Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers



Laser Pointer Store

omg what did i do?

ouch3994

New member
Joined
Jan 19, 2011
Messages
56
Likes
7
Points
0
so i have recently (and perfectly sucessfully) assembled my first 445 laser, and it only had an output of around .9 watts....

the driver is a Lm317 with a resistor array to give me 1.3 ohms, with the entire thing heatsinked.

my input power was 6 AA batteries, and it all worked fine untill i noticed the batteries began to get low, and the laser was becoming less and less powerful.

i wanted a source of power from a wall adapter, so i used an AC to DC adapter that had an output of 12volts, at 1000ma.

i plugged it in and tested it. and now the diode isnt shining, and i assuming i killed it. However i have no idea why this killed it, when my multimeter showd that the voltage was exactly 12volts, and as i have read, that is within the limits of the LM317 driver.....
PLEASE SOMEONE HELP ME AND EXPLAIN WHAT JUST HAPPENED. THIS BROKE MY HEART! :yabbem:
 

Joined
Feb 5, 2009
Messages
199
Likes
10
Points
0
Was your LM317 in voltage regulated mode or current regulated mode?
Assuming you were in current regulated mode, Vin = 6AA = 9V [assuming no voltage drop]
Iout = Vref/R1, Vref = 1.25v R1 = 1.3 ohm, Iout = 960ma +/- 5%
Vout = Vin - Vdrop, Vout = 9V - 2V = 7V output [realistically about 5.5V]

When you switched to 12V input the current output was the same but Vin = 12V with no chance of voltage sag from load, because unlike a battery, AC-DC regulators have much less series resistance than batteries particularly the alkaline varieties. 12V - 2V = 10V, which is about twice the voltage output compared to the 6AA.

Overvoltage is the most likely culprit that may have killed your diode, the LM317 despite being a very inefficient is a diehard chip.
 

Leodahsan

New member
Joined
Aug 7, 2010
Messages
2,088
Likes
76
Points
0
if LM317 were on constant current, then no problem. Post us ur schematic...
 

ouch3994

New member
Joined
Jan 19, 2011
Messages
56
Likes
7
Points
0
My original plan was to use 2X 18650 batteries, but i got impatient with having to wait for them to come in the mail, along with all of the resistors i ordered, so i have a huge array of 5 X 10ohm resistors and a 25ohm potentiometer.
when the potentiometer was turned to the lowest resistance, the measured total resistance was 1.3ohms across the ADJ and OUT of the LM317 chip.

attached is a mspaint schematic of my rig,,,
 

Attachments

Joined
Nov 7, 2008
Messages
5,728
Likes
281
Points
0
It sounds to me like you've been the victim of a transient voltage spike. Unregulated wall-warts will let one of those loose from time to time when they get plugged in or if power is interrupted for some reason.. the protection found in most of the LM317 circuits you'll find around here work much of the time, but not necessarily all the time..
 

Bluefan

New member
Joined
Aug 15, 2009
Messages
1,482
Likes
57
Points
0
You're were already lucky the first time, there's an error in your schematic. The out of the LM317 is directly connected to the laser diode. The LM317 tries to keep ~1.25V over the out and adjust pin, but the adjust pin is only connected via a resistor to the output, so it will be at the output potential. This way the lm317 will open fully in an attempt to get teh output higher then the adjust.
What you need to do it to connect the laser diode after the resistors. This way the current out of the output pin will flow through the resistors, creating a voltage drop. The adjust pin and laser diode are then connected after the resistor network. The adjust pin is now lower in voltage than the output by the voltage drop of the resistors.

What you need to change in your scheme is thus the laser diode connection, it's at the wrong side of the resistors, the current now doesn't flow through the resistor.
Your resistor network is risky, the potentiometer can short the output and adjust. If the potentiometer can ta ke a large current, wire it in series with the other resistors.
 
Joined
Oct 16, 2008
Messages
378
Likes
1
Points
0
uhh no silicone diodes or capacitors?

i'm still waiting on my 15266 batteries and 1 ohm resistor

you shoulda just waited, like i have my 445nm, host and everything, and its tempted to try it but yeah .... i didn't wanna blow the $100 i've spent and from previous experience

btw what does 5 x 10ohm resistors in parallel give you a maximum current of?
 

anselm

New member
Joined
Nov 22, 2010
Messages
2,470
Likes
80
Points
0
You're were already lucky the first time, there's an error in your schematic. The out of the LM317 is directly connected to the laser diode.
Hmmm, he might have just mislabeled the pins in his MSPaint.
In the "real life" LM317, the middle pin is the output and the left pin is the
adjust. My crystal ball says he killed it because he used a wall wart and
no capacitors in his circuit.

btw what does 5 x 10ohm resistors in parallel give you a maximum current of?
5x10Ohm in || is 2Ohm.
1.25V/2Ohm=0.625A
 

ouch3994

New member
Joined
Jan 19, 2011
Messages
56
Likes
7
Points
0
Thanks for the help guys! Oh and if you have any schematic suggestions, could you postem?
 

Cad602

New member
Joined
Feb 20, 2011
Messages
130
Likes
4
Points
0
Dang I did the same thing. Sort of. 'cept my circuit had the cap and diode. Also, I was using a 3.9 Ohm resistor w/a 25ohm pot and previously everything was fine...

But I just finished this project board that has a transformer stepping down the voltage to 12v then a zenar diode giving me DC....well for some idiotic reason I decided to use this 12v but I meant to cut it down to 9v first and dammit I guess I killed my second A140. Anyone have some A140's in the $30 range for sale? I'll take three more! JK I'm not made of money but I need another!

Gr...I feel you pain dude...But I'm just stupid, you were unlucky...

Matt
 
Last edited:

Bluefan

New member
Joined
Aug 15, 2009
Messages
1,482
Likes
57
Points
0
Without capacitors the LM317 can oscillate and kill the diode, with a noisy switching power supply this can probably happen more easily.
 

Bogart

New member
Joined
Oct 18, 2009
Messages
62
Likes
4
Points
0
i used an AC to DC adapter that had an output of 12volts, at 1000ma.
When a wall wart is rated like this, it means that it produces ~12V when operating with a full 1000mA load. At a lower load, or no load, the voltage is often quite a bit higher, maybe even by 50%.
 
Last edited:
Joined
Feb 5, 2009
Messages
199
Likes
10
Points
0
Most likely that your diode died because you had no capacitors :(
uh... what? If he was using a 12V linear unregulated regulator hooked up to a LM317 in current source configuration I still say overvoltage or a transient surge like electrofreak have pointed out may have been the culprit. As with the LM317, Vin decoupler cap is needed but load end cap is optional.

The thing with AC "unregulated" regulators is the output is usually overspec until you draw the amount of current relative to its prescribed output before the voltage will begin to sag down to a more "believable" level. In addition theres usually a huge spike when the load is connected as opposed to an closed circuit at the point of energization.
 




Top