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Red Diodes, power and efficiency

Arc

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Hey, I have recently been making a few lasers in my shed using diodes from dvd drives and since I'm new to lasers and electronics I wanted to ask a few questions regarding electrical power as well as some specific questions about red laser diodes, sorry if this post is a bit long but I don't want to leave out any important info. Please feel free to just skip to question 4 as that doesn't require reading the paragraph below.

I have always been fairly certain that Power = Voltage X Current
Now here are some details about my laser. I am currently using Samsung 24x DVD Writer desktop drive diodes (originally extracted from brand new DVD drives) and these diodes have a big rectangular laser housing with an unusual look very much resembling a camera lens, they look much bigger than any other dvd laser diode I've seen. I've extracted a few different types and selected the Samsung diode since it stands up to the most current before dying. I am running my laser device as follows:
I'm using a home made driver utilising an LM317T, a few resistors, a pot, a capacitor and 2 N4004 diodes. I got the driver schematics from a youtube video. I am powering my home made device with 2 Li-Ion Cells in series giving 8.4V when fully charged. I have set the current of each driver to 400mA. There are several laser diodes on this device (all Samsung) and a driver for each diode with drivers connected to power source in parallel. I have checked and each diode receives 400mA @ 7.2V during operation as supplied by their drivers. I have been using this device for over 3 months now on a daily basis as it actually serves as a very useful tool which does a far better job than any alternative I have found. I would estimate that it is on for 200 seconds per day with duty cycles under 10 seconds. The diodes don't get hot nor warm at any time during operation.

Now with the above in mind here are my questions.
1) How much power is each diode drawing. Is it simply, 0.4A x 7.2V = 2.88W ???

2) I would like to have an idea of the light power output of this laser device. From what I have been able to gather the red laser diodes tend to be the most efficient. To my best knowledge these Samsung diodes have a conversion efficiency of anywhere from 20% to 30%. Please someone correct me if I'm wrong. Assuming the efficiency is 25%. Would the total light output of each diode simply be, 0.25 x 2.88W = 720mW ??

3) Do you think that I'm pushing these diodes too hard with 400ma @ 7.2V ?
So far after 3 months consistent use they are just as bright as the day I extracted them from brand new drives.

4) Do DVD 650nm red laser diodes put out only or mostly red light or are other wavelengths significantly present in the light output of the diode? I am mostly concerned about infrared being present and wouldn't mind if higher frequency visible light were present in the output but I'd still like to know if it is.

Thanks so much in advance for any responses.
 
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ChaosLord

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1.) Not quite. Red and IR diodes have low voltage drops, anywhere between ~2.2 - 3.5V, IIRC. Have you measured the voltage across the diode?

2.) Not sure about actual efficiency, or this particular diode, but you're defiantly not getting anything like 720mA. DVD burner diode, from what I've seen, put out ~250-350mW at 400mA, more or less. However, due to variance between diodes, calculating output power from electrical input is a very rough estimate.

3.) It's definitely not dropping 7.2V, but this depends on the diode. Considering that it's lasted this long, it appears the current is fine. As long as there is some kind of heatsinking, duty cycle is fine and could even be increased.

4.) Doubtful. Some are red/IR combo diodes, but they must be wired to emit both. IIRC, they also may not be able to emit both at the same time. If the diode has 3 leads, it may be able to emit ir, but not if you only connect 2 leads. Diodes do often emit trace wavelengths other than their rated wavelength, but not significant enough to notice.
 
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DashApple

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You are feeding the circuit 7.2-8.4V so the 0.4A x 7.2V = 2.88W will be the overall power consumed .

The red laser diode will only drop around 2 - 2.5V so power to the diode will be around 800mW - 1W .

The circuit is linear based so the laser diode will drop a voltage at 400mA and the rest is dropped over the LM317 regulator and current set resistor .

If you where to assume a 900mW power input to the diode , output at 25% Eff would be 0.25*0.9 = 225mW (0.225W) ( ideal world )
 
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Arc

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Ok since the diode only drops a couple of volts does this mean that it will perform just as well when the battery voltage is low at 7V compared to fully charged 8.4V? Or will the light output decrease as the battery voltage decreases?
Also these diodes have 4 leads but I'm only using 2 of them. Is it likely these 4 lead diodes are capable of IR?
And lastly, do you think it would be safe to turn up the current if I run even shorter duty cycles? I was thinking of trying to go as high as 600mA. Each diode has a nice big heat sink.
 
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I think the output will be the same, because, usually, the LM317T needs 3.0 ~ 3.5V more than the voltage dropped by the LD, so, 2.5V+3.5V = 6V, this will be the minimum voltage that the driver will need to supply 400mA of current.

I've disassembled a couple of DVD-RW, those LD's with 4 leads are IR/Red combos, most of then are frame type LD's (small, slim and black). If you wire it correctly, it can emit both wavelengths at the same time, but each one will need it own driver!

You can try, but some diodes will die instantly if pushed above 500mA, since those diodes are unmarked, we can't find the model number and the current vs power graph.
 
