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I have a 1k pull-up resistor on an input line of a microcontroller on an automotive part.

The signal is coming from another product in the vehicle which is meant to deliver 12V. In this case one might think that the current will be 12mA through the pull-up and hence the power will 144mW so maybe a power rating of 250mW would be appropriate for the resistor.

However, in automotive, there can be very large transients and if the signal becomes noisy or the other product which communicates with our device has a defective or badly designed filter (over which I have not control), it's possible there could be some large transients on the signal line.

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Independent studies by the Society of Automotive Engineers (SAE) have shown that voltage spikes from 25V to 125V can easily be generated [ref], and they may last anywhere from 40ms to 400ms. The internal resistance of an alternator is mainly a function of the alternator rotational speed and excitation current. This resistance is typically between 0.5Ω and 4Ω (Figure 2).

Does anyone know if the power rating of pull-up resistors in automotive systems are typically rated higher than normal?

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  • \$\begingroup\$ Depending on where the device sits and how that 12 V is generated, internally or straight from the 12 V battery, it needs to pass load dump test which will put 80 V on the 12 V rail via a known impedance, so yes, you may need to rate them for said surge voltage. \$\endgroup\$ – winny Apr 5 '17 at 10:28
  • \$\begingroup\$ If your resistor is connected directly to a microcontroller, the other end should never be connected to anything that's even close to 12V, let alone any surges. You need to show us a schematic of exactly what you have in mind. You also need to proofread your post for typos. \$\endgroup\$ – Dave Tweed Apr 5 '17 at 10:41
  • \$\begingroup\$ @DaveTweed I can't release the schematic as it's not mine. The microcontroller input is 12V for the signal line, (0 - 12V), with a 12V pull-up. I edited the the question, so hopefully it's a bit clearer. \$\endgroup\$ – SeanJ Apr 5 '17 at 11:11
  • \$\begingroup\$ @winny we have a load dump requirement on the input power but the line in question is a signal wire coming from another product in the vehicle. My concern is that if the non differential signal wire picks up electromagnetic interference or the actual product fails there could be transients on the wire. \$\endgroup\$ – SeanJ Apr 5 '17 at 11:19
  • \$\begingroup\$ Automotive tolerance includes (from memory, but not limited to) indefinite -12v (reverse battery) and +24v (jump starting from a truck), and a short period (sub second?) of 120v (load dump), engine critical must work down to 5v, others should work to 10v, or is it 8v? It's all searchable somewhere, a good answer would link to it. \$\endgroup\$ – Neil_UK Apr 5 '17 at 11:24
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Nasty voltage spikes on a car "12 V" system are real, but their average power is small.

During normal operation, your resistor could dissipate (13.6 V)2/(1 kΩ) = 185 mW. In theory, the next common size of ¼W would cover it. However, cars get hot. Take a look at the datasheet of a ¼W resistor to see how the power dissipation capability is derated with temperature.

Short spikes on the "12 V" line don't add much average power, but do represent some fixed energy per spike that the resistor has to absorb effectively instantaneously. The power dissipation capability of the resistor is mostly irrelevant. It's the thermal mass that decides how much the resistor will heat for a given energy spike.

Cars are a harsh environment. A ¼W resistor doesn't leave much margin, even at a convenient ambient temperature. I'd go with the next common larger size, which is ½W.

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  • \$\begingroup\$ Hi Olin, many thanks for the feedback. I've edited the question to better reflect what I was trying to say. Basically, I have another regulated product in a different part of the car connected to my product. It is meant to supply 12V (not alternator voltage). My question is whether the potential for issue in vehicle would require a higher power rating on the resistor. \$\endgroup\$ – SeanJ Apr 6 '17 at 7:17

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