How to connect multiple luxeon LEDs together?

[schissler]

Doing a sort of project and bought 2 dozen Luxeon K2 bulbs, my question is how would I got about connecting them all together to a 12v source?

Parallel with a specific resistor on each the best way to do it?

What specs do I need to know when going to buy the resistors?

 

[Illum]

Resistors puts out alot of heat, but I suppose it could do here.

Luxeon K2 has a foward voltage of 3.4V [typ] at 350ma.
Resistance is a function of voltage and current, how much current do you want the LEDs to see? Assuming your 12V power supply can handle it, each K2 is capable [under correct heatsinking methods] of being driven to an amp.

By 12V I assume your doing this by an AC-DC power supply, not the automotive voltage yes? cuz Automotive voltage is more like 14.4V nominal with occaisional [startup] 18V spikes

 

[Luminescent]

Doing a sort of project and bought 2 dozen Luxeon K2 bulbs, my question is how would I got about connecting them all together to a 12v source?

Parallel with a specific resistor on each the best way to do it?

What specs do I need to know when going to buy the resistors?

Not in parallel, in series.

For 12v operation I would string three in series with a resistor like this.

(12V - )----(-K2+)----(-K2+)----(-K2+)----(RESISTOR)----( + 12V)


Note the plus and minus polarity markings above as LEDs are polarized devices that will only work if connected properly.

In this case the total drop across the K2's should be approximately 3.4 volts for each K2 x 3 = 10.2 volts total across the three LEDs

This leaves 1.8 volts that needs to be dropped by the RESISTOR.

Just devide the 1.8 volts by the current needed in Amps to get the value of this resistor using Ohms Law.

A good safe operating current for the K2 would be 350 mA.

For this current level with the 1.8 volts that needs to be dropped by the resistor in the above series string, the dropping resistor calculation would look like this:

1.8volts / 0.350amps = 5.142 ohms

The closest standard resistor would be 5 ohms 2 watt

TWO x 1 watt 10 ohm resistors in parallel could be combined to form the equivelent of a single 5 ohm 2 watt resistor.

/(10 ohm 1 watt)\
\(10 ohm 1 watt)/

FOUR x 22 ohm half-watt resistors in parallel could also be combined to create a single 5.5 ohm resistor.

I mention the option of combining these alternate values, because the one watt 10 ohm and half-watt 22 ohm resistors are often more readily available (they should even be available at Radio Shack)

To see what effect fudging the resistor slightly to 5 ohms or 5.5 instead of 5.142, work the formula backwards.

1.8volts / 5ohms = 0.36amps (360 mA)

1.8volts / 5.5ohms = .327amps (327 mA)

These are small differences since the resistor values are typically plus/minus 5 to 10 percent anyway.

Of much greater concern if you are not using a regulated 12 volt source is the fact that seemingly small voltage changes can cause much greater current changes than you might expect.

For example, as someone has already been mentioned, the nominal "12 volt" automotive systems can easily see peak voltages of 15 volts or more.

The reason that this is an issue is because the LEDs only show small increases in VF as current increases leaving a disproportionate change across the dropping resistor, drastically changing it's voltage drop and resulting current.

For example, let's look at the example above with a 5 ohm resistor, but assume that the input voltage goes up to 15 volts:

With greater current the LED drop increases, but only slightly, so let's assume that we now have three K2s in series dropping 3.6 volts each -

3 x 3.6 = 10.8

As before, any voltage that doesn't drop across the three x K2 LEDs will drop across the dropping RESISTOR.

So taking the 15 total volts available and subtracting the 10.8 volts across the three K2s we get -

15 - 10.8 = 4.2 volts

So for a 25% increase in voltage on the input side (12v -> 15v) we see the voltage across the dropping resistor more than doubles.

By the same Ohms Law formula as we used above, we can see that the current also more than doubles.

4.2volts / 5 ohms = 0.84amps (840 mA!)

The K2's could survive this, but you would need a damn good heat sink, and the dropping resistor is now wasting 4.2volts x 0.84amps = 3.5 watts

So you would probably want to go with a 5 watt rated resistor.