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Arc

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I see. So based on that I must be pushing it close to its limit already. This brings about a few more questions for me.

1) Do you think if I go to 450mA for example the light output will be significantly greater, or since I'm close to the limit, won't there be much of an increase in output for the 50mA jump.
I would like to get more out of them but if it reduces their life just for a tiny performance improvement then it wouldn't be worth it.

2) Would the infrared wavelength burn wood and dried green plant matter better than the red given that they are both running near their max output? Note that I can not opt to use both wavelengths simultaneously since the requirement for 2 drivers per diode will not allow me to house the device in such a small host. But since there are several diodes I could set a few of them to emit infrared and the keep the others red so the device doesn't become invisible and highly dangerous.

3) To emit red light I use just the top 2 pins, top being +ve and the one below that is -ve. When I extract the diode however, the middle two pins are soldered together (see image below), the middle two pins being the -ve pin I use for emitting red, and the one directly below that. I unsolder them such that I can use only the top two. Do you think that perhaps leaving the middle two pins connected would emit both wavelengths simultaneously. Or that using only the top pin and 3rd from top pin would emit infra red?

4) And finally if I did use both wavelengths at the same time would that allow the diode to put out more energy before blowing. I.e. 350mA driver for the red wavelength and another 350mA driver for the infrared. Or would they need to be reduced?
 

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Hi! :yh:

1) Do you think if I go to 450mA for example the light output will be significantly greater, or since I'm close to the limit, won't there be much of an increase in output for the 50mA jump.
I would like to get more out of them but if it reduces their life just for a tiny performance improvement then it wouldn't be worth it.
I generally use 400mA for those diodes from DVD-RW, since is a unknown diode and we don't have a output power vs current graph, I'll suggest that you keep the 400mA, you should get a good lifespan, you could get more power with 450mA, but your diode can not last that long.

2) Would the infrared wavelength burn wood and dried green plant matter better than the red given that they are both running near their max output? Note that I can not opt to use both wavelengths simultaneously since the requirement for 2 drivers per diode will not allow me to house the device in such a small host. But since there are several diodes I could set a few of them to emit infrared and the keep the others red so the device doesn't become invisible and highly dangerous.
Most IR diodes from CD/DVD drives are only 150mW~170mW, the red ones will be around 250mW, I had a good time burning stuff with IR and the result was a bit disappointing, it will burn better black things, colored things will not burn easily, and you will notice that the focus point capable of burning will be very small. For example, it can cut a line of 0,2mm of thickness in a insulation tape, and you will see a yellow glow from it, since IR is heat, this glow can be called fire, at 250mW your red diode will perform much times better and will be more visible.

3) To emit red light I use just the top 2 pins, top being +ve and the one below that is -ve. When I extract the diode however, the middle two pins are soldered together (see image below), the middle two pins being the -ve pin I use for emitting red, and the one directly below that. I unsolder them such that I can use only the top two. Do you think that perhaps leaving the middle two pins connected would emit both wavelengths simultaneously. Or that using only the top pin and 3rd from top pin would emit infra red?
The center pins soldered together are both "-", the top pin are the "+" from the red LD, and the bottom one are the "+" of the IR diode, unless you solder the bottom pin, the IR LD will not light up.

4) And finally if I did use both wavelengths at the same time would that allow the diode to put out more energy before blowing. I.e. 350mA driver for the red wavelength and another 350mA driver for the infrared. Or would they need to be reduced?
Unless you have a good heatsink, you can use both wavelengths at same time. For the red, use 400mA, and for the IR, use 150mA. The output will sum, 250mW+150mW = 400mW, but since are different wavelengths, it will not focus at the same point.
 
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Arc

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Ok thanks so much. That's been some great information that I couldn't find after hours of searching.
Just a couple more things.

Ok so I'm very interested in trying to use the infrared/red simultaneously. Perhaps I'll build another device with less diodes that utilizes this ability.
So,
1) Is the infrared light going to be emitted in the same direction as the red (although focal point may be different)? I'm quite sure the answer is yes but furthermore will this make it as safe as a pure red laser since the red light makes it very visible hence keeping clear of the red beam will keep you clear of the IR beam? Or will it not be as safe as a pure red laser because the infra red light is highly dangerous for some other reason. For example if a reflection from the 250mW of red light poses no danger for a given scenario (i.e. position and orientation of eyes and laser device relative to reflective surface, reflectivity of that surface, focal setting of lens), would it be safe to say that the IR light reflected in that same scenario also poses no danger?

2) I always have the red light focused very close to the lens hence the light spreads out very fast. At this focal setting will the IR light be focused closer to the lens or further from the lens as compared to the red light focal point? And will there be much distance between the 2 focal points? Also is the IR wavelength completely invisible or does it emit a dim red light? (just so i know for testing)

3) You mentioned that I should get a good life span for the diode with the current set at 400mA. How many hours would you consider to be a good life span? Up to 8000 hours perhaps or is that too optimistic? Note that I run very short duty cycles, max 15 seconds but usually under 10seconds. Also the diodes have decent heat sinking and don't even get warm during operation.
 
