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| | #1 (permalink) | |
| Apex Techie Wannabe | Hey folks, I love this site. Lots of great info. I am getting a new laptop to mod. It is a CLEVO D900T. I want (among other things) to put LEDs around the perimeter on the bottom. Finding places to insert the LEDs is not really my problem, but where to wire them up for power and ground. I figure a power lead on a circuit board will have to be used. Has anyone done this or know of anywhere I can get some info? | |
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| | #2 (permalink) | ||||||||||||||||||||||||
| Apex Techie Wannabe |
First off, nice laptop. Though I'm waiting for the D900K myself. Anyway, your best bet would be to hook them up to the back of a USB port for the 5V you want for your LEDs. Look up Spiike's LED mod on Google. He put LEDs in a Dell laptop and has a step by step walkthrough of the process. | ||||||||||||||||||||||||
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| | #3 (permalink) | |
| http://www.kpsec.freeuk.com/components/led.htm I use around 200 to 270 Ohms for the resistor. Use the USB port as mentioned in the above post. | ||
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| | #4 (permalink) | ||||||||||||||||||||||||
If you dont even know what LED's he will be using and what their rated at, how can you possibly suggest what size of resistor to use? HERES A GOOD LINK with basically the same concept in mind. He used a usb cord to make a flashlight, and he goes through the steps to find out exactly what resistor you will need and how to set it up. EDIT: Nice link though Nerdz, that explains it fairly well. | |||||||||||||||||||||||||
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| | #5 (permalink) | ||||||||||||||||||||||||
| Apex Techie Wannabe |
Yea, The D900K is awsome! Spiike's LED Mod is exactly the info I was looking for. I am going to try to poke the LEDs through the bottom of the case. It may or may not be possible. If not, I may try to use electroluminescent film/wire. That stuff is pretty cool. Last edited by LinquE; 09-December-05 at 08:50 AM.. | ||||||||||||||||||||||||
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| | #6 (permalink) | ||||||||||||||||||||||||
LEDs all require about the same current. Their voltage ranges from 1.5 to 2v or even 3V. Its called standard, and I have done many tests of many different color LEDs. Those values gave the best brightness over longevity. | |||||||||||||||||||||||||
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| | #7 (permalink) | ||||||||||||||||||||||||
I dont care how many tests youve done bro, Ohms LAW is the preferred method for a reason. Leds are rated at certain amounts of current for a reason, and that reason isnt so you can just throw some resistors on. An LED can last for up to 10 years if the proper voltage is applied, why wouldnt you use Ohms law if you can achieve this? E= I x R......its as simple as that bro. Your link was awesome however you might need to go through it instead of random spouting off crap. | |||||||||||||||||||||||||
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| | #8 (permalink) | ||||||||||||||||||||||||
Ok lets get technical. Now since all BLUE LEDs drop around 3V, we will let V led=3V. Let VCC=12V, and Iled=30mA So, 12/30mA=400Ohms But we didnt compensate for the voltage drop so, 12v-3=9V 9v/30mA=300Ohms. Closest Value 270 or 330 Now lets use 270ohms and find current. 12v/270 ohms = 44mA Now lets compensate for Voltage Drop 9V/270Ohms= 33mA Now lets use VCC=5V 5/30mA= 166.7Ohms. Closest Value=180 Ohms Now lets use 270Ohms 5/270Ohms = 18mA (A bit small) Now lets compesate for voltage drop 5-3=2V 2v/30mA= 66 Ohms. (off hand I dont know closest value) Lets use 270ohms 2/270=7.4mA (small yes, it won t be bright) I caqn say that on 5V the best Resistor to use is 100Ohms. (look at an LCD to parallel port circuit) Now lets Calculate For Green, Red and Yellow LEDs. Since Vled ranges from 1.5V to 2v, lets use 2v as out voltage. Let VCC=12v. Lets led ILED=30mA First we will compensate for voltage droped. 12-2=10V 10v/30mA=333ohms. Closest value=330Ohms. now lets use 270 ohms 10v/270ohms=37mA Now lets use 5V for VCC and have the same values as above. Lets compensate for voltage drop. 5-2V=3V 3v/30ma= 100Ohms Lets use 270ohms now 3v/270= 11mA Now you may wonder, well what if Vled=1.5v Compensate for voltage drop 5v-1.5=3.5V 3.5v/30mA=116.67ohms. Closest Value=100ohms now lets use 270 3.5V/270= 12.9mA So what if we used 200Ohms Use 9V for VCC and I led=30mA 9v/200=45mA Lets use 5v and compesate for 3v voltage drop 2v/200=10mA Now lets use 200 ohms with 2v drop.VCC=12V 10v/200= 50mA Now let use 2v drop on vcc=5v 3v/200= 15mA Lets use 1.5v drop on 12v and let R=270 12-1.5v=10.5v 10.5/270=38mA Lets use that same drop on 5v and let R=200 5-1.5=3.5 3.5/200=17.5mA And you say I dont use Ohms law... | |||||||||||||||||||||||||
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