Light Characterization for VL53L8CH VCSEL

[JGarrettW]

Hello,

I am trying to do some ray tracing simulation for your VL53L8CH in a specific environment. I was wondering if you have any data beyond the "Field of Illumination" diagram for your VCSEL emitter (I assume you make your own emitters in-house). An "IESNA TM25" or "SPEOS" ray file would be amazing, but even just a maximum output power in lumens, watts, candelas, etc. would be helpful. That way I can at least know what the percentages in the "Field of Illumination" diagram mean. Also, if you have any detailed CAD on your chip that would allow me to place the emitter in the correct location on the sensor, that would be helpful as well. I have recreated it from your datasheet, but that doesn't have details on how deep the VCSEL is or what shape it is.

Thanks,

Garrett

[Anne BIGOT]

Hello Garrett

ST is providing Field of Illumination datas only.

Anyway, you can find CAD design on st.com with a 3D STEP file with cones. 
For the VL53L8CH, you can find it here. Choose SamacSys model and you will be able to download the 3Dfile.

Hope this help,

[JGarrettW]

Do you at least have what maximum power the field of illumination is referring to? 70% doesn't give me enough information to know how much loss to expect in certain environments.

[John E KVAM]

Knowing what the sensor can do in the presence of sunlight is a really tricky question. 

If the sun is directly behind the target, yet somehow is not shaded by the target, the loss is pretty high depending on the angle of the sun. Worst case is having the sun behind the sensor and illuminating the target. 

So, it's not only the Lux level, but the angle of the sun. 

The reason is we have an array of Single Photon Avalanche Diodes. And every photon that hits them gets counted. 

the sensor works because everything except the 940nm light gets filtered out.

But there is a non-zero reset time. 

So, if the sun - which generates a huge amount of 940nm light - saturates the max value the SPADs can receive, there is no bandwidth for the real signal. 

Oddly, if the sun is 90 degrees to the sensor, the results are better. 

But all of this depends on the reflectivity of the target. One wants a reflective target to get a good signal, but that reflectivity also reflects the sunlight. So, the sweet-spot is kind of hard to find.

Best I can tell you is to experiment a lot and see if it works for you. 

- john

[JGarrettW]

This is a zero-sun environment. What I am simulating is loss in an optical system (fiber optic cables, mirrors, etc.). For that, I need to know the initial power sent out by the VCSEL, the wavelength (940 nm), and the characteristics of my equipment.

[John E KVAM · Accepted solution]

I don't have any open data on that. What concerns us is the amount of energy that can pass through a 7mm aperture at a distance of 10cm. The 7mm is about the opening in your eye, and the 10cm is about the closest point you can focus. 

(We did not pick these numbers; the FDA and European heath organizations did.)

The trick was to not exceed the danger point, assuming the energy coming through that aperture was focused by your eye into one spot on the retina. 

It is this power that defines Class 1, and we are about, "20 times below the AEL of Class 1".

The output power of the VCSEL hardware itself, is kind of a closely guarded bit of competitive information. 

It's not something we can post on a forum. 

Sorry about that.

- john

[JGarrettW]

Ah, I understand the need to keep it guarded. Thanks for explaining it to me. That at least gives me a rough ballpark for simulation.