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Thread: driving your LED(No resistors allowed)? Read this

  1. #11

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    my LV green in my MHS hilt is really darn bright, im anxious to see when i get the green seouls from X if those turn out a lil brighter or not. at any rate, as he said, my runtimes on batts will increase. if i understood it right, i have a 700 puck drivin that LV right now, and the new seoul wants 1000, so it wont be maxed, it will be undersriven a lil bit, so i shouldnt blow the led, but it may turn out being just AS bright as the LV is, but greater batt life in the end.



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  2. #12
    Council Member Novastar's Avatar
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    As to what X is saying, and also the thing about over-driving... for what it is worth, I am here to confirm all that to newcomers as well!

    YES, you can "overdrive" (within reason) an LED by giving it more than its listed max continuous current. Within REASON. That does not mean doubling the current, lol. Although I've seen that work too!

    YES, that will generate more heat. Do-Clo said it on the Vs. A LOT more heat.

    YES, in *general* it will get brighter... but some LEDs will differ on how much more you get--if you can even detect it visually.

    YES, even if you cut in half the lifespan of a particular Luxeon LED by overdriving as above, you will almost NEVER see it die within your lifetime. It would still have about 2000 days of usage at 24 hrs./day, 7 days a week. Gee, is that enough???!!?!!

    YES, for heaven's sake, X-wing is right about the current thing! JUST GIVE AN LED "what it wants" regarding current!

    YES, if you're going "direct drive" (without a resistor) you need to be very careful, and match the battery voltage with the forward voltage of the LED. If you can't, you'd better use a resistor.

    And yes, in a way--with Ultra's board... it *MIGHT* have been better if the voice described both the "luxeon style" AND the current being driven... but X-wing paints it out very clearly. In fact... you could technically make a saber staff with an Ultra:

    * Two LEDs wanting 700 or 750ma max cont. current
    * Select Luxeon K2 Blue/Green/Royal B/etc. (1500ma)
    * If you wire it correctly, there you have it.

    You could even do two LEDs wanting 1500ma max cont. current, but I wouldn't recommend it, as the wiring would 1/2 your runtime.
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  3. #13
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    YES, in *general* it will get brighter... but some LEDs will differ on how much more you get--if you can even detect it visually.
    http://youtube.com/watch?v=Cgo48tNSUZg

    Amen! I don't think I ever want to bother with a K2 at 1500mA. I haven't tested it with all my LEDs but Rebels it makes a visible difference from 700mA to 1000mA.

    Mostly I made the topic to make people think. On a driver the volts of the LED and batteries isn't a problem as long as you have enough. To quote the Ronco commercials "SET IT AND FORGET!".
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  4. #14
    Sith Acolyte DACOTA's Avatar
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    Yeah, voltage is like water pressure, you just need enough to push the electrical current.

    Nice topic X-wing. And that was a good quote too.
    Last edited by DACOTA; 02-13-2008 at 05:28 PM.
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  5. #15

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    A correction needs to be stated. Xwingband said...
    "Can I run _______?" Is irrelevant! If it's an LED and you match the current volts don't matter.
    That's not really true.

    One example would be using a Green Luxeon V. The forward voltage is around 6.85v at 700mA of current. If you use a luxeon LED like that, a Plecter/Ultra/Buckpuck to drive it, and only give it a 4-Cell AA battery (4.8v) then you WILL NOT GET 700mA to your LED.

    The specific measured input voltage from the battery at all times must stay ABOVE the voltage the LED requires to keep that constant current to the LED. If you are using a red luxeon III and give it a 2.4v battery for power... it will not get even close to 1.5A.

    Simply put, an LED does not care what your driver is. If you want to push 700mA through a green luxeon V, your voltage will have to be somewhere around 6.85v to the LED TERMINALS. Period. If your battery was only putting out 5v, then what you need is a charge-pumping circuit that can upconvert to the required voltage.

    Crystal Focus boards, Buckpucks, and the Ultraboard do not perform voltage up-conversion. They simply maintain current if and only if the battery supply voltage is over what your LED demands.
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  6. #16
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    Eandori, I'm trying to hit a much lower level of thinking. I know that's true, but it's not my point of posting the topic... I was addressing the idea that people suddenly think the playing field has changed completely because the LED isn't a III or whatever.

    I see constant questions of: "Can I run a ______ on _______?"

    Well...

    Can you:
    A) give it the volts it wants? If yes:
    B) can the driver give the current it wants? If yes:

    YES!YES!YES! You can use it.

    Example I see most often: Can I run a Seoul P4 on and Ultraboard? (or swap for any combo of Rebel, V, K2, Cree, and CF board)

    Now let's look at it:
    Can you give it above 3.8V or so... Yes. Can the Ultraboard give 1000mA... Yes (the III white/green/blues setting which is more correctly 1000mA).

    I also saw people fretting thinking that if they used a V setting on another LED it'd fry it... I'll just lay it out, that's stupid if you think about it.

    What happens when the batteries get lower than fresh on a driver? Does it dim? No, it will keep the current going as long as it has "enough". What enough is, is another ball of grease that you are talking about. I'm trying to combat the absurd idea that say you hooked up an Ultrasound with a III and put it on the V setting it'd fry because it's going to send 6.8V... NOOOOO! It will send whatever volts the LED wants at 700mA (V's current).
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  7. #17
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    I want to seperate this from the previous one...

    An LED is a diode...it has a resistance. A resistance is volts per ampere.

    Each LED is slightly different. Thus the binning...

    SOOOO... I'm trying to convey the constant current idea. A constant voltage driver is absurd for an LED because if you take the above you'll realize that if it sent a fixed voltage the amps (current) would be everywhere.

