Jeff,

Here is the datasheet for the 1206:

http://media.digikey.com/pdf/Data%20She ... 150CKT.pdf
Just putting them in series and adding the forward voltage drops together is not a very good design technique because you can't really guarantee the amount of current that will go through the LEDs. Here is a better way to design what you want to do:

The max DC current allowed is 40mA, but the specs are all centered around 20mA, so that is where the LEDs are really intended to operate. This will also give you a safety margin in case of voltage variations that are common in automotive applications.

While a car's battery is nominally 12 volts, it falls well below that when the starter is cranking, and goes well above that when the alternator is charging the battery. The voltage can typically run up to 15V while the engine is running. For design purposes, I design automotive circuits to survive at least 18V. In fact, for circuits that go into production, I add a schottky diode such as a 1N4148 to cause the circuit to survive someone putting jumper cables on backwards, and I add an 18V TVS (transient voltage suppressor) to suppress voltage spikes generated from things like the battery getting unhooked while the alternator is spinning.

For a 1-off circuit such as the one you want to make, you don't really have to add those protection devices, unless you can't afford to make another circuit should one of these events happen. TVSs are kind of hard to find, and the 1n4148s aren't really needed, since your circuit is made from LEDs (diodes). I just include them FYI.

For the basic circuit you want to make, I would design for a 15V supply, with 20mA running through each diode. The forward bias voltage at 20mA is 2.3 max. So for a circuit with one diode, you would subtract 2.3V from 15V to get 12.7V. From Ohm's law, V=I*R, or R=V/I we know that the current limiting resistor you will need is R=12.7/0.020=635 Ohms. So you would want to pick a current limiting resistor that is the next size up from 635 ohms. To check the wattage you will need, you use the power equation P=V*I=12.7*0.020=0.254W. So to be safe, you should use a 1/2W resistor.

If the voltage is lower than 15V, you are safe. If it goes up to 18V, you will get 18-2.3=15.7 across that 635 Ohm resistor, which gives you 25mA, which is well below the 40mA max.

If the light from one diode is not enough, you can run more than one in series. For example, if you run two in series, the calculations would be as follows:

Subtract 2*2.3=4.6V from 15V to get 10.4V. To get 20mA through the LEDs, divide 10.4V by 0.020A to get 520 ohms.

For 3 LEDs in series, subtract 3*2.3=6.9V from 15V to get 8.1V. To get 20mA, divide 8.1 by 0.020 to get 405 Ohms.

For 4 LEDs in series, subtract 4*2.3=9.2V from 15 to get 5.8V. To get 20mA, divide 5.8 by 0.020 to get 290 Ohms.

For 5 LEDs in series, subtract 5*2.3=11.5V from 15V to get 3.5V. To get 20mA, divide 3.5 by 0.020 to get 175 Ohms.

For 6 LEDs in series, subtract 6*2.3=13.8V from 15V to get 1.2V. To get 20mA, divide 1.2 by 0.020 to get 60 Ohms.

I would not go above 6 LEDs in series, since your required current limiting resistors will fall below 0 Ohms, which is not possible. If you need more than 6, you will need to run them in parallel.

To run them in parallel, you multiply the amount of current used by one LED by the number of LEDs you will use. For example, for two diodes, you would do the following calculations:

Multiply 20mA by 2 to get 40mA or 0.040A. You will get the same 2.3V drop across all LEDs in parallel, so to get the current limiting resistor, you will divide 12.7V by 0.040 to get 317.5 Ohms.

For 3 LEDs in parallel, you would divide 12.7V by 0.060 to get 211 Ohms, and so on. To calculate the resistor power dissipation, you multiply the voltage by the current. For 3 LEDs, you would calculate 12.7*0.060=0.720W, so you should round up to a 1W resistor. While the wattage of the resistors in the circuit where the LEDs are run in series goes down, the wattage in circuits where the LEDs are run in parallel goes up, so you need to recalculate the wattage for the current limiting resistor for every case.

So you will need to run some tests to see how much light you want in order to see how many diodes you will need.

You will also need a substrate upon which to mount those SMD devices. They are really tiny. I'd recommend that you hack up a prototype such as this one from Radio Shack:

http://www.radioshack.com/product/index ... Id=2104052
Or this one:

http://www.radioshack.com/product/index ... Id=3173937
For SMD semiconductors, you will want a steady hand and a very small soldering iron. I use this Radio shack unit:

http://www.radioshack.com/product/index ... Id=2062728
Be sure to use rosin core solder, not acid core solder.

HTH