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DIY Homemade laser diode driver

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yuip

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Okay, I have a few questions. :p

In the circuit diagram, the resistor is 4 ohms. You used a 10 ohm resistor in your circuit because you had it close by, but does it make any difference. I am guessing that it doesn't because you included a 4 ohm resistor in the diagram, linked to a 15 ohm resistor at DigiKey, and a 15 ohm resistor isn't included in the RadioShack pack.

So does the value of the resistor matter much?

Pardon my ignorance, and thanks again! :)
 



pseudonomen137

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Daedal said:
Connect the laser diode and make sure the capacitor is SOLDERED to the laser diode pins, be it through wires or directly on the back of the laser diode. If the laser diode happens to disconnect from the capacitor and then reconnect while the capacitor is charged, then that’ll be worse than the spike from the battery itself! It will literally dissipate all the charge almost instantaneously to the laser diode and fry it because it has very little internal resistance… :(
Minor detail, but can you please explain why this is? I would think that when the cap is in the circuit it would only charge up to the voltage that's already on the diode. Then even if the cap were then directly hooked up to the diode alone, wouldn't it still apply, at most, the same voltage to the diode and thus not kill it? Or am I just completely confused here :-/
 

yuip

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From what I've been told the main reason for the capacitor in very simple LD circuits (just a cap and resistor) is too smooth out in voltage spikes and give the diode a "soft start".

You are right, it seems that the cap would only charge to the voltage on the diode. I think what Daedal is saying is that you can't just connect 2.8-3.0 volts at 250-300mA directly to the diode, since many of the DVD burner red diodes have a threshold of only 100mA according to the spec sheets I've seen.

The way I understand it is that the capacitor slowly charges and slowly outputs more and more current, so that the laser diode doesn't die a quick death by the high current. If the LD became disconnected from the cap after running for a little bit, and was suddenly reconnected, the normal current you want to run the diode at would kill it.


Wow, I just realized I was talking in circles. :-[

Then again, I have no idea what I'm talking about, and probably am wrong.
 

Gazoo

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I found a seller on e-bay that has some of the SMD's we would need if we wanted to make the circuit smaller. The problem is what would we mount the SMD's to? I think I am going to skip trying this. The parts are extremely small and a pain to work with. I have in the past, but building a circuit from scratch would be very difficult...I would think. I suppose it could be done by super gluing the parts to a small board, then wire them up with small wire. But we would have the problem of heat sinking the SMD version of the LM317.
 

Hemlock_Mike

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The LM317 in this circuit is a current source. If the load (LD) disconnects, the Cap will charge to about 6 volts. Reconnect to the LD and you might have an LED :( The cap at 6 volts might blow the LD. (It's happened to me )

Mike
 

pseudonomen137

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Hemlock Mike said:
The LM317 in this circuit is a current source. If the load (LD) disconnects, the Cap will charge to about 6 volts. Reconnect to the LD and you might have an LED :( The cap at 6 volts might blow the LD. (It's happened to me )

Mike
Oh, so the problem is not with the charged cap temporarily disconnecting, but with the diode itself disconnecting?
 

Daedal

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Gazoo said:
My suggestion would be to change that statement to something like as follows:

"How?
To regulate the voltage we will use a voltage regulator and set it up in voltage regulation mode and limit the supplied voltage. And we will use a capacitor to ensure no voltage spikes reach the diode."

Again you have done an outstanding job with your write up and in fact I plan to try it myself. I am looking forward to whatever SMD parts you can find. We do know there is a smaller variation of the LMT317 that can handle up to 500ma's.

BTW, do you know how warm the LM317 gets in your circuit? I mean does it feel hot?
Gazoo, thank you very much for your suggestion, although we are NOT using the LM317 to regulate the voltage. In this circuit I am using the capacitor to dull away the spikes from the battery and the silicon diode to provide polarity protection.

As about the LM317T, it doesn't get anywhere near hot at that supply. A smaller one would definitely start feeling a little warm, but this one feels to be at room temperature even after about 10-15 minutes.. :)

Thank you;
DDL
 

Daedal

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yuip said:
Okay, I have a few questions. :p

In the circuit diagram, the resistor is 4 ohms. You used a 10 ohm resistor in your circuit because you had it close by, but does it make any difference. I am guessing that it doesn't because you included a 4 ohm resistor in the diagram, linked to a 15 ohm resistor at DigiKey, and a 15 ohm resistor isn't included in the RadioShack pack.

So does the value of the resistor matter much?

Pardon my ignorance, and thanks again! :)
I am very sorry about yet another blunder int he post... I will fix it immediately. Please understand that I did this really late at night... :-[

As about the resistor, if you use a 4 ohm resistor, then you limit the maximum voltage to 312.5 mA. If you use a 10 ohm the maximum that will be passed through will be 125 mA. Therefore, yes it makes a huge difference. I will fix the links momentarily.

