Homemade Laser Driver with LM317

This is a simple and cheap homemade laser driver.

The LM317 is capable of up to 1.5A,

which makes it possible to use it for >1000mW laser diodes.

Build your own homemade laser driver

Used Parts

LM317 Voltage Regulator (rs-components)
1N4001 Diode (Reverse Voltage Protection)
10µF Capacitor
Resistors (Value depending on diode current)
Potentiometer (only necessary if you want to change the diode current)
Switch
Fuse

All parts together cost me around 2€. (~2.5$)

Calculate Resistors

The Resistor is to be calculated as follows:

Resistance = 1.25V / max. Current to Diode

The 1.25V is the Reference Voltage of the LM317.

Example:

maximum diode current is 250mA:

1.25V/0.25A = 5 Ohm

The Variable Resistor can be used to control the current to the diode.

The Current can be calculated with this formula:

I = 1.25V / Resistor + <Variable Resistor>

Insert the momentary Resistance of the potentiometer here.

61 Responses to "Homemade Laser Driver with LM317"

Lincoln says:

March 10, 2020 at 3:20 pm

Hi Mr. Langeder. Excuse me for my not soo good english, ’cause i’m from Brazil. I have a doubt, please. Your drive is very interesting about, current out, resistor values, LM 317 reference voltage calcutator, etc…
I made this drive last weekend, and a doubt is about the out voltage. When i used two batteries of 1,5 v, the out voltage is 2,69 volts. but when i use for example 9 v , the voltage out goes to aproximate around 8,48 volts. Can it cause damage the diodo laser from dvd rw 16 x. The diode works with 2,2 – 2,4 volts.
Thanks,

Reply
1. Michael Langeder says:
  
  March 17, 2020 at 12:10 pm
  
  Hello Lincoln!
  
  I understand your concern.
  
  The important thing about this circuit is, that it is a constant current, not a constant voltage supply.
  The diodes need a regulated current, because when you simply apply a certain voltage, the diodes will take whatever current they can get from that supply.
  
  In this case, make sure to apply a test load (resistor and rectifier-diodes in series – to simulate a laser diode)
  and measure the output current.
  
  ALWAYS discharge the output capacitors, connect the output load and only after that, connect the input power supply.
  This needs to be done, because when there is no load connected to the output, the circuit will still try to deliver the set current, thus increasing output voltage until the maximum is reached (8.48V in your case)
  If you would connect the diode now, the capacitor would discharge through the diode with a high current and most likely kill it in the process.
  
  I hope this helps clarify things a bit.
  
  Reply
  1. Lincoln says:
    
    March 17, 2020 at 1:37 pm
    
    Thanks very much Mr. . You clarified my doubt perfectly about voltage and current out. And the important detail about capacitor discharge that i didn’t know. You are great !!
    
    Reply
GARRETT says:

March 20, 2019 at 10:58 pm

Can I use a 10k potentiometer

Reply
1. Michael Langeder says:
  
  September 6, 2019 at 1:41 pm
  
  Yes, but it also depends on the current you want to set.
  
  The LM317 has a reference voltage of 1.25V, with 10kOhms the curcuit would output 1.25mA with the potentiometer maxed (neglecting the -necessary- fixed value resistor),
  using the formula for calculating the current stated above: I = 1.25V / Resistor +
  
  The fixed value resistor is necessary, because if you min the potentiometer, its resistance would be near 0Ohms, causing the current to rise considerably – I= 1.25V/near0 = alot A
  
  Reply
2. Stuart says:
  
  September 6, 2019 at 5:22 pm
  
  You could but it won’t be very precise. Also there are different types of pot.
  Linear and Logarithmic.
  Linear (Lin) or Logarithmic (Log) tracks. With linear potentiometers, the resistance between one end of the track and the wiper varies at a constant rate as the slider is moved along the track. In logarithmic types, the change in resistance is much less at one end of the track to the other.
  Also you can use a multiturn pot for more precision.
  
