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Message from discussion High Power UV LED Driver

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From: Mike Harrison <m...@whitewing.co.uk>
To: london-hack-space@googlegroups.com
Subject: Re: [london-hack-space] High Power UV LED Driver
Date: Thu, 04 Oct 2012 13:29:08 +0100
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>> Hoping someone can help me... I'm working on a DLP resin based 3d =
printer.
>> I have a modified LED Pico projcetor which I have removed the LEDs =
from and
>> replaced them with a UV LED. Currently I have a 5W LED on it for a
>> proof-of-concept. It works, but I need to upgrade to a more powerful =
LED
>> for cure times to be at all reasonable.
>>
>> I'm looking at using the CBT-120-UV-C11-G382-22 UV LED from Luminus
>> Devices, which is explicitly designed for this purpose. It is an =
insane
>> (and expensive) diode that is ideal for my purposes... With 11 Watts =
of
>> actual optical power output it is one of the brightest LEDs on the =
market.
>> The problem is, I need 18amps at 3.7 volts to drive it.=20

Probably a bigger problem is how are you going to cool it?  50 watts from=
 a small emitter is an
awful lot of heat to remove from a very small space.

>So... My question is, how can I get this?

This is in the same power ballpark as high power laser diodes, so may be =
worth looking for any
appnotes  in this area to get some ideas.=20

The PCB layout at these power levels is going to be extremely critical.=20

Bear in mind that a constant current supply is little different from a =
constant voltage one - the
only difference is how the feedback is sensed.=20
Most voltage regulators can be used as CC sources by putting a =
current-sense resistor in the
negative rail and using the voltage across this as the feedback instead =
of the divided output
voltage.=20
The main difference with SMPS chips aimed at LED applications is that =
they're designed for a low
feedback reference voltage (typically around 0.2V)  to minimise losses in=
 the sense resistor - most
voltage-oriented chips use a 1.2V feedback reference. You can use an =
amplifier to boost the feedback
voltage form a low-value reference resistor, but you need to keep an eye =
on loop stability issues -
at least your load isn't going to change suddenly so you can probably =
just damp the crap out of it
to keep it stable.=20
=20
One approach that may work, avoiding the layout issues is to use a 5V =
mains PSU that has remote
output sensing, and drive the sense input from an amplified voltage =
across a low-value sense
resistor. If done properly, the input sense will use a differential =
amplifier - if you delve inside
you may be able to find it and increase the gain to avoid the need for an=
 external amp.=20

If mains PSUs scare you, a similar approach would be to do the same with =
a brick type 5V DC-DC
converter module.=20

>> Ideally, I'd like to run off of 5V. Using a regular resistor to power =
it
>> from 5V I need about 0.08 ohms capable of at least 25 watts (I think).=
 That
>> is going to get _Very_ hot and is simply an insane waste of power. =
Thats
>> assuming my power calculation is correct.... Do I use the full 5V for =
power
>> dissipation in the resistor, or the 1.3 volt difference between the =
diode
>> voltage drop and the power supply? If it's the full 5V then I actually=
 need
>> a 90 watt resistor....

Bear in mind the Vf will probably vary with temperature, so you =
absolutely do need active current
regulation.=20

If you do want to roll your own, the LT3743 may be worth a look:
http://cds.linear.com/docs/Datasheet/3743fc.pdf