22X Samsung driver only class I?

Asherz · Nov 19, 2009

Hi guys,

I have a quick question, I accidentally purchased the wrong optical drive a few weeks ago, and it uses an IDE? connector instead of a SATA cable. This means I have no use for it so I thought why not rip it to shreds and take the diode

I checked the warning label on the back quickly, and it says:

"Class 1 laser product"

Does this mean it's a weak laser inside? I don't really understand because it says it's a 22X DVD writer, there for I assumed it would be one of the powerful 200mW red diodes?

The model number is "DVD Writer Model SH-S222"

Cheers,

Ash.

Edit:

Just found this snippet of information:

"This class includes high-power lasers within an enclosure that prevents exposure to the radiation and that cannot be opened without shutting down the laser."

I gather that this laser is a high powered diode, but because it is in an enclosure it is there for rated Class 1?

brtaman · Nov 19, 2009

Yes Asherz your presumption is correct.

Due to the fact that the diode in contained within an enclusure, the DVD drive itself is labelled as a Class I product.

Hope you get yourself a nice 5,6mm diode and not the dreaded plastic square or another exotic (useless) package.

Asherz · Nov 19, 2009

Cheers Brtaman

In the process of ripping it to shreds now, I can see one of the little 5.6 beauty's now but getting the sled of the rail is proving to be a pain, there's a very stubborn screw holding one of the poles down

Update: Just got the sled out, gonna get to work on that now

Laser_Ben · Nov 19, 2009

Class one means there are interlocks to prevent any radiation from leaving the case. You could put a 10kW laser inside a sealed enclosure and it could be Class one.

Asherz · Nov 19, 2009

Ok, thanks Laser_Ben

Um, I've got everything out now, got an empty sled, 2 rare earth magnets which are firmly glued onto a piece of metal, 1 mirror and 2 lenses.

diodes

As far as diodes go, I've got what looks to be an open can 5.6mm red in a heat sink which is not going to be fun to remove and where you would normally find the IR diode I instead found the heatsink but with a small thin chip saying "C68A" and on the other side there is a small aperture? hole but I have never seen any kind of diode like this if it is :S

it's about 1mm thick.

I don't know if this red diode heat sink is just gonna twist of, might have to gently cut into it.

Asherz · Nov 19, 2009

After a pretty shoddy attempt at removing the ribbon from the diode I'm nearly ready to see if it survived it's ordeal.

I hooked up my DDL driver, feeding it around 10.5V with a test load of 4 1N4001 diodes and a resistor.

The only resistor I have spare is a 4.7ohm resistor, so I got it all hooked up, measured the voltage across and did the small equation I=V/R. I'm getting 1.09V across the resistor so 1.09/4.7 gives me something like 0.231?

So does that mean I'm only getting 0.231A, there for I'm actually get 231mA? cheers

EDIT: just set it at 0.94V which in theory using the above gives me 200mA, I'm gonna discharge the cap hook it up to the diode and see if its working

brtaman · Nov 19, 2009

Asherz said:
After a pretty shoddy attempt at removing the ribbon from the diode I'm nearly ready to see if it survived it's ordeal.

I hooked up my DDL driver, feeding it around 10.5V with a test load of 4 1N4001 diodes and a resistor.

The only resistor I have spare is a 4.7ohm resistor, so I got it all hooked up, measured the voltage across and did the small equation I=V/R. I'm getting 1.09V across the resistor so 1.09/4.7 gives me something like 0.231?

So does that mean I'm only getting 0.231A, there for I'm actually get 231mA? cheers

Hey Asherz,

Good to see, that it is a 5,6 package diode.

With the LM317 chip/DDL driver, determining the output current is quite easy and in general does not require a "test load". You can just hook up your DMM across the +/- effectively shorting it, to get a current reading. Or just set it up based on its reference voltage.

The LM317 has a referance voltage (Vref) of 1.25V

So with the basic equation:
I=V/R

I=1.25 V/4.7 ohm

I=0.266A or 266mA

266ma is a very safe current for a 22x diode (is it LOC?). For a good ~420ma go with 3ohm resistor.

BTW are you sure that you haven't extracted the 780nm diode?

Best Regards,
brtaman

Asherz · Nov 19, 2009

I've extracted everything there is and that is the only 5.6mm diode, it's open can as well. I hooked it up to my driver and it's not even giving me an output. Owell, I wasn't really expecting it too work lol.

Cheers for your help, not really sure why I'm not getting an output. Would it be LED'd if I overheated it? So that means I should still have a small amount of light coming out.

The pin config I did, was the pin which is attached to the case of diode without any gaps on the outside on the left, the other pins I used the top as positive and bottom as negative.