For nice stable operation in a car, with no brightness or power flucuations, I would use a simple three terminal low dropout 12 volt linear regulator IC in front of the dropping resistor to insure a stable 12 volt supply.

A simple one amp 12 volt linear regulator IC costs only a dollar or two, and each one could easily supply three parallel connected .3 amp 3 x K2 LED strings (.9 amps total). With 3 series LEDs in each of these strings, and three strings in parallel on each 12 volt regulator, that's 9 LED's total. Duplicate this setup two or three times and you could easily handle 18 or 27 K2 LEDs.

Or just find a single 3.5 amp rated 12 volt regulator IC and then you can put all 4 parallel 3 x K2 strings on this single regulator (24 K2s total)

Remember to pay attention to the heat sink and input/output capacitor requirements needed for stable operation of the voltage regulator. This need not be complicated, these regulators are ridiculously simple, just a input capacitor, the device itself (bolted to a good heatsink), then an output capacitor. Just be sure to keep the capacitor leads short and make sure to check wheather you can safely bolt the regulator right to the heatsink, or wheather you need an insulating washer and things should be fine.

This would get you a LOT of lumens with fairly reasonable LED drive efficency (60% to 70% efficiency) and very stable light output.

If your 12 volt source is fairly stable, then a simple 5 or 6 ohm 2 watt resistor in series with each 3 x K2 devices in series should be fine by itself.

 

[Popsiclestix]

Isn't that resistor usually there to limit current instead of just dropping voltage. It seems to me it should be specced to be able to take the whole 12V across it in case of a thermal runaway situation.

First, a story:

I actually did something like this 2 years back for blinkers (except I think it was 3 strings of 9 LEDs) on a solar vehicle team I worked on.

We were hooking up to the 12V supply on our BMS system (consisting of a NiCad battery pack). When I hooked up the LEDs to our supply and left the blinkers going for about 15 minutes they started going very dim.

I hooked up my DMM in current measuring mode to the blinker and found that the whole thing was pulling several hundred mA (on 5 mm LEDs!)

It turns out that some strings had a markedly different Vf so a lot more current was running through one string than the other.

The string that was pulling in most of the current had gotten so hot that it's effective resistance was close to insignificant.

Even one bad LED can trigger this situation. If one LED starts having thermal runaway, its resistance will drop. This will increase the current to the other LEDs in the string, then they may also begin to fail. Pretty soon, the resistor will be all that stands between a short-circuit situation.

Getting back to your situation. At 5 ohms for each string and 12V applied across each string:

P = V^2/R = 144/5 = almost 30W!

Generally it's not a good idea to hook large amounts of LEDs up in series (or parallel) for that matter without a current-limited source.

 

[TorchBoy]

:welcome: schissler.

Thermal runaway can't happen in a single series string with a suitable resistor in it. You might get the forward voltage dropping a bit, but that's not thermal runaway.

http://ledcalc.com is one place to start for diagrams of what to do, and it'll give you an idea of what power rating the resistors need.

 

[Popsiclestix]

It sure can. Assume a 3 emitter string. Each emitter has Vf 3.6V. Now suppose due to improper heatsinking (bad thermal epoxying job, whatever), just one of the emitters starts overheating. The overheating emitter's Vf will start dropping.

The Vf drop will now put more voltage on the other emitters that are still functioning nominally. Pretty soon, the increased voltage to those emitters will also cause them to fail, dropping their Vf. In the worst case situation, all 3 emitters will have Vf = 0, putting the entire voltage of the string across the resistor.

 

[Popsiclestix]

Time for some real numbers.

12V source. 5 Ohm resistor as described above.

Philips Lumileds claims the K2 can have a Vf @ 350mA that ranges from 2.79-4.23V with a 3.42V nominal voltage. Let's just simplify these to 2.8, 4.2 and 3.4V respectively.

Suppose you bummed out on the Luxeon lottery and got 3x 2.8V

2.8 * 3 = 8.4V

12 - 8.4V = 3.6V

Oops you suddenly have 3.6V across your 5 Ohm resistor. That's 720 mA! It didn't even take any failures for you to overdrive your LEDs by 100% of specification.

Of course, in the real world, 720mA is going to drive up your Vf, but since Vf changes very little with current, it's still in the ballpark of massive overdrive.

