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vl53l0x sensor detection distance inquiry

pcbsmt
Associate II

(A translator was used for this inquiry.)

Hi,

I am using the vl53l0x sensor.
I would like to inquire about the detection range of the sensor.

  1. What value do you get when the wall you want to measure the distance against is tilted? (Refer to image1)
    What point distance do you get if the minimum length between the sensor and the wall is 100 mm and the maximum is 200 mm?
  2. I know that the maximum measurement distance of the sensor is 2M.
    I would like to know the range of the detectable area when the wall and the sensor are 2M away.
    (Refer to image2)

Thank you.

1 ACCEPTED SOLUTION

Accepted Solutions
John E KVAM
ST Employee

The cone of light - or it' Field of View is 27 degrees. 

With a little trig one sees that the diameter of the illumination circle is 1/2 the distance. 

The astute reader will do the math and get 0.44 but it's a question of accuracy. 

Very near the sensor the FoV is wider, and it narrows as the distance increases.

So, go with 1/2 and you will be as good as you are going to get.

- john


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View solution in original post

3 REPLIES 3
John E KVAM
ST Employee

In the first image when you have a slanted wall, you will get the average of all photon return times.

The trouble is that near objects return more photons than a far object. So instead of the mathematical average, you will see a number weighted toward the closer distance. 

Unfortunately, how much of this effect you see will depend on the reflectivity of the wall.

 

As far as a wall at 2M what you get will depend on the reflectivity of the wall. If it's bright enough to return photons you will get 2M and status =0. A less reflective will will give you Staus =2 (low signal). With a REALLY bright target, beyond 4M you will get a status 4. Status 4 means you have a wrap around issue. Wrap-around (also called Aliasing) is a condition where the photons from pulse A get returned after the pulse of pulse A+1. So with a bright wall beyond 4M, you get a Range Status =4 and the correct distance minus the 2M wrap around point. 

If this bothers you, a pin-compatible part (VL53L3CX) is available. It's an L0 part with better mathematics.  


Our community relies on fruitful exchanges and good quality content. You can thank and reward helpful and positive contributions by marking them as 'Accept as Solution'. When marking a solution, make sure it answers your original question or issue that you raised.

ST Employees that act as moderators have the right to accept the solution, judging by their expertise. This helps other community members identify useful discussions and refrain from raising the same question. If you notice any false behavior or abuse of the action, do not hesitate to 'Report Inappropriate Content'

Thank you for your kind reply.

Regarding question 2, I am curious about the detectable area.

The sensor's detection range appears to be gradually expanding in a cone shape.
Then, I would like to know the area of ​​the detectable area from a distance of 2M.
(Part marked with a question mark in Image 2)

 

Thank you.

John E KVAM
ST Employee

The cone of light - or it' Field of View is 27 degrees. 

With a little trig one sees that the diameter of the illumination circle is 1/2 the distance. 

The astute reader will do the math and get 0.44 but it's a question of accuracy. 

Very near the sensor the FoV is wider, and it narrows as the distance increases.

So, go with 1/2 and you will be as good as you are going to get.

- john


Our community relies on fruitful exchanges and good quality content. You can thank and reward helpful and positive contributions by marking them as 'Accept as Solution'. When marking a solution, make sure it answers your original question or issue that you raised.

ST Employees that act as moderators have the right to accept the solution, judging by their expertise. This helps other community members identify useful discussions and refrain from raising the same question. If you notice any false behavior or abuse of the action, do not hesitate to 'Report Inappropriate Content'