# Automotive: how to safely connect the car's power grid to the input of an intelligent switch?

I have an intelligent switch. Let's take this one as an example - it's the one I'm plan on using, unless I find a better one. What I need to do is drive it by the voltage from the car's power grid. There are two complications here:

1. I'd like to get 3.5-5 volts at the input of the switch, but I have no other voltage source to clamp the input to with a diode.
2. Whatever schematic you suggest must withstand voltage pulses up to at least 60V (better yet - 100), and keep in mind the pulses may have reverse polarity as well.

What are the simplest solutions you can think of?

I've been thinking along the lines of a high-resistance voltage divider + a suppressor diode or a Zener diode. But I have no idea if it will work, let alone - work reliably. Please keep in mind I'm not an electrical engineer, and don't beat me up if I ask how your solution works.

• There is a schematic button on the editor toolbar. "I'd like to get 3.5-5 volts at the input of the switch". What does this mean? – Transistor Jan 30 '16 at 15:21
• @transistor: what would you like me to draw, a 3-pole? I have a 12-15 V source with harsh spikes, and I need to get stable and filtered 3.5-5 V low-current voltage from it. – Violet Giraffe Jan 30 '16 at 15:33
• The schematic would show the relationship between the battery, the intelligent switch, the load and whatever is triggering the intelligent switch. Question 1 is just asking how to generate 3.5 - 5 V and a linear voltage regulator would do that. The HITFET BTS 117 you referenced will accept up to 10 V on its control input so I don't understand what the 3.5 - 5 V is about. – Transistor Jan 30 '16 at 15:43
• 1. The battery becomes irrelevant once the engine is running and alternator spinning. There's no battery, there's a noisy voltage source with floating output. 2. I thought of a voltage regulator, but will it withstand the interference I'm trying to shield the switch from? Do you know any automotive-grade voltage regulator? If you do, that'd certainly do the trick. 3. I see no need to approach the maximum ratings. If 10 V is max and 2.5 is the maximum threshold voltage, then 5 V sounds about right to me. – Violet Giraffe Jan 30 '16 at 15:48
• Searching for "automotive transient protection" should give you a lot of valuable information and examples. – JimmyB Jan 30 '16 at 16:20

Are you looking for something like this?

Signal can come from any 12V(nominal, unregulated) circuit in the vehicle.
Vpos can also come from any 12V(nominal) circuit capable of powering your load.

The zener diodes (marked 16v) protect by grounding overvolate spike & "freewheeling" negV transients while the regular/switching diodes block negV transients from your switch/load.

The 10K and 5K5 resistors form a (5.5/15.5) voltage divider to drive your input pin.

• In principle - yes, this is it, but I have a couple questions. – Violet Giraffe Jan 30 '16 at 16:03
• 1. Won't the input suppressor be better placed AFTER the divider? Then the resistor will limit the current flowing through it when the voltage exceeds the limit. – Violet Giraffe Jan 30 '16 at 16:04
• 2. The symbol that you used for the 16v diodes is a "transil", or a suppressor diode - not a Zener diode. Which, to the best of my knowledge, is actually more fitting here than a Zener. But, how will it work in this circuit? As far as I can see, any voltage spike will cause the diode to open, effectively creating a short circuit and possibly blowing a fuse (the whole circuit will be fuse-protected). It might be the less evil, but I would like to preserve my fuses, too, if possible. Also, I think I don't need to protect the load, just the input, as the switch is "fully protected" itself. – Violet Giraffe Jan 30 '16 at 16:06
• @VioletGiraffe The transistors marked 16v are supposed to be zeners (sorry, that's a fault in my diagramming software on my phone). A zener diode protects from overvoltage spikes by "collapsing" reliably at a preset voltage (16v, in this case) applied to the diode in reverse, effectively "clipping" any voltages that would otherwise go over the zener's set voltage. -- As for the load protection diodes (top-right corner), if you trust your switch's internal "full protection" you can remove them without affecting anything else. – Robherc KV5ROB Jan 30 '16 at 16:21
• Thanks. If those are Zeners then I should certainly remove the top one because no Zener can handle 3 amps. Besides, my load is protected, and I trust the switch's protection as well. – Violet Giraffe Jan 30 '16 at 17:33

simulate this circuit – Schematic created using CircuitLab

Figure 1. Half-supply divider.

• R1 and R2 provide a potential divider with the junction being half supply. This will vary between 6 to 7 volts on battery or alternator. This is well below the 10 V maximum input of the BTS117.
• C1 forms a filter to stabilise the voltage and absorb any transients.
• Current, I, through R1 and R2 is given by $I = \frac {V}{R} = \frac {12}{20,000} = 0.6 mA$.

