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Old 03-23-2016, 06:40 PM #1
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Default Need some explanations.

Hello, im pretty new at electronics and got couple of questions. I have read many times different guides how to build laser driver. But theres still something i dont get it. So
1) Why do i need 5 ohm resistor?
2) Why do i need capacitor? (its necessary for giving consant power?)
3) I know i dont have to use potentiometer but if i wanna do it then as i read i have to use 100 ohm one but if i would use 1000 ohm it would be too big resistance? How to calculate to understund it?

Btw i still thinking of building 1w laser what needs approximatly 1-1,2 volts. So i need to use lm317 and it gives out 1,25 volts and if i need to calculate resitance then i just need to 1,25/1= it gives 1,25 ohms. why 5 ohms? I just cant get it.

Thanks.


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Old 03-23-2016, 08:29 PM #2
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Default Re: Need some explanations.

1) The resistor value across the ADJ and OUT pins sets the current. The regulator will try to ensure 1.25V across these pins, so the output current will vary according to the resistance value. Ohm's law. 1.25V/resistance=current.

2) Capacitors in this circuit are to prevent oscillations. The LM317 usually won't oscillate even without caps, but the caps ensure stability.

3) The potentiometer resistance adds to the base resistance. 1000 ohms is too much, because the current would be 1.25mA across this resistance. This is too low for stable operation, and it is too low to drive any laser diode anyway.

Quote:
Originally Posted by Tamm View Post
...thinking of building 1w laser what needs approximatly 1-1,2 volts.
Those don't exist.

Quote:
Originally Posted by Tamm View Post
i need to use lm317 and it gives out 1,25 volts...
No. Laser diodes are operated with constant current, not constant voltage. If you provide X volts input, the LM317 will automatically vary the voltage from 0 to X-3.5V to source the necessary current to the load. For example, 9V input will have an output voltage of 0-5.5V and the current will be whatever you program with the resistor.
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Old 03-24-2016, 12:35 AM #3
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Default Re: Need some explanations.

Sorry, but it still makes the questions, please understund that english isnt my native language and because of that its a little bit harder to understund and express myself. Thanks a lot for answering.

1)I understund you. So how do i calculate resistance? I can use ohms law if i know laser diode current. 1,25/5= 25 mA is that correct? How do i know 25mA is enough current for 1w laser diode and wont burn it?

2) You said 9v batter can vary 0 to 5,5 volts. Lets say i put 12v battery, so it varies 0 to 8,5volts? How i can ensure that it stays for example between 0 to 3 volts example. 8 volts is a lot and diode would burn. Its confusing connection between current and voltage and resistance. I just know that if i would put too high input then lm317 will overheat. And in this case i could use heatsink?

Its late i dont know. Still confusing. Sorry.
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Old 03-24-2016, 12:56 AM #4
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Default Re: Need some explanations.

General case use is just confusing you, so choose a laser diode first and we can go from there.
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Old 03-24-2016, 04:20 AM #5
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Default Re: Need some explanations.

I hope this doesn't add to your confusion ...

As Cyp explained, laser diodes operate by using constant current sources (=drivers). To get the optical power you want, you must set/build the driver to deliver an amount of current that produces the desired optical power. To find out how much current is required to produce the desired optical power for a given diode, see this website from DTR . He shows pictures of a lot of diodes under different test conditions for input current. You can see the voltage, current, and resulting optical power for a variety of input currents.

Just click on the laser diode of interest and scroll down to see if he took pictures. If he did, then the readings are on the equipment.

Your other questions about how to build a driver using an LM317 are also answered well by Cyp. As Cyp explained, the LM317 is designed so that you can choose how much current it 'pushes' by setting the resistor value between the ADJ and OUT pins. It also looks like you understood his answer but your arithmetic is incorrect. Using a 5-ohm resistor will produce 250mA of current, not 25mA (1.25/5=0.250). Go to DTR's website, look up any laser diode you're interested in, then scroll down to find what optical power is produced with 250mA. If it produces the optical power you want, use the 5-ohm resistor. If the output power is too low or too high, find the test picture which produces the desired optical output power, look at the current required to produce it, then adjust the resistor value using Ohm's Law to get the desired current out of the LM317.

