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   My CPU is located a distance from my workstation and I need to
manufacture extensions for my JFP1 and JFP2 cables, specifically for
the HDD LED and Power LED.  I have the pinouts figured out but I need
to know what type of LEDs (voltage, mA, mcd?) I should be shopping for.

 Should I wire these LEDs in series or parallel from the existing LEDs
in the case or should I keep the extended LEDs remote from the existing

LEDs? Can I use multiple LEDs on one circuit?
   Thanks for your help.



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You should be able to wire them in series without any problems. As far as
LED specs, these'll do just fine...

"I don't cheat to survive. I cheat to LIVE!!"
 - Alceryes



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What type did you want?  The reason why there are so many is
because there's so many wants/needs/desires/etc.

The motherboard circuit driving these may have 5V and a
resistor before the LED.  That means you have some
flexibility in the LED, are not forced to use something
particular, but you can't run a Christmas-Tree-made-of-LEDs
either, not without some supplimentary driving to take the

Generically speaking, the forward voltage is going to be
about 2.2V or higher.  mA is not an issue so much because
that's meant for a designer to build the circuit, and you
already have that circuit fixed on the motherboard.  In
other words, if you want to buy an ultra-bright 12K mcd LED,
go ahead, but you're not going to get anywhere near 12K mcd
out of it because there's already current limiting.

To give you a couple of examples, right now on a testbed I
have a fairly typical PCChips manufacturered Shuttle MN31

Plugging an LED with forward voltage of 2.2V into it
(actually slightly lower, 2.18-something but we need not be
THAT precise), the current is 19mA.  That doesn't mean you
need to buy an LED rated for 19mA max in this example, it
only means that the motherboard has already limited the
current to 19mA so the LED should be rated for at least
19mA, but just about anything normal-sized these days is.

Now for example #2, I'll hook up an ultra-bright LED, whose
spec sheet lists the forward voltage as minimum 2.8V,
typical 3.5V, max 4.0 V.

Note I listed 3 voltages above.  When a manufacturer lists
only 1, it'd be the typical voltage but you're not
necessarily guaranteed that voltage as it can vary from lot
to lot unless you're buying precision spec'd parts (uncommon
and costly).  Instead you just fudge a little and shoot for
the typical, making adjustments later if necessary.

Anyway hooking the above ultrabright white LED up to same
motherboard, the forward voltage measures to be 3.2V and
13mA current.  Since this LED was rated for 100mW, it'd
handle 32mA, again the current put through it was below the
max rating, however it was far brighter than the previous
LED even though at 13mA it wasn't even at half brightness.
It would be too bright for the typical power or LED
indicators though, unless you wanted to put it behind some
kind of translucent, diffused panel so it lit a larger
area... but then if that's the look you're going for there
are LEDs already built into small diffusers too, though of
course they cost more.

Seems I didn't really give you the easy answer to your
question.  You can consider the above and make your own
choice but a typical LED choice might be one with roughly
2.0 to 2.8V forward and a current rating of at least
30mA.... ultrabrights generally aren't used for such things
as they might be novel for a moment then become more of an
annoyance than anything, though if you wanted to use a blue
one for instance you could add a resistor inline to lower
the current further to dim it some, though it's still going
to be a focused beam while many LEDs are diffused already.

Not much point to a focused beam on a power indicator if the
LED itself is visible, unless that's just the look you're
going for, since a diffused case keeps more of the
illumination at the surface of the LED and thus it has a
nice even and brighterappearance on the indicator panel.

In summary, decide what you want it to look like and
(considering the typical ranges LEDs come in) get at least
2.0V forward and 25mA current capability.

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Neither, you should hook up one LED to each led pin pair.
No parallel, no series.

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"Remote" meaning what?

Unplug the existing LEDs.
It's not manditory, you can do the math and come up with
your own target for what voltage, current the LED arrays you
want, will need, and go from there if you want.  For example
if you used a handful of resistors and transistors you can
wire in an aux 12V power supply for these LED arrays that's
controlled by the motherboard LED pins only to the extent
that they bias the transistors, not providing the current
for the LEDs.  I'm suspecting this is a bit more advanced
than you want to go given the basic question about the LED

Another theoretical but not so practical option is to modify
the LED driving subcircuit on the motherboard itself, but
that seems even less of a good idea.

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If you just want the brightest light possible, I suggest
unplugging the LEDs in the case and using one ultra-bright
LED for each pin-header pair (each function, like power or
HDD, sleep/wake, whatever the board supports).

The motherboard is already designed to be the complete
solution for one LED and one LED only (as commonly
implemented).  If you want to deviate from that you'll have
to decide exactly what you want to do, how elaborate,
costly, and how much calculating/soldering/testing you want
to do to get this working.  It's not hard if you have the
inclination for it but if you just wanted to do something
fancy with LEDs, I'd think a flashlight might be a lot more


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You've got two good answers, and I'll take a stab at a
few other details.

First, we need some data to illustrate some points. This
is data for some HP (Agilent?) LEDs I snagged a while back.

Part_Num        Tech   Color   Lambda  Vf_(forward_voltage_drop)

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

I took my spare old motherboard, and measured the resistor used
to drive one of the LEDs. This is a simplified view of the
circuit. It looks like Kony's board had a 150 ohm resistor.

          |           current = (5V - Vf) / 220
        220 ohm       let's use a HLMP-K101 LED
          |           current = (5V - 1.8V) / 220 = 14.5 mA
          | +
         LED          Should be detectably bright.
          | -

The first thing to note, is the LED won't light, until the
applied voltage approaches Vf. The current goes exponential on
you after that, so if you connect the 1.8V LED above, directly
to the 5V supply, there is a bright flash and then it's dead.
That is the purpose of the series resistor - it limits the current

So, the LED V-I response is some kind of curve, and the assumption
that it "lights at exactly Vf" is a simplification.

