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chase
02-22-2008, 10:23 AM
So if I were to get a 350mA micropuck and a 1000mA buckpuck and attach it to a green luxIII, it won't blow up right? It would equal out to 1350mA? I want to overdrive my LED a little bit...

Eandori
02-22-2008, 11:57 AM
Hmm... I would really have to pour over the spec sheets a bit to have a solid answer for this. I think it's possible, but I would be a bit worried about a ground loop backwards through the puck-outputs.

If it was to work at all, it would most likely have the output of one puck connected to the output of the next puck, then on that same middle node the load (your device that you are driving) connected to one side. On the other side of that load you have the ground.

So the currents from the two pucks would add together at the 3 way shared node before entering the top of your device (probably the LED).

That should work... but I would not reccomend it until somebody spends time looking over the datasheets and double checking ground loops.

Eandori
02-22-2008, 12:04 PM
After thinking about it a bit more, I don't think a minipuck would work for this. It's high voltage output rail is probably not high enough. You might need to switch the idea over to a 1000mA buckpuck and a 500mA buckpuck. With their input + and - tied together with the supply in a single 3 way node for each.

Still though, don't try this until I give it more thought or somebody else has a solid answer.

I need a new online storage place where I can upload images. So I can show a circuit diagram.

xwingband
02-22-2008, 12:24 PM
Two 700mAs would be better. I'm not sure what might go on with mismatched drivers, but I have seen diagrams for paralleling pucks.

No time to research... gotta jet to a class soon.

Eandori
02-22-2008, 01:45 PM
Two 700mAs would be better. I'm not sure what might go on with mismatched drivers, but I have seen diagrams for paralleling pucks.

No time to research... gotta jet to a class soon.
Totally agree with Xwing on that. Good point.

Again, we are not saying "go for it." yet. We are simply saying that if it can be done... it would probably work like that.

chase
02-22-2008, 02:26 PM
Ok, well...I don't plan on doing it right now, but somewhat soon. My first plan was to get a green K2, and now I'm wondering if getting a 1000mA + a 700mA buckpuck would overdrive it well? Has anyone done that?

Eandori
02-22-2008, 03:00 PM
Doubt anybody has done that... no.

Again though, like Xwing said you might want to stick to matched buckpucks. Two 700mA would yeild in 1.4A and that's barely under the specified max for that LED (1.5A max.) So two 700mA buckpucks would be much better.

chase
02-23-2008, 07:43 AM
well, I don't want to go under the max, I want to overdrive it.

Eandori
02-23-2008, 04:00 PM
1.4 amps (1400mA) is a lot of current through an LED. You said you have a green Lux III right? 1.4amps IS overdriving that LED. It's rated for 700mA to 1000mA. Your current LED is only rated at 80 lumens.

I assume by wanting to overdrive it, you want it brighter. If you want brighter then use a Green Luxeon V at 1 amp (rated 160 lumens at 700mA). That will be much brighter then a green K2 can get. The green K2 is 130 lumens at 1.5 amps.

Or if you want red, get a red Luxeon III and drive it at 1.4 amps (around 150 lumens). That's right next to the max and you won't see much difference past that.

If you want blue, get a Hyperblade. Blue Luxeon is not as bright as other luxeons (only like 30). Still brighter then an MR if built right though.

If you want white (so you can filter purple, blue, yellow, orange, etc.), get a white K2 at 200 lumens as they become available.

There are other luxeons like Seoul ones that are bright but I have no experience with those and I don't know where to order those.

But still, if you insist on overdriving the crap out of a lux III instead of those things then get 2x 1000mA so they are matched still. But again, a Green Lux V at 1 amp will be brighter. Probably somewhere around 180 lumens of green. That's over twice of what you have now.

chase
02-25-2008, 02:31 PM
I already have a green LuxV and I love it, I just don't love the battery space it takes. Since you and xwing said that it wasn't a good idea to have different pucks, I asked about overdriving a green K2. I asked about a 1000mA +700mA buckpuck combo and if it would work well with overdriving a green K2.

