TVS Diode Question

I've been working with Arduino and other circuits for over a year now, and while I've gained a tremendous amount of knowledge, I'm still struggling to understand basic concepts around voltage, current, and power.

Here's what I'm struggling with at the moment:

I'm trying to build protection in a circuit that I'm placing in a vehicle. From all of my reading, from time to time there can be transient voltage spikes in a vehicle's electrical system, sometimes in the neighborhood of 1,000 volts for a few milliseconds. I found a company that makes a TVS diode specifically for use in vehicles, called LittleFuse.

I settled on this model after talking to one of their engineers:
http://www.mouser.com/ProductDetail/Littelfuse/15KE18A/?qs=sGAEpiMZZMuNo3spt1BaV%2FHTyk9HbrguYFEEAym1JrI%3D

My circuit will be drawing less than 50ma during normal usage, and I don't need any more than 5V for the normal operation of my circuit. I need to protect the power input coming from the vehicle battery. The power input is used in two places, basically branching off. First, it goes into a DC/DC 5V buck converter to give me the "clean" 5V for my circuit. The second goes to a 22k/10k voltage divider so I can measure the vehicle's voltage so I can send an alert if the voltage drops too low.

So here's my question: the engineer at LittleFuse told me that I would need to put a series resistor in between the power input from the vehicle battery and the diode/my circuit (the diode will obviously go across power/ground). He said 75ohms will work fine.

What exactly is that resistor doing? The reason for the resistor is that I didn't want to use a larger diode that they make, because it's leads were too big to fit into a breadboard, so he said I could use the smaller 1.5KE, but I would need the series resistor.

He also said that the resistor would be dropping the voltage, so I would need to take that into account when measuring the battery voltage of the vehicle using my voltage divider going into my analog pin.

How do I calculate how much voltage the 75ohm resistor is going to drop when taking that into account with my already-in-place voltage divider? So now it's essentially a voltage divider with 3 resistors instead of two?

If I only put the 75ohm resistor in series with the "branch" that's going into my DC/DC buck converter, a) will the TVS diode still give me "protection" in the circuit, and b) given the 22k/10k resistors (on "branch" b), will I be "missing" anything critical by not also having the 75ohm resistor?

If anyone has time to speak for a few minutes by phone (would be awesome), please PM me.

Thanks!!!

If you use a DC/DC converter, and a 22k/10k divider, you don’t need “close protection” (15volt TVS).
Depending on the DC/DC converter used, a 24volt TVS diode would do.
Then you won’t need a series resistor. Just a (fast) fuse.
Leo…

edit.
A 22k/10k divider dumps <1mA (300uA) into the pin protection diode with 24volt peaks.
I would consider that safe.

Why are you using a switching regulator for 20ma?
A linear one would be cleaner output.
Many automotive applications get by with a choke coil and a capacitor
for surge protection.
Most all of the high voltage spikes are caused by the starter motor.
Many people blow their alternators when jump starting cars because they
don't understand that.
A capacitor and a resistor going into a linear regulator would be fine for the
20ma load. There you can loose a couple volts drop,
The analog in could be protected with a zener of something like 7V placed
part way down the divider so that it is always off with less than 15V at the top of the
resistor divider ( split into 3 resistors ).
Are you planning to put a trim pot or is 5+% what you are targeting for your voltage
reading?
Dwight

So here's the thing: I'm a smart guy, I promise :-). But I'm totally lost by what you said Dwight.

Also, Leo, thanks for the reply: the buck converter and the voltage divider are on different "branches" of the battery input. One is used for powering the circuit, and the other is used to simply measure the battery voltage of the ambulance.

I know what you might say: if I can't understand fully what Dwight is saying, I probably shouldn't be messing with this. But, I've got all the other programming down, all the sensors wired up, everything is working great, the last piece of the puzzle is the circuit protection.

Any chance you can boil it down for me in very simple terms: if all I need is a capacitor and a zener diode, what specs of each, where exactly should they be placed in the circuit.

I already have the 1.5KE TVS diode and the resistors, so if those will WORK, then maybe I can just use that? I'm just trying (desperately) to understand how voltage and current are factored into and calculated in this circuit design.

Thanks guys! Really, really appreciated.

Current flow though a resistor will cause a voltage drop
I*R=E
Figure how much current without the 1.5KE, since it is not
drawing current. Do this by figuring the total resistance.
Resistors in series add.
A schematic of what your doing would help.
Dwight

Schematic view: Image 2016-01-05 at 8.21.11 PM
PCB View: http://cl.ly/0T471G120T2Y

What exactly is that resistor doing?

When the diode conducts due to spikes the resistor limits the current.

dwightthinker:
A schematic of what your doing would help.

Fritzing confused me more. You should have added "hand drawn, with currents and voltages".

Two? buck converters, a FRAM (Google points me towards air/oil filters and old sail boats), a RTC, bluetooth, temp module, and no I2C pullup resistors.
Peak current (what you usually calculate the supply for) might be more than 50mA.

I suppose FRAM is Ferroelectric RAM, dug up from the grave.
Something I experimented with in the seventies (8 bits! of ringcore RAM).

Did you already get it all working together on a breadboard.
Power protection is probably the last thing to worry about.
Leo..

If you worry about pin protection, and want accurate/stable readings, 22k/10k and default Aref is not a very good way to measure a 12volt car battery.
Using Arduino's inbuild 1.1volt Aref and dropping the divider to 1.1volt kills two birds with one stone.
Stable readings and more protection. Because the "top" resistor of the divider can have a higher value.
1:15 (1/16 of the battery voltage) measures to ~17volt.
1k8/27k, 2k2/33k or even 10k/150k could work.
Do add a 100n cap from analogue pin to ground to kill noise/spikes.
The analogue input stays under 5volt with 80volt on a 10k/150k divider.
Above ~88volt the pin protection diodes take over.
Leo..

But the problem with the 1.1 internal reference is it varies significantly between devices so you
would need to measure a calibration value per chip and store in EEPROM. 5V regultators are more accurate
out of the box.

I don't see that as a problem. You should always calibrate a voltmeter.
Onboard 5volt regulator (line/load regulation), USB supply, voltage divider resistors. All can be off.
At least the 1.1volt Aref stays there.
Maybe a good idea to write the Aref value to EEPROM.
I simply write it on the board.
Leo..