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Old 07-December-05, 03:11 PM   #1 (permalink)
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Default Adding LEDs to a laptop

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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Old 08-December-05, 10:22 PM   #2 (permalink)
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Originally Posted by LinquE
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?

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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Old 08-December-05, 10:37 PM   #3 (permalink)
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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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Old 08-December-05, 11:48 PM   #4 (permalink)
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Originally Posted by Nerdz
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.

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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Old 09-December-05, 08:05 AM   #5 (permalink)
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Originally Posted by Blowfish64
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.


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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Old 09-December-05, 11:16 AM   #6 (permalink)
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Quote:
Originally Posted by Sparky
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.

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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Old 09-December-05, 11:31 AM   #7 (permalink)
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Quote:
Originally Posted by Nerdz
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.

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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Old 09-December-05, 01:55 PM   #8 (permalink)
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Quote:
Originally Posted by Sparky
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.

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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