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I am using the L78L33 on my board in TO-92 package; the weather outside is cold (it is used inside vehicles) and I have had it fail in one of six boards. It has worked for some hours then died out, and the vehicle is a new passenger car with its 12V socket doing fine with other equipment (i.e. not causing them any damage). It looks as in the pictures, with non apparent damage to it besides that small cutter-made-like mark. No other parts got damaged.

enter image description here enter image description here

I have just noticed how the datasheet said

If adequate heat-sink is provided, they can deliver up to 100 mA output current.

I am drawing about 40-50 mA max and there is no heat sink installed. Can this be why the part has failed, even in cold weather ? I am thinking more of some bad soldering, but I don't know really...

Hmm, there is one other thing. There is one switch button connected to an INPUT pin of the MCU, and I am always sloppy with the solder joints around that pin. meaning that I do short the designated INPUT pin with one or two surrounding unused pins which are left to be just as the MCU configures them on start-up, so I would think OUTPUT. Could this be causing issues ?

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    \$\begingroup\$ Maybe getting too hot. May also be getting killed by voltage peaks when starting the engine (or other stressfull moments in a car's life.) 12V in a car isn't always 12V. Usually it is closer to 14V, but can jump way high when the starter motor shuts off. It can also jump at other times. \$\endgroup\$
    – JRE
    Dec 29, 2017 at 17:17
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    \$\begingroup\$ Automotive power can experience surges of +/-200V or more, especially when large loads switch on or off. Any circuit that needs to work reliably in that environment must be prepared to deal with this. This means that you can't just connect the input of your 78L33 directly to the 12V bus. \$\endgroup\$
    – Dave Tweed
    Dec 29, 2017 at 17:37
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    \$\begingroup\$ You need to provide a simple schematic of the circuit this device is in. Because I don't know if you have any protection circuits used to protect the regulator IC from power surges normally encountered with 12V auto (like the car starting, or plugging the unit in while the 12V socket is energized) \$\endgroup\$
    – drtechno
    Dec 29, 2017 at 18:31
  • \$\begingroup\$ +/- 200V ? Well I'll be... even in new cars, there is nothing to take out those peaks ? Sounds reasonable, but if this is the case then the other 5 should have been dead by now as well. \$\endgroup\$
    – kellogs
    Dec 29, 2017 at 20:46
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    \$\begingroup\$ Would agree with Dave Tweed's comment. The only times I've seen holes blown in packages were from a latchup event triggered by voltage spikes. Most folks commenting are focusing on overheat, but severe overheating usually causes cracked and burned packages. Craters and holes are typically severe CMOS latchup, events so fast that the package has no time to burn. \$\endgroup\$ Jan 2, 2018 at 22:24

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Something like this would be better in an automotive environment, though not necessarily fully compliant to standards:

schematic

simulate this circuit – Schematic created using CircuitLab

D2 blocks any negative transients. -100V from alternator field decay is one possible source. It also protects against accidental reverse voltage connection.

D1 clamps the input voltage to less than 42V with 32A applied, which implies an input voltage far exceeding the 300V inductive load switch transients and the 125V load dump transients.

R1 controls the current flowing into D1 and drops a bit of voltage so the regulator does not run as hot. A 400ms load dump transient at 125V would cause R1 to dissipate 130W briefly. That's caused by disconnecting the battery while charging at high current. So R1 should be a fusible type capable of withstanding high pulse power. 300V for a few hundred microseconds is a typical transient from switching an inductive load. Similarly 75V for 90ms can occur several times in a vehicle lifetime from ignition with battery disconnected.

D1 is also rated for more than 24V so that 24V jump starts would not burn out R1. 5 minutes is the expected duration.

C1 and C2 are bypass capacitors for the regulator.

U1 is similar to the regulator you picked, but rated for 40V maximum recommended input with a 45V abs maximum rating, compared to 30V for the 78L33. It also draws less Iq so the internal dissipation is a bit (10%) lower, and is in the SOT-89 package with thermal resistance junction to ambient of 125 degrees C/W compared to 200 degrees C/W. With a large PCB area that 125 can be reduced to more like 50-60.

So with 50mA being drawn, input voltage of 13.8V, the drop across R1 is 3.1V, so the regulator sees 10.0V nominally. Internal dissipation is 0.335W, so the temperature rise is 42 degrees C, allowing an ambient as high as 83 degrees C before we hit the 125 degrees C Tj(max). During a 24V jump, the input sees about double that so the dissipation is 0.845W, meaning we would hit the maximum Tj with a 20 degree C ambient. At about 55 degree C ambient the overtemperature protection should come on and the regulator should shut down until the temperature drops 20 degrees C or so. Hard on the chip, but that is an unusual condition. ST claims 55 degrees C/W with 6cm^2 PCB copper area for heatsink, which would be much, much better.

