3D Time-of-flight distance measurement with custom - Universität ...

POWER BUDGET AND RESOLUTION LIMITS 93
contrast Cdemod, which depends on the pixel performance, (usually less than 100%),
have to be considered:
A = Cmod
⋅ Cdemod
⋅PEopt
Equation 4.15
For 100% modulated optical signal we obtain:
demod opt PE
C A ⋅ = Equation 4.16
Thus we can rewrite Equation 4.11:
L background + Npseudo
+ PEopt
∆ L = ⋅
Equation 4.17
8 2 ⋅ Cmod
⋅ Cdemod
⋅ PEopt
For the following example we assume a 100% modulated light source (Cmod=1) with
a total optical power of Popt=700 mW. The light source emits at 630 nm, where the
sensor has a quantum efficiency of 65%. We use a CS-mount lens with a focal
length of f=2.6 mm and a F/# of 1.0. The pixel size is 12.5 µm x 14.4 µm. With a
beam divergence of 50° (LED) we get an image size at the sensor plane of 4.6mm 2
({2.6 mm * tan25°} 2 ⋅π). With optical losses of lens (0.7) and interference filter (0.5)
we obtain klens=0.35. Choosing an integration time of Tint=25 ms we can easily
calculate the number of electrons Ne generated in one pixel by rearranging
Equation 4.7:
⎛
⎞
⎜
⎟
2
⎜ P light source ⋅ ρ ⋅ D ⋅ klens
⋅ QE(
λ)
⋅ λ ⋅ Tint
⎟ 1
Ne = ⎜
⎟ ⋅
Aimage
2 Equation 4.18
⎜
4 ⋅ ⋅ h ⋅ c
⎟ R
⎜
A
⎟
⎝
pix
⎠
Choosing a target with 20% reflectivity we get the number of electrons per pixel,
which now only depends on the distance R in meters (R/[m]):
Ne ≈ 170,000 photoelectrons / (R/[m]) 2 . Hence, for a distance of 5 meters, we will
integrate a number of 34,000 electrons in one pixel. The following overview
summarizes the chosen parameters.