Look at that bulb below : it consumes about 200 mA of current (at 12V DC) and produces considerable heat when lit. Needless to say that its light intensity diminishes over time (or easily burns out completely !).
There is a far better, cheaper and economic alternative...
Step 1 : Get a high intensity LED of your favorite color (blue, white, red, green etc) and a 470Ω resistor (1/4W). This combination consumes about 20 mA @ 12V DC and produces a very satisfying light intensity. You can also try a 330Ω resistor for an even higher current of about 25 mΑ. Don't go any lower because you might lose the LED :)
Step 2 : Place the resistor between the two LED legs as shown in the picture below.
Step 3 : Solder the resistor legs to the LED positive (long) leg as shown in the picture below. Use a small yet sufficient amount of solder.
Step 4 : Cut the resistor legs as shown in the picture below.
Step 5 : Cut the LED leg part between the two solder spots as shown in the picture below.
Step 6 : Get a small length of heat-shrink tubing. A 4 mm diameter will do perfectly.
Step 7 : Cover both LED legs and the resistor with the heat-shrink tubing. Pay attention to the exact length of the tubing; shouldn't be too long (just to cover the resistor). Be careful not to overheat in order not to melt the solder inside !
Step 8 : Cut the LED legs shorter, as shown in the picture below. However, leave the positive leg (with the resistor) a bit longer !
Step 9 : Carefully bend the LED legs outwards, as shown in the picture below. Bend a bit less than 180 degrees, in order for the LED to 'lock' better in the bulb housing.
Step 10 : Place the 'LED bulb' in its housing. Don't forget to check the polarity !!! (former bulbs didn't have any polarity).
You're done !
Enjoy your new LED indicators :
- they consume less
- they light the same (if not brighter)
- they live longer !
Great work Bill! thumbs up from me :)
ReplyDeleteIndeed, it's a great post, you have illustrated a simple but useful way on make a LED Bulb.
ReplyDelete