Hasid Lafre
02-25-2008, 02:38 PM
2 1000 mah pucks. there ya go. overdriving it by around 500.

Eandori
02-25-2008, 04:26 PM
I *think* it would work either way... 1000+700 or 700x2, if it's hooked up right. The pucks should inrease the output voltage until each of them has enough current flowing. But what I worry about is this...

Say one puck is already putting out 700mA, but the other puck is only putting out 700mA and wants to put out 1000mA. I think that puck would keep increasing the common node voltage until it satisfied it's 1000mA, but what does that do to the 1st puck? Does it force it to yank out more current, making it try to pull that node voltage down? Does it force that current backwards through the 2nd puck?

I don't know. I'll simply say it seems a bit risky and I would not connect it like that unless I was willing to fry my parts and buy more. But hey, it's your stuff.

strengthofrage
02-25-2008, 04:46 PM
Arm on Fire used two 700 buckpucks in parallel on his last/current saber if I remember right. I just ordered 2 myself for the same setup to use with the newest MR board (616).

chase
02-26-2008, 10:18 AM
Hasid, have you tried overdriving a K2 that much? I don't want to blow it up and have to order new ones.

Is it even that much different to overdrive a LED? Am I waisting my time trying to figure out this stuff?

Hasid Lafre
02-26-2008, 12:11 PM
Wouldent be any different that the luxIIIs being overdriven to 1.5A or a luxV driven to 1A.

you overdrive the led by half you take half its life off. Its not the current so much that kills the led its the volts that kill it.

Atleast thats how I understand how lux leds work.

Eandori
02-26-2008, 12:51 PM
You were right on all of it but the last statement. It's the current that kills it. But the current is a function of the input voltage... so you "could" say the voltage kills it and be partially correct. It works like this.

The input voltage raises, the active silicon P-N junctions open up the channel to allow more electron flow. As a result, more electron flow occurs (current). The high current heats up the very small traces on the silicon and the wires connecting the silicon to the solder tabs. The small traces heat up enough that they vaporize, severing the connection between teh solder tabs and the silicon LED.

That's how it actually happens. :)

chase
02-27-2008, 06:30 AM
You were right on all of it but the last statement. It's the current that kills it. But the current is a function of the input voltage... so you "could" say the voltage kills it and be partially correct. It works like this.

The input voltage raises, the active silicon P-N junctions open up the channel to allow more electron flow. As a result, more electron flow occurs (current). The high current heats up the very small traces on the silicon and the wires connecting the silicon to the solder tabs. The small traces heat up enough that they vaporize, severing the connection between teh solder tabs and the silicon LED.

That's how it actually happens. :)


You might as well be speaking Japanese. I'll just stick with the recommended mA and puckage and later I'll boost it up. Thanks for the help fellas.

Eandori
02-27-2008, 10:04 AM
Said more plainly....

The voltage raises, then a higher voltage allows more current to flow. The higher current is what destroys the luxeon. So the current "killed" it, but the current could not happen until the voltage was higher.

Hasid Lafre
02-27-2008, 11:51 AM
Ok I see, So in a sence its actually the heat from overdriving it that kills it. What if you made sure that that led stayed as cool as possable?

Ghostbat
02-28-2008, 10:08 AM
Ok I see, So in a sence its actually the heat from overdriving it that kills it. What if you made sure that that led stayed as cool as possable?

The cooler you keep a component the more you can overdrive it. This is the reason for those insane water cooling systems for your computer, so you can drive your processor far beyond what ti was designed to take without cooking it.

In theory you could run some tiny cooling system across your LED, maybe a many-finned heatsink with a fan and vents to keep the air constantly flowing around it, and drive it much brighter.

