OK so the car manual says not to stick items into the car's USB port while starting the car. Like any good engineer(ing student) I tried to see what will happen if I did so, and am now the proud owner of a few dead USB sticks. So yes, the manual wasn't kidding.

Without going through the hassle of sticking an oscilloscope into the USB car ports, I presume that some kind of voltage spike must be going through the USB port when the car engine is started. What causes it? The current surge in the starter motor circuit causing an inductive current pulse in adjacent wiring from the 12V battery to the USB port?

It is somewhat tedious to remember to remove items from the USB port before starting the car, so I was wondering if we could design some kind of surge protector for the USB port to prevent it frying sensitive electronics.

Any comments on my attached circuit diagram? What capacitances and inductances would you suggest, absent any tests with an oscilloscope to quantify the surge size and times? (It is OK to over-engineer it)

Circuit sketch

Would you also recommend similarly protecting the data in/out ports of the USB output, just in case?

Regarding the question about the car model, see a similar warning from a Youtube video on another model of car. Kia USB port video

USB port warning

  • \$\begingroup\$ Maybe add a ferrite bead in series, and I would throw in a 5.1V zener. Any other ideas? Can't say anything about datalines, but they probably depend on vcc, so vcc protection is probably enough. But then again, I'm not the one to know it for sure \$\endgroup\$ – Ilya Jun 9 at 7:36
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    \$\begingroup\$ car manual says not to stick items into the car's USB port while starting the car is a design cop-out. It should be ok to start the car with the USB in or there should be a device that spits out the USB when starting or the car won't start with the USB in place. This is an administrative solution where an engineering solution is required. What kind of car is this so I may avoid purchasing one? \$\endgroup\$ – D Duck Jun 9 at 12:22
  • \$\begingroup\$ That may be specific to that model car, maybe even to that year, rather than a general problem. \$\endgroup\$ – scorpdaddy Jun 9 at 13:50
  • \$\begingroup\$ I would try a few zener diodes in parallel \$\endgroup\$ – Fredled Jun 10 at 7:26
  • \$\begingroup\$ A single Zener, even one with higher power than usual, won't have enough power so more than one are required. Maybe 5 or 6. Then TVS Diode on the data pins. There are TVS diodes designed for USB. You can put TVS diodes on the power line too, but zener are much cheaper and as fast, and you need them in the same number. \$\endgroup\$ – Fredled Jun 10 at 7:33

If "car manual" says so about USB sticks, it is a very poor design. USB ports are fed from 5 V supply, which should be generated from inside of USB-enabled equipment. 12 v to 5 V should provide enough margin to have a stable +5V with a little engineering effort. So making a L-C filter makes no sense.

The problem is likely in the fact that 12V rail fluctuates wildly when cranking the car engine. Here is the cold-crank waveform as assumed by ISO 7637-2 standard

enter image description here

So it is expected that the supply voltage drops down to 3V level and stays at 50% level for up to 20 seconds. It is difficult (costly) to maintain an audio/USB/on-board computer equipment to sustain such deep and long drops. So in most cars (and especially inexpensive ones) the equipment simply is turned off during cold cranking. The problem with USB is likely that "surprise disconnect" of USB host might cause device corruption if there is no sufficient power hold and graceful system shutdown and recovery. A L-C filter of any reasonable size cannot help here. The problem can be solved by having a small secondary (backup) battery that is capable of maintaining quality power to USB-related equipment just for 20-30 seconds, a small local UPS so to speak.

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  • \$\begingroup\$ It's possible that voltage drops to 3 or 6V then jump suddenly above 12V. And the isolated power supply can't adapt fast enough. \$\endgroup\$ – Fredled Jun 10 at 7:25
  • \$\begingroup\$ OK thanks, I learnt something new regarding the voltage drop during start-up. So as I understand it the problem is probably not a voltage spike killing the USB drive, but a corrupted filesystem from a sudden voltage drop during engine start-up? What battery would you recommend for keeping in a car? The maximum operating temperature of lithium batteries is 55C and car interiors can reportedly reach up to 68C in the summer. I've always been worried about having batteries in the car - even have a camping icebox for keeping my laptop inside if I know I will be leaving it in the car, in the sun. \$\endgroup\$ – MeEngineerTrustMe Jun 10 at 10:47
  • \$\begingroup\$ @Fredled, there are plenty of DC-DC converters that are capable to withstand 40-60V input range. Usually operating at 200-kHz- 1.5 MHz switching rate, they are more than fast enough to react to 5 ms edges, as the standard cold-crank waveform implies. More, it is easy to put a simple surge protector >12V at input end of the power supply and forget the problem. Or the same C-L-C filter as in the op question. \$\endgroup\$ – Ale..chenski Jun 11 at 3:36
  • \$\begingroup\$ @MeEngineerTrustMe, there are small lead-acid sealed batteries on the market. All you need is a 150 mA-h battery to hold 5 A for 100 seconds. Also, there are Li-ion cells with -40+150C range. customcells.org/products-development/… Tesla/Toyota are also cars, and operate under the same sunny environment. \$\endgroup\$ – Ale..chenski Jun 11 at 3:48
  • \$\begingroup\$ @Ale..chenski The reaction time is not related to switching rate but by the time it takes for the output voltage to be sensed back and a new frequency being applied. It's never instant. If the voltage goes from 6V to 18V in a very short but very sharp spike (the example of clod-crank waveform is academical theory) it won;t have the time to react. A few microseconds at twice the max voltage is enough to kill a USB stick after a few times if not immediately.You don't need a millisecond. \$\endgroup\$ – Fredled Jun 11 at 22:57

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