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  • \$\begingroup\$ Could I use just one bidirectional TVS and do away with D2 ? \$\endgroup\$
    – kellogs
    Dec 31, 2017 at 17:06
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    \$\begingroup\$ @kellogs No, that would be worse than just leaving D2 out. The unidirectional type as shown will actually clamp negative voltages by itself, but not perfectly (-3.5V at 100A) so you'd have to make sure the regulator didn't die or pass too much negative current along under those conditions. Neither part is well specified for that condition. The resistor might fail or get too hot if the input voltage was reversed for some time unless it was physically large (~3W dissipation at -13.8V in). Some regulators are specified to withstand some negative input- the AP7381 is not one. Eg. Infineon TLE4xxx. \$\endgroup\$ Dec 31, 2017 at 17:15
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    \$\begingroup\$ Generally regulators with MOSFET or NPN pass elements cannot withstand negative inputs. The ones that use lateral PNP pass elements may be able to, but the beta is low and they can be unpleasant in other ways. \$\endgroup\$ Dec 31, 2017 at 17:17
  • \$\begingroup\$ the D1 symbol is for unidirectional diodes, but the datasheet for smcj26ca says it is a bidirectional diode. Is there a mistake with the symbol above ? \$\endgroup\$
    – kellogs
    Jan 2, 2018 at 16:53
  • \$\begingroup\$ @kellogs Yes, will fix it. Should be ..26A for unidirectional. \$\endgroup\$ Jan 2, 2018 at 17:16
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For load-dump info, I suspect that ISO 7637-2 and ISO 16750-2 are two documents you may want to consider.

For protection, as well as distributing the dissipation load, perhaps something along these lines?

schematic

simulate this circuit – Schematic created using CircuitLab

The above is more a concept than a crafted design. I just picked one possible series of TVS you might use. There are others, of course. And I didn't point out which voltage to use there. I'd probably recommend that you examine the details of the 78L33's maximum input specification and find a TVS voltage that keeps you comfortable. I added a couple of small caps, too.

The zener pre-regulator section was just pasted in, from memory. I picked 6.2 V because that seems high enough to satisfy the dropout of your regulator and it will distribute some of the dissipation onto \$R_1\$ and \$R_2\$. I chose two resistors there for two reasons -- (1) if one opens then the other is still there and (2) also for dissipation, in case it matters. You could further complicate that by arranging things so that if a screwdriver shorts one out there is still more, by adding a series resistor, too. Up to you.

The exact values of those two resistors does matter. I picked ones that set up about \$50\:\textrm{mA}\$ each, for a total of \$100\:\textrm{mA}\$. However, the zener I selected is only tested at most for about \$40\:\textrm{mA}\$. So again, you need to think this through on your own end. You may need to set up an added dummy load (and/or add a BJT emitter follower for it after first selecting a 6.8 V zener.) Again, this is conceptual and you need to think it through on your own. I just wanted to call your attention to the idea of adding this kind of section.

The PI filter is mostly for RF. I'm not sure if it is needed and I also don't know how the 78L33 will handle the preceding PI. But it is there to make you at least think about the subject, in case it matters to you.

Finally, your 78L33. You might also consider adding a PNP BJT bypass around it, too.

So:

  1. Input protection section for load dump events.
  2. Possible pre-regulation section.
  3. Possible filter for RF.
  4. Linear regulator, with an optional PNP BJT bypass.

You can pick and choose. But consider including the TVS protection and perhaps the PNP BJT bypass to shift some dissipation away from the TO-92 regulator.

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As mentioned in the comments, the car battery voltage is about 14V while the engine is running.

If the regulator IC's input is directly connected to the supply (Since no schematics are provided I assumed so) then the power dissipated by the IC will be \$P_D = (14-3.3) \cdot 0.05 = 0.54W \$(approx). The IC has a TO-92 package and the datasheet shows that a thermal resistance of 200°C per Watt for it. So the temperature rise of the IC will be \$\Delta T = 200°C/1W \cdot 0.54W = 108°C\$.

What does this value say? If the ambient temperature is Ta=18°C then the IC's case temperature will be Tc = Ta+ΔT = 126°C which is higher than allowed. I think that's the problem.

Maybe you can put a series resistor for dissipating some power so that the stress on the IC gets lowered. For example, if the input voltage of 78L33 would be 8V then the stress would be lower. So, a series resistance of Rs=(14-8)/0.05=120R/1W can be used.

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    \$\begingroup\$ the maths is good, thanks, however the 'accident' took place while the load was and has been 0.2 - 0.25 mA for a while. Also, the datasheet starts with employ internal current limiting and thermal shutdown, so thermal shutdown should have prevented this kind of damage \$\endgroup\$
    – kellogs
    Dec 29, 2017 at 20:49

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