Simply buying LEDs rated as "high efficiency" should be good
enough to avoid the super cheap "dud" LEDs. You want a LED that
can handle the current the circuit will be supplying. In my
sample motherboard above, the current is 15mA, and a 50mA max
LED should do nicely.

Now, let's try to do a series connection for the LEDs.
I'll draw this circuit side ways to save space.

                        +     -     +     -
   +5V --- 220_ohm ------ LED1 ------ LED2 ------- GND
                          1.8V        1.8V
                          Red color   Red color
   current = ( 5V - 1.8 - 1.8 ) / 220 = 6.3 mA

The current has dropped, due to the additive effects of the
two LEDs Vf in series. If the engineer designing the motherboard
knew the user would do this, then he would have put a 100 ohm
resistor on the motherboard. With that resistor choice, and
the two LEDs, the current would be 14 mA again. As end_users,
we don't have the luxury of changing the resistor.

The next experiment, is to put a red LED and a blue LED in
                        +     -     +     -
   +5V --- 220_ohm ------ LED1 ------ LED2 ------- GND
                          1.8V        3.6V
                          Red color   Blue color

Oops ! Now neither LED lights. The sum of the two Vf of the
LEDs is more than 5V. Neither LEDs minimum forward drop is
being met.

Similarly, from this thought experiment, you can see that if
only one blue LED is inserted in the circuit, the current
will not be as great. Whether this results in a brighter or
dimmer perception to the user, really depends on the efficiency
of the LED, and how receptive the human eye is to different

Now, we'll try the parallel LED connection. First we'll
mix a blue and a red LED into our circuit.

          |           current = (5V - Vf) / 220
        220 ohm      
          +------------------+            Oops! The red one
          | +                | +          lights, and the blue
    red  LED  1.8V    blue  LED 3.6V      one doesn't.
   color  | -        color   | -
          |                  |
          |                  |
         GND                GND

Darn! That blue LED again! In this case, the red LED Vf
requirement is met first. If the voltage rises above Vf,
the red LED is in its exponential region -- it hogs all the
current. The node at the top of the two LEDs never even
gets close to the 3.6V the blue LED needs. This "current
hogging" problem will plague us, even when we use two
red LEDs in the circuit, in parallel.

LED manufacturers offer to "match" LEDs. At the factory, certain
lines of LED product are sorted into "bins". Two LEDs of the
same color, will match characteristics to within some percentage.
This is important when building display devices that happen to use
that lousy circuit above. If we order a bag of matched
LEDs, then the intensity of two LEDs in parallel, will be
close to being the same.

Well, what else could we do ? Say we use two LEDs, one being
red and the other one being deep_red. The first one has a
Vf of 1.9V and the second one a Vf of 1.8V. How do I know which
color has which Vf ? The slope of the line that relates forward
voltage to color, happens to have Plancks constant as part of
the equation. Physicists would way "E = h * v" where the v is
pronounced "nu". That is how I can predict that the deep_red
LED has a lower voltage drop than the simple red LED. And,
that also allows me to predict that the blue LED will be a pig
to work with, even without data from HP/Agilent.

          |           current = (5V - Vf) / 220
        220 ohm      
          |                  |
          |                trimming
          |                resistor
          |                  |      
          | +         deep   | +        
    red  LED  1.9V    red  LED 1.8V  
   color  | -        color   | -
          |                  |
          |                  |
         GND                GND

Well, how do we figure out the value of the trimming resistor ?
First, we take the right hand LED out of the circuit.
With the Vf of 1.9V, the current is  (5 - 1.9)/220 = 14ma.
Now, we put the right hand LED and trimmer resistor back.
The trimming resistor has 1.9 - 1.8 = 0.1 volts across it.
We want roughly half the current in each LED.
R_trim = 0.1V / 0.007A = 142 ohms.

(Note - the above calculation and method would not pass the
scrutiny of my electronics instructor. Please forgive me --
I have sinned. Some linear equations with non-linear diode
terms might have satisfied my instructor, but would obscure
the principle I'm trying to demonstrate.)

If we put a 1K ohm potentiometer (variable resistor) as the
trimming resistor, we turn the knob until the brilliance is
matched. We can dial the pot from end to end of its travel,
and nothing in the circuit will burn. If we put the trimming
resistor in the 1.9V side of the circuit, what happens ?
(Hint - the deep red LED wins...)

Since the characteristics of the two LEDs will not be
matched (one is AlGaAs, the other GaP), as the temperature
changes in the room, an inperceptable change in brilliance
will occur.

If we put trimming potentiometers on both sides of the
circuit, we'll achieve some degree of adjustment, but
probably the end_user won't be too impressed with the
results (after all, we split the 14ma current in two,
so at the best of times, each LED is at half the brilliance
it used to have).

Moral of this story ? Disconnect the LEDs in the computer,
and only connect your two remote LEDs. No maths needed,
just buy a couple high efficiency red LEDs. If they don't
light, reverse the connection of the LED to the motherboard
header. That is why the LEDs in my drawings above have the
polarity shown. Anode = plus, cathode = minus. Anode is the
"triangle" part of the LED symbol, cathode is the "bar".
In the datasheet, notice the cathode leg is shorter. The
plastic on the LED body, may have a "flat" spot, which is
also an indicator of polarity.

Datasheet (5989-2809EN.pdf -- HLMP-1321):

Have fun,

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