 

[Luminescent]

It sure can. Assume a 3 emitter string. Each emitter has Vf 3.6V. Now suppose due to improper heatsinking (bad thermal epoxying job, whatever), just one of the emitters starts overheating. The overheating emitter's Vf will start dropping.

The Vf drop will now put more voltage on the other emitters that are still functioning nominally. Pretty soon, the increased voltage to those emitters will also cause them to fail, dropping their Vf. In the worst case situation, all 3 emitters will have Vf = 0, putting the entire voltage of the string across the resistor.

If this happens, it would just be due to an overall bad thermal heatsink design, and not because any single LED in a series string 'hogged' all the power and run itself away thermally. In a series string, by definition, the current is exactly the same through the resistor and all three LEDs. So if one LED heats up more than the other two, it's VF drops and it will actually dissipate relatively LESS than it's neighbors (which should tend to equalize the temperatures between the devices).

As long as you use a dropping resistor that drops a reasonable voltage, then the relatively minor negative Vf temperature coefficient voltage change as the LED heats up won't cause any more of a current rise than a similar small voltage increase on the input (which will be easily accomadated).

In the case of the three LED string I gave as an example for instance, a 100 degree C change in junction temperature would not shift the VF down enough (even if it occurred on all three LED devices) to raise current by as much as the 3 volt shift up to 15 volts from 12 volts that I gave as an example of the kinds of normal shifts you will see in an automotive 12v system.

As noted, the K2s can survive the worst case current even at 800mA and if thermal effects raise that to 900 mA or even 1 amp, that is not going to be an issue unless the heatsinking is not up to spec.

To be on the safe side (or to make up for marginal thermal design on the heatsink), we would normally just derate the operating current point, by using a larger resistor.

With three LEDs and a 10 ohm 1 watt resistor on a nominal 12 volt battery, for example, the operating current would be only 180 mA at 12 volts, and if the system was on charge and the voltage went up to 15 volts, that would only increase the LED drive to perhaps 500 mA.

These are safe numbers, but only if the heatsink is adequate.


Unfortunately, most folks assume that they can just bolt the LED to ANYTHING (even plastic!) and it will be fine, and that's where they run into thermal problems.

Inadequate heatsink design (along with the morons who insist that LEDs can be treated like light bulbs and direct-driven "if you just get the voltage right"), are responsible for virtually ALL cases of so called "thermal runnaway" that I have seen in power LEDs.

Consider the example I gave with three series connected LEDs . . .

If the heat sink is marginal and one of the LEDs gets so hot that it fails and shorts, then the dropping resistor will suddenly see an additional three and a half volts, and the current flowing will more than double making the other two LED's already marginal thermal situation hopelessly worse, till finally all three have failed or the resistor has burned up.

If you want to be on the ultra conservative side (so you can treat K2 Luxeons or CREE power LEDs more like simple 12 volt 'light bulbs) then you should:

1. put three K2s in series with an 8 ohm dropping resistor, with a conservative 3 to 5 watt power rating for the resistor. (if the resistor is not the 'fusable' type that will fail open gracefully in case of a short, then you can add in a 1 amp series fuse).

2. use a good heat sink having at least 9 square inches of area per K2. This may sound like a lot, so please note that I am not talking about a nine x nine inch square, but rather nine square inches total (three by three) for a flat plate design, and this can be much more compact for high density finned design. (if ventilation is poor, you will need a larger heat sink or you will need to increase the value of the dropping resistor spec'ed above to a value higher than 8 ohms to reduce the current and power per LED)

 

[Popsiclestix]

If this happens, it would just be due to an overall bad thermal heatsink design, and not because any single LED in a series string 'hogged' all the power and run itself away thermally. In a series string, by definition, the current is exactly the same through the resistor and all three LEDs. So if one LED heats up more than the other two, it's VF drops and it will actually dissipate relatively LESS than it's neighbors (which should tend to equalize the temperatures between the devices).

As long as you use a dropping resistor that drops a reasonable voltage, then the relatively minor negative Vf temperature coefficient voltage change as the LED heats up won't cause any more of a current rise than a similar small voltage increase on the input (which will be easily accomadated).

