# Electronic – Decomposing Kinect Laser Projector – Is it really safe

kinectlasersafety

I am looking at Kinect Laser and couple of things that are not clear, so I thought I bring up here.
First the known facts:
– Kinect has a laser projector that works at 840nm
– Output power of the laser is 60mW
– The laser projector has a diffuser. I didn't break one and look inside but my guess is laser diode->collimator->diffuser is the path.
– Kinect is a Class 1 safety device.

My primary interest is understanding Kinect's laser safety. For interested parties, the following links discuss some of these stuff and I will not repeat them here.
– A very informative link
– An independent Kinect Laser output measurement

My approach is as follows:
– Assume 60mW output power is correct
– Diffuser efficiency is 50% and therefore 50% of the energy is lost
– Diffuser output projects a rectangle image with angles of 45 degrees on both axis (vertical and horizontal)[Probably the angle is different but good enough for us to make some calculations)
– Diffuser output is 20mm away from the Projector window (this is the glass in front of the laser diode, light output leaves this window and projects the image)

Now, given 45 degree angle, the size of the rectangle at the output of the window is still 30mW spread over the size of the rectangle. (the size of the rectangle is tan(45)*20mm*2=40mm one side, 1600mm2 area of the light rectangle at the window.) Assuming uniform energy distribution, we get 30mW/1600mm2 = 1.875mW/cm2

This looks high to me. I am trying to figure out why it is given Class 1.

Now the questions:

• This light is not collimated, in fact is diverging, if you think of your eye, the eye will focus only a very small portion of this light to the retina. (model for eye is just a lens with focal length of 17mm) Assuming pupil size is 7mm, the area of the pupil is 0.39cm2. So the energy enters the eye would be 18.75*0.39=7.3mW.
If this was a straight line, you would end up blind since the lens would collimate that straight line, however in this case, the light comes with 45 degree angle, therefore eye will only be focusing on a very small portion of this energy properly. I am trying to figure out is this the reason why Kinect denoted a Class 1 certification. Even if I am off by 50%, there is still tons of energy coming out of Kinect laser. I did more checking on this, using the attached pdf, one can conclude the beam from Kinect will have an area of d=ftheta, d=1.7(45 degree in radians)=1.7*785mrad->1.335cm. This is the dia1meter of the spot on your retina. The area is a=pi*d^2/4=1.4cm2 so the retinal irradiance becomes E=7.3mW/1.4cm2=5.2mW/cm2. This still seems exceptionally high to me, even if it is diverging beam.

• My second suspicion is that, Kinect is indeed dangerous however no sane person would stick his eye to the projector and at normal working distance (say 1meter) the energy entering to the eye would be very small (tan(45)*1000*2=2000mm, area is 4m2. 30mW/4m2 =7.5mW/m2=7.5 10e-4 mW/cm2 which is safe for eye. However if this is how Kinect got the Class 1 certification, it is pretty scary, since I am pretty sure some idiot out there will stick his eye to see what is inside.

Any insights would be appreciated..

[Update]
I have done some measurements today.. Here they are:

Using a PD (Osram BPW34FA) and a 10K resistor. I manually touch to the surface of the projector's window with the PD and 10K resistor. I measure voltage across resistor using a scope. The result is approximately 500mV. (I took out the 4mW ambient measurement, it was 504 but when Kinect isn't there it is 4mW, so the delta due to Kinect is 500mW)

Now, 500mV translates to 500uA current on this diode. (500mV/10K=500uA). At 7mm2 (assuming pD is uniform energy absorption, which is a reasonable assumption) per mm2, I get 71uA. The diode's efficiency is 0.65A/W, I simply use this to calculate incoming power per mm2, it translates into 109uW/mm2.

I also measured the angles of the diverging pattern. They are 50 degrees vertically and 54 degrees horizontally. (Use a sony cam in nightshot mode and a ruler. The patter is not perfect rectangle but rectangle is a reasonable assumption)

So, assuming the output of the diffuser is 12 mm away from the window, you get an approximate area of 12*tan(50)= 14.4mm and similarly 17.7mm. So the total area of the beam at the window is 254mm2. Since we established per mm2, this thing emits 109uW, total energy output becomes 27.6mW, which is inline with 60mW laser output with 50% optical efficiency. I didn't break the Kinect to measure the depth of the diffuser output but I guess it could easily be 1mm, in this case the output power translates to 18mW at the window. Anyway it is high and I am still not understanding why it is Class 1. Hope you guys can help.

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.