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FrozenGate by Avery

DIY Homemade laser diode driver

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Voltate (1.25 at the LM317 sense pin) divided by current = Ohms...

Example:

1.25 volts / 1250mA's = 1 Ohm
1.25 volts / 250mA's = 5 Ohms

gillza, I realized that a 1.8 Watt resistor may not be available. But I know that a 1 Ohm 2 Watt resistor is available. So that's why I made the statement that you will need that resistor for 1250mA's...
 
jayrob said:
Voltate (1.25 at the LM317 sense pin) divided by current = Ohms...

Example:

1.25 volts / 1250mA's = 1 Ohm
1.25 volts / 250mA's = 5 Ohms

gillza, I realized that a 1.8 Watt resistor may not be available. But I know that a 1 Ohm 2 Watt resistor is available. So that's why I made the statement that you will need that resistor for 1250mA's...
Click to expand...

Jayrob,

I got what you were doing. You were right. I just missed the W=VA conversion form ohms law and thought that your calculations were incorrect. As I wrote in the previous post: you were right! :)

And about 1.8W I figured that :) hehe

My apologies for that original statement.
 
I think it would be a perfect little driver to make using just an LM317, and a 1 Ohm 2 Watt resistor for a 445 build giving 1250mA's.

I think that 2 X 18650's in series would work as well...

Because at this current, lets say that the diode will be drawing maybe 4.5 volts. But it may be lower. Maybe even 4 volts. So the two batteries will give good regulation down to maybe 3.5 volts each before the output power starts to drop...

But you could always go 3 X 18650's, and heatsink the LM317 to the main sink as well...
 
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hey. i need some help with this please!
(please bear with me im new at this and just a kid)

so i have all the parts for this:

-LM317t
-Laser diode (340mw / 340ma / 660nm)
-3.7 ohm resistor (giving me around 340ma)
-capacitor (16v / 10uf)

what i am getting from this article is that no matter what power i put into the driver, it should always give me the same current and voltage that i need. However, when i connect them all as shown in the diagram, i get almost the same voltage as i am putting in (around 8.7v from a 9v battery) and the current doesnt even show up (most likely because there is to much ma for the multimeter to read.

so i guess my questions are:
1. the diode needs 3v to run. should 3v always be coming out of the driver, or will it be the same voltage coming out as i put in?
2. will the driver keep the output current at 340ma no matter how much current i put in?
3. what is the point of the POT if i have already measured that 3.7ohms is perfect to keep the current at 340ma?

thanks alot for any help. :)
 
3) Pot is used to change resistance on the fly if you want to increase/decrease the current to the diode.
1) The driver will supply the diode with the voltage needed at a particular current assuming that your power source can provide enough voltage both the driver and diode (read previous page, lm317 needs ~3V to function.)
2) The driver will draw the current from the power source, and I guess you need to make sure that the power-source can handle your current draw.

As per your weird readings, how do you test? Did you use the dummy load? Did you use the voltmeter to test the voltage across the resistor in the dummy load?
 
thx
this driver works really well. i might need to get a better module ;) but the driver works fine!
:thanks: :gj:
 
It seems to me that the pinout on the diagram is backwards. It shows the negative on the right and positive on the left. Of course, I am a noob, but from the 5 or 6 diodes I've hooked up, the positive is on the right side, the case pin is on the bottom and negative on left. It all depends on how you look at the diode, of course, and I'm talking about the one in the picture of this thread. ........So, if you place one pin on the bottom, then that's how I determine which is right and left. It looks like a smiley face to me, with two eyes and a mouth. At least, that's how I remember it. To date, I haven't blown up my 445nm 400mW, nor my 405nm, or any of my cheap reds. I wonder if some pinouts are different. I see one on hightechdealz website that shows a LPC815 650nm 20x pinout which uses the bottom pin as negative and the left pin as "not used". So, now I'm getting a little bit confused. But, like I said, I'm only a noob. Peace, out.