In the case of the three LED string I gave as an example for instance, a 100 degree C change in junction temperature would not shift the VF down enough (even if it occurred on all three LED devices) to raise current by as much as the 3 volt shift up to 15 volts from 12 volts that I gave as an example of the kinds of normal shifts you will see in an automotive 12v system.

One would agree that 5mm (several milliamps) LEDs don't need to be heatsinked. Yet I encountered a situation where a 9s constant-voltage configuration drove them into a thermal runaway! (and yes there was a current limiting resistor in the circuit and the LEDs were not being overdriven).

It seems variation in the Vf (and a bit of bad luck) can result in your design being driven to 700mA even with Tjunction at 25C with a circuit calculated for 350mA. Add in a failure and it's pretty much "good game, burn up all your emitters"

As noted, the K2s can survive the worst case current even at 800mA and if thermal effects raise that to 900 mA or even 1 amp, that is not going to be an issue unless the heatsinking is not up to spec.

To be on the safe side (or to make up for marginal thermal design on the heatsink), we would normally just derate the operating current point, by using a larger resistor.

With three LEDs and a 10 ohm 1 watt resistor on a nominal 12 volt battery, for example, the operating current would be only 180 mA at 12 volts, and if the system was on charge and the voltage went up to 15 volts, that would only increase the LED drive to perhaps 500 mA.

These are safe numbers, but only if the heatsink is adequate.

Unfortunately, most folks assume that they can just bolt the LED to ANYTHING (even plastic!) and it will be fine, and that's where they run into thermal problems.

Inadequate heatsink design (along with the morons who insist that LEDs can be treated like light bulbs and direct-driven "if you just get the voltage right"), are responsible for virtually ALL cases of so called "thermal runnaway" that I have seen in power LEDs.

The OP doesn't seem to grasp the basics of LED circuit design, let alone the thermal issues associated with power LEDs, I was merely pointing out that these back of the envelope calculations aren't particularly safe when you're dealing with "2 dozen K2 emitters" in a complicated series-parallel configuration, which has very good potential to start a fire if not driven properly.

If the heat sink is marginal and one of the LEDs gets so hot that it fails and shorts, then the dropping resistor will suddenly see an additional three and a half volts, and the current flowing will more than double making the other two LED's already marginal thermal situation hopelessly worse, till finally all three have failed or the resistor has burned up.

Sort of what I was getting at, and the definition of thermal runaway.

 

[TorchBoy]

It sure can.

It sure can't, and you're about to learn what thermal runaway actually is.

As I said, thermal runaway can't happen in a single series string with a suitable resistor in it. As Luminescent explained a series situation tends to be self-balancing - the opposite of a runaway situation, in which a condition gets more and more unbalanced, and at an ever faster rate. That's what runaway means - it's out of control, and gets more out of control as it progresses.

In a series string even if the Vf of each LED drops enough for the LEDs to conduct so much more current that it becomes LED-threatening, as soon as one LED burns out, any would-be runaway cannot develop; it stops instantly because the circuit is broken.

Thermal runaway is only displayed in parallel arrays of LEDs. When heat causes the Vf of parallel LEDs to drop, the Vf of one LED can drop more than the others, leading to that LED to draw more current than the others. (Remember, series LEDs can only draw the same current. This isn't the case for parallel LEDs.) Because it's drawing more current, that LED will get hotter, causing its Vf to drop further, causing it to draw more current, and so on, until it blows.

Unlike a series arrangement, having one LED blow doesn't stop the current flowing through the other LEDs, but they are now sharing a current intended for more LEDs. This can cause the next LED to blow quicker, the one after that even quicker, and a runaway situation develops. It continues at an ever-faster rate until all the LED blow.

I hope that's all clear now.

 

[Popsiclestix]

As I said, thermal runaway can't happen in a single series string with a suitable resistor in it. As Luminescent explained a series situation tends to be self-balancing - the opposite of a runaway situation, in which a condition gets more and more unbalanced, and at an ever faster rate. That's what runaway means - it's out of control, and gets more out of control as it progresses.

