I'm running this regulator off of a ~12V car battery. I've set it up almost identically to the 'typical application' schematic and layout that the datasheet suggests to generate a 5 V output, except that I'm using a 22 μH inductor.

enter image description here

The problem is that when we run the starter motor, and also at undetermined points in time when the car is driving, the microprocessor on the board dies -- presumably due to some kind of glitch on the regulated 5 V.

I suspect the problem is noise, but I can't get a proper measurement of it while it is on the car. I'm going to try measure it in lab at some point.

I've got 8-9 0.1 μF bypass capacitors at the microprocessor, but apparently it isn't enough. Another problem might also be because the regulator is running in discontinuous mode most of the time. If I were to redesign it should I put some filters at the input of the regulator? Is this something that is typically necessary?

What's the best fix for this? It's too late for me to add signicant hardware to the PCB, but what would be the best thing to do if I redesigned it?

For reference, this is my layout, (C102 is the input cap, D101 is the catch diode, C106 L101 are the output filter network, R102 is a jumper to the 5V plane). The important loop (catch diode/input cap) is at the bottom.


  • \$\begingroup\$ Is \$V_{IN}\$ directly connected to the battery, or do you have some polarity, transient, overvoltage, and ESD protection on that line that's not shown in the 'typical applications' schematic? Put more simply, what guarantees do you have that transients on the battery line are limited to 60 V? \$\endgroup\$ Commented May 15, 2012 at 15:14
  • \$\begingroup\$ I didn't put any additional protection hardware at the input of the regulator, which is a problem, but I don't think that it is the problem here. If it were, the regulator would have gone up in smoke. \$\endgroup\$
    – wjin
    Commented May 15, 2012 at 15:25
  • 1
    \$\begingroup\$ It's possible for transients to break through the regulator when it starts to break down from over voltage without destroying it. But relying on this always being the case would be unwise :-). The automotive environment is an exceptionally hostile one. \$\endgroup\$
    – Russell McMahon
    Commented May 15, 2012 at 15:43
  • \$\begingroup\$ @wjin - You're oversimplifying your environment and the possible problems you'll encounter. Read this appnote, and ask yourself if you've taken into account all those possibilities. I think your best bet will likely be to buy or build a surge and brownout stopper which you can use to supply your board with clean 12V, since you haven't designed for the automotive environment. \$\endgroup\$ Commented May 15, 2012 at 16:07
  • \$\begingroup\$ What is your CAD program and what are those gray crosses with red pads behind them? \$\endgroup\$ Commented May 15, 2012 at 16:51

2 Answers 2


You're likely dealing with either voltage spikes or big dips. I would tend to agree with those who believe that it's more likely a spike vs. a dip that's causing downstream damage.

You should put a high-power rated TVS or other crowbar across the input of your circuit, to attempt to contain big spikes. A DC line filter will also help smooth out high-frequency disturbances.

Ultimately, you may need a crowbar on the output as well, just in case the other protections fail (What happens if the buck goes to 99% duty cycle or the series MOSFET fails?) that's rated big enough to blow your fuse. Um ... please tell me you have a fuse in your design...

Another problem might also be because the regulator is running in discontinuous mode most of the time.

DCM is a good choice of operating mode for this topology, since there's only a type-2 compensation network on the voltage feedback. If the converter transitions to CCM, the added phase boost that a type-3 compensator gives is often necessary to ensure stability.

  • \$\begingroup\$ Problems "when we run the starter" suggests a problem with dips to me...not to say there couldn't be more than one problem. \$\endgroup\$
    – The Photon
    Commented May 15, 2012 at 19:18
  • \$\begingroup\$ If the load being drawn by the starter motor isn't continuous, the sudden changes in current will create voltage spikes via the inductances throughout the power train. \$\endgroup\$ Commented May 15, 2012 at 21:30

You may also be getting dropouts rather than spike ride-through.

If that is all the filter capacitance available on the 12V and 5V rails it won't last long. Quite likely the uCboard has some extra large caps on it.

Holdup time ~~ =~ V x C / I
C = allowable droop, I = load current, C = capacitance.
For 4 Volt input droop T = 4 x 2.2 x 10-6 /I
=~ 10/I uS.
ie at 1A in its 10 uS.
At 100 mA in its 100 uS etc. Not long regardless.
Under cranking a 12V battery can approach 6V.

One less than perfect system which worked well on a taxi system I was involved with long ago was to add a small lead acid battery at the input and feed it via a Schottky diode. Under cranking and misc dropout spikes the small battery provided the supply. It also adds a degree of +ve going spike filtering as well.

A lead acid battery is not a good long term solution as it runs at a diode drop below the main battery and experience shows that this is enough to reduce longevity by a surprising amount.

A supercap at the input would substitute well enough - or just a very large electrolytic. For T= VC/I - say 4C/I:

For playing, a 1000 uF gives you 40 mS at 100 mA in.
A 10,000 uF givs 400 mS. [ ~~~~= in all cases + E&OE]
Note I_12V ~= 0.5 x I_5V.

For +ve spikes a simple inductor input, cap to ground filter will help muchly. In this case the inductor can be in the supply wire so no PCB change. An exra cap can be too.

  • \$\begingroup\$ I didn't get a good trace of the 12V line from a noise perspective, but I'm pretty sure the battery voltage doesn't drop below ~9V (Let alone the 5.5V minimum required by the regulator). \$\endgroup\$
    – wjin
    Commented May 15, 2012 at 15:38
  • 1
    \$\begingroup\$ In a car environment I'd be surprised if the"battery" circuit did not sometimes go well below 9V for short periods "at undetermined points in time when the car is driving," \$\endgroup\$
    – Russell McMahon
    Commented May 15, 2012 at 15:41
  • \$\begingroup\$ When the car is driving, the alternator is also running. The battery drives a 150A starter motor, and the other stuff on power bus doesn't draw more than ~10A, so running the starter motor should present the largest possible load on the battery. \$\endgroup\$
    – wjin
    Commented May 15, 2012 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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