Fast, Quick, Cheap, Good looking LED room lighting (for anyone)

Step 3 Make the LEDs

Now what i love about this system is that you can switch, move and change colours so easily!

How to ready the LEDs...

Bend one corner of the resitor and the + (positive) leg of the LED and hook them together.
Then put on the solder.
Then cut the extra off.

And who said soldering was hard!


Then bend the legs like shown in the picture so that both the legs point downward and make them the same length. about 4mm not long

ok now do that as many times as you want... Its easy...

Comments

[meissler]

I'm a bit confused with the resistors here. Everything is in series, right? So you have a 12V battery, and let's say (just for a clear example) 10 LEDs that can have 1.5V across each. Why do you need a resistor, and why one resistor for each LED? Isn't the voltage across each LED going to be the same without a resistor or am I wrong? Thanks.

[junits15]

Each LED will be receiving 12V when they are wired in parallel. So each LED will need its own resistor in order to work with the higher voltage. It is important to note that the more LEDs that you use the more power they will draw from the battery.

[meissler]

oh crap it is in parallel cause of the 2 cables, didn't put that together. I'm a little confused about the cables though. You use 2 of them: one end of each of them gets hooked up to +/- of the power adapter, but what about the other ends of each wire?

[junits15]

they are left untouched and seald possibly with tape, if they were connected it would just create a short.

[Hycro]

I would just stick a female plug on the end, just to make it a little neater in appearance, plus, you could connect another string, assuming your power supply is capable of it.

[Electronics Blurred]

Actually , you could connect 3 (3.3V) LED's .
that would give you 12 - 9.9V = 2.1 V ( Instead , of a resistor , connect a smaller LED ) (and a small amount of resistance , usually )
It can be seen on http://www.LEDCalc.com/
What i did for my room lightning was 2 LEDs and 1 130 & 120 Ohm Resistor ( I Wanted more brightness , since it's my room , and i could replace led's quickly )
In each line .
Because it's 9V and 350ma From my wall-wart , i placed 17 lines of LED's , (34 LED's )

[errow]

if i used a 6v power supply would i need resitors ?

[Thenwcp]

yes, because resistors do not limit voltage or current very well and will burn out quickly even at voltages as low as 3V

[skrubol]

Running like this is quite inefficient, also your resistance calculation is a bit off. White and blue LED's run at about 3v, so you are burning 9v in the resistors and 3v in the LED. About 25% efficient. Also, for your resistance calculation, for 'v' you should use 'Vsupply-Vf'=12-3=9v. So your LED's are really only using 18mA.
If you want to double or triple your efficiency, you can run 2 or 3 LED's in series with one, smaller valued resistor. 2 LED's with a 200 ohm resistor would get you 30mA. 3 LED's with a 100 ohm resistor will be a bit less stable (LED's will get brighter as they warm up, and small variations in your power supply will show.)

[aarons123]

ok, you could do this, but how would you get the same cool effect XD around 10 cm apart