Hardwiring LED’s

Calculation for resistor:

Car battery voltage: 12V
LED forward voltage: get from LED specs, about 3V
LED current: get from LED specs, about 30mA

Resistor = (12V - 3V) / 0.03A = 300 Ohms
0.03A = 30mA

Power dissipated in the resistor:
(12V - 3V) * 0.03mA = 0.27Watt = 270mW

So, it should be a 1/2Watt, around 300 Ohms resistor (doesn't matte if you don't find exactly 300 Ohms, can be higher BUT NOT LOWER).

Get that resistor from a local electronics supplier (or Radio Shack if you're too lazy to look for one and have extra $$$).

Hope it helps,
TI|CP

OK, let's say each LED is rated at 3 volts. So, you can only put 4 LEDs in series (3 times 4 is 12V), WITHOUT needing the resistor.

If you want to put more, than make series of 4 leds and put those series in parallel. EX: you want 8 LEDs, so make two groups of 4 LEDs and put those two groups in parallel.

Make sure you don't mess the polarity. In series, (+) of one LED goes to (-) of another LED. In Parallel, (+) goes with (+) and (-) with (-).

If you need to put 6 LEDs, then put 4 in series, then form a group of 2 LEDs and a resistor (value = [12-(3+3)]/0.02 = 300 Ohms), and put this group in parallel with the first one (**I assumed LEDs of 3 Volts at 20 mA).

If you're still confused, I can post a picture.

Alex

i understand what you are saying, but i am a visual learner and that would help out greatly

Whatever groupings you choose, try and make them even and the total count even, then you won't need different resistors for each group.

Ex: 9 LEDs is better in 3 groups of 3 than 2 groups of 4 and one single.

Hope it helps (and doesn't confuse you more, :D)

Alex

I don't agree with that picture very much, the first image would be ok if your LEDs matched ok.  The second one has the resistor on the negative side, you need the resistor on the positive side before the LED series.  The third picture has the LEDs in parallel with one resistor, this should be avoided in case LEDs don't have the exact same voltage, one will light and others not and then the one will burn out.

The second image is the most useful, but put the resistor on the other side of the series, on the positive side of the first LED.  The second most effective way is just one resistor and LED, then you don't have to connect the LED legs together if they are far apart.

There's NOTHING wrong with the resistor's placement in second image. It doesn't matter where it is, as long as it drops the remaining voltage (at a given current). It can even be placed between two consecutive LEDs.
The thing is that LEDs will drop 3 volts each, for a total of 9 volts for 3 LEDs in series. There remains 3 volts that should be "dropped" by something. That something is a resistor, calculated using the 3 volts "to drop" divided by the total current through the group.

I assumed that ALL LEDs are same, rated at 3 volts and 20 mA. That way, you can put groups in parallel with only one resistor (the current through the resistor will increase, decreasing its value and increasing the dissipated power).
Of course, if you have mismatched LEDs, then you will have to recalculate the resistor value for each group, and you probably couldn't put them in parallel with only 1 resistor (like in 3rd image).

Alex

My mistake, I was thinking backwards.

As for the parallel, I don't do that because I bought some 4000mcd red LEDs once and didn't have enough so I bought a few more and they ended up being 5000mcd and a different mA, can't quite remember what though.  Anyway, I was working with them and got them mixed up.  They are all red and I was not able to tell the difference, not really on the brightness either so having them in parallel might not be the best decision.  Although they were the same voltage and I ended up averaging the mA rating for the resistorsand they worked fine.

https://www.home.agilent.com/cgi-bin/pub/agilent/expandedresults/cp_ExpandedResults.jsp?NAV_ID=-25324.0.03&LANGUAGE_CODE=eng&contentType=Editorial&entityType=ED30&COUNTRY_CODE=US

If the link works, it's all you've wanted to know about LEDs.. poke around on the site for more info..

Jim

Wow, that website is great. Everyone should check it for a HUGE amount of info :)

Chris, no problem. As long as the current isn't TOO different, you're fine. If it's higher, than your LEDs will be brighter, if it's lower, they will be dimmer. It's just that a LED can't drop too much voltage (because its equivalent internal resistance is VERY small, so if you increase the voltage a little, like 2 volts, the current goes higher exponentially).

-Alex

LOL, yea, I'm 100% with you here, jeepmansro.

The inconvenience with neons is that they break easily, not as long-lasting (as far as I know) and consume much more current (need a bigger alternator). LEDs can be ran over with a truck several times and nothing will happen to them. They last longer and are very little consumers. The problem with LEDs though is that the light is concentrated (focused) and not as diffused (going in all directions) as neons. But that's not a problem, right???