Thank you;
DDL
 

Daedal

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pseudonomen137 said:
[quote author=Daedal link=1185701612/0#1 date=1185701644]
Connect the laser diode and make sure the capacitor is SOLDERED to the laser diode pins, be it through wires or directly on the back of the laser diode. If the laser diode happens to disconnect from the capacitor and then reconnect while the capacitor is charged, then that’ll be worse than the spike from the battery itself! It will literally dissipate all the charge almost instantaneously to the laser diode and fry it because it has very little internal resistance… :(
Minor detail, but can you please explain why this is? I would think that when the cap is in the circuit it would only charge up to the voltage that's already on the diode. Then even if the cap were then directly hooked up to the diode alone, wouldn't it still apply, at most, the same voltage to the diode and thus not kill it? Or am I just completely confused here :-/
[/quote]

The reason for the nice spike from the capacitor in such a case if because basically the capacitor is charged up like a second battery. This second battery has very little to no internal resistance and is not regulated by the LM317, in fact it is directly wired to the LD... thus when the LD works as a discharge venue for the capacitor, the electrons are eager to go through the tiny wire and will instantly fry it. In a case like this the LD won't turn into an LED, the hair-thin wire inside the diode will overheat and snap like a fuse wire. The internals of the LD would all be fine, but it would be disconnected... :(

I am not sure if this explains it any further Pseudo. Sorry if not, and please feel free to ask more if you need;
DDL
 

Daedal

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pseudonomen137 said:
[quote author=Hemlock Mike link=1185701612/15#20 date=1185756896]The LM317 in this circuit is a current source. If the load (LD) disconnects, the Cap will charge to about 6 volts. Reconnect to the LD and you might have an LED :( The cap at 6 volts might blow the LD. (It's happened to me )

Mike
Oh, so the problem is not with the charged cap temporarily disconnecting, but with the diode itself disconnecting?
[/quote]

Mike, thank you very much! This is exactly what I was trying to say.

Pseudo, the capacitor is the danger here. If the LD is disconnected form the Cap and then a second later reconnected, you have a fatal situation. If the capacitor is soldered to the LD permanently, then you could disconnect and reconnect the pair as much as you want, and nothing would endanger the LD. At least that is how it works in theory... ;)

--DDL
 

Gazoo

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Daedal, I am trying to understand why you need the diode. If I am understanding the circuit, the diode effectively is converting the 6 volt battery to a 5.5 volt battery. This is the same thing as adding a half volt to the dropout voltage. I guess what I am asking is wouldn't a resistor with a higher value accomplish the same thing?
 

Daedal

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I have updated the link for the resistor to a 3.9 Ohm resistor offered at Digikey. This would limit the max current supply to around 320mA.

Please all remember that you can ADD resistors as follows:

----====----====----
R1 R2

This is the same as one resistor equal to R1 + R2. Therefore if you wanted exactly 4 Ohms you could put 4 resistors in series.

Also, you can reduce resistors as follows:

--|--====--|-- <-- R1
--|--====--|-- <-- R2

This adds up as follows:

1/R = 1/R1 + 1/R2 and can keep adding up as many as you want. Thus two 10-ohm resistors in parallel are the same as one 5-ohm resistor. The benefit is that in this case the ratings would add up, merging these 2 1/4 watt resistors into one 1/2 watt resistor ;)

GL;
DDL
 

yuip

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Daedal said:
[quote author=pseudonomen137 link=1185701612/15#21 date=1185757164][quote author=Hemlock Mike link=1185701612/15#20 date=1185756896]The LM317 in this circuit is a current source.  If the load (LD) disconnects, the Cap will charge to about 6 volts.  Reconnect to the LD and you might have an LED :( The cap at 6 volts might blow the LD.  (It's happened to me )

Mike
Oh, so the problem is not with the charged cap temporarily disconnecting, but with the diode itself disconnecting?
[/quote]

Mike, thank you very much! This is exactly what I was trying to say.

Pseudo, the capacitor is the danger here. If the LD is disconnected form the Cap and then a second later reconnected, you have a fatal situation. If the capacitor is soldered to the LD permanently, then you could disconnect and reconnect the pair as much as you want, and nothing would endanger the LD. At least that is how it works in theory... ;)

--DDL[/quote]

Oh okay, so my explanation was pretty flawed. ;D

Eh, thanks for the clarification about the resistor value as well. No need to apologize. I can't thank you enough for taking the time to write up this guide in the first place. I think I'll try to make one of my boards with a removable resistor so I can tinker around with driving a diode past 500mA with pretty extreme cooling. ;)
 

Hemlock_Mike

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DDL is doing his best to explain a really basic circuit which has been around for years.  He's telling you that the 4 ohm resistor sets the "MAXIMUM" current from the LM317 and the 10 ohm is the variable part which reduces the current to a lower level.  HINT: Start low.
NOW:  He is working with a specific LD (diode) which has its own parameters so YMMV with what you connect this to.  Each diode has its own forward voltage drop and the "effective" ohms (V/I) is not linear.  You have to be careful at the top end.
A bunch of us, SenKat, DDL, Gazoo and myself among many others, have found this out the hard way   :(    You will be in good company if you cook an LD yourself  :eek:

Mike
 

Gazoo

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Tell me about it. I have cooked at least six so far ;D Your advice is very good.
 
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SenKat

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Daedal - to make it even easier for folks (you have done a GREAT job, BTW) Here is a suggestion - make a basic, non-schematic drawing for folks of what it is you are talking about - label each item, R1, R2, etc....and place a small legend at the bottom....

JUST a suggestion - you have done great work thus far, it may help some folks to visualize the circuit a little better though....
 
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