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FrozenGate by Avery

DIY Laser Torch

I'm pretty sure if you add more voltage to the input of the LM317 when it is configured as a current source ( DDL's circuit ) the voltage across the diode will still be the same.

The only problem I can see is that this extra voltage will have to be dissipated in the regulator and result in a little extra heat.
 





Well, I was talking about the LM317 - obviously using just two bigger batteries without proper "countermeasures" will kill it.

Ah, gonna try it with the LM now...
 
The lm317 can handle about 30v. you will just have to add extra components to adjust it back down to the required voltage.
 
yes8s said:
I'm pretty sure if you add more voltage to the input of the LM317 when it is configured as a current source ( DDL's circuit ) the voltage across the diode will still be the same.

The only problem I can see is that this extra voltage will have to be dissipated in the regulator and result in a little extra heat.
Click to expand...

Right on... :D It is very important to note that anything less than 6 volts may cause the regulator to begin dropping out.
 
To drop the voltage to the LM317 you can add a series resistor between + of the battery and Vin of the LM317. Say if the current through the LD will be 250mA then the current drawn from the supply is also 250mA. And with the supply voltage at 7.2V but you want to drop it to 6V at the LM317... then:

R = (7.2 - 6) / 0.250 = 4.8 ohm.

So this resistor will drop 1.2V across it.

This was just an example, with a 7.2V supply i wouldn't even worry but if you wanted to use something with a higher voltage this will work just fine.
 
It's not a very good idea because you would not be getting the most from the batteries, and while the regulator does get hot with a 9 volt input, it is fine.
 
The extra power supplied from the batteries will be dissipated either way. Whether it is dissipated in the resistor or in the regulator (in the form of heat), in theory, it won't matter. If the regulator is in close proximity with the LD this heat may affect the performance of the LD. But really i know it doesn't really matter... it was just for thought.
 
But what happens when the regulator begins to drop out and then you remove the resistor? The regulator is able to regulate again until the batteries reach a level where the regulator drops out. So are you saying run times would be the same with or without the resistor? I am not meaning to be argumentative, but rather understand.
 
Assuming the battery voltage doesn't sag, the battery run time would be the same. For Vin=9v and Isup=250mA, the input power would be 9*.25 = 2.25W. This is the same whether there is a resistor or not...The battery run time is directly proportional to the power supplied.
 
Hi Can I construct a laser driver with components I can find In copmputere pieces? RESistors ANd capacitors I think yes but what about the voltage regulator?? there is another way to have the same effect??
I have a 16x Sony diode.

I pull off this circuit from a laser driver I had yuo think I can fit with my new diode?
 
Probably not. It's best to start from scratch and build your own. We don't know anything about the driver form the picture you posted. Is it is capable of supplying enough current to the laser diode? Where did you get the driver from?

And please stop posting the same question in multiple threads!
 
As Gazoo said (thank you) please do not multi-post the same thing...

The driver is unknown to us --or myself at least-- unless you provide some more info on it... such as where it came from, some test results, close-up pictures... etc... etc...

--DDL
 
More focused photo I hope will help you and me :)
 

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thats from a 5mw pointer.. no chance mate make your own.. all the part are cheap at maplins(uk) i got all the parts i needed for £4.78 for 3 sets.. for me the most expensive bit is the diodes (4x gb) ( 3x diys 1x fixing a kenom dorcy)

a square 9v will work fine BUT you WILL have to heatsink your LM317 to whatever case you use or make (this can be done with ceramic thermal paste and a isolating thermal pad)
im making one in a square case...


the flashlight on this guide is perfect.. its all i needed.
 
:) :) I finally finished making one of these and works pretty well. It burns un-sharpied red matches, pops balloons, engraves black plastic, etc. I used a smaller surface mount TI LM317 TO-252 package (which is good to 500mA).

I used a 5 ohm resistor (10ohm || 10ohm) which should give me 250mA but when i test it I don't actually get to 250mA - i get more like 220-230mA. I've tried using both rechargeable and non rechargeable batteries:

http://www.dealextreme.com/details.dx/sku.3273
http://www.dealextreme.com/details.dx/sku.5639

When I connect the circuit to a bench supply @ 6-7V it puts out 250mA on the dot. So I can deduce that the batteries are not strong enough.

Does anyone else get similar results?
 
I do... :(

The battery sag tends to get the best of the circuit sometimes... :-/

My next project might have some light to shed on this all... I'm seriously getting tired of waiting for things to arrive in the mail... Not to mention the next solution might have to be SMT ;)

--DDL
 
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