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Old 01-02-2012, 04:06 PM   #1
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Default 38mA out of 2 AAAs?

Hi Guys, I cant find any solution for my problem, so I just try to ask you

I got a weak 10mW red diode and want to put it into a Pen Host I got here. The datasheet says at 10mW the Diode drains about 32-38mA at 2,2-2,5V. Im going to power it with 2 AAAs, so I got a Voltage of 3V... First I wanted to use a flex drive (May be a waste for a 10mW Laser, but I have one here and I dont need it atm... Also I just wanted to build an "less dangerous" laser then my other Lasers )

But then I noticed the Flex Drive starts at 65mA! So I search for any Idea what I could use to power my diode. I would even somehow modificate my Flex Drive, or build some myself, but the Problem here is it must fit my Pen housing, so it should have about the size of the FlexDrive...

Now its up to you! Any Idea how I can power this diode without a resistor (I really dont want to use a Resistor for a weak diode like this, actually I would never power a Diode without a Driver...)
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Old 01-02-2012, 04:58 PM   #2
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Default Re: 38mA out of 2 AAAs?

Power Supply Driver for 5mW 650nm/660nm Red Laser Diode LD | eBay

^^^Here.

Adj from 20mA to 40mA. Supply voltage up to 3v or in your case 2x AAA 1.5v cells.

I bought 1 a wail ago and i can tell you it does what it says it does.

Hope that helps.

What diode is it you are trying to power.??? DO you have a link to where you got it.?

There are also other driver that will work of 2x AAA 1.5v cells at 40mA Just fine.

Like:
3V Laser Diode Driver for 532nm Green 650mA & 200mA-240mA 650nm 660nm Red | eBay

2pcs adjustable current laser diode driver 80-500mA w/TTL | eBay


Those can work too depending on the diode.!!!!

Pay no attention to the tittles on them that say they cant go bellow a certain currant. Both those drivers can go down to Zero no Problem.
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Old 01-02-2012, 08:16 PM   #3
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Default Re: 38mA out of 2 AAAs?

Hi, thank you for the Links!

Im using this Diode: http://www.uocnet.com/pdf/LD/U-LD-65...reliminary.pdf

Looks like I will buy the first one. Also the seller has some interesting filters in his shop

So thanks again!

-MadEye
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Old 01-03-2012, 02:42 AM   #4
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Default Re: 38mA out of 2 AAAs?

looking at that data sheet, there is 2 pin out diagrams, A & C.

"C" is case negative & "A" is case positive, hope you choose "C" if you want to use the most common drivers!

Fiddy.
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Old 01-03-2012, 07:02 AM   #5
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Default Re: 38mA out of 2 AAAs?

you could just use a 25ohm resistor in series with your diode and batteries.

Your laser would just get dimmer as the batteries got low.
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Old 01-03-2012, 09:04 AM   #6
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Default Re: 38mA out of 2 AAAs?

Quote:
Originally Posted by justinjja View Post
you could just use a 25ohm resistor in series with your diode and batteries.

Your laser would just get dimmer as the batteries got low.
Please read the OP again. He clearly stted he does not want to use a resistor or direct drive this laser diode.
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Old 01-03-2012, 03:54 PM   #7
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Default Re: 38mA out of 2 AAAs?

Quote:
Originally Posted by lazeerer View Post
Please read the OP again. He clearly stted he does not want to use a resistor or direct drive this laser diode.
I figured I would throw it out there, as ordering stuff from china on ebay seems to take like a month before it gets delivered,
Personaly, I can never wait that long
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Old 01-03-2012, 04:21 PM   #8
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Default Re: 38mA out of 2 AAAs?

Quote:
Originally Posted by justinjja View Post
I figured I would throw it out there, as ordering stuff from china on ebay seems to take like a month before it gets delivered,
Personaly, I can never wait that long
Tell me about. Ive waited so much for so many different things that iam use to it now and doesn't bother me. But at first i use to go nuts.!
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Old 01-07-2012, 01:25 AM   #9
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Default Re: 38mA out of 2 AAAs?

This is very tricky since the voltage leaves zero to little room to work with. You could try a single FET current source to work with - perhaps running two or three of those in paralel. The other choice would be to boost the voltage first and do the regulation afterwards, making the build much more complex.
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