  Reply
Aman says:

November 9, 2018 at 1:05 pm

Sir,
Can we use 9v battery for Power up laser diode using this circuit.

Reply
1. Michael Langeder says:
  
  September 6, 2019 at 1:34 pm
  
  This depends on the current you want to supply the laser diode with.
  9V should be fine for most diodes, but a typical 9V battery can’t sustain a current of i.e. 1A
  
  Reply
Marco Burgman says:

June 22, 2017 at 4:22 pm

Is there any video which shows this design?

Reply
1. Michael Langeder says:
  
  June 26, 2017 at 11:28 am
  
  No, I only took photos.
  Is anything unclear?
  
  Reply
  1. Marco Burgman says:
    
    June 26, 2017 at 12:13 pm
    
    Not anymore, I understand it now. But I do have another question, my Laser diode is 300mW if I calculated it right, I need 9 ohm, but is it a problem if I do 10 ohm?
    
    Reply
    1. NetBUG says:
      
      April 23, 2018 at 5:16 pm
      
      Your diode will work at 95% power. That’s great, it’s more safe in case of any issues.
      Always try to use current below recommended if you cannot match the exact calculation.
      
      Laser diodes die very quickly if you exceed the maximum power.
      
      Reply
Eric Tyler Bettencourt says:

May 10, 2017 at 6:18 am

Would this driver work for a 2w laser?

Reply
1. Michael Langeder says:
  
  May 19, 2017 at 9:30 am
  
  It depends on the diode current. The LM317 can only supply 1.5 Ampere
  
  Reply
  1. Eric Tyler Bettencourt says:
    
    May 27, 2017 at 7:05 am
    
    The data sheet states a max of 1.8A, the resistor values I punched numbers for was for 1.7A. I thought the LM317T supplied an excess of 1.5A?
    
    Reply
    1. Michael Langeder says:
      
      May 27, 2017 at 12:51 pm
      
      Thats interesting. I had a closer look at the datasheet:
      
      http://www.st.com/content/ccc/resource/technical/document/datasheet/group1/a0/db/e6/9b/6f/9c/45/7b/CD00000455/files/CD00000455.pdf/jcr:content/translations/en.CD00000455.pdf
      
      Page 6 Table 4: The maximum output current I_O(max) ranges from 1.5A min. to 2.2A typical, depending on the batch your LM317 is from.
      The 1.5A is the guaranteed value for I_O(max).
      If you are lucky, it can supply 2.2A or maybe even a little more.
      
      You should also consult Figure 3 on Page 9 of the datasheet for your use case.
      Junction temperature and the input to output voltage difference also effect the maximum current you can draw.
      
      Reply
      1. Stuart Bassett. says:
        
        May 27, 2017 at 3:18 pm
        
        With a large heatsink and a fan it is possible to push the LM317 a bit more than you should, but I don’t know how long it would last.
        
        Reply
      2. Eric Tyler Bettencourt says:
        
        May 27, 2017 at 9:35 pm
        
        Ahhhh misread on my part, thank you for the insight my friend!
        
        Reply
Ava Hedayatipour says:

April 14, 2017 at 6:34 pm

Hi,
Thanks for all your explanation, it was extremely helpful!
I have the circuit and it works well, I am always monitoring my current and use the minimum allowable current to drive my laser so I am pretty sure I am not over driving it, the circuit works fines for days but sometimes it burns my laser very slightly, in the sense that the light gets dimmer!
I found out that if I don’t discharge my capacitor and chance the laser diode it would be burnt, do you have any similar experience?

Thanks!

Reply
1. Michael Langeder says:
  
  April 18, 2017 at 6:06 pm
  
  The capacitors (I assume you are refering to the capacitors right at the output) have to be discharged prior to connecting the laser diodes.
  When the circuit is powered, but not connected to a laser diode, the capacitors will be charged to approx. supply voltage. When the diode is connected then, the capacitor will immediatly discharge, hence damaging the laser diode through over-current.
  Adding a discharge resistor in parallel to the capacitor can help avoid this problem.
  