If the heat sink is marginal and one of the LEDs gets so hot that it fails and shorts, then the dropping resistor will suddenly see an additional three and a half volts, and the current flowing will more than double making the other two LED's already marginal thermal situation hopelessly worse, till finally all three have failed or the resistor has burned up.

This is thermal runaway in a series string, quoted directly from Luminescent. Heating up LED causes more current to flow through the string, causing it to heat up some more. The very definition of a thermal runaway.

In a series string even if the Vf of each LED drops enough for the LEDs to conduct so much more current that it becomes LED-threatening, as soon as one LED burns out, any would-be runaway cannot develop; it stops instantly because the circuit is broken.

NO! LEDs are DIODES, and diodes can (and will often) fail SHORT. They'll still continue to function as perfectly good conductors with little to no power dissipation. This will continue until they experience physical destruction of the wires. At that point, hopefully you'll just blow the resistor or maybe the wire bonds on the LED, but all the other LEDs in that string have been destroyed because they've had to bear the increased current. And if you've mounted the device on something flammable, it'll catch on fire.

 

[VidPro]

when i run many and many series parellel, with resistance, i use way to much wattage of resistance, so the resitance itself doesnt change.
and i use lots of LEDs run lower , as opposed to less run harder, by doing that the things seem to last forever.
havent had a string blow out using that methodolgy.

i think if your going to have everything run PEAK drive constantally you might have a greater need for current control, but then you often lose more power, and have more chances of failure with more unnessisary junk in there as opposed to light output items, plus you then only get normal lifetime of the light output item. if the money is put into more output items run lower, with simpler curcuit system it will run longer today and tomorrow TOO.
and these things are more efficent and less of a heat problem when run lower anyways. Where does the heat go in containment when you have driver losses too?
put your money where the light is

as far as 5mm led, throw them out use a few high powered in the same place. i have a device that ran a 5mm replaced with a cheap high powered run low, and it puts out more light and hasnt failed in 10+ years, wheras the 5mm lasted mabey 3, before waning down in output.

when your talking 12v though, you gotta ask is it ALWAYS 12v or is it automotive 12v, or battery operated, or magnet transformer unregulated, or failing switching power supply, because that isnt always 12v even if it says so on the label

 

[Popsiclestix]

when i run many and many series parellel, with resistance, i use way to much wattage of resistance, so the resitance itself doesnt change.

Resistors don't change resistance (much) unless you blow them up.

Unless you're using a lightbulb as a resistor, to provide additional illumination.

You just gave me an idea to use lightbulbs as a circuit element. Because filament resistance increases with temperature and the device almost inevitably fails open, one could use it to control LEDs? :naughty:

 

[VidPro]

Resistors don't change resistance (much) unless you blow them up.

Unless you're using a lightbulb as a resistor, to provide additional illumination.

You just gave me an idea!

yes again why waste power the right size light bulb will also offer up some minor current control too.
and yes resisters go freaking nuts when people use way to small of sizes for thier applications, and then not enough cooling exists. then it becomes the Double Carbon resister

 

[Luminescent]

Time for some real numbers.
12V source. 5 Ohm resistor as described above.

Philips Lumileds claims the K2 can have a Vf @ 350mA that ranges from 2.79-4.23V with a 3.42V nominal voltage. Let's just simplify these to 2.8, 4.2 and 3.4V respectively.

Suppose you bummed out on the Luxeon lottery and got 3x 2.8V

2.8 * 3 = 8.4V

I have worked with dozens of high power white LED's, including Luxeons, Rebels, and CREEs, and have never seen one with a Vf below 3 volts at 350mA, but just for the sake of argument, fine, let's assume 2.8

12 - 8.4V = 3.6V

Oops you suddenly have 3.6V across your 5 Ohm resistor. That's 720 mA! It didn't even take any failures for you to overdrive your LEDs by 100% of specification.

Of course, in the real world, 720mA is going to drive up your Vf, but since Vf changes very little with current, it's still in the ballpark of massive overdrive.

Wonderful that you now seem to be able to calculate this stuff, glad to see that my explanation wasn't totally wasted.

Just a few quick points . . .

