Bi-Colour leds, anyone know how to wire them?

ElectroBlaster

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I apologise but this is something new to me!

I can wire leds, I know resisters pretty much for lowering current but never messed with leds that can emit more than one colour.

I basically want this:

Colour 1 for power, always on.

Colour 2 for drive activity, only when drives are reading.

Can this be done? I am currently sat on google trying to get the gist of what needs doing.

This will be going into an old C64 Breadbin case (orignal board is fubard). I will remove the original led and put in the bi-colour one, keeps it neat and tidy then.

EDIT: I have found these http://cgi.ebay.co.uk/ws/eBayISAPI.dll?ViewItem&item=170710148796. If these were red instead of clear then I will be a very happy chappy lol
 
It depends on what type of multi LED you chose. For simplicity I'd advise you to use a duo-LED. These have 3 pins and are essentially two LEDs in one casing. They usually have a common cathode and will allow you to have three colours in total (by illuminating LED 1, LED 2 or LED 1+2).
The current limiting resistor is calculated as with a normal LED. Both LEDs should have their own separate resistor.

Bryce.
 
So I will need a resister to limit them and stop things going awry?

I bascailly want a mini-itx board to power the led and also report drive activity.

Or

I could power one side of the led directly from a psu and then just connect drive activity to the board...

Just need to figure out the value of resistors I need.
 
Both LEDs can be seen as separate entities. The resistor depends on 3 values you need to know:

1 - The voltage (or signal voltage) that will power the LED (Vcc).
2 - The voltage the LED requires (Vled)
3 - The current the LED will require (Iled)

2 and 3 can be found in the datasheet of the LED.

Then just use R = (Vcc - Vled) / Iled
to calculate the values for each resistor.

So if the power LED is being supplied by a 9V source and the LED requires 2.2V @ 20mA (typical for green LED). Then the resistor needs to be:

R = (9 - 2.2) / 0.02 = 340 Ohms

If the activity LED is being driven by a 5V signal and the LED needs 2.0V @ 20mA (typical for red LED). Then the resistor needs to be:

R = (5 - 2) / 0.02 = 150 Ohms

However, you'll find that the 20mA given on the datasheet usually illuminates the LED to its maximum, so I usually use 10mA in my calculations... You only want to see it lighting and not be blinded by it.

Bryce.
 
Thanks :D

Really good info.

Same thing with the brightness! I dont want to be blinded by it, just enougth so it tells you things are on and doing the job.

I would like to use the mini-itx board header for this job if possible.

EDIT: I just browsing the tech docs for the board I have and it seems it can drive a dual-led by default! It states the pins can be connected to a single or dual coloured led. Also the the voltage seems to be 5v which I assume is the same as all pc motherboards?
 
Back on track with this!

Got some more info:

http://www.alldatasheet.com/view.jsp?Searchword=L-59EGW

  • LED, 5MM RED/GREEN, 3MCD/8MCD
  • Bulb Size: T-1 3/4 (5mm)
  • LED Colour: Red, Green
  • Luminous Intensity / Colour: R 3mcd, G 8mcd
  • Viewing Angle: 60°
  • Forward Voltage / Colour: R 2V, G 2.2V
  • Forward Current / Colour: R 30mA, G 25mA
  • LED Mounting: Through Hole
  • Lens Shape: Round
  • Wavelength / Colour: R 617nm, G 568nm
  • MSL: -
  • SVHC: No SVHC (20-Jun-2013)

And my board is:

http://www.intel.com/content/www/us/en/motherboards/desktop-motherboards/desktop-board-dn2800mt.html

Just trying to suss out what voltage is coming from the motherboard headers. I read something about this board being able to drive dual-colour leds... but unsure what voltage its pumping out?

EDIT: The manual states, Front Panel Header of 9pins - HDD Activity Pull-Up resister (750) to +5V. Nothing stated for Power Led other than it can drive a dual-led but I now assume that is for power alone?

:)
 
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click on "interactive layout" then hover over "17" on the picture;)

youll see what i mean.:)


do you have the board already?
 
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Motherboards normally have the resistors on board for the LED headers so all you need to do is to attach an LED to the pins, however, you can put a resistor in series with it if you want to reduce the brightness even further.

If two LEDs are arranged like this on the header:
+ - + -
|-| |-|

then you can simply connect a 3-pin dual LED on the left three pins:
+ - + -
|---|
(apologize for bad ascii gfx :))
 
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