Remember you're dealing with 3 variables here; voltage, current, and output power. Mostly people use Li-Ion batteries which are nominally 3.7V each (4.2V fully charged). Some laser diodes need more than 3.7V, some need less (note that I'm being a little general here, the LM317 needs a little more voltage on its input than it produces on the output, often referred to as headroom). So depending on the laser diode you want to use, you will have to drive the LM317 with one or two batteries to generate enough voltage to turn the diode on and produce the desired current to deliver the desired optical/output power. For example, one battery is enough to drive the LM317 for some red laser diodes while blue laser diodes typically need more voltage so you'll need two Li-Ion batteries to be able to turn it on. You can't drive the laser diode with current if you can't turn it on.

Pretty wordy, I get it. But is it clear enough?

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Old 03-24-2016, 07:38 PM #6
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Default Re: Need some explanations.

First, i think i have choosed diode. Thought this one:
1W 1000mW Laser Diode Blue 445nm to 18 5 6mm | eBay

Next i really wanna thank you, today woke up with clear head and im gonna understund it better, at least i think that. And also a lot thanks for this link this was really helpful. I didnt see any 1w blue laser but i found green one. What i saw was that he needed aproxximately 5 volt and 1,6-1,7 amperes. Is that correct? I think what makes me confused is reading/calculating units. Or that is right what i wrote? So 1,6 amperes is 1600 mA and in this case i dont get how you can use that 250mA current because its so few?
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Old 03-24-2016, 09:04 PM #7
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Default Re: Need some explanations.

That's a 1.2A diode, so you need a 1ohm 2W+ resistor to set the current. 1.25V(ref)/1ohm= 1.25A. A potentiometer is not necessary. You can verify the current is 1.25A by checking the voltage across this resistor when the circuit is active. For 1ohm, 1V = 1A (ohm's law). A voltage of 1.25V across this resistor means the circuit is in regulation as it should be.

Your input will need to be 9V to account for the minimum dropout of the regulator. 12V could also be used, but will create more heat in the IC. It needs to have a heat sink bolted to it. If the heat sink gets too hot to touch, use a bigger one. Do not use a 9V battery, as these cannot supply enough current. Use 3 lithium cells in series, or use a 9V DC power supply capable of at least 2A.

Do not worry about the voltage output or the maximum current rating of the power supply. The LM317 will ensure 1.25A goes through the diode if you've followed the standard schematic.
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Old 03-24-2016, 10:38 PM #8
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Default Re: Need some explanations.

Yes, thanks, but its easiest way to just tell me what to do. I want to understund what im doing otherwise its pointless. Let me make an example to check my understunding.
Lets take this one for example:

405nm 50mW CW Violet Blue Laser Diode LD SLD3232VF Sony New | eBay

Is that correct in this case i need 1.25v/0.050mw= 25 ohms. So i need 25 ohms worth resistance. And i know theres no 25 ohms resistance. So theres two main choices what i figured out: first is simplest, just add 10+10 series and two 10+10 paralleel so it makes 25 ohms. or just 10+10+1+1+1+1+1 but here is 7 resistors instead of previous four. I didnt get it first what did you mean by 2w+ but then i found this:

basic - When to use a 1/4 Watt resistor vs 1/2 Watt resistor - Electrical Engineering Stack Exchange

So i did calculationns P=0,05 square mutiplied with 25 ohms i get 0,0625 watts and so 1/4 watt 25 ohms resistance would be enough. Did i get it right?

And for a last thing i would to know why you said 9 volt would be the best for my 1w laser diode. Input should be at least 3v because of lm317 and then i read i need about 1,2 additional voltage and some for laser diode but i dont know how much. I saw from that link what OVNI posted that 5volts takes 1w laser so i really need yes 9v battery.
But lets say i dont know know much voltage takes 1w so how do can i calculate it ?

Sorry for this long text but thats the only how i can express myself. I really appreciate your help.
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Old 03-25-2016, 12:36 AM #9
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Default Re: Need some explanations.

Quote:
Originally Posted by Tamm View Post
Is that correct in this case i need 1.25v/0.050mw= 25 ohms.


That's like saying 0.4liters/17tarantulas=275 decaffeinated banana pies.

It has been mentioned several times here that laser diodes are CURRENT driven, and you're continuing to ignore current entirely. I don't blame you, I blame the language barrier. I just don't think I can help if we're not speaking the same language

I wish you luck.
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Old 03-25-2016, 01:10 AM #10
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Default Re: Need some explanations.

