8
\$\begingroup\$

TLDR:

What are possible failure sources for a schottky diode? Overheating, overvoltage, overcurrent (leading to overheating) anything else?

Long version

I am currently measuring our new prototype device. It works fine, does not draw more current than expected and stays relatively cool (hottest point on the PCB is 50°C at the buck converter.)

In front of the buck I have a schottky diode for reverse voltage protection:

enter image description here

I left the device running for some minutes, current draw on the 3V3 rail was below 500mA, 24V current must have been below 100mA. Then I saw a small lightning, like ESD, and the device stopped working (no current draw). On closer investigation I found a small hole in the middle of the schottky diode package, and the diode was open circuit.

I would rule out ESD, as the device was lying on the desktop and nothing touched it. The device was powered by a lab power supply. Any ideas why the component failed?

\$\endgroup\$
7
  • \$\begingroup\$ Reverse voltage breakdown could also be a possible source of problems. For that to happen, your device would need to increase the voltage behind the diode, so that'd be a bit strange. \$\endgroup\$
    – Arsenal
    Commented Aug 6, 2020 at 11:04
  • 1
    \$\begingroup\$ @Arsenal would it be strange? Because I see a large inductor in there, and inductors + changing currents = voltages. Inductors + Capacitors + periodic excitation at the right frequency range = resonance... \$\endgroup\$
    – Marcus Müller
    Commented Aug 6, 2020 at 11:13
  • 1
    \$\begingroup\$ @MarcusMüller well, I guess that could happen. But then probing with a scope should make that problem visible? \$\endgroup\$
    – Arsenal
    Commented Aug 6, 2020 at 11:15
  • 1
    \$\begingroup\$ I have not analysed in great detail, but the common mode choke looks like it does not perform any filtering on the 24V - Gnd circulating currents produced as part of normal operation. It looks useful to stop EMC getting out onto the supply rails generally, but if you are expecting it to provide an LC filtering arrangement, you would be better to have a single L and keep the GND continuous. \$\endgroup\$
    – elchambro
    Commented Aug 6, 2020 at 23:34
  • \$\begingroup\$ The Schottky diode is under rated both in current and peak reverse voltage. A SMD 1N4007 would have survived. \$\endgroup\$
    – user105652
    Commented Aug 7, 2020 at 1:35

4 Answers 4

Reset to default
21
\$\begingroup\$

Any ideas why the component failed?

The maximum repetitive peak forward current for the BAT54J is specified as 300 mA in the data sheet. Take note: -

enter image description here

Simple math about the converter

To achieve 3.3 volts on the output of the buck converter requires a duty cycle of about 0.1375 assuming the converter operated in CCM. In other words: -

$$\text{Duty} = \dfrac{3.3\text{ volts}}{24\text{ volts}} = 0.1375$$

If the load was 500 mA, the power out would be 1.65 watts and that power has to be passed through the diode from the 24 volt rail in 13.75% of the time.

So, the average current from the 24 volt rail is 1.65 watts ÷ 24 volts = 69 mA but, given that this average is delivered in only 13.75% of the time, the peak current would be: -

$$\dfrac{69 \text{ mA}}{0.1375} = 502\text{ mA}$$

And this exceeds to peak stated in the data sheet.

Just rough and ready calculations of course.

A proper estimation would need more information such as: -

  • The duty cycle of the buck converter because if it's working in DCM the current peak demanded by the converter may be several times that estimated if the duty is significantly smaller.
  • The ESR and ESL of the 10 uF capacitor (it could be a show-stopper). Cheap capacitors are not going to be cost effective in this part of the circuit and can vastly increase the diode current.
  • More information about the CM choke for instance, what self-resonant frequency it has and how closely coupled the windings are. This can also be a show-stopper.
\$\endgroup\$
4
  • 2
    \$\begingroup\$ If the front-end filter does its job ok, then the diode current should be fairly flat, close to dc with some ripple. However, considering the unusually-large common-mode inductor value of 4.7 mH, I would recommend to take a look at the voltage at the diode cathode while the Schottky is temporarily replaced by a 1N400x. An oscillation with a lethal voltage for the diode is likely to happen. Damping is therefore necessary and the BAT54 (this is not a power diode) be advantageously replaced by a MBRA140 for instance. \$\endgroup\$
    – Verbal Kint
    Commented Aug 6, 2020 at 12:41
  • \$\begingroup\$ @VerbalKint I take your point entirely but there are practical considerations such as the reliability of the 10 uF capacitor at high frequencies and its ESR or ESL. We don't know anything about the choice of that capacitor. We also don't know how good the CM choke is regarding the coupling or whether it's actually any good at the buck converter switching frequency. With a winding inductance of 4.7 mH (unusually large as you say), it may be well beyond its self-resonant frequency and ineffective. It wouldn't surprise me if it's actually a 4.7 uH CM filter. Possibly worth a simulation.... \$\endgroup\$
    – Andy aka
    Commented Aug 6, 2020 at 13:38
  • \$\begingroup\$ The buck converter could be operating at close to BCM and the peak current it draws may be rising to 1 amp over the on-part of the duty cycle. I can see a peak diode current of over 300 mA if the 10 uF had an ESR of 1 ohm and the CM choke was 4.7 uH. If the ESR was 2 ohms, the diode current would be peaking at a little over 450 mA. I guess the OP will have to deliver some information. \$\endgroup\$
    – Andy aka
    Commented Aug 6, 2020 at 13:44
  • \$\begingroup\$ Of course, if the unit is operating in DCM the peak inductor current could be higher than 1 amp because the duty cycle might be somewhat less than 0.1375. If it's running around half this duty cycle, the peak inductor current could be 2 amps and the diode current could peak around 800 mA \$\endgroup\$
    – Andy aka
    Commented Aug 6, 2020 at 14:10
6
\$\begingroup\$

The BAT54 not very conservatively rated for that application, both in current and PIV (only 30V). Without even picking up my calculator I would have started with the SMT version of a 1N5819. I'd also feel better with (C31) much higher. We don't know what the normal mode inductance of that choke is but it's not zero.

However, that does not really explain the failure mode (open with destruction of the package) which sounds more like a short circuit was caused. The diode going short should have just disabled the reverse polarity protection. You may find that one of those 5 capacitors has shorted or has an intermittent short due to cracking.

\$\endgroup\$
1
\$\begingroup\$

I think i had to explain, how a buck-converter works! There is a high-side FET inside which switch the input to the output for a specific time (Dutycycle). So if there were 500mA at the output, there are also 500mA at the input, at the time, the switch is on. And because we have to charge the choke at the output, we need some more ampere! So it is important, how much ripple you have on the output choke. The same ripple you would have on the diode at the input (for the time the switch is on). If you would touch the diode when you start, you would feel very fast how hot they gets! In this case the diode with 300mA repetitive peak forward current is much too small.

\$\endgroup\$
0
\$\begingroup\$

You really need to attach oscilloscope (properly) to the various input nodes - both before and after the choke - and see what happens there. The input capacitors to the switching regulator are way too small, and the choke you chose is way too big - it will have microhenries of differential mode inductance (the "unwanted" or "non-ideal" inductance) and it may be generating enough inductive kick during transients to be quite destructive. It's a wrong part for the job - you don't need millihenries of differential inductance in this application.

\$\endgroup\$

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