Help a newbie with a problem

Hi all. First post and I really need some help if someone can assist. A little background on this problem. I got a laser some time ago and unfortunately the laser emitting diode is toast. The original setup was with a 635nm red laser with two power levels with an output of <1mW on low and <3mW on high. This was on a 3VDC system being powered by a CR123A battery. I can not replace the original laser diode because it is not available, thus I want to upgrade the LED to what I can find.

What I want to do is to replace the laser diode with a 630nm 10mW laser but the replacement laser diode operates on a 6VDC system. I would like to keep the original driver board but because I am increasing not only the voltage to 6VDC, I am also increasing the output of the laser diode to around 3mW on the low side and around 8mW on the high side.

I think the small driver board that I want to keep due to perfect size for the application I need it in will obviously have to have some components replaced on it because of going from the 3VDC system to a 6VDC system and because the output of the laser will be higher, it will require more amperage to drive it.

Now I am not a electronics guy by any means of the word. I can solder some items and have a basic understanding of some of the electronic components. It is my understanding that a laser emitter has 2 circuits basically. One is the laser driving circuit and the other is the laser monitoring circuit. I was able to do a schematic of the circuit but I am unsure of the values of some of the components. Specifically, I am unsure about the two transistor drivers. One is marked as 6CW with the number 82 lying on its side and the other one is marked 5AW with the number 60 lying on its side. From what I have been able to ascertain, one of these is a NPN while the other is a PNP and they both are rated at 45V 500ma according to the documentation that I have found here:

6CW: http://www.s-manuals.com/pdf/datasheet/b/c/bc337,_bc817,_bc817w_series_nxp.pdf

5AW: http://www.s-manuals.com/pdf/datasheet/b/c/bc327,_bc807,_bc807w_series_nxp.pdf

I have no idea what the 82 or 60 numbers mean or are in reference to. The other component that I can not find any information on is the DC decoupler but I do not think that this will be an issue on the upgrade of the board.

I have a schematic that I did of the driver board and although I think that it is accurate, I am unsure about the symbology of the laser diode itself.

Here is the schematic:



It appears to me that the high circuit utilizes the 13kOhm resistor only while the low circuit used both the 33kOhm and 13kOhm resistors in series to control the "monitoring" side of the laser unit.

What I am thinking is that I need to replace the resistors on the driver board because I am upping the voltage to 6VDC and upping the output of the laser from <1mW on the low side and <3mW on the high side to 3mW on the low side and 8mW on the high side. So I am thinking that I will need to replace the 33kOhm and 13kOhm resistors to increase the power to the laser. What I am unsure about is whether I need to replace the 1.2kOhm and 560 Ohm resistors as well.

So here is the dilemma and what I need help with.

Is this even possible? And if it is, how do I calculate the values of the new resistors? Is there more that I have to do than just replace the resistors to work with the new laser emitting diode (LD630P10)? Also, do they even make a 6VDC battery with the same dimensions of the 3VDC CR123A that originally was used with the burnt out laser? Are there other components that I will need to replace as well?

Like I said, I am not well versed in the operation of the lasers and I appreciate any help that can be offered to get me out of this mess that I am in right now. I really want to keep the current driver board because of the size of it and the application that it will be used in.

Can I do the changes that I need to this new laser work at the output that I want?

I can't help too much, but I can say that not every diode will work with every driver. diodes all have different limits on how much current they can take without burning out. If you want to use the same driver, you will need to find out what it's output is and make sure it is compatible with any diode you end up choosing. As far as a 6 volt battery the size of a cr123, good luck. there are 3.7 volt 16340, these are very nearly the same size as cr123, but only .7 (up to 1.2) more volts

these red diodes only need like ~2V. it's the new driver that wants 6V to give the ~2V to the diode. the new and old driver are 2 different types.
it's likely, tho not certain, the new diode will work with the old driver as it is.

cheech226 said:

these red diodes only need like ~2V. it's the new driver that wants 6V to give the ~2V to the diode. the new and old driver are 2 different types.
it's likely, tho not certain, the new diode will work with the old driver as it is.

Click to expand...

I measure 5.4VDC at the LD side of the emitter. And no, the old driver board did not operate the new LD.

Holy crap, I wish I had this noob's CE skills when I came into this hobby...Hell, I wish I had them now! +1 for doing your homework and being concise.

I've been puzzling over this for a while now, and I've come to the conclusion it doesn't work. You must have recorded something wrong. You should fix your battery (it looks like a ground at the moment). As the circuit stands, more light means more current through the bottom resistors which will shut off the PNP (which looks backwards, at the very least) and serve to increase the current through the main driving NPN at the top there. This is the opposite of what the feedback mechanism should do. More light should result in less current.

Are you sure that PNP isn't an NPN? Because that would make a lot more sense.