Automotive Controller LDO Failures

Question

I've designed an automotive lighting controller for two LED pucks, powered from the vehicle's ignition with a headlight signal input. The circuit uses a UA78L05AC 5V LDO to power an ATTINY 8-bit MCU.

Here's the simplified schematic (decoupling capacitors are not shown, but there are 1uF caps on the LDO and on the MCU):

I've experienced two failures where the 5V LDO that powers the MCU burns up. The circuit includes series diodes on both the power input (SS1H10 Schottky) and signal input, which should protect against negative transients. However, I suspect high-voltage positive transients might be the culprit.

Questions:

1. Given this design, what could be causing the LDO failures? How can I improve the circuit's protection against automotive transients, particularly high-voltage spikes?
2. Are there specific automotive-grade components or design techniques I should consider to increase reliability?
3. What additional diagnostics could I perform to pinpoint the exact cause of failure?

Any insights on hardening this design for the harsh automotive electrical environment would be greatly appreciated.

Answer 1

Yes, transients are likely an issue. UA78L05AC has an absolute maximum 40V input rating. Many of the transients defined in ISO 7637-2 or ISO 16750-2 exceed this, at impedances and time scales that will progressively or instantly damage it, even if it doesn't apparently overheat.

Hot-plugging may also be an issue, or similar issues triggered by transients. You don't mention what type or size of bypass capacitors are being used, but ceramic are a common choice; see:
Correct Placement of Series Ferrite Beads to Avoid DC Disconnect During Power Cycling
for more information. The linear regulator won't exhibit negative input resistance as a switching regulator can, but the ringing and C(V) curve effects remain important.

With the polarity protection diode, most of the negative-going transients can be ruled out; this leaves the positive ones. In particular, test pulses 2a, and 5a or 5b, are the most onerous to your system.

You might simply decline testing with pulse 5a/b, because it's an infrequent event; it does happen in real situations, but aftermarket devices regularly skip it, and the user might consider a perspective of aftermarket devices being noncritical and thus failure is acceptable in extreme circumstances.

That leaves pulse 2a, which is best protected against using a TVS. A SMAJ24A ought to suffice. This also partly addresses overshoot due to input ringing, though an electrolytic is still suggested there. This TVS will get cooked easily during load dump (test pulse 5a/b), failing shorted, for which a fuse should be provided (and should in general anyway).

You might also want to use a regular rectifier for the polarity protection diode; 1N4004 will withstand any negative pulse in the series just fine.

Answer 2

With the incomplete schematic, it's hard to tell, but here are a few suggestions:

1. Check the maximum current draw from 5 V+, and consider heat-sinking for the 7805 linear regulator. Remember that it must dissipate ~9 V (from a battery being charged to 13.7 V down to 5 V) at that current, and heat-sink appropriately.
2. Check that there is no other path from 5 V+ than the MCU and the 10 kΩ resistor.
3. Add a resistor of a few ohms in series with ignition, and a 15 volt Zener diode and 100 µF or larger capacitor across the regulator input. Size the resistor based on maximum current to be drawn -- the larger the resistor you can use, the safer for the Zener and regulator.

Answer 3

You can add an R/C low pass filter at the input side.

Since you only need some mA for the ATTiny MCU, you can use e.g. 100 Ω in series and 10-47 µF at the 78L05 input.

If you don't have an oscilloscope to show such transients, a TVS diode in series with a LED and, say 100 Ω, will produce a short flash at such a voltage spike. The human eye can see surprisingly short light pulses.

Answer 4

Bipolar Linear regulators are thermal and current-protected so it is wise to assume the thermal fault is due to a transient fault.

An unlikely abrupt 0V input (gnd) will cause a Vio reverse voltage fault. An abrupt open input cutoff may cause a rapid high side voltage flyback with a negative output Vo = Ldi/dt (also dependent on C) with output conducting I and wire inductance (<10 uH/m).

The classic solution is a reverse Schottky Diode on the output shown by TI on 8.2.2.7 Polarity Reversal Protection. You can verify this transient's different wiring scenarios and minimize L or ESR(C).

Answer 5

In reality, (automotive) 12 V can vary from –14 V to +35 V for extended periods of time and experience voltage spikes with extremes ranging from +150 V to –220 V Source: https://www.analog.com/en/resources/analog-dialogue/articles/protecting-and-powering-automotive-electronics-systems-with-no-switching-noise.html

So that UA78L regulator needs protection against overvoltage conditions. A current limiting resistor and TVS diode would be a good start.

You also have no capaictors on the regulator, this would help with transient stability and help smooth out short voltage spikes.

The third thing that could be killing a linear regulator is heat it doesn't look like the design is drawing much heat if the full 100mA isn't being used. But if the full 100mA is being drawn, then you have 12-5V *100mA which is 0.7W. Seeing that the worst case thermal resistance would be somewhere between 10 C/W to 20C/W that would be a 7 to 14C temp rise. Cars can get up to 80C and the part can survive up to 125C (minus ~15C if the part is hot) so it's likely not heat that is killing the regulator.

Answer 6

Are there specific automotive-grade components or design techniques I should consider to increase reliability?

Back when I was working in automotive, the standard regulator was the LM2940. This has built-in protection against pretty much everything that a user can do wrong to the power supply on a piece of automotive electronics. There may be better alternatives these days, but that's a starting point.

Answer 7

I use 18 volt TVS diodes in front of 78M05 voltage regulators in the ignition system of my 1974 camper van. No regulator failures.

I have had a manufacturer fitted fuel gauge stabilzer 7806 regulator explode in the dashboard of a Ford back in about 1987..

Answer 8

I see that there is no reverse current flow bypass diode across the regulator. If Voltage is cut and the 12V drops to 0 faster than the 5V side, then the capacitor on the 5V side will provide reverse Voltage across the regulator. I doubt this is causing the issue, but technically you're supposed to have it to stay within the absolute maximum ratings. To avoid needing the bypass diode, the 12V tap would need to be moved over to the anode side of your reverse polarity protection diode. The 12V relay circuit would be what is bringing the 12V side down faster than the 5V side.

edit: Also, if the diode across the relay coil isn't properly connected, when the power is cut, the entire current of the relay will be forced to flow through the regulator in reverse direction.

Answer 9

For good resistance to voltage transients, and low current requirements, a shunt regulator is sometimes a good choice.

For example, suppose 10mA and 5V +/- 10 or 15% is enough for your circuit and you need it to operate down to 10V. You could use a 510Ω resistor with good pulse withstand capability and a 5.1V 1W zener diode.

Below is a simulation with input voltage from -100V to +150V.

The blue trace shows the instantaneous power dissipation of the resistor, the red trace the power dissipation of the Zener diode.

Answer 10

Use a regulator designed for automotive duty. Then you don't need to worry about protecting it - it will operate just fine with nothing more than wire impedance as a form of input transient current limit. 7805, as you have learned, will not survive automotive environment. Protecting it from this environment is best avoided if you can just buy a part that includes all the protection needed and is designed specifically for automotive use. For a beginner, that's what I would suggest.