Jon Fleming said:
My son's computer has an Asus A7N8X-E Deluxe and one SATA drive in a
case that has no hard drive activity LED. I've got a software
activity indicator installed but I'd certainly prefer a hardware
activity indicator.
What is the appropriate voltage for an LED to hook up to the MB
terminals? Is there any chance that such an LED will indicate SATA
activity?
This is what a typical LED driver circuit looks like.
+5V
|
|
Current
Limiting
Resistor
|
|
X
<--- LED connects here
X
|
__|/ (Output drive
|\ pulls to GND)
|
|
GND
First of all, pick up a red LED at RadioShack. Connect it to
whatever two pins are intended for an IDE drive light. If
it does not light, simply reverse the terminals on the LED
and it should work. Since the motherboard should already have
a current limiting resistor in the path, no additional
protection for the LED should be needed. (If playing with
LEDs at home, always insert a resistor in series with the
LED, to limit the current flow.)
As for the SATA chip, be aware that not all storage interfaces
are tied into the LED circuit. It appears the SIL3112 doesn't
have a convenient LED output drive, and some tricks are used
to generate an output. Asus may not have bothered to hook
up the LED drive from the SIL3112, to the IDE LED pins.
The second link here, will download a PDF document containing
some suggestions from Silicon Image.
http://12.24.47.40/display/2n/kb/article.asp?aid=10064
http://12.24.47.40/utility/getfile.asp?rid=1023
*****
The rest of this post is just FYI.
The intensity of the LED is a function of current flow.
The motherboard has a current limiting resistor, and no
more than 5volts divided by R can flow. A rough current
limit might be 0.010 amps (10mA) for the LED.
Each LED has a forward voltage drop. In fact, the forward
voltage drop (Vf) of a LED, is a function of the output wavelength.
(You can actually derive a value for Plancks constant, by
plotting forward voltage of colored LEDs versus their
dominant output wavelength.)
HLMP-K101 AlGaAs Red 637nm 1.8volts
HLMP-1321 GaP Red 626nm 1.9volts
HLMP-1401-E0000 GaP Yellow 585nm 2.0volts
HLMP-1521 GaP Green 569nm 2.1volts
HLMP-K640 GaP Green 560nm 2.2volts
HLMP-DS25-R0000 InGaN Blue 470nm 3.6volts
HLMP-KB45-N0000 GaN Blue 462nm 4.0volts
OK. Since the intensity of the light output is proportional
to current flow, and the LED does drop some voltage, we
need to modify the equation to work out the current. In
the same circuit at the top of this posting, there is 5V
used to power the circuit. The voltage across the resistor
is 5 minus Vf, so the higher the Vf, the less voltage there
is across the resistor. Ignoring the driver device for a
moment, the current flow is (5 - Vf)/R.
Taking some examples, say the current limit resistor is
330 ohms. Plug in a red LED, which drops 1.8V. The
current flow is (5-1.8)/330 = 9.7mA . Now, plug in a
blue LED, which drops 3.6V. The current flow is
(5-3.6)/330 = 4.2mA . Eventually, if the Vf of the
LED gets close to the 5V we are using to drive the
circuit, no current will flow, and we won't get any
light. This means, when a designer picks the current
limiting resistor, the designer usually has a particular
color of LED in mind, like a red one.
The choice of 5V drive is not an accident. Many LEDs have
a reverse voltage rating of 5V, and that is why if you
insert it the wrong way in the circuit, it won't be damaged.
If the circuit was powered by 12V, it might be possible
to exceed that reverse rating, so if the drive circuit
uses the higher voltages, you would want to make sure the
LED is fitted the correct way, the first time. The 5V
used on the motherboard, ensures that the LED can take
a reversal.
So, when making LED indicators, make sure to find out what
maximum operating current it can use, only use a portion
of that rating, then work out a resistor value to limit
the current flow to that rating. 10 mA is a good
starting value for the current flow, if you don't have any
data on the LED. If we were using 12V to power a red LED,
then (12 - 1.8)/0.010 amps = 1020 ohms would be a good
current limiting resistor. A 1K resistor is the nearest
standard value. The power (heat dissipating) rating for
the resistor is V**2/R, which in this case is 10.2*10.2/1020
= 0.102 watts, and a 1/8th watt or a 1/4 watt resistor
could be used.
LEDs are available with extremely high current ratings.
When I was a kid, I bought one that could handle 2 amps,
and it made an amazing toy. The best modern LEDs now,
like some of the Lumileds, take about 0.7 amps, so that
LED (which I destroyed in the process of playing with it),
remains a champ in terms of the amount of current it could
take.
Paul
