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FrozenGate by Avery

DIY Homemade laser diode driver

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Ericab, thank you very much for your kind words. If you would like to use the Dorcy mini to make a burning laser using these diodes, the Dorcy already has a regulated circuit that is sufficient to run the diodes for quite some time. I have made 2 of them and many here have done this same mod.

http://www.laserpointerforums.com/forums/YaBB.pl?num=1181635652

This is a link to a full guide with video even of the entire process by Kenom. He has done a superb job at explaining the process and I would recommend you check it out.

GL;
DDL
 





My light take 2 x 18650's = 3.6v
They are charged at 4.2V and they say the voltage can be anything up to that.
How would i use these to run the diode with Daedal's Circuit?
Thanks
Jonno
 
Inefficiently, you can add a couple 1N4001 diodes and drop the voltage to make up for the voltage hike. Or you can try and make this circuit with a pair of CR2 batteries and wait a little while longer for me to actually get the parts I ordered for the new circuit (I hope they work out) and supply the new circuit with just one of these batteries :)

In the meantime, I really would say the best power source for this thing is a reliable 6V source.

--DDL
 
I forgot to mention that 1N4001 diodes will drop the voltage by 0.7V each. So, if you have 3.6V on the battery, use 2 total to drop the voltage of the 2 batteries back down to 6V (closer to 5.8, but it's still fine because we're not using the full current threshold on the LM317). If the batteries are putting out 4.2V, then you would have to use 4 diodes (it'll still be fine at 5.6V as well).

GL;
DDL
 
All I could find at Radio Shack was a 1k-Ohm pot...they didn't even have the 25-Ohm that was suggested. Will this pot work, or should I work up an order to Digikey (I don't just want to order a $0.50 part)?
 
1K pot will not work here. You'll need to get one form DigiKey, or find another supplier. :-[

GL;
DDL
 
Can I ask a q? Will a 1N1004 diode work? Datasheet here
h tee tee p://physics.gac.edu/~huber/classes/phy270/SpecSheets/1N4004.pdf
The only diff I see is that the 4004 is rated for reverse current. Someone comfirm its OK to use?
And will any cap do?
 
I was informed that the 1N1004 WILL work because it is of the same series. I used them when making my driver, don't fully quote me though wait for someone like Daedal to confirm this.
 
Any diode 1N400X will work :)

As about the capacitor, try to get the biggest cap that will fit in your application (47uF - 300uF is fine). As for the voltage rating on the capacitor, the higher the better :). At a minimum, make sure it is rated at more than twice the supply voltage.

GL;
DDL
 
Hey DDL,

Is the voltage drop in your circuit so high that you need to drive it with 6V to get ~3 at the LD? I was planning on powering it with 2xAAs, but I need to change that to 4xAAs or 2xCR123s if I need more voltage.
 
Woo! Good news everyone who looked into making the driver Daedal helped me with! I finished it yesterday now that I finally got a new diode. It's working great, a tiny bit more power would be nice but it's still putting out over 100mW I would guess, but I'll take that in exchange for having it last longer from not being so over powered. Well, have fun making lasers everyone! ;D ;D ;D
 
Okay I know that the resistors are used as a feedback loop between the vout and adj lines to keep the current regulated but what I am not sure about is if the LD load goes through the resistors.

I want to build a test board that can deliver 400mA max. If I use a 3 ohm resistor speced at 3R 1% 0.6W. That would be 1.25/3 = 0.416mA
I then want to add a 10 ohm 18 turn trimmer that is rated at 0.5w. This would give me very fine control between 0.09mA to 0.416mA.

Now this is where I need help. My ohms law is a bit rusty but w = v * a. So 2.5v (LD voltage) * 0.416mA = 1.04 watts. That is over twice the rated wattage of the resistors  :-/ Now if the current is supplied via the adj line then I am good.

I hope this makes sense  :-/
 
You have to calculate the power going through the resistors to rate them - not the LD.  

Use the series resistance in the Out-Adj loop. If we use the 3 ohm resistor only:

P=I^2 x R
P=0.416 x 0.416 x 3
P=0.519W (therefore, you should really use a 3/4 or 1W resistor).

Paul

xgeek said:
Okay I know that the resistors are used as a feedback loop between the vout and adj lines to keep the current regulated but what I am not sure about is if the LD load goes through the resistors.

I want to build a test board that can deliver 400mA max. If I use a 3 ohm resistor speced at 3R 1% 0.6W. That would be 1.25/3 = 0.416mA
I then want to add a 10 ohm 18 turn trimmer that is rated at 0.5w. This would give me very fine control between 0.09mA to 0.416mA.

Now this is where I need help. My ohms law is a bit rusty but w = v * a. So 2.5v (LD voltage) * 0.416mA = 1.04 watts. That is over twice the rated wattage of the resistors  :-/ Now if the current is supplied via the adj line then I am good.

I hope this makes sense  :-/
Click to expand...

EDIT: Oops. I forgot a 2 above in the formula but the calc was done correctly. Now corrected.
 
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