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1) Is the infrared light going to be emitted in the same direction as the red (although focal point may be different)? I'm quite sure the answer is yes but furthermore will this make it as safe as a pure red laser since the red light makes it very visible hence keeping clear of the red beam will keep you clear of the IR beam? Or will it not be as safe as a pure red laser because the infra red light is highly dangerous for some other reason. For example if a reflection from the 250mW of red light poses no danger for a given scenario (i.e. position and orientation of eyes and laser device relative to reflective surface, reflectivity of that surface, focal setting of lens), would it be safe to say that the IR light reflected in that same scenario also poses no danger?
Yes, it will be emitted at the same direction, the focal point will be very close, but not the same, if you don't look at the red then you will not look at IR too.

2) I always have the red light focused very close to the lens hence the light spreads out very fast. At this focal setting will the IR light be focused closer to the lens or further from the lens as compared to the red light focal point? And will there be much distance between the 2 focal points? Also is the IR wavelength completely invisible or does it emit a dim red light? (just so i know for testing)
If you will burn things, I'll suggest that you set the focal point away from the lens, since smoke can get in it. The focus point of IR will be away (search for chromatic aberration) from the red one. At 150mA, you will see IR as a very dim light, just be careful with it.

3) You mentioned that I should get a good life span for the diode with the current set at 400mA. How many hours would you consider to be a good life span? Up to 8000 hours perhaps or is that too optimistic? Note that I run very short duty cycles, max 15 seconds but usually under 10seconds. Also the diodes have decent heat sinking and don't even get warm during operation.
Yes, it will vary from diode to diode, I think 5000h will be fine, it represents more than 200 days of continuous operation, you said that runs at 10 seconds duty cycle, it will last even more! Red diodes are very proud to COD (catastrophic optical damage) when overloaded.

My have a small heatsink and I run it up to 15 minutes at 415mA and never had problems with it. Heat, noise and spikes are the bigger enemies of a LD.

Only to confirm, your diode looks like this one? That one I extracted from a Samsung SH-S182 drive.



Your LD should be in a metal housing, this one is just the bare diode.

:beer:
 
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Arc

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Yeah its in a metal housing so I'm not sure exactly what the bare diode looks like. Here are a couple of pics though of the assembly. Also is that metal housing going to provide much heat sinking?
 

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This housing is just a enclosure, it don't provide any good heatsinking, you will need found a way to attach it in a heatsink, here is the way I did it.




There's a thing that you need to do before using it, you can see a small black piece holding a diffraction grating?

You will need to remove it, because when you focus the light using a lens, it will split the beam in a line with various little spots and you will loose output power!
 
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Arc

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Ok thanks for the advice. When I get another one of these diodes then I'll try and do that. I think keeping under 10 sec duty cycle should be alright still with just the housing, they don't seem to get warm until 20 seconds. When you mention the beam being split by the diffraction grating. The raw diode output without lens has an oval shaped beam with lots of horizontal thin lines running through it. Is this what you mean?
 
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Ok thanks for the advice. When I get another one of these diodes then I'll try and do that. I think keeping under 10 sec duty cycle should be alright still with just the housing, they don't seem to get warm until 20 seconds. When you mention the beam being split by the diffraction grating. The raw diode output without lens has an oval shaped beam with lots of horizontal thin lines running through it. Is this what you mean?
Now I understood, you're running the LD only with the housing, if the LD isn't getting hot, then there's no problem.

Yes, the horizontal lines are from the diffraction grating, if you focus it with a lens, every horizontal line will be a small spot with a bigger one in the center.



Remove it and you should get a smooth oval raw output.
 
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Arc

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Oh yes I recognize that pattern. Yeah I saw that too when I tried focusing it to a point with a collimating lens. Thankfully my application does not use that kind of focus and I don't get that effect with the lens I'm using I have all the light focused close to the lens. However the power loss is still there I assume?
How much power (%) do you think I lose with the diffraction grating there?
And is it very difficult to remove the diffraction grating without damaging the LD?
 
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Oh yes I recognize that pattern. Yeah I saw that too when I tried focusing it to a point with a collimating lens. Thankfully my application does not use that kind of focus and I don't get that effect with the lens I'm using I have all the light focused close to the lens. However the power loss is still there I assume?
If you don't need to collimate the beam in a single point, I think you will not loose much power, but since is more one obstacle between the LD and the lens, remove it will only improve the things.

How much power (%) do you think I lose with the diffraction grating there?
Is hard to guess, it will dislocate some power and put in a vertical line, the power density will not be the same, since I don't have a LPM, I can't answer.

And is it very difficult to remove the diffraction grating without damaging the LD?
No, is very easy, you will need a knife or something sharp, just insert it between the black plastic piece and the housing, it must come off easily without causing any damage to the LD.
 

Arc

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Hi again :) What is the plastic piece that I must take off, is it on the front of the housing of on the rear side where the pins/leads are? I attached a few photos.
See the second photo is this correct what I have removed there? Or should I remove something from the back as seen in the 3rd photo?

Thanks again for all of your useful advice, it is very much appreciated.
 

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