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I'm continuing to troubleshoot a broken SMPS.

At this point, I'm just trying to understand what components are broken, working under the assumption that it saw a voltage spike as a result of input ringing.

Here's what I'm seeing when I scope the switch pin (10uS/div, 1V/div):

enter image description here

The FB pin is sitting at 200mV, output is sitting at 800mV, input is steady at 11.8V.

Given this information, is it possible to guess what is broken? Is it the SMPS IC, the inductor, both?

Circuit for reference: enter image description here

Switch for reference: http://www.ti.com/lit/ds/symlink/tps62125.pdf

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  • \$\begingroup\$ Maybe I'm just looking at it wrong. This is the voltage waveform of the SW pin? Is it just me, or am I not seeing any evidence of capacitance? The voltage came through but didn't have to decay? \$\endgroup\$
    – user39962
    Apr 15, 2015 at 4:57
  • \$\begingroup\$ @SeanBoddy can't speak to the bulk of your comment, but can confirm this is the SW pin. \$\endgroup\$
    – kolosy
    Apr 15, 2015 at 5:00
  • \$\begingroup\$ Troubleshooting board level is something I'm not great at. The voltage changed, with respect to ground so the SW pin is not explicitly grounded. Neither is the output. But the voltage isn't being stored - so - bulk capacitance isn't available, but it can't be shorted either. I think your inductor is broken or disconnected. I'll let someone better than me mull that one over, I don't feel sure. \$\endgroup\$
    – user39962
    Apr 15, 2015 at 5:07
  • \$\begingroup\$ I recommend doing more probing. Is this SW signal periodic? At what rate? What do \$V_{IN}\$ and \$V_{OUT}\$ look like during this event? If you have a current probe, what does \$I_L\$ look like? Also: What type of inductor is it? What type of capacitors are you using? There's a lot of interplay in switching converters, and it's exceedingly difficult to tell what's going on just by looking at a single voltage. \$\endgroup\$
    – Zulu
    Apr 15, 2015 at 6:35

3 Answers 3

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If you can reasonably guess that something is broken due to electrical overstress, the most likely candidate is the integrated circuit.

It is typically the power switches and control circuits that break. The most common mechanisms are over-heating and over-voltage breakdown. And especially considering your previous question, which suggests that you had \$V_{IN}\$ overshoot, over-voltage breakdown is quite likely. When semiconductors experience over-voltage breakdown, they suddenly absorb a lot of power (often in a snapping, positive-feedback fashion), which causes damage.

As for the inductor, essentially the only way you could electrically break it would be to over-heat it until it melts. The solder will melt long before that happens.

I suggest you swap out the IC and see if it works again.

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  • \$\begingroup\$ I know this is the most likely solution, intuitively... I still wish I knew why the output capacitance isn't coupled to the SW pin. \$\endgroup\$
    – user39962
    Apr 15, 2015 at 16:58
  • \$\begingroup\$ @Sean, if by "output capacitance" you mean parasitic capacitance on the switch node \$C_{SW}\$, you're right: when the inductor ceases to conduct, you should see some ringing as the energy stored in \$C_{SW}\$ resonates with \$L\$. It seems here that the ringing is very heavily damped. Perhaps the bottom switch has developed a resistive gate-to-drain short (this could explain why the bottom FET isn't turning on), and this resistance is damping the ringing? We'd have to dig deep (much deeper than a single waveform) to find out for sure... \$\endgroup\$
    – Zulu
    Apr 15, 2015 at 17:09
  • \$\begingroup\$ so swapping out the IC had an interesting effect. At first, it turned on, and started at 2.5V, slowly ramping up to 3.3 (slowly being over the course of 30 seconds or so). I then swapped out the inductor, which had the opposite effect - on power on, the system starts too hot (3.8V), and then slowly drops to 2.5V or so (again, 30s+ to do so). The effect seems to have some sort of memory - disconnecting and reconnecting briefly doesn't affect the decay timing. Any other ideas other than scrapping the whole thing? \$\endgroup\$
    – kolosy
    Apr 17, 2015 at 15:10
  • \$\begingroup\$ Don't give up! When it's all over, you'll appreciate your amazing new soldering skills. You still haven't answered my questions of what type of inductor or capacitors you're using (these are important). You also haven't shown any new waveforms. Additionally, switching converters are very sensitive to layout; information on your layout could be helpful. \$\endgroup\$
    – Zulu
    Apr 17, 2015 at 17:00
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I'm going to guess that R17 is open.

The waveform that you show makes it seem as if both the chip and inductor are working fine. So now you need to find out what the internal reference voltage for that chip is.

If I were home in front of my computer, I'd go to the chip website and look up the datasheet. But I'm not, so I'll calculate it from the voltage divider network.

Output voltage is supposed to be 3.3V. Feedback voltage divider is 1.8M & 575K. A little bit of math shows that the reference voltage is 0.8V. That happens to be exactly the voltage that you are getting at the output of the power supply.

Therefore, R17 is most likely open.

Note: it is possible that R16 is shorted but that is less likely.

[Edit]

Upon reading the comments and re-reading the question, it looks as if the FB pin is supposed to be sitting at 800 mV but is currently sitting at 200 mV. That means either of two possibilities:

1) the chip is in current-limit.

2) the chip is defective.

Not sure where I'd go from here.

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  • \$\begingroup\$ Unfortunately, the resistor is fine (checked it with a DMM right now). You're right, though, Vref is 800mV (datasheet is at the bottom of the post) \$\endgroup\$
    – kolosy
    Apr 15, 2015 at 4:15
  • \$\begingroup\$ Still very suspicious that the voltage on the FB pin is exactly Vref. Somewhere, somehow the output voltage is shorted to the FB pin. You can test that by removing R16. Under normal circumstances, that will cause the output voltage to swing up to V_BATT, so disconnect the load and any of those capacitors that can't handle 11.8V. If the output is still 800mV, then the pins are shorted internally on the chip. Or there's a hidden solder splash on the outside you haven't found yet. \$\endgroup\$
    – Dan Laks
    Apr 15, 2015 at 5:01
  • \$\begingroup\$ Well then, R16 might not be what you think it is? \$\endgroup\$
    – gbarry
    Apr 15, 2015 at 5:02
  • \$\begingroup\$ @DanLaks, isn't the feedback pin at 200 mV right now? Not that that makes a bunch of sense either. \$\endgroup\$
    – user39962
    Apr 15, 2015 at 5:27
  • \$\begingroup\$ @SeanBoddy, ah, you're right. I must have gotten the numbers confused after reading through the question and Dwayne's answer. Thanks for the correction. \$\endgroup\$
    – Dan Laks
    Apr 15, 2015 at 5:29
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Okay, I'm going to take a stab at this. Don't laugh, I don't do this much.

200 mV on the FB pin is consistent with the given feedback divider, provided 800 mV on the output.

Based on the waveform on the SW pin, it would appear that battery voltage is being made available, and there is nothing attached to that pin to store it, so something in the 3.3 V rail is disconnected such that battery voltage cannot make it to the bulk capacitor.

Under the condition that VOS outputs Vref - I don't know why it should - with sufficient current to sustain the feedback resistor divider current, the most likely single point failure is that L2 is broken or disconnected anywhere in the line before it joins the VOS node.

I cite as evidence that it would appear that the SW peak to peak is receiving battery voltage - briefly - and it appears that it is not connected with any significant source of capacitance. Take an oscilloscope reading from SW to VOS to confirm.

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