Figure 2. Extract from BTS117 datasheet.

• Normal operation: The datasheet shows that normal 'on' current on the input is typically 30 µA. This is 1/20th of the current through R1 so the voltage on C1 won't drop dramatically.
• Current limit operation: If the device goes into current limit this is signalled on the control-pin (rather cleverly) by increasing the input current to 120 µA. This is 1/5th of the 600 µA passing through R1 so the voltage on C1 will drop by about a fifth. It will still be above the 'on' threshold.
• Thermal shutdown: If the device goes into thermal shutdown the input current will increase to 2200 µA (2.2 mA). The most R1 can supply will be about 1 - 1.2 mA so I'm not clear what will happen at this point.

Update: R1 - C1 will have a time constant given by $\tau = R \cdot C$. I've increased C1 to 10 µF so that the time constant becomes $\tau = 10~kΩ\cdot 10~µF = 100~ms$. That should be plenty for typical transients.

Seeing as you have raised the point about internal resistance, I've added C2 to bypass any really high frequency noise that gets through.

• Thanks a lot. I was wondering what those "current limitation" and "thermal shutdown" currents were, and you explained it! Do you trust the capacitor to reliably filter transients, though? Do you have a recommendation on the type of the capacitor (in regards to its internal resistance, for example)? Shouldn't I at least add a serial input diode before the voltage divider? – Violet Giraffe Jan 30 '16 at 17:24
• The datasheet shows how to detect current limitation and thermal shutdown by measuring the voltage drop on the input feed resistor using an analog input pin on a micro. I've updated the answer to address your capacitor questions. If you mean a diode pointing down in series with R1, then no. It will pass transients as well as the supply current. – Transistor Jan 30 '16 at 17:44
• It will pass forward transients as if it wasn't there, but at least it will block the reverse polarity transients, no? I mean, it basically costs nothing (as long as it doesn't get burned). And thanks for the answer. Your solution is neatly simple. In fact, I thought about the very same thing, but I had no idea if the capacitors will, in fact, do what's expected from them here. – Violet Giraffe Jan 30 '16 at 18:03
• You're welcome. Just to clear up some of the comments on the OP: the battery is still relevant as it's in circuit when the alternator is running and greatly affects the stability of the vehicle's electrical system. Any circuit schematic would have helped even with a question mark for the part you are unsure of. For example, I added SW1 as I assume you want to switch the device whereas your OP reads as though you will leave 5 V on the input permanently. I had to read several times. The better the question, the less guessing we have to do and, probably, the better the answers you will get. – Transistor Jan 30 '16 at 18:20
• Yes, of course you're right. The switch you added is exactly right: it's the ignition key switch. And no, I can't just omit the intelligent switch and power my load directly from the ignition switch. Asking a good question requires knowing half the answer, they say. And the better you explain which aspect of the question you don't understand, the better I can clarify it! I mean, it's a 2-way road. – Violet Giraffe Jan 30 '16 at 18:33

Something like an off-the-self voltage regulator like this one can produce 5v with an input anywhere from 8-45v, however it may start to get a little toasty at 45v.

In order to protect the regulator from the harsh transient voltages present you should use a TVS(Transient voltage suppression) diode. They come in bidirectional types so if the voltage spikes to high or low they will turn on.Putting it simply they are like a zenzer diode with a fast reaction time.

Example of such a diode

• I think Violet Giraffe's concern is with voltage transients from, say, the car's starter motor kicking in/out for the 60-100v +/- spikes concern. – Robherc KV5ROB Jan 30 '16 at 15:20
• @RobhercKV5ROB: exactly! So no, even if there is a well-regulated line somewhere in there, I have no access to it. – Violet Giraffe Jan 30 '16 at 15:28
• While I don't know a ton about automotive wiring, I would think the 12v source that is present on the cigarette lighter 'jack' already deals with with the transients. Otherwise even things like phone chargers would have to be able to deal that transients. – Dave851 Jan 30 '16 at 15:28
• @Dave851: no, you would be wrong to assume that, and if you do - your devices would die after some time. And yes, most phone chargers do not account for the automotive specifics, and they die from time to time (not instantly, ofc.). But some chargers actually are designed correctly to withstand the transients. – Violet Giraffe Jan 30 '16 at 15:29
• My mistake, like I said I don't know to much about cars. There is actually a diode to deal specify with transient voltages however, I'll update my answer in a sec. – Dave851 Jan 30 '16 at 15:34