JR said:
No its on in all its dim blue glory. I checked polarity both directions.
The other way gives nada.
When Asus put the power supply and resistor for the LED, they assumed
a certain forward voltage drop across the LED. The blue LED has the
highest forward drop, while a red LED is much lower.
Here is the circuit. It should be powered from +5V, so that if the
user reverses the wires to the LED, the LED won't be damaged. The
LED datasheets I've looked at allow up to -5V to be applied to the
LED with no damage. So, it is safe to use the +5V supply to power
a LED.
|\ |
----------- | \ |
+5V_mobo --------| Resistor |----X---| \|---X----- Ground
----------- | /|
| / |
|/ |
The two "X" are the header pins on the motherboard.
LEDs operate from current. The more current, the brighter they get.
The forward drop of the LED as it operates is called "Vf".
LED_Current = (5 - Vf)/R
For a red LED, the forward drop is say 1.8V. Say that Asus plans for
a red LED with a current of 20ma, the resistor used would be 160
ohms.
A blue LED has a forward drop of 3.5 to 4.0 volts. Reworking the
numbers and using 160 ohms for the resistance, gives 6 to 9ma.
So, as the color of the LED moves "up spectrum", the intensity
will drop.
A different resistor value would fix it. Or using a red LED would
make it brighter as well.
HTH,
Paul