JoshuaG
New member
Thanks for the heads up, just placed my order for the XP-E's. Can't beat $5 a pop! Now if I could just find the royal blue's for $5 I would be in business!:beer:
DIY LEDs - The write-up
- Thread starter JohnL
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Thanks for the heads up, just placed my order for the XP-E's. Can't beat $5 a pop! Now if I could just find the royal blue's for $5 I would be in business!:beer:
Jay1982
New member
Alright I got my shipment from Cutter, I've just tried quickly searched for how to adjust the internal trimpot on the Meanwell eln-60-48-P. Do I turn it clockwise or counter clockwise to turn the power all the way down?
Also, I got XPGs and XPEs. The cool whites have yellow emitters, correct? While RB have white?
Also, I got XPGs and XPEs. The cool whites have yellow emitters, correct? While RB have white?
widmer
Drug Enthusiast
In the interest of full disclosure, I've just noticed that those XP-E's are one of the lower bins in terms of efficiency (XP-E = 100 lumens @ 350 mA while XR-E = 107 lumens @ 350 mA). It puts them at almost exactly as efficient as the XR-E, bearing in mind that the XP-E has a slightly lower forward voltage at the same current than an XR-E (XP-E = 3.5v @ 1000 mA while XR-E = 3.7v @ 1000 mA).
So it is essentially the same as buying a standard XR-E Q5 for $5, although optics will take a little guess and check. Overall a good deal IMO.
widmer
Drug Enthusiast
If you are referring to the actual die as you look at it closely, then yes.
Add them to your cart, that is the easiest way I have found. Price is AUD but after fees etc usually works out to be pretty close.
Jay1982
New member
They were pretty cheap price wise. But they took a 5 weeks to get here and I have 3 big screw ups on my order. If I was you I'd look elsewhere and save yourself the headaches. Emailing them doesn't work very well. The only way I get a response is by phoning them, talking to a rep (after being placed on hold for 10-15 minutes). And then having to call back 2 days later because they never sent the email they promised they would send.
Which brings me to another question about items I recieved:
I ordered 40 degree optics for my XPGs and XPEs. They sent me these optics that are labeled:
MMR20085
CTP/X5480
101196R2 - 0.8mm Frosted wide 20mm optic (stamped on the package with Carclo).
So are these the actual 40 degree optics? Or is frosted something else I didn't order?
Which brings me to another question about items I recieved:
I ordered 40 degree optics for my XPGs and XPEs. They sent me these optics that are labeled:
MMR20085
CTP/X5480
101196R2 - 0.8mm Frosted wide 20mm optic (stamped on the package with Carclo).
So are these the actual 40 degree optics? Or is frosted something else I didn't order?
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Jay1982
New member
Thank you
JoshuaG
New member
Jay1982
New member
Sure enough after looking at the carclo website I recived completely different lens than what I ordered....
Can someone please tell me if they'll fit on to both my XPEs and XPGs. And what degree optics they are. I can't read the data sheet properly (too many acronyms).
http://www.carclo-optics.com/opticselect/intranet/optics/details/index.php?id_optics=39
Is there a severe drop off from the regular non-frosted optics? If they are pretty much the same I'm just going to say screw it and use them.
Can someone please tell me if they'll fit on to both my XPEs and XPGs. And what degree optics they are. I can't read the data sheet properly (too many acronyms).
http://www.carclo-optics.com/opticselect/intranet/optics/details/index.php?id_optics=39
Is there a severe drop off from the regular non-frosted optics? If they are pretty much the same I'm just going to say screw it and use them.
CJO
New member
Cut and paste at its best! Actually, it's you, der_wille_zur_macht and a few others who we should thank- you are the people who supplied the information :thumbsup:
CJ
tahiriqbal
Sialkot
They should fit fine for both XP-G and XP-E LEDs. You need a dab of glue to have them mounted on your LEDs or PCBs. The lens you got from Cutters has poor efficiency comparing to http://www.carclo-optics.com/opticselect/intranet/optics/details/index.php?id_optics=86 . They also seems to be extra wide viewing angle with 84% efficiency which is pretty good at the moment for XP-G LEDs by Carclo.
I am lucky enough to have Carlco in my backyard, I have received numerous free samples from them and Johnathan is your main man for any further information. For my own DIY project I have gone with these http://www.carclo-optics.com/opticselect/intranet/optics/details/index.php?id_optics=28 lens, they are really tiny and look very neat with good efficiency output for the time being. I am wondering what people thoughts are regarding ETG's 60 degree lens for XP-G LEDs, are they Clear, Frosted, Elliptical, Ripple or what? If anyone got them on hand please could you kind post your views and pictures of the lens, if possible?
Jay1982
New member
Thanks tahiriqbal! Yeah I noticed they were based in the UK. Man I love England. I was there last summer for the first time. I'm going back in August, I simply love it there.
So onto my optics, are they still 40 degree optics? I just talked to cutters and they said they are going to look into my "problem." Hopefully I can get them to ship these out quicker than the first time.