Of course at some point you have to wonder if the extra gear you have to build in, the increasing size of the hilt, and the dramatically reduced battery life is worth the amount of increase :)

chase
02-28-2008, 10:25 AM
Fantastic. How about just using some of that heatsink gel or whatever it's called? That should help.

Hasid Lafre
02-28-2008, 02:24 PM
Should work.

Eandori
02-28-2008, 04:20 PM
Even with a top notch, copper air cooled heatsink you will still fry the LED with too much current. While it's true that "heat" fries the small wires it's not the kind of heat you are thinking of.

The heat you are thinking of comes out of the actual LED silicon itself. Each P/N junction dissipating some power, next to other P/N junctions dissipating power. Overall adding heat together that's getting sinked into the heatsink.

The heat I'm talking about is individual wires that are really thin having too much current and frying. Imagine trying to run 100amps through a very narrow wire the size of a hair... the wire will heat up and disintigrate. If that wire was connecting to a heating element (like on an electric stovetop) then you are not going to stop the wire from disintigrating by putting a heatsink on the stovetop coils.

Does that make sense?

xwingband
02-28-2008, 06:53 PM
Yeah, there is the limit to the heat of the diode and even with a great heatsink you may be screwed. There is a thermal resistance in the star itself so it might be like car traffic... if it's too much at the diode the star won't be able to get the heat out. Then a heatsink isn't going to help no matter how big if it's not getting the heat.

Novastar
02-29-2008, 01:22 AM
Those are EXCELLENT ways of pointing out the "bottleneck" the LED runs into.

It isn't *always* solely about just getting a bigger, better heatsink for over-driving.

This is actually my main concern with the future of the single LED... unless there is a total architecture change... there IS a ceiling. X-wing alluded to it as well. And it makes sense. It's like... expecting to get 70 volts out of a coin cell. Um... on what PLANET?

Maybe I'll visit Area 51 and ask the aliens what to do. They should know. After all... I think that one alien named XZLKHZ*Z^ZONK discovered back in the 50's was the one who gave Lucas the idea of SW in the 1st place... ;)

Ghostbat
02-29-2008, 10:54 AM
Fantastic. How about just using some of that heatsink gel or whatever it's called? That should help.

A little bit yes. Thermal paste is recommended to deal with even the amount of heat generated by the LED running at recommended levels so it's not going to let you get much hotter by it's lonesome.


Even with a top notch, copper air cooled heatsink you will still fry the LED with too much current. While it's true that "heat" fries the small wires it's not the kind of heat you are thinking of.

The heat you are thinking of comes out of the actual LED silicon itself. Each P/N junction dissipating some power, next to other P/N junctions dissipating power. Overall adding heat together that's getting sinked into the heatsink.

The heat I'm talking about is individual wires that are really thin having too much current and frying. Imagine trying to run 100amps through a very narrow wire the size of a hair... the wire will heat up and disintigrate. If that wire was connecting to a heating element (like on an electric stovetop) then you are not going to stop the wire from disintigrating by putting a heatsink on the stovetop coils.


Well... that depends on how GOOD a heatsink you are using :)

I am speaking almost entirely in the abstract as I am very dubious that you could get appreciable results from any additional heat sinking you could fit in a hilt. Luxes are rated pretty close to their failing point unlike traditional LEDs (which have a lot of wiggle room).

But say you had a mass of ultra thin fins in a circulating liquid cooled by a nitrogen tank or something functionally similar. You could suck off a LOT of heat and run the LED way harder than it was ever intended. Of course only having a heatsink coming off the back of the component wouldn't be able to deal with the increase fast enough to save some parts on the front of the component, so you will need to run coolant across the face as well. which will of course suck up a lot of the escaping light.

The question is, would an increase of complexity and cost of, say, 500% cause an increase in brightness to match? Unlikely... And now you have a tank of liquid nitrogen, a pump, and a battery on your back with a hose feeding to an 8 pound hilt... not to mention an LED who's brightness you can only actually see part of :)

Eandori
03-04-2008, 06:24 PM
No, you still missed what I said and what Xwing said.

