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This question already has an answer here:

Background: I am building a laser that is running off of an M140 M-Type 2+ W 445 nm laser diode and an adjustable constant current driver.

Here are the specifications for both the driver and the diode as stated by the seller.

DRIVER
8 - 12V Operation
Adjustable Constant Current Output up to 5A (preset to 500mA)
Suitable for Laser Diodes up to 7W+
0 - 5V TTL/PWM Input Control
0 - 150kHz Frequency
2x 12V Fan Driver
Delayed Startup - <10 Seconds
Includes 2-pin JST-XH Crimp Connectors
Driver is ON by default when powered. 

DIODE
This listing is for ONE Brand New Nichia M140 M-Type Japan 445nm 2+ 
Watt 5.6mm Laser Diode mounted in 12mm x 30mm Metal Housing Module 
w/ G-2 Glass Lens, ~6" 22AWG Silicone Leads. The diodes are  
meticulously extracted from a new laser diode array. Recommended   
current is 1 to 1.6 Amps. A constant current driver must be used to 
power the diode. Do not connect directly to battery or unregulated 
power supply! 

As of now, I have the diode in module with a lens and leads, the driver, and a heat sink which is currently on its way.

Problem:

What I need to do now is supply power to the driver. Since the driver is operational up to 12V, I am considering running 3 18650 batteries rated at 3.7V in series. This will give me a total voltage of 11.1 volts. My concern, however, is that I am unsure if the current of these batteries running in series matters. I know that the current remains the same, however, does the input current need to match the preset current on the driver (500mA)? Does the input current need to be more? If so, how much more? What is a safe amount of current that can be driven to the driver so that it does not break down. I am having trouble understanding how a constant current driver steps down the incoming current without messing up the driver. I apologize if these questions seem silly, but I am genuinely interested in this project, and I would like to power it properly and safely.

Also, how can I calculate the output power of the laser diode? If Watts is equal to voltage * amps, and the driver is sending 500mA to the diode with 11.1 volts, the output power SHOULD be 5.55 Watts. However, this doesn't sound right since the diode is rated at 2 Watts AND since the seller recommends a current between 1.0 to 1.6A. Is my calculation wrong?

Note: My major is in chemical engineering. I am very unfamiliar with electronics. I have tried my hardest to learn everything there is to know in order to successfully build a high-powered laser. If there is anything wrongly stated, please correct me.

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marked as duplicate by brhans, Anindo Ghosh, Warren Hill, Elliot Alderson, Dwayne Reid 14 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ "the current rating of a power supply must be at least what the device wants but there is no harm in it being higher. A 9 volt 5 amp supply is a superset of a 9 volt 2 amp supply, for example." This was very helpful from the post. Thank you. Am I allowed to keep this post up as I am also concerned with other methods that can power my laser, and since other questions are remained unanswered by the thread you referred me to? Thank you. \$\endgroup\$ – Andrew J Padilla Jan 6 at 16:46
  • \$\begingroup\$ You can edit your question to remove the part that was answered by the old question. \$\endgroup\$ – The Photon Jan 6 at 19:15
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The laser diode looks kind of like an LED, but you probably want to be driving the thing with a constant current source set to somewhere around 1-1.3A or so (1.6A would be really pushing it), which your laser diode driver can do (There is probably a trimmer to set the diode current).

The current mode driver will draw only however much current it needs from the power supply that might be either slightly more then the laser current setpoint (Linear driver), or less (switched mode driver or switching pre regulator), but in any case it will only take what it needs.

At 1.3A the diode will probably be dropping about 5V or so (But this is temperature and phase of the moon dependent, which is why laser diodes of this sort are current driven), and so the INPUT power will be about 5-7W, of which somewhere between 1 and 2W will exit as coherent radiation, the rest will come out as heat, so you will need heatsinking to cool the diode assembly.

If your driver is a linear current regulator then the laser driver will be dissipating your (11V supply - the diode voltage (say 4.5V)) = 6.5V * the diode current, so will also potentially need a heatsink (There is a reason big lasers use switched mode drivers).

Now a word about safety: 2W of laser is class 4 (in the old money) which means that both specular and diffuse reflections from surfaces are hazardous to vision (Never mind the direct beam). Bolt the thing down to an optical bench and use appropriate ANSI certified eye safety gear (and shielding with interlocks so that a stray beam cannot blind anyone ELSE) and you are all good, but please, nearly nobody needs a handheld class 4 laser. Gear at this power is NOT a toy, no matter what some seem to think.

Finally, get yourself over to www.photonlexicon.com where the real laser nerds hang out, FAR better for serious advice then here.

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