Oh sorry no it wasnt language barrier problem, its late and i looked wrong. I get what u mean by current. At first i saw 0,75 mA but also noticed that was maximum and then i looked headings and i read 0,50 mA but it was actually mW. My mistake. And as i can see i made one huge mistake more maybe. Is 75 mA 0,075 Amps or 0,75 Amps? It should be 0,075 Amps? 1,25/0,075= ca 17 ohms. Correct now?
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Old 03-25-2016, 07:36 AM #11
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Default Re: Need some explanations.

Quote:
Originally Posted by Tamm View Post
I didnt see any 1w blue laser but i found green one.
Actually, there are a lot of blue laser diodes that are able to produce 1W of optical power. It doesn't have to say "1W" in the name of the diode. All it has to say is it can produce > 1W. For example, I pulled these pictures from DTR's website and highlighted where to look for current, voltage, and output power.

NUBM44 Diode: 900mA produces 1.056 Watts





NDB7875 Diode: 800mA produces 1.109 Watts


Just find the picture that has the desired optical power, then design your driver to produce the amount of current shown in the picture. Does that help?
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Old 03-25-2016, 01:53 PM #12
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Default Re: Need some explanations.

Thanks OVNI it doesnt help much but im still impressed that green laser used 5v and blue laser uses only about 4 voltage to get about 1 w output. 1 whole voltage is a lot i think. But i didnt get answer to my questions:
1) Was i correct? i mean that 17 ohm calculation.
2) I can see see voltage and current from your webpage. But can i calculate it somehow?
3) I havent asked it yet but lets say im using potentiometer. How do i know which potentiometer should i use? Can i calculate it? 100 ohm is the best as i can see but why not 10 ohms or 1000 ohms?

Thanks for your time.
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Old 03-25-2016, 11:15 PM #13
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Default Re: Need some explanations.

Can you please point me to the diode you're interested in and remind me how much optical power you want? I'm not following this thread very well. Then I'll try and answer each of your questions one by one.

EDIT:
I'll answer based on this diode on eBay: 405nm 50mw CW Violet/Blue Laser Diode LD SLD3232VF SONY New

1) Was i correct? i mean that 17 ohm calculation.

First, I would use the typical value, not the laser diode's maximum value. Second, the text in the eBay auction states the maximum current is < 75mA (75mA=0.075A) however the datasheet provided by the link in the auction lists the maximum operating current at 65mA. From the datasheet the typical operating current of 55mA produces 50mW of optical power. Therefore the resistor value needed using an LM317 is 1.25V/0.065A~19 ohms.

So yes, your calculation of 17 ohms using 75mA was done correctly - only I recommend that you use 55mA instead of 75mA for the reasons stated above.

2) I can see see voltage and current from your webpage. But can i calculate it somehow?

I am not aware of a model, calculation, program, etc. that computes optical power out as a function of input current. That doesn't mean one doesn't exist. Try Google. I would guess that the manufacturer has extensive models based on their own internal research and development. Maybe try looking through a laser diode physics textbook. I've also noticed some diodes input current/output power relationship appears pretty linear so all you would need are a couple points to create your own equation/expression.

I use pictures taken from other people, e.g. DTR's website, or conduct a test myself.

3) I havent asked it yet but lets say im using potentiometer. How do i know which potentiometer should i use? Can i calculate it? 100 ohm is the best as i can see but why not 10 ohms or 1000 ohms?

As you probably already understand, using a potentiometer with an LM317 allows you to vary the current into the laser diode thus varying the output power. Be aware, however, potentiometers are 0 ohms on one end and whatever value it is on the other. You don't want to accidentally (or allow someone else to accidentally) allow too much current into the laser diode by allowing 0 ohms as it will destroy the diode. A way to fix a maximum current value but also allow variability is to use a fixed resistor and a pot in series with each other.

The first step is determine what range of input current you want. The datasheet doesn't give a minimum current but lists typical at 55mA and max at 65mA. For the sake of this calculation, let's pick a minimum current of 45mA. The fixed resistor determines the max current and is calculated to be 1.25V/.065A=19 ohms. The range of the pot is determined by the difference between the resistance values for minimum and maximum current. Calculating the resistor for minimum current we have 1.25V/0.045A=28 ohms. So 'ideally' you want a pot that goes from 0 ohms to (28-19)=9 ohms so you select a 10-ohm pot. Anything higher than a 10-ohm pot will simply produce less current thus producing less optical power. It's a design/requirements choice.