First, if you did succeed in DOUBLING the current from 360 to 720, from the K2 spec sheet we can estimate that the Vf will increase increase about .2 to .3 volts which is NOT trivial, so your real current increase would be slightly less. I agree that it would not be a lot less though, because the device will heat up a bit, and that takes the VF very slightly the other way. Even so, the resistor will still limit the current enough to insure very stable operation at slightly less than 700mA (and NO the operating current will not "run away").

Second, if you are going to presume to give advice, you should perhaps at least check the spec sheets for the K2, before pronouncing 720mA "massive overdrive" on a device that can quite easily handle an amp or more.

From the Philips Lumiled K2 Luxeon spec sheet -

Industry Best Lumen
Maintenance —50,000 hours life
at 1000 mA with 70%lumen maintenance

However, as I stated above, to use the device at these high current levels, you need to be VERY careful about the heatsinking if you want to maintain that specified 50,000 hour life, and use of the device with a simple resistor hookup dictates that we should be more conservative and derate the device.

Lumiled has all kinds of pretty derating curves, but long story short, the reccomended 5 - 8 ohm resistor should be on the conservative side with any kind of reasonable heat sink, and with reasonable assumptions about Vf and input voltages.

As I also said above, resistor circuits are subject to fairly wide variations due to fairly small changes in operating conditions in the 12 to 15 volt range, so you might indeed see some fairly significant variations.

The simplest way to check your final real world operating conditions is to actually measure the current with a meter on each string during assembly and setup, and then check the resulting temprature rise on the back of the heatsink just under the K2 device with either a contact probe or IR thermometer (I got a very nice little non-contact IR thermometer for 9 bucks on sale, so this need not be a big investment).

Then you can easily adjust the operating point if needed by simply changing the resistor value. In a automotive setup you will want to keep the LED current on the low side with the engine off when the alternator is not charging, and be sure to double check that the current and temp don't go too high with the engine running.

Power LEDs are pretty tough, so if you can keep the worst case heatsink temperature directly under the device below about 80C then things should be fine.

 

[Luminescent]

yes again why waste power the right size light bulb will also offer up some minor current control too.
and yes resisters go freaking nuts when people use way to small of sizes for thier applications, and then not enough cooling exists. then it becomes the Double Carbon resister

Light bulbs work great as cheap dirty PTC (Positive Temperature Coefficient) current limiting devices. In this mode, they don't produce a usefull amount of light, and don't expect perfect current regulation, but if you pick a suitable bulb, the current regulation will indeed be several times better than a simple dropping resistor.

A 12 volt bulb operating at only a few volts drop will run just about forever, and won't fail if the LED string shorts, but it may not be optimum for current regulation since it won't heat up very much at only a few volts (during normal operation when most of the voltage is dropping in the LED string).

So, for something like the above mentioned 3 x K2 series string of LEDs, a 6 volt bulb would be better, and would still last almost forever, but it can be a little tricky finding a bulb with just the right internal resistance because the resistance changes as the filament heats up.

Remember, light bulbs disapate watts primarily through IR and visible radiation, not mainly through conduction or convection like a regular resistor, so you have to take this into account when mounting. This can actually be a GOOD thing. If the IR is trapped inside the enclosure, it will heat up, but if the IR is allowed to pass outside, then the bulb will run a lot cooler overall than an equivelent resistor (because the heat is going elsewhere).

 

[TorchBoy]

diodes can (and will often) fail SHORT.

Where do you find these diodes that often fail closed circuit in a LED driving situation? Have you really considered the implications of that? If it happened often then whole parallel arrays would often go dark because one LED is shorted out. Do you really see that often?

And even if it does happen on occasion, for a runaway you would still need all the LEDs in the series to fail one after another, not quite at the same time, by all failing closed circuit. Quite literally exponentially more unlikely.

 

[TorchBoy]

I ...have never seen one with a Vf below 3 volts at 350mA, but just for the sake of argument, fine, let's assume 2.8

That really may be a bit too low. There are a few CPF threads that discuss the bandgap for blue photons and the absolute minimum voltage isn't much below that, about 2.75 V. So you wouldn't get a huge amount of light out of a real-world white LED running at that voltage.