So onto my optics, are they still 40 degree optics? I just talked to cutters and they said they are going to look into my "problem." Hopefully I can get them to ship these out quicker than the first time.
tahiriqbal
Sialkot
Thanks, I am glad you enjoyed your time here in England. Our summer has just started to kick inn and its lovely out there and today is the hottest day so far
Not sure which optics are you aksing about, the one you got or the one I have suggested? It seems that Cutter's are making lots and lots of blunders and messing up orders. Anyone has any comments regarding ETG's 60 degree lens?
spamreefnew
New member
ok im gonna give this a shot. but one issue, my tank has a foot print of 385 square inches BUT.. with my black eurobracing the exposed surface area of water is only 225 square inches. if i go buy the rule of 1,3 watt led per 10-20 sq" of surface area i can buy the 24 led kit and have all the power i need right? what should i do for optics? i was thinking because i will have more power than i need with the 24 leds that i could get away with the optics mostly used on an 18" deep tank when my tank is 24" deep ,,,or should i go with the recomended optics and just use real wide optics on the outtermost leds to try and get under the eurobrace????
DeepPirate
New member
Heat-shrink tubing?
Heat-shrink tubing?
Question for the experts...I plan on using some solid 24AWG, pretinned, PVC insulated wire to connect the LED(s) to other LED(s) and some stranded wire, consisting of similar specs, to connect the driver(s) to the LED string(s). Should I also be using a Heat-shrink tubing that provides a moisture barrier? Or, is there something to protect the various connections...ie LED to LED & Driver to LED.
Heat-shrink tubing?
Question for the experts...I plan on using some solid 24AWG, pretinned, PVC insulated wire to connect the LED(s) to other LED(s) and some stranded wire, consisting of similar specs, to connect the driver(s) to the LED string(s). Should I also be using a Heat-shrink tubing that provides a moisture barrier? Or, is there something to protect the various connections...ie LED to LED & Driver to LED.
TheFishMan65
New member
LED to LED is usually soldered to a pad, and the heat shrink will not cover it. Best option here is an acrylic shield to keep down salt spray.
Don't remember the connection on a Meanwells, but if it is wire to wire I would use heat shrink and be safe.
Don't remember the connection on a Meanwells, but if it is wire to wire I would use heat shrink and be safe.
GEORGEDOPE
New member
Reguarding Parallel strings wouldn't using a resister instead of a fuse be better. Like the drawing below. If one LED blows then the current to the other strings would not increase. And if 30 3W LED's are on 1 Meanwell would that mean I'd need a 100W Meanwell. Like the CLG-100-48 that has the 0-10V dimming?
Solution 3: 10 x 3 array uses 30 LEDs exactly
+48V R = 22 ohms
R = 22 ohms
R = 22 ohms
The wizard says: In solution 3:
"¢each 22 ohm resistor dissipates 10780 mW
"¢the wizard thinks the power dissipated in your resistors is a concern
"¢together, all resistors dissipate 32340 mW
"¢together, the diodes dissipate 73500 mW
"¢total power dissipated by the array is 105840 mW
"¢the array draws current of 2100 mA from the source.
Sorry the drawing didn't copy. It would be 3 strings with 10 LED's per string . But do you see what I mean? doesn't this make sence, or don't I understand resistors.
Solution 3: 10 x 3 array uses 30 LEDs exactly
+48V R = 22 ohms
R = 22 ohms
R = 22 ohms
The wizard says: In solution 3:
"¢each 22 ohm resistor dissipates 10780 mW
"¢the wizard thinks the power dissipated in your resistors is a concern
"¢together, all resistors dissipate 32340 mW
"¢together, the diodes dissipate 73500 mW
"¢total power dissipated by the array is 105840 mW
"¢the array draws current of 2100 mA from the source.
Sorry the drawing didn't copy. It would be 3 strings with 10 LED's per string . But do you see what I mean? doesn't this make sence, or don't I understand resistors.
TheFishMan65
New member
Pardon me while I think this through.
Looks like you are driving 700 ma per string. The voltage drop across the resistor will be 15.4 volts (.7 * 22) and the power will be 10.78 watts (15.4 * .7). Given that the Meanwell can supply 48 volts this leaves 32.6 for the LEDs or 3.26 per LED. Sounds about right.
Now a string get opened the the meanwell will want to continue to drive 2.1 amps so each string should get 1.05 (which may blow an XR-E). So at 1.05 amps the resistor drops 23.1 volts and leaves about 25 for the LEDs (not enough voltage get the LEDs to one amp). So the the system will balance with the Meanwell supplying 48 volts and whatever current it takes to get the right drop across the LEDs and resistor. Yes this would work.
BUT think efficiency. Yo have 10 LEDs at 3 watts each (maybe at .7 amps they probably don't use 3 watts), that is 30 watts. The resistor in normal operation was dropping 10.78. So at best you are wasting 25% of your power. Seems like a waste to me. Efficiency is one of the main reason I am going LED.
PS what tool were you using?
Looks like you are driving 700 ma per string. The voltage drop across the resistor will be 15.4 volts (.7 * 22) and the power will be 10.78 watts (15.4 * .7). Given that the Meanwell can supply 48 volts this leaves 32.6 for the LEDs or 3.26 per LED. Sounds about right.
Now a string get opened the the meanwell will want to continue to drive 2.1 amps so each string should get 1.05 (which may blow an XR-E). So at 1.05 amps the resistor drops 23.1 volts and leaves about 25 for the LEDs (not enough voltage get the LEDs to one amp). So the the system will balance with the Meanwell supplying 48 volts and whatever current it takes to get the right drop across the LEDs and resistor. Yes this would work.
BUT think efficiency. Yo have 10 LEDs at 3 watts each (maybe at .7 amps they probably don't use 3 watts), that is 30 watts. The resistor in normal operation was dropping 10.78. So at best you are wasting 25% of your power. Seems like a waste to me. Efficiency is one of the main reason I am going LED.
PS what tool were you using?
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