ok i havent slept much this week, had a funeral and stuff to deal with, and all my electronicks knowledge went bye-bye .... i just got 3 3000mcd red led's to light up the interior of my bro's case, i was wondering if i could be lazy adn just hook one 470 ohm resister to the 12v lead and not have to deal with heatshrinking all the resistors to the led's individually , or do i HAVE to do each led with its own resistor ....
The answer is....insufficient data. You need to consider the power dissipated in the resistor. For the single LED case (I gonna assume Vf = 2V Id = 20 mA for your red LEDs), P = (12 - 2) * 0.02 = 200 mW A commonly available 1/4 Watt resistor will handle this. For the three LED case, add the currents: P = (12 - 2) * 0.06 = 600 mW So if you use a 1/4 W resistor, it will fail. The cost-effective way to go is use three resistors, one per LED. Or you may want to consider using +5V for your supply: R = (5 - 2) / 0.06 = 50 ohms P = (5 - 2) * 0.06 = 180 mW Oooh, look there, you can use a 1/4 W resistor. This is why I include the power calculation in the LED calculator
leds are 1.7v typ(2.4max) 20ma, resistors are 470ohm 1/2watt 5% ... i am seriously electronics impaired.
oh DUH .... yes i am a retard .... so what i would need to do it the lazy was is this. , or am i still retarded?
Well, to me, lazy would be using the parts you have. So three of those 470 ohm resistors, and you're not exceeding the ratings on anything, and you don't have to go back to that awful place you linked to. Another good trick to keep in mind: if you need a 1W, 470 ohm resistor, use two 1/2W 1kohm resistors in parallel (actually 500 ohms, but resistors have a tolerance as you noted). You can add the power dissipation ratings of resistors in parallel. A nice trick when you don't have the right critter on hand... (I still think you should use +5V supply.)
i would, but the bus already runs 12v only, on/off for my 92mm intake, 40mm radeon fan, and my 6' red EL wire. and to me, lazy = less soldering/heatshrinking.