  Reply
Nguyen Thanh Tung says:

December 11, 2016 at 10:16 am

I did follow this, and test use led, voltage’s led are 8V… if use for LD can destroy LD?
Vcc 9,3V, R: 3,3ohm VR obout 5ohm

Reply
1. Michael Langeder says:
  
  January 29, 2017 at 12:20 pm
  
  Using an LED to test the circuit won’t work.
  Here’s a simple nearly-equivalent circuit:
  http://i.imgur.com/VPc2V2Q.jpg
  You can use this circuit to replace the LD for testing.
  
  Reply
Krasi says:

December 7, 2016 at 10:24 am

Hi, can the circuit be powered with 12V? I want to power red 200mW module from ebay… It does not have other specs – just power and wavelength. If I understand the principle correctly, it’s the current going through the laser diode that matters, right?

Reply
1. Michael Langeder says:
  
  January 29, 2017 at 12:02 pm
  
  12V is fine for this circuit.
  Yes, the diode current has to fit the diode.
  Try to find a data sheet for the used diode.
  If you can’t find any, you’ll have to slowly increase the current until you get a satisfying result. (But take care not to overheat it)
  
  Red laser diodes have a forward voltage of approx. 2.2V
  The operating current of a 200mW (output) diode would be somewhere between 100-150mA.
  I’d suggest starting with 50mA though.
  
  Reply
kimju says:

December 2, 2016 at 9:40 pm

Hello could you tell, me how can i do to increase it up to 4000mw ?

Reply
1. Michael Langeder says:
  
  January 29, 2017 at 12:27 pm
  
  With the LM317 IC its not possible.
  But there are other circuits for this purpose.
  Generally I’d suggest using a pulsed driver for high currents. (The LM317 is a linear regulator)
  There are many pre-assembled boards on ebay that are not too expensive.
  
  Reply
Aidan Ayoung says:

November 10, 2016 at 12:18 am

I can replace the 1N001 diode with a 1N004 diode, right?

Reply
1. Michael Langeder says:
  
  November 19, 2016 at 3:34 pm
  
  Yes. The 1N4001 has a maximum reverse voltage of 50V,
  the 1N4004 has 400V.
  Here’s a datasheet: http://pdf1.alldatasheet.com/datasheet-pdf/view/90515/VISHAY/1N400X.html
  
  Reply
Olafur says:

August 29, 2016 at 5:16 pm

Hi, I have laser that I need to build a driver for.
It is 650nm red 5mv and the output power is 2-5 mW and the power supply I have is 5v

Can you tell me how I can build similar driver as seen above ?

Thank you so much in advance : )

Reply
1. Michael Langeder says:
  
  September 21, 2016 at 3:58 pm
  
  The only thing you would have to change is the resistor, which sets the current.
  for a diode wih 2-5mW I would guess that the current would have to be be around 1-2mA
  
  Hope that this helps
  
  Reply
  1. olafur says:
    
    November 5, 2016 at 9:59 pm
    
    Thank you. It seems to work : )
    
    Reply
Viswa says:

May 18, 2016 at 8:18 am

I need a circuit which provides 1.7V and 37mA output. Please suggest.

Reply
1. Michael Langeder says:
  
  June 11, 2016 at 12:23 pm
  
  Does it have to be constant current or constant voltage?
  What load will you connect and whats the supply voltage?
  
  Reply
Stuart Bassett says:

February 17, 2016 at 3:56 am

Hi. I noticed in the photos of the driver circuit you have 5 resistors in a row between the pot and the regulator.
Was this to make up a certain value or for some other reason.

Thanks.

Stuart.

Reply
1. Michael Langeder says:
  
  March 2, 2016 at 5:11 pm
  
  I needed a 5 Ohm resistor, but only had 1 Ohm resistors at hand, they are just in series.
  