You also want to select the proper wattage for both the fixed resistor and the pot. Since power is (current) x (current) x (resistance), or (voltage) x (current), use the value that produce the largest power. In this case we have 0.065A going through a 19-ohm resistor which will dissipate 0.08 watts (or ~1/12th of a watt). The pot goes from 0 to 10-ohms but is in series with the fixed resistor so it's power dissipation is less than the fixed resistor. So a common choice for resistor wattage (fixed or pot) is 1/10th which meets your requirements. Some 'derate' their operational values which means choosing a value that is more conservative like 1/8th Watt or 1/4 Watt parts.

Let me know if you have any questions or find something that isn't calculated quite right, I'm off to dinner and will proof when I get back.

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Default Re: Need some explanations.

Tamm, it was nice talking to you on the phone briefly today, your evening. Do you really need or want to build your own driver when you can buy a diode already mated with a driver with wires already soldered on to it from DTR LPF (google search it to find his online store)?

Do you have a host to hold and heat sink the diode with batteries in mind yet?

Did you know that for the exact same output power green 520-532nm is about ten times brighter to the eye than either 450nm blue or 650nm red?
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Old 03-27-2016, 02:07 PM #15
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Default Re: Need some explanations.

These guys are really trying very hard to help you understand but some things just don't seem to be clicking in your understanding. Do you absolutely need to start with building a driver on your own? Why not start simpler and work your way up the the more complex. I would hate to see you blow diodes and components when you are just getting into it.
If I oh really want to build your own they really NEED to know what diode you are working with as it would GREATLY decrease the confusion as some things you mentioned about it don't make sense. Please help them help you and provide the information they are asking for
Best of luck with your build

I repped you guys for spending the time to try and help.
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Old 03-28-2016, 08:44 PM #16
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Default Re: Need some explanations.

Thanks to all of you guys. Its been busy last days. First OVNI. Your last explanation was very good, now im getting it at least i think that. I got only one question for you. You said that if i use potentiometer i dont want to anyone set it to 0 ohms. I undertund that gives highest current. But i was just wondering if i put in in series to 1 ohm resistor in that 1w 445nm diode case then its resistance adds to my 1 ohm. But 1 ohm shows to be enough for 1w laser, so if i set it 0 so theres only my 1 ohm resistor and its enough, in this case i cant kill my diode, am i right?

Alaskan, it was nice to talk with you too. Actually i can understund about 90% what you are saying, some words just dissapeared. But your question is really big question. Yes i thought to build one myself and in this case i know i cant fit it into host problably because that driver would be just so big. And i thought first power supply from my home electricity not batteries. But yes which would be better to do one yourself or buy one almost finished. Im now considering making 2 lasers, first gonna buy and make handheld and another making totally myself.
And yes i know that green light over 500 nm is the best visible to human eye. But i actually didnt know that difference would be like 10 times. And yes i could make also green laser but the problem is 520 seems pretty exotic. 445nm is only one and 405 nm too which is easy to get from ebay. I put this link again what i thought first:

1W 1000mW Laser Diode Blue 445nm to 18 5 6mm | eBay

445nm is only one with 405nm what diode i can get easily from ebay for very good price. My favourite would be 473 nm but it seems to be one of the most expensive. Also 520nm seems to be very rare. I cant buy mostly from your given webpages because the postage cost would be just too high. If i pay like 40 bucks for postage then it makes buying from there pointless. I checked eagles protective glasses and what i found. They cost 40 buck and postage is another 40 bucks. That makes it pointless. Best postage prices are usually in ebay. I wanna also thank for your offering to assist me.

Pman, yes, i m rally thankful for these guys. I think making one would be more fun than just buying and also helps to understund what im doing. And i also hope too i wont blow up any diode, im gonna make test for driver first so i hope it ensures i wont blow it. And i have already read many guides and i think im aware of those problems what might come. I also watched whole day guides from youtube what difference has paralleel and series resistors. I think i understund a lot better now. I can also put some links if someone wants to learn from them too. This video was really helpful and i recommend this guy to anyone, explains everything so well. Thanks again

https://www.youtube.com/watch?v=CZgqGTxL9cA
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