Electronic – Protection against automotive power supply hazards

I can't speak to SAE J1113, but for SAE J1455 (12-V heavy truck, where the loads should be larger) the load dump is defined as a 100 V peak with about ~0.6 s fall time and ~0.6 Ω impedance, which is a pain to live through.

The two broad methods to survive are either

• Disconnect yourself and let it pass:

Which is usually preferable and cheaper. Load dumps are in a class of faults that many devices are not expected to operate during (unlike coupled inductive transients), so unless you're some critical device (ABS, ECU), you're allowed to shut down and reset when you see a load dump.

Very broadly speaking, to do this you could have a Zener diode on your input, where once it breaks down and starts conducting, switches some pass transistor to disconnect yourself entirely. Obviously your pass transistor will have some voltage rating, so selecting a TVS is still needed (see following), but it won't have to clamp anywhere near as much voltage, energy, and power.

• clamp the whole thing.

This is also quite possible with TVS as you mention, and then it really depends on how hard you want to clamp it. If you're fine with 75 V coming through, I think I've seen 500 W SMC's used. If you want it like almost nothing ever happened, you can do as I've seen and use (2) 5 kW 5KP22CA TVS in parallel. They alone can clamp the entire load dump themselves; I've tested a pair that survived (5) 100 V dumps in a row, about 10 seconds apart between each.

The math behind it is somewhat hazy to me, as the figures provided on the datasheets don't seem as though they were meant for transients any slower than 60 Hz. The 5 KW rating is for a 1 ms pulse, which is obviously just 5 J.

The peak energy it dissipates will be (100 V - 24 V)/0.4 ohms * 24 V = 4560 W, but this will decay roughly exponentially to nothing with a tc of about 300 msec. If we just call that a triangle (very conservative), it's 0.5 * 4560 W * 0.3 s = 684 J. If we extrapolate the Figure 1 rating curve on the 5KP datasheet, it suggests that a 100 ms pulse can have a maximum power rating 1000 W, or total energy of 100 J, and even more energy if we smear it out further, so we're in the ball park with 2 of them in parallel and tests seem to bore it out.

Littelfuse 5KP-series TVS datasheet, Figure 1

If you wanted something better, I'd come up with an equation for the curve and give it an asymptote at the maximum steady-state dissipation (8 W...though that might not make a difference), then do some integration with that over your pulse to see how much of the rating you use up :P

First pass:

1. Your schematic is confusing. What is the relationship between JP1-1, VIN, BPOS and the battery voltage?

2. You should have the fuse on the positive input 'ahead' of your protection circuitry (on the battery-side of things). You never know how things may fail in real life.

   Given the orientation of D2, the battery feed should be JP1-1, correct? If so, and if you have reverse battery, your unidirectional TVS will conduct big current (it will indeed act like a diode) and melt things, and your PTC is powerless to help.

   You may want to consider the bidirectional TVS 5.0SMDJ30CA (notice the extra C) so that it will clamp regardless of polarity.

   If VIN is the battery feed, your circuit won't work as D2 will be blocking all the time.