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FrozenGate by Avery

DIY Homemade laser diode driver

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Mountaineer said:
Here are the pics of my circuit.  I have it running a 5mm, 2.6V, 28mA LED.

The first one is just of the circuit powering the LED.  The battery is on the right out of the picture.

The second is of it the voltage measured on my DMM with the leads at the LED.  It's showing 2.339V (the battery is quickly dropping in voltage it was at about 2.45V last time.)  Measured like this, the current is 57mA.

The third picture is of the current being measured straight from the circuit without being connected to the LED.  Like this, it's at 57mA and 4.63V.

I figure I'm supposed to measure it with it hooked up to the LED?!?!  If that's right, then I'm getting ~41[ch8486] (2.339V/.057A) and 133mW (2.339V*.057A).  Does the rest of the circuit besides the resistor create about 19[ch8486] of resistence (41-22) :-??  If anyone has any questions/suggestions that could help me out on this or if you need more pictures, let me know, I need all the help I can get.  I can answer your questions/post more pics in a couple of hours if needed (I have to get some mergers & acquisitions homework done - ouch! [smiley=cry.gif])  Thanks in advance.
Click to expand...

You've got it part right.  The LED is a 5mm LED, period.  Its voltage and current are dynamic and are dependant on each other.

It looks like a 22 ohm resistor so,if you supply the LM317 with enough voltage overhead, the LM317 will adjust its output current so that the voltage drop between the Out and Adj pins equals 1.25V.  1.25V/22 ohms = 56.8mA

The way you measured current is not the way to do it. There is a reason why it worked out this time, but I caution you that this method will only work in certain circumstances (in others, it will blow the fuse in your multimeter!).

There are several other ways to measure current, but the BEST and easiest way in this case is to measure the voltage drop through a known resistance.  You already have the resistance (22 ohm resistor). Measure the voltage drop on it and calculate the current (I=V/R).

Another way is to insert the multimeter in SERIES with the load.  Use the highest current setting on the multimeter (and the correct jacks) to minimize the meter's effect on the circuit.  This method works OK to check battery current by using the meter in current mode to complete a light/laser's tailcap circuit.
 
Mountaineer, your circuit appears to be configured properly. Although, with 22 Ohms of resistance you are only permitting 56.818mA through, which is what you got there on the meter.

As chimo said, you do not measure current across the load. You measure the voltage like that and that is ok, but to measure the current you have to make the current forcefully flow through the DMM.

I suggest testing the circuit again with a reliable 6V supply, and using 2 of the diodes you have there. The diode you have there will not read more than the voltage that it needs to operate, is you use 2 of them, then the voltage reading is well past that of the LD and you get a better picture/idea of your supply voltage to the diode. As about the current, I suggest using a 5-Ohm resistor for use with one of the GB diodes :)

GL;
DDL
 
Thanks for the help guys. Would a 4.7ohm resistor give me 266mA (1.25V/4.7ohm), and is that too much? What would the mW be then? I'm still trying to figure out all the math and stuff involved in figuring this out. Kinda makes a guy's eyes want to cross... Sorry if all this is beating a dead horse, but I'm just not quite getting it yet. Thanks again chimo and DDL [smiley=dankk2.gif], now it's time for me to take my helmet off and go to bed, I've got to catch the short bus early in the morning haha ;).
 
If you're going to be using .25 or .5 Watt resistors from RS, I suggest using 2 X 10-Ohm resistors in series. Just twist them onto each other. That'll give you double the rating and half the resistance :) That's what I use ;)

On the other hand, I think you'd be alright with 266mA. That should get you anything from 150mW to about 250mW ;D That's a big range and it all depends on the temperature of the diode.

GL;
DDL
 
I wish I was good at drawing... ;D But here is what I use.

The only resistor I use is a 1 ohm resistor connected in series with the diode. This is used to measure current. Measuring voltage across the resistor will give the same readings as measuring current, and will be safer as there is less chance of ruining the diode on account of accidentally discharging the voltage built up in the capacitor..... into the diode.