OK! SO NOW I SEE AFTER POSTING THIS, THAT IT SHOWS UP ON A DIFFERENT PAGE THAN THE ONE I THOUGHT I POSTED IT TO. THAT MEANS THERE IS NO PRINTOUT PICTURE THAT I WAS REFERENCING! OH WELL, I GUESS I'LL FIGURE IT OUT IN TIME.
 
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uttershock said:
It seems to me that the pinout on the diagram is backwards. It shows the negative on the right and positive on the left. Of course, I am a noob, but from the 5 or 6 diodes I've hooked up, the positive is on the right side, the case pin is on the bottom and negative on left. It all depends on how you look at the diode, of course, and I'm talking about the one in the picture of this thread. ........So, if you place one pin on the bottom, then that's how I determine which is right and left. It looks like a smiley face to me, with two eyes and a mouth. At least, that's how I remember it. To date, I haven't blown up my 445nm 400mW, nor my 405nm, or any of my cheap reds. I wonder if some pinouts are different. I see one on hightechdealz website that shows a LPC815 650nm 20x pinout which uses the bottom pin as negative and the left pin as "not used". So, now I'm getting a little bit confused. But, like I said, I'm only a noob. Peace, out.

OK! SO NOW I SEE AFTER POSTING THIS, THAT IT SHOWS UP ON A DIFFERENT PAGE THAN THE ONE I THOUGHT I POSTED IT TO. THAT MEANS THERE IS NO PRINTOUT PICTURE THAT I WAS REFERENCING! OH WELL, I GUESS I'LL FIGURE IT OUT IN TIME.
Click to expand...

This is what you need to know.
19757-pin-out-diagram-lpc-815-pin_out_001.jpg
 
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Silvershot: Thanks for clearing that pin layout up before I hook up my LPC815 diode! Wait a minute...brb......I just remembered, that I tested this LPC815 the other day, because I wrote a comment about not caring for the dot on the wall it produced. After looking at the soldered pins, the middle(-) and the left(not used) are soldered together. I guess that's why it worked. Anyways, I'm not going to undo it. I've learned a long time ago, "If it works, fiddle with it till it don't". I think I'll use that as my quote from hence!
 
I made this driver using a 3.9 ohm resistor to limit the driver to 320-321 mA. I also used a 100 ohm single turn potentiometer. The laser I have for this driver is a LPC-815 and from the information nikokapo posted about diodes I want to run this laser at 200mA for now. I have a radioshack meter [50µ (250mV) and 250m) under DCA and I also have a Simpson model 260 which I can test the mA on my dummy load.

Which would be better to measure the mA?
 
Kmssd said:
I made this driver using a 3.9 ohm resistor to limit the driver to 320-321 mA. I also used a 100 ohm single turn potentiometer. The laser I have for this driver is a LPC-815 and from the information nikokapo posted about diodes I want to run this laser at 200mA for now. I have a radioshack meter [50µ (250mV) and 250m) under DCA and I also have a Simpson model 260 which I can test the mA on my dummy load.

Which would be better to measure the mA?
Click to expand...

Either one, just use the correct setting range on the DMM.
 
You can use any of those.
You can put the multimeter in the 10A range and measure directly the output
of your driver, OR you can use your "dummy load" and either measure current
directly using the 10A range, or measure the voltage drop across the resistor
on your dummy load. Using Ohms law you can then determine
the current that is flowing.
 
10A as in amps, the Simpson only has 500mA, 100mA 10ma and 100 microamps. Then the radioshack seems to only have 50µ amps (.00005 amps) and 250 mA.

Edit: Using the radioshack in 250mA DCA I connected the red and black lead to the + and - of where the laser would go (respectively) and I set the potentiometer so the meter read 4 (looking at the 0 - 5 scale) which I then multiplied by 50 (top of the scale x 50 = 250mA). So this is what I think is 200mA?
 
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