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Old 03-22-2008, 04:25 AM #1
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Default current issues with DIY open can

Well I finally finished my DIY open can red laser. Its not at full power yet because my batts are not providing enough voltage, but its still BRIGHT. and yes, I have glasses for it! so far I've lit several matches unsharpied (red head) and I've popped 2 balloons from quite a distance. It can also cut electrical tape and etch into wood and plastics that aren't red or white.

This is the second DIY I've done, and Its kinda ironic because I successfully made a blu-ray before I made this one, and I had more trouble with this one than with the blu-ray. I killed one open can before I got it to work. I think it was the circuit, I had alot of trouble with the circuit.

But anyways, *I'm going to go take some pictures right now with it compared to my 5mW greenie


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Old 03-22-2008, 04:33 AM #2
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Default Re: FIRST DIY RED SUCCESS

HURRY UP WITH THOSE PICS. ;D
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Old 03-22-2008, 04:54 AM #3
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Default Re: FIRST DIY RED SUCCESS

lol

well I don't have a good camera, and I'm not that great of a photographer, but here's what I managed to get


the actual laser


spot comparison ith my Galileo 5


beam comparison with my Galileo 5

I'll try to get some better pictures later
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Old 03-22-2008, 05:11 AM #4
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Default Re: FIRST DIY RED SUCCESS

Nice, I know your batteries aren't giving it full power, but what current are you driving the open can at?
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Old 03-22-2008, 02:58 PM #5
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Default Re: FIRST DIY RED SUCCESS

right now its at about 280mA. with the correct batteries, it should be at 420mA. That seems kinda like a big difference, so I'm going to check the curent again after my batts get here in the mail. I might even go do some more extensive measurments on the circuit right now just to see if its correct as of right now.

at 280mA, the laser is only at about 150mW lol. in the end it SHOULD be 250!

edit: alright chido, I'm gonna need some help with this. If my open can is running at 280mA, and I'm using an LM317 and a 3 ohm resistor, how would I set up Ohm's law to get me the voltage I need? whenever I just plug in .420 and 3 to I and R, it comes out with 1.26 as the voltage. I think its cause I'm not taking the LM317 into account. is that right? and how would I set this equation up?
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Old 03-22-2008, 06:42 PM #6
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Default Re: FIRST DIY RED SUCCESS

Quote:
Originally Posted by Abray
at 280mA, the laser is only at about 150mW lol. in the end it SHOULD be 250!

edit: alright chido, I'm gonna need some help with this. If my open can is running at 280mA, and I'm using an LM317 and a 3 ohm resistor, how would I set up Ohm's law to get me the voltage I need? whenever I just plug in .420 and 3 to I and R, it comes out with 1.26 as the voltage. I think its cause I'm not taking the LM317 into account. is that right? and how would I set this equation up?

I thing the LM317 put out 1.25. Use this value in place of 3
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Old 03-22-2008, 07:06 PM #7
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Default Re: FIRST DIY RED SUCCESS

Quote:
Originally Posted by Abray
edit: alright chido, I'm gonna need some help with this. If my open can is running at 280mA, and I'm using an LM317 and a 3 ohm resistor, how would I set up Ohm's law to get me the voltage I need? whenever I just plug in .420 and 3 to I and R, it comes out with 1.26 as the voltage. I think its cause I'm not taking the LM317 into account. is that right? and how would I set this equation up?
You don't need to worry about voltage when using the LM317 circuit, the diode will take whatever voltage it needs, just think current.

Look at GAZOO'S canned response:

The LM317 is no mystery and very easy to work with. The following calculations always apply since it uses 1.25 volts for its reference voltage, and ohm[ch8217]s laws don't lie.
*
To calculate the resistor needed for a given current, take 1.25 and divide it by the current. So say you want to drive a SenKat diode with 250mA. 1.25 divided by .250 = a 5 ohm resistor.
*
Another way you could do this is to take 1.25 and divide it by the resistance. 1.25 divided by 5 = .250.
*
Next you will want to calculate the wattage of the resister needed. We know 1/2 watt resistors are common for use with the regulator. But to figure it out, simply take the 1.25 and multiply it times the current. 1.25 times .250 =.3125 watts.

The rule of thumb for the voltage going into the regulator is it should be 3 volts more than the voltage going to the diode. A SenKat diode running at 250ma's will have about 3 volts across it. Therefore a minimum if 6 volts is needed.
I recommend 6 NIMH batteries or 2 RCR123's for use with Daedal's driver.
*
This is why you need at least 8 volts to run the blu-ray. You will find when you have it hooked up, the voltage across it will be appx. 5 volts. But since it will only draw ~38mA, you can run it with a 9 volt battery.
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Old 03-23-2008, 07:02 AM #8
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Default Re: current issues with DIY open can

Why can i see my open can diode beam at night but when i take a picture with my camera phone, i cant see anything? How do you people get the beam to come out so well in the pictures? I'm using a Nokia N73 3.2MP Carl Zeiss lens camera to take my shots.
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Old 03-23-2008, 07:18 PM #9
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Default Re: current issues with DIY open can

i use a 1.3mp 4y/o camera that takes 32mb cards its that crap and it comes out fine
:P

i have one of these made by jayrob on its way... cant wait
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