What Olin says technically (as almost always :-)). Also, as he says - you need to be clearer in describing your need if you want a good answer.
Lots of circuits here plus various pictures (this is just a Google image search - but useful). Be aware that the quality of the offerings will range from very good through rubbish. Caveat Emptor.
Here are a few samples from above. No guarantees !!!
This is a superb LASER resource - hav a good look at what it offers. Sam's LASER FAQ is a long time well known site with much good information. Here is his Diode LASER power supplies page.
From the above, see at the end of this post "Care and feeding of LASER diodes".
Commercial LASER Diode Driver PCBs. These LOOK like they could be modestly priced if they chose to make them so - but they are probably horriobly over priced.
This is for information only - worth a look. It shows what goies into a commercial LASER diode driver block diagram - scary stuff.
From Sam's LASER Diode page - reformated:
Care and feeding of LASER diodes.
The following must be achieved to properly drive a laser diode and not ruin it in short order:
Absolute current limiting. This includes immunity to power line transients as well as those that may occur during power-on and power-off cycling. The parameters of many electronic components like ICs are rarely specified during periods of changing input power. Special laser diode drive chips are available which meet these requirements but a common op-amp may not be suitable without extreme care in circuit design - if at all.
Current regulation. Efficiency and optical power output of a laser diode goes up with decreasing temperature. This means that without optical feedback, a laser diode switched on and adjusted at room temperature will have reduced output once it warms up. Conversely, if the current is set up after the laser diode has warmed up, it will likely blow out the next time it is switched on at room temperature if there is no optical feedback based regulation.
Note that the damage from improper drive is not only due to thermal effects (though overheating is also possible) but due to exceeding the maximum optical power density (E/M field gradients?) at one of the end facets (mirrors) - and thus the nearly iSnstantaneous nature of the risk.
The optical output of a laser diode also declines as it heats up. This is reversible as long as no actual thermal damage has taken place. However, facet damage due to exceeding the optical output specifications is permanent. The result may be an expensive LED or (possibly greatly) reduced laser emission.
I see a fundamental units problem in your math. You are using the symbol m which stands for milli (10^-3). The actual energies are µ (10^-6). You are off by a factor of 10^-3. Since most keyboards do not have a symbol for µ (micro), people often revert to an m (milli), and end up confusing the units. Hey, they both start with the same letter, right? :)
Additionally, there is the question of "wall plug efficiency". While a diode may 60mW, even military grade lasers only have a 10-15% wall plug efficiency. Thus that means an output of around 6mW. Then any optical element will reduce power by around 50% for each element. Assuming at most 2 optical elements (nu-naturally low number), that means the output can at max be 1.5mW.
In the answer you quoted, use this paragraph as a point of comparison:
For the sake of comparison, sunlight is one kilowatt per square meter and perhaps 5% of that is near infrared i.e. 700 to 1000 nanometers. Just going outside will expose you to much greater power densities of SWIR than the Kinect.
Also, remember that even though the generator is 60 mW (yes, I used the correct units), there is a series of diffusers, optics, and such so that at the very extreme of the exit aperture, the power density is <25 μW (again, note the symbol). The series of steps required to get at the 60 mW generator would indicate a willful intent to cause self harm, and be beyond simple mechanical failure.
Your initial assumption is incorrect.
My approach is as follows: - Assume 60mW output power is correct - Diffuser efficiency is 50% and therefore 50% of the energy is lost
The diffusers and optics reduce the power to <25 μW at the aperture. Run your math with that figure and you'll have an accurate representation.
Best Answer
The 1N4001 diode is included for reverse-power protection. As the linked site says, the diode "protects LD if batteries are inserted the wrong way round".
In normal operation, the diode is reverse biased and will have very little effect on the circuit operation.
But if you accidentally miswired something, or hooked up your power backwards so that you were trying to push current the wrong way through the laser diode, the 1N4001 diode could save the laser from being destroyed (assuming your power source has enough internal resistance that it doesn't just blow up the 1N4001 and the laser).
Laser diodes are generally optimized for efficient light output rather than ability to withstand high reverse voltages, so protection circuits like this are often needed to improve the reliability of laser diode circuits.