DIY Laser Torch

[LarryQ]

OK...Now I see that to get the voltage drop the rectifier diode must be in series with the positive lead to the LD..

So...what was throwing me was in the circuit, the rectifier diode is shown in parallel along with the capacitor!

So..One goes across the Laser diode leads, and one must go inline to the Laserdiode to get the voltage down from3.1 volts.

I think I am having a hard time simulating a laserdiode with the regular diodes I have.

I went through my Box O Stuff, and found an LED Light/Pushbutton...

Wiring ONLY the Led light, I now see 2.4 Volts 230 Mw current.


Now if I could only be sure this Simulates a LD accurately!!!!

Larry

 

[Daedal]

Given 5 ohm and 6 volts in...once thru the LM 317.
I plaved the stock diode that came with the MXDL as a test..

I am seeing 150Ma and 3.1 volts...

I added a Zeiner diode to this circuit thinking it would lower the voltage by .7 volts.

But I testeed it and the voltage remained 3.1????

Also the current seems a bit low.....

Is this simply that the stock diode is doing a poor job "Simulating" a LD???


THis mod will have a nice big alum heatsink, to keep the diode cool.

Larry

not all diodes have a .7V drop

--DDL

 

[Daedal]

OK...Now I see that to get the voltage drop the rectifier diode must be in series with the positive lead to the LD..

So...what was throwing me was in the circuit, the rectifier diode is shown in parallel along with the capacitor!

So..One goes across the Laser diode leads, and one must go inline to the Laserdiode to get the voltage down from3.1 volts.

I think I am having a hard time simulating a laserdiode with the regular diodes I have.

I went through my Box O Stuff, and found an LED Light/Pushbutton...

Wiring ONLY the Led light, I now see 2.4 Volts 230 Mw current.


Now if I could only be sure this Simulates a LD accurately!!!!

Larry

Larry, to (as close as possible) simulate a laser diode, use 4 X 1N4001 rectifier diodes in series instead of the LD. At .7V drop each, you have a 2.8V drop across them, which is the most common voltage drop I see on LD's like this. A laser diodes, or any diode for this matter, is a current drain. It simply drops the voltage going through it by some amount and then passes all the current across. If it manages to get too much current (which is loves) it gets too hot and then dies! With the diodes wired in series you get a perfect idea of what a LD will do when hooked up... except the whole coherent light thing

If you want to go one step further, you can use regular LED's. They are the same thing, but they will have a higher V-drop. I would use 3 or 4 of them to make a nice big V-drop and then test to make sure that the voltage, across the whole bunch, is what it needs to be. If the voltage across all that many diodes is too high, then your supply voltage is too high, if it's low, then it'll simply be low. Using a single LED with, say, 2.6V drop would only simulate the circuit up to 2.6V. If you then add another one in series you should notice that he voltage drop is a maximum of 5.2V, but if you get more than 3.5-4V then you should try cutting it down. The current will always be the same regardless of what diode you use, it's the voltage + current that make the difference. They have to be just right

Current still gets the higher precedence of the two though

--DDL

 

[LarryQ]

Ok...looking good!!!

However, I was wondering one last thing....

In the picture above, where the center leg of the LM317 is cut off, and the resistor goes from where the heatsink tab was to the Adj leg of the LM317.

It was mentioned that the lead to the Laser diode connects to the Adj leg.

I then asked if a .5 watt resistor would be enough, and it was answered YES....given that not all the voltage/current would be going through the resistor.

How can this be???

It looks like ALL the current HAS to pass through that resistor if you connect the lead to the laser diode to that adj leg.

I need to take a refresher in electronic theory!!!

I managed to fit a single 5.1 Ohm 1/2 watt resistor Alongside the LM317 (rather than over of under) in that tiny housing, and I hate to have to take it apart to change it to (2) 1/2 watt 10 Ohm resistors now that the whole thing is closed up.

But only having that 1/2 watt resistor in there has me concerned!

Larry

 

[a_pyro_is]

Larry,
I guess I'm still not understanding, :-? please forgive me. You said earlier that you're supply is 6V, and that you're getting 3.1V out of the circuit at 150mA.
At which points are you measuring the voltage?

Again, sorry I'm a bit slow on the uptake sometimes. :-[

My understanding is that the power rating of a resistor needs to be equal to a safety margin (say 120%) times the voltage drop across the resistor, times the current through the resistor.

