Swapping indicator LEDs

[korrelate]

Bphiphany: now that looks like a real meal! I've got to sit down, let it sink in and digest all this info. Many thanks!

[frogamic]

I switched the LEDs on my Filco from blue to green to match my Deathadder, I really wish they'd just put normal resistors on the main board, there's no shortage of space in that area. The greens are much brighter than the old blues, I'm thinking I should actually solder resistors directly onto one of the legs of each LED, but I can't really be bothered right now.

[bpiphany]

You can always resort to the state-of-art black Sharpie solution =)

[korrelate]

I actually don't remember if I measured the voltage drop before, only the resistor values I think. So I just whipped out my Fluke...

Total USB voltage 5.05V
Filco blue LEDs 2.55V, resistors 10kΩ, current 0.25mA
Filco red LEDs 1.97V, resistors 510Ω, current 6mA

I only did one measurement per color. The resistors are 5% I guess so it will vary a bit. Different colors have different efficiency I suppose.

Hey bpiphany,

I'm trying to derive the specs for the red leds in your example above. I'm using the fact that the sum of the drops has to equal the total V (I) input.
If I assume that input is 20mA @ 5.05V here's what I get for FV and FC:

FV:
Resistor 3.08V
LED 1.97V

FC:
Resistor .006A
LED .014A

And in my case, I'm going to replace the red LEDs in my Filco MJ2 (100% layout) with green less. So this means I need to find a green LED with FV around 2V and FC around 14mA.

Am I on the right track here?

Cheers,

K

[OldDataHands]

Am I on the right track here?

You've got the voltage split correct, but the same current flows through both the LED and the resistor.
bpiphany indicated that the LED is driven with 5.05v over a 510Ohms resistor.
Ohms law: I = E / R :applies like: I = (5.05v - 1.97v)/510Ohms ~= .006Amps
This 6mA current flows through both the resistor and the LED.

You'll need to pick a green LED, then substitute its forward voltage (instead of the 1.97v) into the above
equation to find out how much current will flow through it.

If for some reason you really wanted to drive the LED at 20mA (you don't though, it'll be too bright) then
you'll need to change the resistor to get the right mix.

[korrelate]

Roger that ODH, thanks for the feedback.