 

[TorchBoy]

when your talking 12v though, you gotta ask is it ALWAYS 12v or is it automotive 12v, or battery operated, or magnet transformer unregulated, or failing switching power supply, because that isnt always 12v even if it says so on the label

Back on topic - to cope with automotive variations it might be more sensible to go for a couple of DX's buck driver 4 packs in the new design (details). The output current would need to be to toned down (change the sense resistor) or use a single driver for multiple strings with balancing resistors.

 

[Luminescent]

Back on topic - to cope with automotive variations it might be more sensible to go for a couple of DX's buck driver 4 packs in the new design (details). The output current would need to be to toned down (change the sense resistor) or use a single driver for multiple strings with balancing resistors.

I must have missed something, or are those buck regulators only rated to 9 volts?

Nor could you just put two or three of these in series like you could series connect two 6 volt light bulbs across 12 volts. Voltage regulators have a very non-linear negative resistance characteristic (as voltage goes down current goes up). So putting two in series will cause them to immediately go into huge voltage swing oscillations and destroy each other.

Even with a switching regulator, you are lucky to get as high as 75% to 80% efficiency.

Using my previous example, with a 12 volt input and only 1.8 volts across the dropping diode, a simple resistor driver is 85% efficient.

In a car with the more likely 14 volts input, and a realistic 3.4 volts each across the K2s, the total LED voltage drop becomes 3.4 x 3 = 10.2 and then the overall efficiency becomes 10.2 / 14 = 72%

This is NOT a bad overall efficiency rating.

As noted, the power regulation is poor, so the current and therefore the light output will vary a bit, but the eye is non-linear when it comes to sensing illumination levels, so even a two-to-one difference in light output as the input voltage swings will not look as drastic as it sounds.

This also happens with incandescent light sources in cars, where a 20% drop in input voltage can lower the usable light output by 50% as the filaments temperature drops and more of its blackbody radiation shifts farther down towards the infrared.

As I mentioned above, if you want to run the chips harder and stabilize the output, then you can use a linear voltage regulator chip to regulate the drive.

This can be done by running a variable regulator chip like a LM317 in a constant current mode or [much more simply], by using a automotive rated three terminal low dropout 12 volt regulator chip like a LM2940-12 as a 12 volt 'pre-regulator' for each LED/Resistor string.

Or as also noted above in another post, if you are not wedded to the absolute requirement to have a solid-state solution, you can also get improved regulation of the LED drive current by simply using a 6 volt 4watt to 6 watt light bulb as a dropping resistance for each string of three series connected K2 LED's.

For example, consider the case where a 6 watt 6 volt bulb is in series with three power LEDs dropping 10 volts with a worst-case car input voltage of 16 volts.

In this case we are assuming that the drop across the three series connected LEDs is 10 volts, so the drop across the light bulb is 16 - 10 = 6 volts. Since this is the bulb's rated voltage, we can be sure that the current will be 1 Amp @ 6volts in accordance with its 6 watt rating. 1 Amp would be high for a normal Luxeon class power LED, but is fine for high power parts like the K2 if the heat sink is adequate (for lower power parts, or to derate the K2 so a smaller heatsink can be used, just use a proportionally lower wattage 6 volt bulb).

A light bulb current regulator won't give the rock stable current regulation of a solid state regulator, but it will be noticeably better than a simple resistor.

For example with that worst case shift from 12 volts to 16 volts in the example above, the dropping resistor will see a three-to-one change in voltage from 2 to 6 volts (after the 10 volts dropping across the LEDs is subtracted), and therefore the drive currrent would also change by about three-to-one. A properly selected light bulb current regulator could hold this down to two-to-one or less. With a more resonable 12 volt to 14 volt input voltage swing, a simple resistor could cause a two-to-one current change, but a light bulb based dropping resistance could keep the change down to an almost unnoticeable 30% to 50%.

Remember, the efficiency of ALL these solutions is still in the 75% to 80% range with the normal range input voltages of 12 to 14 volt, and would still be a reasonable 68% even with a 15 volt input.

So I would go with a simple resistor if the input voltage is reasonably stable, and make the switch to a light bulb as a dropping resistance, or a solid state three terminal voltage regulator IC only if you need more stable output, or need to insure stable operation over a wider voltage input range.