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I'm making a project where the Raspberry Pi will be mounted inside a closed box, with buttons and LEDs on the front panel.

Is it possible to have panel-mounted LEDs that replicate the status of the FDX and LNK LEDs on the Pi itself, so they blink etc?

I'm willing to give up 2 GPIO ports if needed. Perhaps some sort of daemon that runs in the background analysing the network traffic and controlling them?

  • 4
    Have you considered using flexible lightguides? – joan Jan 25 '15 at 14:50
  • I'd de-solder existing leds and would use wires to the front panel. In my opinion that would be most practical solution and no drivers needed. – Bungee75 Mar 7 '16 at 7:22
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Assuming a Pi B+ here (might be adaptable to other models though).

I'd simply suggest to go for a doubling of the existing LEDs. Either i) simply solder a thin wire to the each cathodes of the LEDs or ii) desolder the LEDs on the PCB - and still solder the wire to the pad of the cathode. See schematics model B, Rev 2.1 page 3. The external LEDs would be wired in the same fashion. With resistors of 1k at 3.3V the max current per LED will be less than 3.3mA. Adding additional LEDs in parallel keeps the total current below 10mA (for the 3 LED case - PCB, jack, and external).

Checking the datasheet of the LAN controller LAN9512, pp. 10, 11 and 20, we find that the pins' buffer type (for nFDX_LED, nLNKA_LED, and nSPD_LED) is:

OD12 Open-drain output with 12mA sink

Well, seems that even a three LED solution (on board + jack + external) would be in spec with the LAN9512's buffer circuitry.

  • Actually - are the other GPIO pins (5 I thought?) on the LAN9512 device "available" (with a spot of soldering and some thin wires) - a few extra output pins might be useful for some people.... – SlySven Aug 4 '16 at 13:58
  • I do not know. Sounds like good question in its own right. – Ghanima Aug 4 '16 at 14:01
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I think you will be hard pressed to get the results you want by paralleling the existing LEDs - with a 1K series resistor and a 3.3V supply you really need to take into account the 1.5Vforward for Red or more for other colour LEDs - the best (lowest voltage / current combination I could find on CPC - a division of Farnell who are also an Official UK RPi supplier) is this Red one. With a 1.6Vf to drive this with 1mA from 3.3V you need a series resistor of: (3.3 - 1.6) / 0.001 = 1.7 KOhms - so you'd use a 1.8 KOhms and have a little lower current. The same calculations for this Green one has a Vf of 1.9 and can use 2mA so needs: (3.3 . 1.9) / 0.002 = 700, so you'd use a 820 Ohms. However nether of these are going to be blinding bright - the red is a minimum 2mCd and the green a minimum 2.3mCd at the quoted 1 or 2 mA current (viewing angles 55 to 60⁰) respectively!

A better bet would be to check the circuit (if possible) and try and add a transistor stage or two with a series resistor in its base wired to the existing LED circuit and a more powerful LED in the collector circuit running between the 5V and 0V rails - this may not enough as if the LED is wired between the 3.3V supply and pulled low to ground via the integrated circuit pin to light the LED than a second stage will be needed to invert the "low-voltage" = "LED on" level. Indeed whether each LED circuit is active high or low may determine which of an NPN or a PNP transistor circuit is best to use.

I would comment that I will have to do something like this myself very soon for my Home Automation project both because I too will be putting the Pi into a box and because the UPiS module I will be using with it actually sits over the LEDs on the Model B I am going to use and that has it's own - completely different - LED status indicators as well in the same XY position!

  • I maintain that the datasheet promises 12 mA at this output pins which should be fine for a parallel set of low current LEDs (with appropriate resistors). So I'd think that a transistor stage is not necessary. Still a helpful answer that should not stay unvoted ;) – Ghanima Aug 4 '16 at 9:13

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