  Reply
  1. Stuart Bassett says:
    
    March 2, 2016 at 9:04 pm
    
    Ok, thankyou, I thought that may be the reason.
    
    Reply
  2. Zach Wald says:
    
    June 2, 2016 at 10:30 pm
    
    Did you mean 5 ohms then in the instructions? Because 1.25/.25 is 5 not 50. Or is there some conversion
    
    Reply
    1. Zach Wald says:
      
      June 2, 2016 at 10:53 pm
      
      Also what fuse are you using
      
      Reply
      1. Michael Langeder says:
        
        June 11, 2016 at 12:18 pm
        
        I am using a standard fast melting fuse,
        which actually isn’t proper for a laser diode, (It is still waaaayyy too slow to protect the diode, it might only save the LM317 from overheating in some cases)
        but I went with the cheap “solution” anyways.
        
        A proper protection circuit is adviced,
        although (so far) it works fine for me without one.
        
        Reply
    2. Michael Langeder says:
      
      June 11, 2016 at 12:13 pm
      
      I should write 5 “Ohm”, not 5Ohm, sorry for the confusion!
      
      Reply
Jacob Smith says:

November 30, 2015 at 6:53 pm

Could I leave out the fixed resistor, and just use one pot instead?

Reply
1. Michael Langeder says:
  
  January 7, 2016 at 11:43 am
  
  The fixed resistor is in there to make sure that the current does not get too high.
  If you turn the poti all the way down, its basically 0 Ohm, and the diode will get fried.
  
  Reply
chirs says:

March 9, 2015 at 10:17 am

Hi Michael, thanks for your share great project. However, may i ask you can i use this circuit to switching on/off laser diode around 10~20kHz ?

Reply
1. Michael Langeder says:
  
  April 30, 2015 at 12:23 pm
  
  Not as it is. I think there might be a way to modify the circuit, but I never tried.
  
  Reply
David W says:

January 26, 2015 at 2:09 pm

Hi, I want to power a laser diode that has typical operating voltage of 1.8v and current of 350ma. Does this circuit only operate at 1.25v?

Reply
1. Michael Langeder says:
  
  January 28, 2015 at 7:17 pm
  
  No, this curcuit operates with a constant current,
  the voltage depends on the resistance of the load and will be around 1.8V for you, if the current is set to 350mA.
  To calculate the value of the resistor, you only need the operating current of the diode,
  since the voltage will be according to it.
  
  The 1.25V is the reference voltage of the LM317 IC.
  
  Reply
David says:

January 18, 2015 at 2:43 am

Thanks for the article. It looks very easy but I can not seem to make it work. I left out the fuse and the pot but I dont see how that would make it not work. I am not an electronics person but I can use a soldering iron. I have tried to make it like numerous times and each time when I test it with a multimeter it still shows almost 9 volts coming through. Im using 2x10ohm resisters.
Please help.

Reply
1. Michael Langeder says:
  
  January 18, 2015 at 8:52 am
  
  Testing just with multimeters might not give you good measuring.
  I use a dummy laser diode (which is 6 N4001 diodes and a 1 Ohm resistor all in series) to test the curcuit.
  It simulates the behavior of a blue laser diode pretty well. I just connect that instead of the laser diode, so I can see if it works (by measuring the output current) without putting the real laser diode at risk in case the driver doesn’t work properly.
  
  I hope that helps!
  
  Reply
  1. David says:
    
    January 18, 2015 at 3:42 pm
    
    Thanks for the reply.
    Im new to all this stuff so could you tell me the components for testing a red laser driver circuit? I have a 400mW, 2.35v, 205mA red laser diode Im making the driver board for.
    Can you go into more details of how to use the testing circuit please. Sorry for so basic questions, I just have never made stuff like this.
    Thank you very much for all your help.
    
