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DIY LED-Illuminated Sleep Chamber
Bulldozer2003 writes "'Finally something both nerdy AND sexy engineers can do.' It sounds like an oxymoron but this guy took a cue from The Vos Pad and decked out his own dorm room bed with Light Emitting Diodes. They're even fully adjustable 'allowing me to create every color of the rainbow.' Total cost, according to him in an email: 'Around $25, the LEDs cost me about $0.25 a piece in bulk, and the potentiometers cost about $6 a piece from digikey. I got the LM317 voltage regulators as a free sample from Texas Instruments. Lots of companies will ship you free samples, its a good deal for college students.'"
Reminds me a lot of this guy's projects. He made some damn cool things out of LEDs there, complete with howtos.
the link wasn't working so I'll try again!
Pot signing out
Jaj
Most LEDs operate at a relatively low current (~20mA) and voltage (~3V). This amounts to maybe .06 Watts (60mW) per LED. It looks like he has 5 panels of 4 LEDs and a 6-LED reading lamp, from the pictures; this makes 26 LEDs, consuming around 1.5 Watts in total. This is 1/40th of the power consumption of a single 60-Watt light bulb. If we say that electricity costs $.06/kWh ("US Federal Average"), then it would cost approximately $.09 to run these lights for 1000 hours.
The kinds of LEDs will probably have different operating characteristics than those I have in my head (like those UV LEDs, which are higher frequency -and energy- than I'm used to).
Whatever it is, it will not exceed the power output of the wall wart he's using.
Interesting! This guy's project basically connects a dimmer switch each to red, green and blue LED strings. The colors sorta mix, sorta producing colored light, but as you can see in his pictures there are major fringing effects (multicolored bands of light). The howto on this page, suggested by another poster, gives a much cleaner result.
The link above uses a microcontroller and pulse-width modulation to vary each color's intensity, producing a much more even color effect.
Now, of course, I want to redo the apartment with them. Eternal lighting with no more power consumption than a couple of flashlights...yum...
You can use any sort of LED you like - It basically comes down to:
a) The colour you want.
b) How bright you want it.
c) How big you want it.
You shouldn't connect LED's straight to a battery.
You need a series resistor (In between the battery and the LED). The value of the resistor is calculated using
R = (Vb-Vf)/I.
Vb = Battery Voltage
Vf = Forward Voltage (On Data Sheet)
I = Current (On Data Sheet)
so for this 5mm white LED using a 9V battery Vf = 3.6, I = 30mA => R = 180 Ohms.
Make sure that you connect the LED the correct way round (The long lead should be connected to the '+' terminal on the battery, the short lead to one end of the resistor and the other end of the resistor to the '-' terminal on the battery. If you get it the wrong way round it don't worry the LED won't blow up, the LED just won't work!
Good Luck.
Sig (appended to the end of comments you post, 120 chars)
Don't use batteries, they pollute the environment something shocking. Use a mains adaptor instead, possibly even an old phone recharger or similar. Note that the voltage ratings quoted on the label are only for show and don't reflect what your trusty AVO will indicate.
Measure the output voltage, subtract the forward voltage of the LEDs {add together if wiring several in series} to get the "excess" voltage {guess 2V if you don't know it, 3V for blue or white diodes} to find out the excess voltage.
Select a suitable resistor to give about 20mA of current, it is not critical that this be spot on or anything, using the formula R = V / I -- where V is the excess voltage that needs to be dropped, I = 0.02 {because we said 20 milliamps but the formula wants whole amps} and R will be in ohms. Now resistors are made only in certain values, usually multiples of 10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68 and 82, so choose the nearest; eg. if your power supply is putting out 4.9V and your LED wants 2V, you have to drop 2.9V at 0.02A so R = 2.9 / 0.02 = 290 / 2 = 145, so use a 150 ohm resistor.
Finally, work out the power rating required, by multiplying the voltage dropped across the resistor by the current through it. In this case, P = 2.9 * 0.02 = 0.058 watts, so a 0.25W resistor will do fine.
Each room will need a separate switch and resistor, but if you have the volts available you can put multiple LEDs in series from the same switch. Watch the current consumption doesn't exceed the power supply rating: as you get close to it, the voltage will start to fall and the LEDs will get dim. But power supplies of this kind must be short-circuit-safe by law, so you won't burn your house down even if you do actually overload it.
Je fume. Tu fumes. Nous fûmes!