I am powering a PIC32MX microchip using a 3.3V Linear Regulator (LM2937), supplied from an RBO40 which I fitted in an effort to protect against adverse conditions such as load dump. As an aside, I've just ordered a Murata buck converter (RS-Online part 796-2141) to replace the linear regulator.

The PIC32MX is connected exactly as per the datasheet with the recommended capacitors within a few mm of the power supply pins.

The PCB is mounted inside my classic car and controls various systems such as PWM dimmed lights, gauges, switches etc etc. Whilst not safety critical it would be desirable if the PIC would remain running during engine start, which it doesn't (always resets). A breadboard-ed thru hole version of the PIC does continue running during engine start on my regular (modern) car, so I wondered if perhaps the classic has a more aggressive starter motor or alternator, or weaker battery that is contributing to this issue.

I'll try the buck converter instead and see if that's any different.

Otherwise should I consider any particular rating of bulk capacitor perhaps? Before or after the voltage regulator? I've googled the issue but not got anything definitive, just confusing myself. Thanks in advance.

  • \$\begingroup\$ s/bulb/bulk/ ...? \$\endgroup\$ May 13, 2020 at 14:20
  • 2
    \$\begingroup\$ I do believe you normally fit series resistors and TVS diodes or MOVs from the line to GND to clamp the line and protect against a load dump. There is a linear regulator right after and the 12V from your car is so much higher than 3.3V that an extra large series resistor won't hurt things. It will just take some of the voltage drop so your regulator won't have to and will increase the transient power at which the TVS can clamp at. electronics.stackexchange.com/questions/483959/… \$\endgroup\$
    – DKNguyen
    May 13, 2020 at 14:41
  • \$\begingroup\$ LM2576 with a SK56 diode before all components on the input, disconnect the pcb for the first seconds and turn it on with a button or automatically 5 sec after power on. \$\endgroup\$
    – CFCBazar
    May 13, 2020 at 18:52

1 Answer 1


New theory:

Keep in mind a significant voltage sag during engine start is normal. For many years, automobile engines did not have electric starters because it was impossible. The starter motor would weigh 500 pounds. The battery would weigh even more. What changed is the realization that they were sizing based on the continuous rating, and they didn't need to. Electromagnetic equipment (transformers, motors) can be insanely overloaded for short time limits, because of the "thermal inertia" of the copper and iron. Now the motor gets dramatically smaller and the battery is optimized for surge start. (at the expense of deep cycling).

Series wound DC motors make torque proportional to current. Voltage is irrelevant. Therefore the starter-battery system was optimized to deliver the necessary current even if it means significant voltage drop! (and the ignition and fuel injection system must be ready for this.)

So must you. Prepare for the nominal 12V to drop to as low as 4 volts.

I don't see that being a huge problem, as long as you are using a bucking DC-DC converter. It should cheerfully accept anything as low as setpoint voltage.

First theory: (may help others with similar problem)

Because you have it wired into a circuit that is "Hot in run", like the fan blower.

The system is being switched off by the ignition switch when you move it from "Run" to "Start".

Generally a key has 4 positions:

  • Off - really off.
  • Accy - turns on auxiliary loads such as car radio
  • Run - turns on all normal loads
  • Start - Only turns on loads essential to starting the engine, e.g. the ignition.

Most circuits are "hot in Run" meaning they only are energized in Run mode, e.g. the fan blower. They are shut off during engine starting to conserve power.
Some circuits are "hit in Accy and Run" such as the radio. Again, off during engine starting.

So you are on one of those circuits. You need to move your device to a circuit that is hot in Run and Start. There are precious few of those - the ignition being one. Beware: Line noise.

  • \$\begingroup\$ Thank you all. The pcb is always powered during start. The problem manifests itself even with the voltage regulator connected directly to the battery. I will keep working on it. \$\endgroup\$
    – steve9876
    May 14, 2020 at 10:52
  • \$\begingroup\$ Thank you for the updated answer - makes sense to me. I've swopped the linear regulator for a buck converter which does sometimes still drop out during cold cranking - but the datasheet states it has a minimum input voltage of 7 volts, so this makes sense. I've found another buck converter module (TPSM84203) that states 4.5 volts as the minimum so I think I'll try that next. Your answer led me to reading some good TI articles on cold cranking. ICs are available that switch to boost mode and maintain output down to 2.5 volts input, but I don't want to disappear down this rabbit hole just yet! \$\endgroup\$
    – steve9876
    May 16, 2020 at 10:42

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