    Reply
    1. Michael Langeder says:
      
      January 18, 2015 at 4:16 pm
      
      To do some basic testing for a red diode, I’d use 4 N4001 (~0.7V*4=2.8V) and a 1 Ohm resistor.
      I would solder them together (all in series) on a small piece of pcb and connect it (instead of the real laser diode) to the driver curcuit. Then I measure the current through the dummy laser diode, which should be pretty much 1.25V/5Ohm=250mA, with 2x10Ohm in parallel (I assume), which by the way is too much for your diode.
      
      Since the curcuit is designed to opperate as a constant current source, using 4 N4001 and using 6 N4001 should give almost the same results (The color of the laser doesn’t matter for the current, only for the voltage)
      You could even short curcuit the output to the laser diode, and only have 250mA (or whatever you set it to) flowing from the driver.
      
      Reply
  2. David says:
    
    January 18, 2015 at 8:52 pm
    
    Thanks again for taking the time to help me.
    1. Should the line of diodes and resister be connected to the negative or positive out put from the driver?
    2. Everything I could find said 2 x 10 ohm resisters gave the driver 150mA. Is that right?
    Thanks again.
    
    Reply
  3. David says:
    
    January 18, 2015 at 9:29 pm
    
    Opps. Ment to type 250mA not 150mA.
    
    Reply
    1. Michael Langeder says:
      
      January 18, 2015 at 11:33 pm
      
      1. To both ends, between positive and negative
      2. How are the two 10 Ohm resistors connected?
      In parallel they would give 5 Ohm, resulting in 250mA,
      in series they would give 20 Ohm, resulting in 62.5mA.
      If you want 125mA, you would have to use a single 10 Ohm resistor. (1.25V/10Ohm=0.125A=125mA)
      
      Besides that, you might want to get a bit familiar with electronics before working on lasers. Not only are they really dangerous if handled improperly (giving it practically no space for experimenting), but also laser diodes break easily if anything is done wrong (including overcurrent, overvoltage (even small peaks), minimal reverse voltage, and so on) So having a good understanding of how each component of the curcuit works and behaves makes it much easier, cheaper, more fun and safer.
      
      I’m not saying that you can’t do it, but that it’s most likely better to first get some experience with not so dangerous stuff where you can just try and see what works and what doesn’t work, without putting that much at risk.
      
      Reply
ANAND AGRAWAL says:

January 11, 2015 at 11:13 am

Kindly tell me. If this circuit would work with IC-7805 instead of LM317. Reply Asap.

Reply
1. Michael Langeder says:
  
  January 13, 2015 at 1:12 am
  
  No, that doesn’t work. The 7805 is a fixed voltage regulator, meaning, that it doesn’t have a feedback pin (which is needed to sense the voltage on the shunt resistor – that gives a measure of the current)
  
  Reply
Tony G says:

February 22, 2014 at 7:00 pm

There’s one problem: since the current flows via the potentiometer, if the potentiometer is not designed for over 1.5W, it will overheat and burn out.

Reply
1. Michael Langeder says:
  
  February 22, 2014 at 7:48 pm
  
  Of course you need to use appropriate parts. (The fixed value resistor needs to be “big” enough as well)
  
  But be careful:
  If you have (lets say) 125mA using a potentiometer set to 10 Ohm,
  you will have a heating power of ~156mW.
  If you use an linear 100 Ohm potentiometer,
  that power needs to be dissipated on 1/10 of the potentiometer. (since its set to 10 Ohm)
  The potentiometer would then have to be capable of dissipating 1.56W in total.
  
  It’s usually better to use the maximum current of a potentiometer than the power to find the right parts.
  
  I hope thats helpful!
  
  Reply
Trigate says:

February 16, 2014 at 1:45 pm

Thanks!!!
Really clean and easy to understand shematics.

Reply
1. Michael Langeder says:
  
  February 16, 2014 at 4:30 pm
  
  You are welcome!
  
  Reply
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Homemade Laser Driver with LM317

Build your own homemade laser driver

Used Parts

Calculate Resistors

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Build your own homemade laser driver

Used Parts

Calculate Resistors

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