I guess you could just try it while monitoring the resistor temp. :-/

 

[LarryQ]

I am measuring the voltage and current as it goes into a Luxeon Star LED (Simulating a Laser diode).

If you look up 10 posts to the one by Rog881, i have my LM317 setup the way he has it pictured.

I placed the positive wire going out TO the laser diode on the ADJ leg

It seems that if I tap on the laser diode feed at the ADJ leg...not only is the voltage that goes into the feedback of the LM317, but all the voltage and current that goes out to the Laser diode passes through that resistor.

I always have this trouble with electronics...I think way to 2 dimensional......

Larry

 

[a_pyro_is]

Sorry for the edit, I didn't want to double post. I guess I'm not quick enough.

 

[LarryQ]

Ok..

I will also require 2.4 volts and 300Ma in it's final form..So I believe 1 watt is where I will need to be....

THo I still need to understand why the wire lead that goes to the laser diode doesn't get soldered the where the heatsink was cut off (vout) rather than at the Adj leg.



Larry

 

[a_pyro_is]

Don't think of it as connected to the Adj pin, as much as connected to the Vout by way of the 5 Ohm resistor. The Adj pin just happens to also be connected to the resistor and is a convenient place to connect the load/LD.

 

[rog8811]

Confusing init .....If you think of the adjust pin as a "sense pin" that senses the output does that help?

Regards rog8811

 

[rog8811]

@ Larryq....Having looked back over the thread, you said.....

So...what was throwing me was in the circuit, the rectifier diode is shown in parallel along with the capacitor!

That diode is across the output but reverse polarity, it is a safety device to protect the LD, just in case you connect the supply up the wrong way round....

Regards rog8811

 

[a_pyro_is]

I thought the 317 was already reverse polarity protected... unless it's supposed to be soldered directly to the LD like the capacitor, so that you can remove the LD/Cap/Diode assembly from the power supply without destroying it.
lol funny how when you talk something out you see something you had missed.

 

[Gazoo]

Well I have accidentally hooked up the input to my 317 wrong and nothing happened and nothing was damaged. I have even hooked up my LD wrong and nothing happened. So I don't really think the silicon diode is necessary.

 

[rog8811]

unless it's supposed to be soldered directly to the LD like the capacitor

That is certainly what I have done, the diode and the electrolytic are inside the Aixiz housing with just the flying leads emerging...

Regards rog8811

 

[Daedal]

Ok...looking good!!!

However, I was wondering one last thing....

In the picture above, where the center leg of the LM317 is cut off, and the resistor goes from where the heatsink (Vout) tab was to the Adj leg of the LM317.

It was mentioned that the lead to the Laser diode connects to the Adj leg.

I then asked if a .5 watt resistor would be enough, and it was answered YES....given that not all the voltage/current would be going through the resistor. It is recommended to use a higher rating resistor than .5 Watt, but .5W will be enough.

How can this be???

There is a voltage drop (constant) between the Vout and the Adj of 1.25V. All the current is going to be running through the resistor, but to reach the maximum capacity of a .5 Watt resistor you need to have 400mA current.

It looks like ALL the current HAS to pass through that resistor if you connect the lead to the laser diode to that adj leg.

I need to take a refresher in electronic theory!!!

I managed to fit a single 5.1 Ohm 1/2 watt resistor Alongside the LM317 (rather than over of under) in that tiny housing, and I hate to have to take it apart to change it to (2) 1/2 watt 10 Ohm resistors now that the whole thing is closed up.

But only having that 1/2 watt resistor in there has me concerned!

The resistor you have in there should be enough, but I personally recommend changing it out only for the safety of the circuit/diode. When the resistors heat up to extremes their resistances change drastically. I tested that and heated one of the 1-Ohm RS resistors with 1Amp current for 3 seconds. It turned black, smoked and a resistance check afterwards gave me about 50-Ohm resistance!! The opposite could be true if I started with 100-Ohm and burnt it, I might have ended up with 10-Ohms... but that is speculation ;D

Larry

Answers in red

--DDL

 

[Daedal]

Well I have accidentally hooked up the input to my 317 wrong and nothing happened and nothing was damaged. I have even hooked up my LD wrong and nothing happened. So I don't really think the silicon diode is necessary.

You lucked out

There is a good damage potential from hooking any diode in reverse polarity

--DDL