Duck eagle, what is it? any link?
About myself, i gave up to buy Pure, the only few Pure in stocks, comes with not the right backlit color or right switch i want.
I choose Pro over Pure.
http://duck0113.tistory.com/81
It's a custom board by Duck113. Fully programmable, aluminium case, backlit. Not sure when they'll become available or the price. Edit - apparently elton5354 will run a GB in May for the Viper and Eagle, cost is likely to be around 200 dollars: http://geekhack.org/index.php?topic=55427.0
I bought a Pure with green backlight, but have desoldered all the LEDs and will be putting in white LEDs to replace them (when they arrive)
At least it came with the switches I wanted (Browns). I prefer the layout of the Pure to the Pure Pro, especially the larger right shift and 1.5x modifiers.
How easy is it to just replace the LEDs with white ones? I have a Pure Pro in the mail, but I could only get it with blue LEDs and I want white ones if I use the LEDs at all. Wouldn't I need new resistors etc?
Blue, white and high brightness green LED's often have similar voltage drop and current characteristics, so quite often you can just swap them. Orange, red, yellow and low brightness green LED's often have lower voltage drops. As long there actually are resistors in the circuit it should be safe enough to swap them out. If they're too bright or dim you can replace the resistors to taste.
I'm using some low current (10mA IIRC) white LED's to replace the greens in my Pure.
You can calculate the current going through the resistor and LED by using V=IR, where V is voltage (in Volts), I is current (in Amperes) and R is resistance (in Ohms). For instance, if the original LED's had 3.2v drop and used 270 Ohm resistors:
V
resistor = V
total - V
LED = 5v - 3.2v = 1.8v
V
resistor = I
resistor*R
resistorso I
resistor = V
resistor / R
resistor = 1.8 / 270 = 0.00666' A = 6.7mA
and I
LED = I
resistor, so I
LED = 6.7mA
If you use a new LED with only 3.0v drop, but the same resistor, the current will rise:
V
resistor = V
total - V
LED = 5v - 3.0v = 2.0v
I
resistor = 2.0 / 270 = 0.00740' A = 7.4mA
It's not a big difference and in keyboards it's rare that the LED's are given anywhere near their max current anyway, so it's mostly a matter of adjusting the resistors if the LED's are too bright or dim for you. White LED's most often have the highest voltage drop AND the highest current rating and therefore it's quite safe to switch TO white LED's.