Why would a switching power supply fail with a battery supply?

Question

I have this power supply circuit I have been using for a couple of years:

I have put it on a lot of boards, and ranging input voltages up to 50 volts (as long as the capacitors are rated that high). I had never been able to get it to fail, and I even load tested it drawing the maximum off both rails, and it shutdown before it failed. However, I had always powered the input from an AC/DC converter.

I recently began using a board with this supply on it in a battery powered application (24 volt - 2 12 volt lead acid in series). Everything worked fine for months. I was charging the batteries separately using a car battery charger. I then bought a 24 volt charger, which seems to work well. After charging the batteries (there are two supposedly identical systems) I powered up and one of the power supply circuits blew on the board. In the second system all was fine.

To summarize my troubleshooting from that point forward: that one board that didn't fail on the original power up after charging is the only one that I can get to work on this charged battery system. I have replaced the LT3988 and tested it in the lab off a lab supply (up to 40 V) and AC/DC brick, and it works perfect. I take it back to the fielded unit and plug it up to the battery (only the supply connected) and the switcher immediately smokes. I populated an entire new board, the switcher works on a lab supply, and immediately smokes on these batteries. The one that works, works on both systems...but I cannot recreate it.

So...what problems can you guys think of when using a battery supply that are not encountered on an AC/DC converter? I can think of a lot of more problems powering a switcher with a switcher, so I need your help.

Edit: When powering this circuit from an AC/DC brick, I usually have the brick plugged into AC and just plug the barrel jack connector into the power input on my board. When powering from a lab supply, I obviously just press the button. When powering from the batteries there is a switch in the line between +VDC and my board.

Answer 1

The LT3988 data sheet mentions the use of ceramic capacitors at the input:

A final caution is in order regarding the use of ceramic capacitors at the input. A ceramic input capacitor can combine with stray inductance to form a resonant tank circuit. If power is applied quickly (for example by plugging the circuit into a live power source), this tank can ring, doubling the input voltage and damaging the LT3988. The solution is to either clamp the input voltage or dampen the tank circuit by adding a lossy capacitor in parallel with the ceramic capacitor. For details, see Application Note 88.

You might want to add some extra input impedance to your source.

Answer 2

The available current is much higher from lead-acid batteries, and it could have a very fast rise time.

I would expect problems if the C37 4.7uF is a tantalum. They are very intolerance of high currents and will tend to produce fireworks, burning, smoke, small explosions etc.

If the chip is failing, that's more of a mystery. Maybe some extra inductance somewhere. A larger aluminum electrolytic cap in place of C37 and perhaps a series polyfuse (you should have a fuse somewhere as the batteries can produce enough current to melt wires and cause a hazard as a result).

Edit: Another possibility is that there is a layout issue that is resulting in transient reverse bias of the chip. If the incoming power return is fed into the wrong side of the circuit, for example. I don't like that the schematic is 'flipped' left-right.

Answer 3

It's likely that the start up current is causing the issue. In addition to other suggestion to deal with the switcher's own current at startup, consider increasing the value of C38 and C39 to increase the soft startup time. This will ramp the voltage, and thus current, on the output up over a longer period of time, reducing the startup current.

If you can, connect an oscilloscope in the field with a newly built circuit. Measure the input voltage and current, and the output voltage and current while connected to the load, and from the AC power supply to the battery power supply. I expect the difference between the two will become immediately obvious, and the solution won't be far behind.