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FRONT PANEL LED SPECS
Hello,
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. Bobby |
#2
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FRONT PANEL LED SPECS
"Bobby" wrote in message oups.com... Hello, 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. Bobby You should be able to wire them in series without any problems. As far as LED specs, these'll do just fine... http://www.directron.com/ec074.html -- "I don't cheat to survive. I cheat to LIVE!!" - Alceryes |
#3
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FRONT PANEL LED SPECS
On 29 Nov 2005 03:26:07 -0800, "Bobby"
wrote: Hello, 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. 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 load. 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 motherboard. 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. Should I wire these LEDs in series or parallel from the existing LEDs in the case Neither, you should hook up one LED to each led pin pair. No parallel, no series. or should I keep the extended LEDs remote from the existing LEDs? "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 specs. 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. Can I use multiple LEDs on one circuit? 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 rewarding. |
#4
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FRONT PANEL LED SPECS
In article .com,
"Bobby" wrote: Hello, 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. Bobby 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. +5V | current = (5V - Vf) / 220 | 220 ohm let's use a HLMP-K101 LED resistor | current = (5V - 1.8V) / 220 = 14.5 mA | + LED Should be detectably bright. | - | | GND 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 flow. 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 series. + - + - +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 wavelengths. Now, we'll try the parallel LED connection. First we'll mix a blue and a red LED into our circuit. +5V | current = (5V - Vf) / 220 | 220 ohm resistor | +------------------+ 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. +5V | current = (5V - Vf) / 220 | 220 ohm resistor | +------------------+ | | | 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. http://www.home.agilent.com/cgi-bin/...OUNTRY_CODE=US Datasheet (5989-2809EN.pdf -- HLMP-1321): http://www.home.agilent.com/cgi-bin/...LENT_EDITORIAL Have fun, Paul |
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