mikeinaus wrote:what ones did you guys buy? is there something special im supposed to do for them? there not even 1/2 as bright as the regular bulbs.
The first thing you have to know is that "super white" doesn't mean "super bright".
The LEDs you bought are not designed to be brighter than the bulbs they replace. They're designed to be whiter, more reliable, and consume less energy.
Then, you have to consider that, while incandescent bulbs emit light in (pretty much) every direction, LEDs are more or less directional, depending on their design.
The link Mi|<E provided is great, if you know what you're looking for.
So, if you look at your bulb specs, you will see that, for example, an 1157 bulb high filament emits 32 candela, or 402 lumen.
http://www.run-n-lites.com/bulbspec.asp
which means that, if you look at Mi|<E's link, if you were to use the first of the "5 mm LEDs" listed, you'd need 21.33 LEDs to get the same brightness over a 120° viewing angle (like, for example, for a brake light), while, if you only needed to cover a 30° viewing angle, you'd hit the jackpot with the 5th LED, of which you'd need only 2.
Then, if you go down to the "5050 SMD LEDs" list, using either one of the 3 first ones (depending on the color you want), you will only need 6 LEDs to be a little brighter than the bulb.
Once you chose your LED, click on the part number. You will get the LED specs.
What you're looking for is the "forward current" and "forward voltage" values. For example, for the last LED I used above (5050-CW6000), the forward current is 3x20mA=60mA ( the "3x" is because this is an RGB LED, which means there's 3 LEDs in one capsule), and the typical forward voltage is 3.2V.
Armed with those values you can calculate the resistor you need to connect to the LED, so you won't burn it. For that, take your battery voltage, plus some extra for safety (I normally use 15V), and subtract the forward voltage. In this case, 15V-3.2V=11.8V.
Now, take that voltage and divide it by the current in Amps (60mA=0.06A), and you will have the value for your resistor, in Ohms. In this case, 11.8V/0.06=196.66 Ohm, or, to round it up to the next HIGHER value, 200 Ohm. Then, you multiply the voltage by the current, and you get 11.8Vx0.06A=0.708W. Resistors come in 1/4, 1/2, 1,2,4 W, etc., so, in this case, you're left with a 200 Ohm, 1 Watt resistor. Solder that resistor to either the anode or cathode (or, in this case, the 3 anodes or cathodes), and connect to the 12V battery. If you put the resistor in the anode, connect the resistor to the positive, and the cathode(s) to the negative. If you connected the resistor to the cathode, invert the polarity.
Finally, you can add the forward current of all your LEDs (providing they all have the same values, of course), and calculate only one common resistor for all of them. The downside of that is that, if one LED burns, depending on the total number of LEDs, sooner or later, they all burn.
Hope that helps. Sorry for the long post.
