22X Samsung driver only class I?

[brtaman]

How did you get 831ma? Where is the 1.93v from? The Vref of the LM317 is 1.25V. Do you have the driver set-up exactly as the DDL driver? If so, then 2.35 will equal a maximum output of 520mA.

I don't like using pots on my lasers as they can be very finicky, IgorT convinced me of this a long time ago. Do you want the laser to be adjustable while in action, is this the reason for the pot?

If no, then I would suggest going for a 4.7ohm x2 in parrallel and 1ohm x2 in parallel in series (2.35ohm+0.5ohm), this will give you 2.85ohm which will mean the diode will get ~440 mA, should give great output and still be safe. That is unless you simply find a 3ohm resistor somewhere.

 

[LarryDFW]

Ashertz;

Is it a long open can Red diode ??

The Samsung 22X have good specs for their DVD burners.

Make sure you have good solid connections to the Red laser diode when you have it powered up.

I have been testing a lot of Red diodes lately ...
and even with a good power supply,
any loose connections can ruin the diode.

LarryDFW

 

[butthead]

I just saw that this drive is still available, but only the SATA version.

Since the last post it's been more than 1,5 years, does anybody know if the present revision still uses this oh so much desired 5,6 mm diode?

 

[qumefox]

Just order an LPC-815 sled from someone here (or elsewhere online) that sells them. It will be cheaper than buying a whole drive and will have a 5.6mm LOC in it.

Also on the subject of the potentiometer listed above. I know this is an ancient thread, but I think this information is relevant in case someone else stumbles upon it. Most pots are available in two tapers. Linear, and audio. The one discussed above sounds like an audio taper pot. You'll see 90% of the useful adjustment on one end with an audio taper. You need to make sure you use linear taper potentiometers for stuff like this... And again.. this isn't directed at the OP.. since I know this is an ancient thread.. This is directed at anyone else who may stumble on this.

 

[butthead]

Bought a few of these drives today, but bummer!! There's no such thing as a 5,6 mm diode inside.

It's a small one, which is mounted in a square piece of cast aluminium.

I had a laser diode mount, which I ripped out of a prehistoric HP laserprinter, and it gives a nice sharp beam, @ 220 mA aimed at a black piece of plastic it burns a wee bit, so will try again tomorrow with some more current, see whether it will have some more impact.:beer:

 

[newbie noob]

man ! i got the same samsung driver model ! it got only one red laser diode with low light !
could it be that the burner didoe is the same as the irda diode ?? that cant be ! i know !

would it be safe to power it with a 9v battery ??

 

[qumefox]

It's quite possible to have both the 660nm and 780nm dies in the same package, so that very well could be the case. ANd no. laser diodes require current regulated drivers to operate properly, Just hooking a 9v to one is instant death.

 

[newbie noob]

sry bt i ment hooking the 9v to the driver ,
another question , i bought like 20 lm317 and none of them shows the correct 1.25 ref voltage ! wat the hell for god sakes ?
how do i test if the lm317 was faulty , i bet that not all of them are faultys , is something wrong with the circuit ? it is power with an dc power supply
plz help am stuck in here !

 

[qumefox]

They don't 'put out' 1.25V.. What they do is change their voltage output until the adjust pin sees 1.25V.. How the adjust pin is tied to the output voltage depends on how your using the regulator. as a constant current source, you'll have it connected to the opposite side of a current sense resistor on the Vout pin. The value of this resistor determines how much current it outputs.

 

[newbie noob]

They don't 'put out' 1.25V.. What they do is change their voltage output until the adjust pin sees 1.25V.. How the adjust pin is tied to the output voltage depends on how your using the regulator. as a constant current source, you'll have it connected to the opposite side of a current sense resistor on the Vout pin. The value of this resistor determines how much current it outputs.

some of you seniors are weird and i really cant understand wat ur problems are ! when i ask any question i don't get the answer that i seek , everyone here knows that information u provided so no need to prove ur acknowledgment level
i didnt mention that the ic isnt showin a 1.25 vout ! , i said the reference voltage ! the voltage between vout and vadj ! , shouldnt that be 1.25 all the time ? whether iam putting a 55 kohm or 200 kohms it would still be 1.25 right ? a simple answer is all iam asking for ,
what could affect that change ?

 

[aryntha]

some of you seniors are weird and i really cant understand wat ur problems are ! when i ask any question i don't get the answer that i seek , everyone here knows that information u provided so no need to prove ur acknowledgment level
i didnt mention that the ic isnt showin a 1.25 vout ! , i said the reference voltage ! the voltage between vout and vadj ! , shouldnt that be 1.25 all the time ? whether iam putting a 55 kohm or 200 kohms it would still be 1.25 right ? a simple answer is all iam asking for ,
what could affect that change ?

Look,

The way you're phrasing your questions, with lots of 'netspeak' contractions (ur, u, wat) and run-on sentences, no capitalization or punctuation, in "walls of text" -- these make it very hard to extract what you're trying to get at.

If you're looking for a simple answer, spell out what you're trying to convey simply. I read the above paragraph a few times and am really honestly trying to understand the issue, but it's kind of hard to decipher.

Qumefox knows what he's talking about. Elucidate what you want to know succinctly and clearly and i'm sure folks will do their best to help. We can't just be in your head and understand what you're seeing, and you aren't doing the best job of relating it to us.