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ArcticMyst Security by Avery

Could a DDL driver supply a 2.5A current?

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Hey guys, I'm in the process of building my first IR labby and its recommended Current for the 2 Watt Diode is 2.5 Amps. I am questioning if two .5 ohm resistors in parallel would work with the driver to make a 2.5 A current, or if it would just fry the board, any suggestions?

~Jeremy
 





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The LM317 is only good for 1.5A, you need a LM350 which can go up to 3A. Also your resistor need to be rated for 4W and be 0.5 ohms, not 0.25.
 
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Yes you could use 4ea 2ohm 1watt resistors in parallel that would give you a .5ohm 4watt resistor.


The LM317 is only good for 1.5A, you need a LM350 which can go up to 3A. Also your resistor need to be rated for 4W and be 0.5 ohms, not 0.25.
 
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Like FML said, those are only capable of 1.5A. You can use multiple LM317s for more current though. I'm not sure how to wire them up, but IIRC it might be just as easy as wiring two of the circuits up in parallel.
 
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I just went to my local Radio Shack and could only find my silicon diodes, pcb's, and lots of solder. I'm in the process of finding the 4: 2 Ohm resistors along with the LM350 and a pot. Could anyone please lead me in a direction to find the resistors and the lm350? I can't find anything on them anywhere.
:thanks:
~Jeremy
 
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So the pot would kinda be rendered useless with all the resistors?
Sorry for the noobish questions, I'm not used to working with diodes of this power..

~Jeremy
 
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At this power the pot probably wont hold, and you don't need it to be adjustable anyway since you know what current you want.
 
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I don't really like using pots on most of my DIY driver builds either because it just makes things more complicated, if you do the calculations right the current will be perfect when you use the right resistors. With a high power IR laser diode you won't ever be adjusting current so it will just be a waste of space. Just calculate the resistors to the current you need and then you are set, plus you don't have to spend time setting the current.
 
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Thanks for all the input guys. This will be my last question I swear.
Would a 9v battery or 2 3.6v CR123 batteries work better with this diode?

~Jeremy
 
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A 9V battery is nowhere near capable of this job. Also, 9V is a little high, as the IR diode run at it's rated current will probably drop 2.2 to 2.5V. The linear regulator would have to dissipate the rest as heat. 2 x 18650 would be a better choice for current handling. If the regulator heats up a lot, try a putting a 3A or better diode between the batteries and the regulator to drop the voltage. Another alternative that I have used is 1 x 18650 and two DX disk regulators in parallel, you would need a 1400mA and a 1050mA for your diode. These are negative side regulators, which can simplify construction of a handheld, since the IR diode is positive ground. Let us know how your project goes.
 
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Well, ATM I just got my parts in, but my PCB isn't set in Parallel or Series, so making it so is a difficult task. I'm pretty sure I'm screwing up everything... this isn't working well.

~Jeremy

7133_188063191039_704276039_3935106.jpg



9925_188063686039_704276039_3935108.jpg
 
Last edited:

Benm

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You could source 2.5 amps using two DDL circuits in paralel, but they would sure be burning a lot of power. Since the full current goes through the setting resistor, pots would be very difficult to use since you'd need the ceramic wire-wound kind to handle the power dissipated.

The DX disc regulators mentioned might be a better source - they just consis of AMC7135 chips, each delivering about 350 mA to your load. You would need 7 of them to make it work, but it sure would. The chief benefit is that these are low drop out regulators. Using these, you can easily power your pump diode from a single (3.6V) lipo battery.
 
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You are Unreal Man --- A 9 volt battery can't put out 0.5 amps !!! You need a power supply or a big gel cell.

Mike
 
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well the second picture is for my fan.....

The first one is waiting for the 18650's to come in the mail... along with the battery pack

~Jeremy
 
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Also, make sure that the rating of your switch can withstand 2.5 amps at whatever voltage you're going to use. Most switches will handle it, but a low quality switch could burn out.
 




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