IC destruction.
Now how to select an adequate diode?
You have to look for thermally adequate package. As the AbsMax is usually valid for 25degC, but you will most likely operate it higher, you have to calculate with the power dissipation and thermal properties, as the current would be limited by the maximum junction temperature.
Now when going to non-Schottky, you have to count with about 1V forward voltage drop to calculate the power dissipation and usually that is all. The non-Schottky have usually the leakage in order of uA even at the maximum operating junction temperature, so are of no real power dissipation contributor. But these diodes tend to be rather slow, so you have to look for "Ultrafast" diodes, with the reverse recovery time up to about 50ns (real maximum).
When the 1V yield to high power dissipation, you have to go for the Schottky. But with many of these, the reverse leakage is by far not negligible (in fact with many types it is the associated power dissipation and not the real electrical breakdown, what limit the max voltage rating), so you have to look for the types, where the leakage can not cause more than 10..20% of the power dissipation room available at expected equipment operating temperature (not the dissipation you desire to operate the diode), maximum rated junction temperature and expected thermal resistance. Higher figure mean asking for runaway troubles, don't forget only 10degC double the leakage...
Achieving this need either very low leakage, or a good heat sink even for low currents.
So check mainly following datasheet parameters: Leakage at maximum temperature, thermal resistance figures, maximum rated junction temperature.
For the current capability, calculate the power dissipation
Ptot = 0.5V*RequiredLoadCurrent + MaxInputVoltage*LeakageAtMaximumTemperature
Calculate the junction operating temperature:
Tj = Tmaxambient + Rthja*Ptot
and compare with the max rated Tj, if higher, choose other diode.
The calculation takes a lot of margins, but as many parameters you may only guess (the Rthja, it strongly depend on the PCB) and the leakage is so sensitive on the resulting temperature, such margin could be eaten by the inaccuracies pretty quick...
