Electronics > Projects, Designs, and Technical Stuff
Laser diode efficiency
Zero999:
--- Quote from: james_s on February 12, 2019, 06:30:56 pm ---I don't see how you're gonna mark steel with even a true 15W visible diode laser, most of the energy will just bounce off. 15W far IR from a CO2 laser might do it. In order to mark something, it has to absorb the energy, most metals make pretty good reflectors at visible wavelengths.
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Metals are even more reflective at infra red, than visible. The reason metals reflect light so well is because they conduct electricity. As a general rule, at lower frequencies, there are less losses, so more reflectivity. To melt steel the frequency needs to so high, the free electrons in the metal can't move fast enough, so deep UV might do.
james_s:
I've seen steel being cut by a CO2 laser but I don't know what the power was, quite a big one though. According to this https://fslaser.com/Applications/Metal a 90W CO2 laser can mark steel directly without using a special coating although they mention a fiber coupled laser is better I'm not sure what wavelength fiber laser they're referring to. CO2 is 10,600nm so very far IR.
Either way a 15W laser isn't gonna engrave steel unless maybe with a marking coating.
ajb:
--- Quote from: duak on February 12, 2019, 06:25:17 pm ---I prepared a few of the laser safety reports and applications to the CDRH to get approval to sell our equipment in the US. There are a lot of misconceptions about what Brightness really means in physics. (hopefully I have the correct terminology) Brightness is the power intensity received divided by the solid angle subtended by the emitter. In other words, how much power received per area per the apparent size of the source. This is very important for laser safety. For an unfocussed system, ie. no lenses, it doesn't matter too much if the power falling on a surface comes from a point source or from something much larger. For a focussed system, such as the eye, the lens can concentrate the power in a much smaller area. A laser has a very small exit aperture and the solid angle it subtends is very small. This means that the power intensity where an image has been formed is very high and can cause damage. Laser beams usually have small divergence and so any spot formed is also quite small and can also be a hazard.
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"Brightness" can mean a lot of things, but is typically a photometric description rather than a radiometric one, since the apparent brightness of a source depends on its spectral content in conjunction with the spectral sensitivity of the human eye. In physical terms it refers to the total luminous flux as well as some combination of the emitting or receiving area and emitting or receiving solid angles.
Generally laser safety considerations are based on irradiance (incident *power* per unit area), without regard for the emission or reception angles (except insofar as the emission angle determines the irradiance as a function of distance). The angles do have an impact on the risk of injury, but in most applications it's not worth splitting hairs about. A multi-watt beam with an initial diameter of a couple of mm and only a few mrad of divergence obviously maintains a rather high irradiance for quite some distance. (Pulsed laser safety measurements must consider pulse energy and repetition rate, which is not irradiance per se, but still the angle is not generally a significant factor.) You could certainly produce a more accurate hazard model with a more sophisticated optical model, but no one wants to be responsible for the empirical aspects of that study. . .
--- Quote --- I don't know what sort of sensor could accurately measure laser radiation that can ablate metal. Perhaps a defocussing lens to spread the beam out and a larger area sensor? Large area silicon photodiodes have a fairly well defined response to radiation of various wavelengths and produce a photocurrent proportional to incident power so it should be possible to build a power meter with available parts.
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Power sensors for up to 120kW are available, but obviously not cheap! A lot of the magic is in the absorber, with high power sensors using cone-shaped deflectors and annular absorbers to spread the incident power out over a large area. A crude measurement can be made using an integrating mass and a temperature sensor, as long as you can calibrate the absorbance and thermal mass (here is a very DIY example). The trouble with photodiodes is the damage threshold is usually low and poorly specified. You can send only a fraction of the beam to the diode (reflective sampler or attenuator), but then you have to be able to figure out what that fraction is!
--- Quote from: james_s on February 12, 2019, 06:30:56 pm ---I don't see how you're gonna mark steel with even a true 15W visible diode laser, most of the energy will just bounce off. 15W far IR from a CO2 laser might do it. In order to mark something, it has to absorb the energy, most metals make pretty good reflectors at visible wavelengths.
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CO2 won't mark plain metals well (at least not until the ~100W level), they're too reflective even at 10um. I believe most marking lasers are UV fiber (~300nm?).
--- Quote ---Regarding pointers, in the US I believe the legal limit has always been 5mW but nobody seems to enforce that. You can easily buy pointers from China that are well over 100mW.
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Basically, but this covers products, not necessarily components.
james_s:
You can buy fully assembled ready to use laser pointers in the hundreds of mW range all over ebay and various direct from China sellers. Totally illegal in the US but I've bought a few and never had any issues getting them. Fortunately I'm sensible enough to not go pointing them around carelessly but a lot of people aren't.
jolshefsky:
When you're talking efficiency, you're asking about power-in versus power-out, right? So for practical accuracy, P[in] = P[out] + P[loss], and P[loss] is going to be heat, so why not measure the electrical power input and the thermal power output (e.g. heat sink dunked in known quantity of water; measure temperature change over time; calculate power) and subtract?
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