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FrozenGate by Avery

DIY Homemade laser diode driver

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joniverson;

I am glad you liked the circuit and hope it works out for the best for you. In regards to using a resistor as a current limiter for such a circuit, you have an inaccurate cap that is a variable of the heat and other resistances in the circuit. This may be the resistance of the LD, the capacitor, or the batterys' internal resistance. All of which play a role in the current control. The capacitor is an excellent dissipater for voltage spikes, but current spikes are not something anything this simple can control. I cannot testify for how long a circuit of cap + resistor would hold up for a diode, and depending on the power source it may not be a constant at all.

I honestly urge everyone to control the current going through any laser diode.

GL;
DD:
 





Again, my question iterates my lack of knowledge on this subject...

I have pored over the PDF on the LM317T regulator and think I finally understand how it works, mostly-- really confused me for a while-- but I still have 2 questions;

Can I use a single pot instead of a pot/resistor? The only candidates in the RS grab bag were a 417 ohm pot that goes down to 15.6, (prolly gonna use that one) and a 48 ohm pot that goes down to ~7.5. Before I actually tested them, I didn't realize most pots would bottom out that high. :-?

Second, how the heck does the LM317 regulate current? The PDF makes it sound strictly like a voltage regulator, and the pack its in doesn't say a thing about voltage. I think I am getting confused with Ohm's Law... Current goes down when ohm's go up... But then your reducing current AND voltage. I guess the diode doesn't need much voltage to run... But iunno-- I will have to wire up the circuit then take readings I suppose, cant guess.

At least I can borrow my dads multimeter from work-- I have no idea what 2/3rds of the settings are! And only that cause it has build in memory storage and infrared download settings! ;D ;D

Thanks so much guys for educating a poor fellow like myself! :) Am I allowed to blame my ignorance on age? ;)
 
The LM317 will try to maintain 1.25V between the Vout and Adj pins. It will output whatever current (as long as it is within the device's specs) through the resistors to maintain a the 1.25V delta between the pins. Since you know the voltage, you pick a resistor value to obtain the desired current. These three terminal linear regulators have been around a long time. They are not overly efficient but they are quick to set up and easy to use in a pinch.

You can use it without the extra resistor but the resistor is there in case you accidentally turn the pot to 0 ohms and put the device's full current through the laser diode. The 4 ohm resistor limits the max current to 313mA (1.25/4) if you turn tehpot to 0 ohms.

Paul

Xenodius said:
Again, my question iterates my lack of knowledge on this subject...

I have pored over the PDF on the LM317T regulator and think I finally understand how it works, mostly-- really confused me for a while-- but I still have 2 questions;

Can I use a single pot instead of a pot/resistor?  The only candidates in the RS grab bag were a 417 ohm pot that goes down to 15.6, (prolly gonna use that one) and a 48 ohm pot that goes down to ~7.5.  Before I actually tested them, I didn't realize most pots would bottom out that high.  :-?

Second, how the heck does the LM317 regulate current?  The PDF makes it sound strictly like a voltage regulator, and the pack its in doesn't say a thing about voltage.  I think I am getting confused with Ohm's Law... Current goes down when ohm's go up... But then your reducing current AND voltage.  I guess the diode doesn't need much voltage to run... But iunno-- I will have to wire up the circuit then take readings I suppose, cant guess.  

At least I can borrow my dads multimeter from work-- I have no idea what 2/3rds of the settings are!  And only that cause it has build in memory storage and infrared download settings!  ;D ;D

Thanks so much guys for educating a poor fellow like myself!  :)  Am I allowed to blame my ignorance on age?  ;)
Click to expand...
 
Okay, cool-- Thanks for the info-- The PDF said to use it with 2 resistors, and I was like, WTPF? Why not one pot? ;D

You saved me a LONG time there ;D
 
Seems like you guys got it covered... ;)

The reason for the resistor, just as Chimo said, is to limit the maximum current regardless of what the pot reads. Some pots have the insane ability to drop from 10 or so ohms to 0 by simply turning it a hair's width! That could be something bad to a diode... ;) I try to make it as fool-proof as possible :P

GL and let us know how it turns out Xenodius;
DDL
 
Daedal,

Thanks for answering my questions. I think I'm going to try building your circuit, but I do have a couple more questions if you don't mind.