    So to make sure you are giving a constant current (brightness) the driver will change the voltage to give what the LED wants. That's why a III on a "V" setting works, or a III on the K2, etc...
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  8. #18
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    Thanks, that's helpful.


  9. #19
    Council Member Novastar's Avatar
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    Since this topic is about driving an LED with no resistors, or at least akin to that... let me confirm what X is saying:

    Let's say we have:

    If... Driver + LED...
    Then Batteries can be nearly any voltage (within reason) that the board can support... and you then set the current to where the LED is best at for you. THIS is basically what X means by "voltage doesn't matter". Within REASON. You cannot light up a Lux III Green with 1v. That is just stupid. Just like you can't light up a Lux V with 3v. that is also very gay. BUT... you can give BOTH of those 7.2v with a driver, control the current, and BINGO. No problem.


    If... *NO* driver + LED... (maybe even no resistor!)
    Then Batteries should be AS CLOSE TO the LED's forward voltage as is possible, and preferably a *tad* bit below. This way, you will not need a resistor (or only need a very tiny one), and you won't be "burning away" any power. The resistor acts as a certain "choke point/gate" for the batteries, and will only allow so much juice to make it through to the LED. So where does that extra energy go? Essentially "nowhere". It gets used up as heat, and you never really "see" it via the light. It's a big waste. So... if you're going to do "direct driving"... or throw in a resistor... try your BEST to match things up perfectly if at all possible!

    Best examples I can think up (and have used successfully without issue and LED problems):

    * Lux III or K2 "Light Side" [Green, Cyan, White, Blue, Royal Blue]
    * 3.6v or 3.7v Li-Ion Cell
    * No resistor.


    Why no resistor? Because the Lux III/K2's fwd v is around 3.8v or 3.9v. PERFECT for the 3.7v cell, which initially "slightly overdrives" when the charge is full at first, and the cell spits out 4.2v or so. In a short time, we get closer to 3.9v, which is the most ideal brightness... and later, we drop to 3.7v for the major duration of the cell's charge cycle. Once we drop to 3.4v or so, the saber is nearly useless, and it's time to replenish the cell's juice!

    In my thinking--an absolutely PERFECT setup for a very simplistic saber where you do not have/wish to have a driver.

    * Lux III "Dark Side" [Red, Red-O, Amber]
    * 3v Alkaline, or even 3v Li-Primary (as X-Wing has used!!! Very smart!!)
    * No resistor.


    Again, why no resistor? Because the Lux III's fwd v is around 2.95v. PERFECT for the 3v Li-Primary cell, or two 1.5v Alky cells... which drive the LED almost exactly perfect. In this case, no real overdriving is occurring, and if you count 0.05v as over-driving, you have some serious brain issues that NO one or no thing (not even a brand new CF) could ever fix!

    Besides, as X has mentioned (and many others)... LEDs are not a perfect science. LED X will be "best" at 0.15v higher than the same type of LED which is called LED Y. The binning of LEDs also figure into this, simply because of the nature of creating them, causing different shades of color (measured in nm) and so forth. You never TRULY know the LED's "best" voltage and current "sweet spot"... unless you test it 100% on some kind of spectrometer or dedicated power supply with a bunch of dials to help you truly nerd out over it.

    BUT... the specifications that are made about LEDs are (for the mostpart) about as good as an estimation as anything. Trying to milk out an extra 5 lumens by dorking around with an LED *MIGHT* be of value... but I seriously doubt it is worth it due to the time invested and results realized. So your LED is now 5% brighter than a similar saber next to yours. So what. Can you really detect 5% with your eyes? Not a chance.

    Anyhow. More blabbing from me... in hopes that people get the understanding on it. Even though it isn't the WHOLE story... you get what you need, just like X said.
    Last edited by Novastar; 02-28-2008 at 05:24 AM.
    ~~ GREYTALE NOVASTAR (Writer, Director, Choreographer, Sound Designer, Actor, Saber Designer, Vocal Artist)
    ~~ Balance of Power, EP I: "Into The Lion's Den"
    ~~ Balance of Power, EP II: "Ashes of The Phoenix"
    ~~ The Crystal Focus Sound CD Compendiums... are HERE! ~~
    ~~ Nova & Caine's Staged Combat System... comin' SOON!
    ~~ Crystal Focus Wiring Guide

  10. #20

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    I wasn't saying anything about hooking up a lux III on a board with a setting for lux V. Yeah that will fry the LED.

    What I said was if the battery input voltage drops below the required voltage of the LED, then you won't hit your desired current. That includes when batteries get low.

    So sticking to the luxeon V example. You connect a 8.4 volt (rated) pack. It actually ends up measuring out to around 10.5v with no load, but when you yank 1 amp out of it the pack drops to around 9v under load while fully charged. Your LED uses around 7v forward so you have around 3v of drop at 1 amp for 3 watts dissipated in the board, and 7 watts dissipated in the LED.

    Now over time that battery starts to die. Say it drops all the way down until while driving the LED as hard as it still can, the 8.4v rated battery is actually only pushing out 6v. Now.. your LED will not be getting the current it wants. Probably somewhere around 500mA instead of that 1A.

    I had a 7.2v pack hooked up to my last saber with a CF board driving a green lux V. It would be fully bright at full charge. Then after like 15 min my saber would dim a few notches (down to what the green K2 puts out) but still stay "on" for another 90 min at least.

    I understand what you are saying about making it simple. But you can't remove that part of it. It's just not true. Your battery (while under load) must remain above the required voltage of the LED or you will not get the current.
    Edwin Tracy (Eandori)

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