Ah well, the pin soldering wit the wires went well, I think removing the ribbon and my driver is what let me down. I might just buy a lock and a pre-set driver and do it that way lol.

brtaman · Nov 19, 2009

Asherz said:
I've extracted everything there is and that is the only 5.6mm diode, it's open can as well. I hooked it up to my driver and it's not even giving me an output. Owell, I wasn't really expecting it too work lol.

Cheers for your help, not really sure why I'm not getting an output. Would it be LED'd if I overheated it? So that means I should still have a small amount of light coming out.

The pin config I did, was the pin which is attached to the case of diode without any gaps on the outside on the left, the other pins I used the top as positive and bottom as negative.

Ah well, the pin soldering wit the wires went well, I think removing the ribbon and my driver is what let me down. I might just buy a lock and a pre-set driver and do it that way lol.

Hmm are you sure you got the pin config correct? From your explanation I can't quite make it out.

The pin without the black o-ring (connected to case), is the negative of the diode.

Here is a picture, got it from this topic: http://laserpointerforums.com/f51/pro-tips-w-spyderz20x6-building-your-first-diy-laser-40157.html

Also, a LED'd diode will not output anything in some cases, while in others quite a bright output of incoherent light is noticeable.

How did you extract the diode from the heatsink (dremel, hacksaw?), these diode have microscopic wires connecting to the die, which just a tiny fragment of heat sink could well rip.

Asherz · Nov 19, 2009

Ah right that would explain, it got the pin config incorrect. I'll change it and see if it's still going lol.

and I didn't extract it from it's heatsink, I left it there as it's too thick to do anything with it.

Thanks again, +rep

Asherz · Nov 19, 2009

Woooo!

Got it working, thanks very much

I had the pin out wrong in the first place. I now have a very nice and bright red laser

Now I just need to get a 1 ohm resistor in there, think I'm gonna max the pot out now and should get me around 260mA.

Thanks again, Don't no how to get this diode outta the heatsink though it's well an truly stuck in there lol.

brtaman · Nov 19, 2009

Asherz said:
Woooo!

Got it working, thanks very much I had the pin out wrong in the first place. I now have a very nice and bright red laser

Now I just need to get a 1 ohm resistor in there, think I'm gonna max the pot out now and should get me around 260mA.

Thanks again, Don't no how to get this diode outta the heatsink though it's well an truly stuck in there lol.

1ohm resistor? That will give you an output of 1.25 amps from the driver!

For the forum standard of 420mA, go with 3ohm on the 317.

Glad to hear it is working. Extracting the diode really depends on the heatsink on hand, I have had to use everything from a dremel to pliers to a hack-saw etc. Some heatsinks just "pop", while with some it takes a lot of time and patience.

PS: No problem, glad I could help.

Asherz · Nov 19, 2009

Yea I know I'll get 1A+, but with the 4.7ohm I can only get upto around 260mA. So with the 1ohm I can lower the pot too like half way and get 400mA

brtaman · Nov 19, 2009

Be very careful with that pot. :yh: Pot's tend to not be linear and you could see well over 500ma much sooner than expected. Has happened to me before, so I felt I should warn you.

It is always much better to be extra careful with the diode, then to have an expensive led on hand.

Since based on this thread I am sure you can solder, may I suggest scavenging from an old component/s, I have found an unbelievable amount of useful stuff through re-cycling of old components.

Do you have another 4.7ohm resistor on you by chance? In parallel it would equal 2.35 ohm, which would limit current to 530ma, much safer when combined with the pot.

Asherz · Nov 19, 2009

very good point, Don't know why I didn't think of that

I do have another 4.7 on hand, My current circuit is all soldered together on a small board so I'll have to just solder it over the top of the current one and that should work.

I'll give that a try tomorrow

I think I'm gonna order a LOC in a aixiz module then use my driver on that. Then I can get it in a little labby enclosure

Thanks again for all the help.

edit:

Just did what you said, got the 4.7ohm in parallel with the current resistor, so divided by 2 that gives 2.35ohms and 1.93V/2.35 = 821mA

I'm gonna set it as 0.82V which should give me 350mA

Asherz · Nov 20, 2009

Hi guys,

another quick question as I didn't want to open a new thread, I've got it all running put I've decided to put the test load back in as I got a new 9V battery and I just realised my pot is ridiculously sensitive, it's a 100ohm single turn pot and it seems like the last 1/4 is the only section that is not 0, I get to about 1.4V as the lowest then it just jumps to 0. It's really annoying.

Is there anyway to sort this out without buying a multi turn? cheers.

22X Samsung driver only class I?

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