I notice in the picture that the 47 uF cap is a tantalum. I didn't notice this at first and bought a 47 uF electrolytic instead. Can the electrolytic be used instead?

Also, I want to use this to drive one of the Sony diodes at 2.5V at say 160 mA. I see you have 6 v input going into the circuit. I know current is affected by adjusting the pot, but what does the voltage do? If I'm going to be using these 2.5 v diodes, should I reduce the input voltage to 2.5V or leave it at 6?

Thanks,
Jon
 
joniverson, you can use electrolytic, and I actually think they are better at their job in this circuit.

As about the voltage, the input is going to have to be 6V. The LM317 requires 3V for itself, so you're left with 3V, and then the silicon diode takes about .7V, so you're left with about 2.3V going into the LD. I haven't ever encountered an LM317 that uses more than about 1.5V to itself, and I have run this circuit as low as 4V and it worked just fine, but I suggest 6V as input, and the LD will be very happy. :)

GL, and keep us posted;
DDL
 
I built the circuit today without trouble. The only thing I lacked (and still do) was the 100 ohm pot, so I started started at a fixed 50 ohms with resistors and reduced accordingly while monitoring current. I ended up at around 165 mA, 2.56 V. Not sure what this is in mW, but easily burned through electrical tape. Even after running the Senkat diodes for over a minute at this power, they just start to get warm.

One thing I did want to note, in case it hadn't been mentioned before, was that the case of the LM317T is NOT at ground, but does have a voltage. I could "feel" a very slight shock every time I applied power, so I quickly noted the case not being at ground from the data sheet. A quick mica insulator/ nylon screw solved the problem. I'm not sure what this could have done to the LD had I let it this way, but I have read of "fingers" modulating diodes in similar circuits.

Note that the aforementioned paragraph applies only if you mount the LM317 on a conductive chassis (metal). If using plastic, you would never have noticed the voltage unless you measured the LM317 casing itself.

Thanks for what appears to be a fine circuit.

Jon
 
I'm glad you got it working joniverson!! ;D :D :D

This is great news! Also glad to hear that you managed to power up the diode and get it fully operational! This is really exciting! :D :D (notice all the exclamation marks?)

Ok, I'm sorry if I had not mentioned it before, but the metal heat sink itself is hooked to the Vout line, as noted in the data sheet on the right side, and middle, of page 2. :)

Hope to hear from others who get their circuits working :)

--DDL
 
joniverson said:
I could "feel" a very slight shock every time I applied power
Click to expand...

That is very odd. With that low of current and voltage I am very surprised you felt a shock. What are you using to power the circuit with?
 
I'm very sensitive to mild shocks, but I can detect them even though low voltage DC.

Daedal, another quick question. I wanted to see what would happen to the circuit once I recharged the batteries. I did notice a slight increase in current. Is this normal even though using the LM317 as a current regulator?

Thanks,
Jon
 
How slight is slight? I personally tested it with a wall-adapter and with different kinds of batteries and it was all the same.

Could you explain please? :-/

--DDL
 
After recharging the batteries and powering up, I noticed about a 10 mA increase in current. After it's engaged for a few seconds, the current starts to drop slightly, maybe 5-10 mA. I am wondering if the 1/4 watt carbon resistors I am using in place of the 100+4 ohm might be heating and changing value slightly. If so, it might explain the current change.

Would the freshly charged batteries, having a higher voltage, also slightly increase the current?

Jon
 
I highly doubt it...

If you are using a predefined value resistor, say 2.5 ohms, try putting a few resistors in parallel to distribute the load... For example, if you put 2 5 ohm resistors in parallel, you now have a 1/2 Watt 2.5 ohm resistor... :)

GL;
DDL
 
The resistance of two equal resistors in parallel is divided by two. If you put two 1/4 watt 2.5 ohm resistors in parallel, you end up with a 1/2 watt 1.25 ohm resistor. Here is a calculator:

http://www.1728.com/resistrs.htm
 
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