Let's imagine your LED runs 1.5 amps fine. At 3 amps it fries with a "normal" heatsink.

Even with your liquid nitrogen system I would expect that LED to pop if you gave it slightly more then 3 amps. Do you understand? You are not going to fix that problem with more heat sinking!

take a 24 gage wire and hook it up to the terminals of a car battery. It will get red hot and fry. Now cool a similar wire ANY WAY YOU WANT and hook it up to the car battery again. You will see that wire fry again.

It's a case of heat not being piped out of the unit fast enough. The heat channel is too narrow. To describe it like eletricity, you are not going to get a 100amp DC current through a 100 Ohm resistor if your battery is only 9v. Make the wires connecting the battery solid gold if you want. It's not fixing the PROBLEM.

The problem here is small traces where current flows in the Silicon and at the PN junctions. Heat will NOT transfer out of that silicon and into the LED heatsink fast enough for you to stop it from burning out.

Does that mean future LED's will not handle more current? No! But they will be designed differently. Instead of traces on the silicon dye being 1-15 mils wide they will be many times that size. Which allows more current to flow with less heat generated. In short... the luxeon LED will need to be beefed up in design phase. You won't fix this with any amount of liquid nitrogen.

EDIT: Adding some links for your reference/understanding.

http://web.mst.edu/~gbert/basic/thermo.html
Thsi page discusses the basics with some examples

http://www.educypedia.be/education/mechanicsthermo.htm
This page has links to many other more specific topics on the subject.

http://forums.thecustomsabershop.com/showthread.php?t=3555
This is a thread I created talking about the parts that apply to what we do here. Trying to simplify it for you.

chase
03-07-2008, 10:27 AM
Ok, how about this....Green LuxIII with 1000mA buckpuck vs. Green K2 with no puck. Who is brighter?

Eandori
03-07-2008, 10:43 AM
Ok, how about this....Green LuxIII with 1000mA buckpuck vs. Green K2 with no puck. Who is brighter?

This question cannot be answered in it's current form.

How much current is flowing through the Green K2? If 5mA then the Lux III is brighter. If 1.5A then the Green K2 is brighter. I'm going to assume you were asking about Green Lux III at 1 amp, and Green K2 at 1.5 amps. If that's the question... The answer is right on those websites.

Green Lux III at 1000mA = 80 lumens
http://www.luxeonstar.com/item.php?id=387&link_str=196::198&partno=LXHL-LM3C

Green K2 at 1500mA = 130 lumens
http://www.luxeonstar.com/item.php?id=1803&link_str=330::1433&partno=LXK2-PM14-U00

If you wanted to compare the K2 to the Lux III at 1 amp each... then the K2 should be brighter. The Green Lux III is rated at 80 lumens for 1 amp, and the K2 is rated at 75 lumens at 700mA. So at 1000mA it should be brighter, somewhere between 75 and 130 lumens (probably around 100 lumens at 1000mA).

Another quick note. It does not matter what is providing current to an LED. 1000mA from a buckpuck is EXACTLY THE SAME as 1000mA from a resistor setup. If you measure the current from both to be the same, it is the same!

What changes is over time the buckpuck will remain at it's rated current as the battery drains. A resistor will slowly pass less and less current as the battery drains.

Sorry to be so specific on this stuff. But I must be, the devil is in the details!

chase
03-08-2008, 03:38 PM
It's great that you give details. That's why I ask these questions. If I had the money I would do these experiments myself and wouldn't even bother to ask questions. Thanks for the help though. Looks like I'll be getting some green K2's. I've got about 3 sabers I'll be putting them into. I've got a luxV green, and I love how bright it gets, it's just the battery pack is so big I have no room for anything. The royal blue K2's I have just don't do it for me, they are pretty dim compared to green. I plan on running the K2's with the 4 AA battery pack with a 1000mA puck (seeing they don't make a 1500mA puck).