How much power do I need to run 50 uv leds

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diesalweasel

Joined Aug 28, 2010
36
Hi guys as the post says I'm doing my self a little project I've seen 50 uv leds and I was wondering how much power I would need to run all of them together.
I'm building my self a big uv light as I have started to repair mobile phones it for the loca glue which is used in between the glass and the lcd display. I'm going to convert an old flood light. I mite not even need 50 mite get away with 20 or 30 leds. But for how cheap they are about £3.70 it not going to break the bank lol here is the eBay link which will give you the full power spec this is the item number for ebay as it wouldn't let me insert a link

290736745166

Thanks in advance regards James
 

MikeML

Joined Oct 2, 2009
5,444
each led is 20 to 30mA , so 50 x 30mA = 1.5 Amps max.
Not if he puts groups of them in series.
I'm guessing that for a uV Led Vf is about 3.3V at a current of 20mA, so the theoretical total power required is 50*3.3*0.020 = 3.3W.
10 parallel strings of 5 each would only require 10*0.02 = 0.2A
Min input voltage, allowing at least 2V across a current ballasting resistor, is 2 + 5*Vf = 2+5*3.3 = 18.5V, so I would be looking for a 18 to 22Vdc Wall-wart plug-in power supply.

You need to accurately know Vf before we proceed further....
 
Last edited:

Marcus2012

Joined Feb 22, 2015
382
Hi :)

LOCAs should have a datasheet available somewhere telling you the range of wavelength needed to cure the glue and data usually in the form of a graph showing you the radiant energy density required against % reacted glue. As radiant energy density is measured in joules and 1 watt is 1 joule per second it's not too hard to convert. For example the datasheet below for a LOCA requires 2000mJ/cm2 and 0.2mm glue thickness to that should be around 2W/cm2 for a second. But as the relation of Joules to Watts is a function of time you can use a lower wattage but it will take longer to cure. I think I remembered that right it's been a while since I applied any of this so might be worth getting some to check lol but hope that helps. This may be easier to achieve with a conventional bulb as opposed to LEDs but the datasheet actually advices sources and specific LED wavelengths.

Intensity (mW/cm2) = Radiant energy density (mJ/cm2) / Time (seconds)
 

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crutschow

Joined Mar 14, 2008
23,544
Hi :)

LOCAs should have a datasheet available somewhere telling you the range of wavelength needed to cure the glue and data usually in the form of a graph showing you the radiant energy density required against % reacted glue. As radiant energy density is measured in joules and 1 watt is 1 joule per second it's not too hard to convert. For example the datasheet below for a LOCA requires 2000mJ/cm2 and 0.2mm glue thickness to that should be around 2W/cm2 for a second. But as the relation of Joules to Watts is a function of time you can use a lower wattage but it will take longer to cure. I think I remembered that right it's been a while since I applied any of this so might be worth getting some to check lol but hope that helps. This may be easier to achieve with a conventional bulb as opposed to LEDs but the datasheet actually advices sources and specific LED wavelengths.

Intensity (mW/cm2) = Radiant energy density (mJ/cm2) / Time (seconds)
You need to include the efficiency of the LED to convert from input power to radiant power.
 

John P

Joined Oct 14, 2008
1,763
It is a type of glue which is cured (i.e. made to turn solid) when exposed to ultraviolet light. He's working on repairing the displays on cell phones.
 
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