In series with the resistor I have a RS 25 ohm 3 watt rheostat. This allows me to adjust the current as needed. Of course this will not fit into a flashlight but is a nice way to begin experimenting.
 
Mountaineer said:
Thanks for the help guys.  Would a 4.7ohm resistor give me 266mA (1.25V/4.7ohm), and is that too much?  What would the mW be then?  I'm still trying to figure out all the [highlight]maths and stuff involved in figuring this out.[/highlight]  Kinda makes a guy's eyes want to cross...  Sorry if all this is beating a dead horse, but I'm just not quite getting it yet.  Thanks again chimo and DDL [smiley=dankk2.gif], now it's time for me to take my helmet off and go to bed, I've got to catch the short bus early in the morning haha ;).
Click to expand...
I dont know if this would help but i thought id post it. you probably alrdy know all of this and the questions on there... http://sol.sci.uop.edu/~jfalward/seriesparallelcircuits/seriesparallelcircuits.html it helped me a lil bit but i am a noob!  :)
 
daedal i hooked up the diode to the circuit but it shines dim but i put and LED on and bam light up the blue LED when i added the 5 ohm did it need to be in series or paralel
 
Well, I came to this thread to complain (nah, rather ask for help politely) because my circuit doesn't work.

Problem was, I thought "1-2-3" on the circuit on p.1 referred to legs 1,2,3 on the LM317. Unfortunately, with the LM317 T, this is not the case, instead of "in-adj-out", it is "adj-out-in". Now gotta go and see if it works :)

Just if that helped anyone.
 
Good point. As many times as I have looked at that circuit diagram I never noticed it.. :P
 
last week I decided to not be such a noob  ;) i started figuring out how this works and I finished the driver last night!
but today i realized i didnt use 2.2ohm resistors, i put in two 2.2k ohm resistors in a series; 4400 ohm! :o
i kno i have to replace these guys but just want to kno how would that have effected the laser diode
 
philguy said:
Well, I came to this thread to complain (nah, rather ask for help politely) because my circuit doesn't work.

Problem was, I thought "1-2-3" on the circuit on p.1 referred to legs 1,2,3 on the LM317. Unfortunately, with the LM317 T, this is not the case, instead of "in-adj-out", it is "adj-out-in". Now gotta go and see if it works :)

Just if that helped anyone.
Click to expand...

Philguy, if you had looked at the schematic and then at the picture, you would have noticed that the pins on the LM317 are arranged as such. They are always arranged such that the heatsink is Vout.

GL and I hope you managed to get it working;
DDL
 
Mountaineer said:
Would a 10uF 16V capacitor handle the spikes OK? I don't figure it will since the recommended capacity is almost 5X that, but I just thought I would make sure. The reason I'm asking is because I'm trying to find a tiny capacitor to fit into the Aixiz module that I can get at the local Radio Shack. Would this one work?:

http://www.radioshack.com/product/i...ssion=1&numProdsPerPage=100&parentPage=family

Thanks!
Click to expand...

Pretty sure it will. It might not be as effective as, say, a 100uF, but it'll numb some of the spikes and dull them enough to make them safe for the laser diode. :)

GL;
DDL
 
toked323 said:
daedal i hooked up the diode to the circuit but it shines dim but i put and LED on and bam light up the blue LED when i added the 5 ohm did it need to be in series or paralel
Click to expand...

Toked, what voltage are you supplying the circuit, and what voltage does your LED need?

Do you have a V and I measurement? Try and get a 1-Ohm resistor in series with the diode as Gazoo did and let us know what the voltage across it is. It should be outside of the circuit and should pretty much act as one of the LD pins.

Also, do you know if the LD itself is still ok? Measuring the voltage across the diode itself and then across the resistor should give you an idea what this thing is taking. If it is in fact drawing ~2.8V and 250mA (because of the 5-Ohm resistance) and it's still dim, you may have a COD diode there :(

GL;
DDL
 
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