Driver question

[Leodahsan]

What voltage do I need for a 1W (XJ-A140) laser diode?
are 2 18650 cells with a LM317 circuit enough to pump enough power?

thanks in advance.

 

[BMWG650GS]

I recently bought a 1 watt 445 diode from pulled from a Casio I was not told if it was an A140 OR AN A130 but I do have these test current values for the diode I bought

200 ma -- 23 mW* --* 3.7 Vld
300 ma -- 125 mW -- 3.8
400 ma -- 244 mW -- 3.9
500 ma -- 340 mW -- 4.0
600 ma -- 450 mW -- 4.1
700 ma -- 550 mW -- 4.1
800 ma -- 633 mW -- 4.15
900 ma -- 720 mW -- 4.2
1000 ma -- 820 mW - 4.2
1210 ma --1000 mW- 4.3

This is the driver I plan to use http://laserpointerforums.com/f42/diy-homemade-laser-diode-driver-26339.html I bought the parts at radio shack today. I am relatively certain this driver would work for ether model diode! I plan on using 4 C cell batteries. I hope this helps

 

[qumefox]

Judging by those numbers, it was probably an a130 diode.

And to the OP, your input voltage would always have to stay higher than the voltage the LD wants, so yes, two 18650's (or pretty much two of any rechargeable single cell Li-ion's) would work.

 

[Leodahsan]

> BMWG650GS
thanks for the information, very very useful.
I am studying eletronics so, LM317 as positive regulator with 2 18650's 3,6V WILL BE ENOUGH to feed a 1 watt diode?
> qumefox
7,2volts will work...? with an a130 or a140?...
I'm new to Class IV lasers, i'm accostumed to 3B...
thanks in advance...

 

[qumefox]

The LM317 circuit your playing with is what they call a 'buck' regulator. Meaning it cuts the source voltage down to what it needed by the load to be to be at the amount of current it's set for.

For the circuit to work properly, your supply voltage always has to be at or higher than the voltage the load wants.

So going with the above posted numbers for that diode, if you wanted to run your diode at 1000mA, the forward voltage would be 4.2v. That means your supply voltage to the regulator would have to stay higher than 4.2v to keep the diode running at 1000mA. If the regulator input voltage sagged below 4.2v, a buck only regulator would source less voltage and current as the battery voltage dropped. Meaning it would get dimmer and dimmer as the batteries died.

 

[Leodahsan]

The LM317 circuit your playing with is what they call a 'buck' regulator. Meaning it cuts the source voltage down to what it needed by the load to be to be at the amount of current it's set for.

For the circuit to work properly, your supply voltage always has to be at or higher than the voltage the load wants.

So going with the above posted numbers for that diode, if you wanted to run your diode at 1000mA, the forward voltage would be 4.2v. That means your supply voltage to the regulator would have to stay higher than 4.2v to keep the diode running at 1000mA. If the regulator input voltage sagged below 4.2v, a buck only regulator would source less voltage and current as the battery voltage dropped. Meaning it would get dimmer and dimmer as the batteries died.

what driver circuit do you recommend for me?
i'm going to use a heat sink... dunno how much current to apply on the diode :S

according to BMWG650GS' table...

...

200 ma -- 23 mW* --* 3.7 Vld
300 ma -- 125 mW -- 3.8
400 ma -- 244 mW -- 3.9
500 ma -- 340 mW -- 4.0
600 ma -- 450 mW -- 4.1
700 ma -- 550 mW -- 4.1
800 ma -- 633 mW -- 4.15
900 ma -- 720 mW -- 4.2
1000 ma -- 820 mW - 4.2
1210 ma --1000 mW- 4.3

...

I want at least 1210 ma (1 watt). More only, if my heat sink would withstand the power...
thanks in advance, and sorry for bad english.

btw, LM317 is a 'positive terminal 3 leads linear regulator'

 

[Hemlock_Mike]

BMWG ----
Funny thing -- Those are exactly the same readings I got testing my first 445 diode on 6/11/10! I am amazed! I posted that here on 6/12.

HMike

 

[BMWG650GS]

Hemlock_Mike those numbers where provided by the e-bay seller who soled me my 445 1 watt. are you e-bay seller quijalicious?:beer:

 

[qumefox]

More than likely the ebay seller just copied the specs from this forum.

And leodahsan.. I know what an LM317 is. I'm just saying that the regulator circuit your talking about is only capable of lowering the voltage (bucking) from source to output. Therefore it is a buck regulator. The alternative is a boost regulator (which is only capable of raising the voltage above that supplied), or better yet, a boost/buck regulator that is capable of doing both.

Most people build the LM317 buck regulators because they're really simple. Building a boost or boost/buck regulator is considerably more complicated. If you want to go those routes, your better off buying them. Examples are the microboost and microflex drivers from dr.lava.

I also have to ask.. why do you need 1 watt? Do you have any previous laser experience? Do you own safety goggles rated for 445nm? 1W is a lot of power. It's advisable that you read up on the dangers of class 4 lasers to both yourself and others (and property, unless you like having random burnt spots in your walls) before messing with one.

 

[midias]

If you are going to run it at 1.2A for 1W output make sure to heatsink the LM317 and remember the tab has voltage it is connected to pin 2.

 

[qumefox]

He also should remember that the current required to get 1W varies from diode to diode. Just because one hits it at 1200ma, doesn't mean the one he ends up with won't take 1300ma to do so.. Or 1100ma.. The only way to know is to use a LPM and set the current accordingly.

 

[Leodahsan]

And leodahsan.. I know what an LM317 is.

sorry, didn't mean to offend you.
And... yes, I have build many lasers (including class IV IR) and have all the appropriate protection for many of the nanometers (that includes 445nm), many at 3.0OD and 4.0OD... thanks for caring

and... I don't know what a boost driver is... I do not speak english very well so... I don't understand :S someone send me some topics about it?

and... I don't have access to an LPM anymore... because of that I need the necessary current to safely run the laser

thanks in advance.

 

[chipdouglas]

The LM317 circuit your playing with is what they call a 'buck' regulator. Meaning it cuts the source voltage down to what it needed by the load to be to be at the amount of current it's set for.

For the circuit to work properly, your supply voltage always has to be at or higher than the voltage the load wants.

So going with the above posted numbers for that diode, if you wanted to run your diode at 1000mA, the forward voltage would be 4.2v. That means your supply voltage to the regulator would have to stay higher than 4.2v to keep the diode running at 1000mA. If the regulator input voltage sagged below 4.2v, a buck only regulator would source less voltage and current as the battery voltage dropped. Meaning it would get dimmer and dimmer as the batteries died.

missing important info...

remember if talking about the lm317 in a driver config that the driver itself needs a little over 2 volt. so you will take that into consideration as well. not just the 4.2 need by said LD.

michael