Cause of diode failure in step-down switching mode power supply

Question

In a basic switching step-down LM2675 voltage regulator design (as shown in the picture from the datasheet below), what could possibly be the main cause of a short-circuited diode failure (D1 in the picture)?

A little background:

The company I am working for designed a special power supply as part of a larger product a few years back, that accepts 24V as input and uses an LM2675 to produce 5V output. The type of the diode is MBRS340 (Shottky, 3A, 40V).

Some devices (with these power supply units) have been returned by the customer and all units have the same fault: All Shottky diodes are short-circuited.

Additional to the circuit in the picture there is also a poly fuse to limit the output current to 0.9A, so a secondary short-circuit should not destroy the diode. Usually the output load is less than 100 mA.

The strange thing is that the rest of the circuit (on the 5V part) does not get damaged, so it remains working after replacing the faulty diodes.

The waveform across working diodes looks like this while the input voltage is 24V, so no unusual high voltage spikes visible.

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Waveform:

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Waveform zoomed in:

_{Original image source}

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Do you have any experience with these kind of faults?

Answer 1

After repairing a returned unit, take a very careful oscillogram across the Schottky with a short 10:1 scope probe (directly connect the tip and use a very short ground wire) and no bandwidth limiting on your good quality (100MHz or better, digital storage preferred) scope.

An example of a short probe, borrowed from elsewhere on EE.SE:

What you're looking for is high voltage (a spike, or ringing) on the leading edge of the voltage waveform. Any time you switch an inductive load, there's the potential for high voltage - unless you carefully look for it (i.e. with a short probe and a good scope) you won't see it. Your layout will have a great influence on whether or not there will be nastyness across the Schottky, so measure to be sure.

If you do end up seeing something ugly, you may need to introduce an RC snubber across the Schottky to keep the diode from avalanching.

Answer 2

This is just a sanity check...

Your output power is stated as 5 W and because the switcher operates at 260 kHz, the energy transferred per cycle is: -

\$\dfrac{5\mathrm{~W}}{260,000} = 19.2\mathrm{~\mu J}\$.

To liberate this energy from a 47 uH inductor requires a "charge" current of: -

\$\sqrt{\dfrac{2\times19\times 10^{-6}}{L}}= 0.905\mathrm{~A}\$.

This doesn't sound too much for a 3 A diode capable of non-repetitive peaks of 80A.

The switcher is quite capable of switching down to 0% duty cycle so I can't see this has any bearing on things. The polyfuse might be an issue - were these found to be OK? Did you test any of them?

I suspect you may have a bad batch of diodes or there are some other considerations you are not aware of. Last week I had several 400 V diodes go pop on me and they were under no stress at all. Maybe it's a bad week for diodes!

Answer 3

Shottky diodes leak more than standard diodes.Your design uses a 40V type which on the face of it is OK.The 40V diode you chose will have a lower foward drop than say a 1ooV diode so you would think that things are fine and what is the point of having a diode that has a volt rating higher than the chip.If you take a shottky diode and apply a reverse voltage to it thats below its rating from say a lab supply you will see a small reverse leakage current .If you externally heat the diode by whatever means you will see the leakage current rise .According to semiconductor physics it will roughly double for every 10 degree rise.The extra dissipation due to this increased leakage current at elevated temperatures can be significant .You can now see that there is a mechanism where the diode could thermally runaway and die.Sure as temp increases the Si foward drop goes down but the bulk resistance goes up .You must take thermal runaway seriously when using shottkies.If you are old enough to have cut your teeth on germanium transistors you will be fine .Remember that the band gaps are in the same ballpark.So for your problem is your diode cool ? .If you can heat up your SMPS externally by whatever means and the diode dies prematurely you will have a clue .Remember that the heatsinking or thermal impedence of the diode will have to be lower than you think .Using a higher voltage diode could be a quick get out of jail card .

Answer 4

It will be overvoltage spikes that are causing the Schottky shorts (many years experience in SMPSU). If you look at the waveform with a higher bandwidth scope you will seen them. 100 MHz not really good enough, suggest 500 minimum to see all. Use snubber as suggested or higher voltage rating.

I realise this is very late, but may help others, who find the thread as I did.

Answer 5

Your input is labeled as 8V to 40V. Your shottky diode will see the input voltage during switching. If your input can really go to 40V then your schottky should be 60V rated or at least 50V rated.

The switch current limit is also 1.5A. Seems like your customer must be putting more than 24V at the input. Shottkys need a healthy voltage margin or they will reverse leak and get hot.

Answer 6

I think the issue its with your poly fuse 0.9A is probably the holding current of the PTC fuse. The actual tripping current for might 2 or 3 times higher as seen on this datasheet Also its important to note on the time to trip they specified a fault current of 1.25 which is 5 times greater than the hold current and its what I have normally seen as a guideline for the trip current. The most likely scenario is that with a 0.9A holding current your fault current will be 4.5A however the LM2675 will be limiting the current using its internal current limit to 2.2A max which will never trip the PTC. So under a fault the diode will be dissipating the most of the power and if there is no good heating sinking for the diode it will most likely fail while the LM2675 will remain unaffected because of its internal thermal shutdown