6.1 Operation of Photodiodes
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the optical power level as this photon flux passes through an incremental distance dx
in the semiconductor is given by dP(x) = −α s (λ)P(x)dx, where α s (λ) is the photon
absorption coefficient at a wavelength λ. Integrating this relationship gives the power
level at a distance x into the material as
P(x) = P in exp(−α s x)
(6.1)
Figure 6.1 gives an example of the power level as a function of the penetration
depth into the intrinsic region, which has a width w. The width of the p region
typically is very thin so that little radiation is absorbed there.
Example 6.1 If the absorption coefficient of In 0.53 Ga 0.47 As is 0.8 μm
–1 at 1550 nm,
what is the penetration depth at which P(x)/P in = l/e = 0.368?
Solution From Eq. (6.1)
P(x)
P in
= exp(−α s x) = exp(−0.8x) = 0.368
Therefore
−0.8x = ln 0.368 = −0.9997
which yields x = 1.25 μm.
Example 6.2 A high-speed In 0.53 Ga 0.47 As pin photodetector is made with a depletion
layer thickness of 0.15 μm. What percent of incident photons are absorbed in this
photodetector at 1310 nm if the absorption coefficient is 1.5 μm
–1 at this wavelength?
Solution From Eq. (6.1), the optical power level at x = 0.15 μm relative to the
incident power level is
P(x)
P in
= exp(−α s x) = exp[(−1.50)0.15] = 0.80
Therefore only 20% of the incident photons are absorbed.
Drill Problem 6.1 An InGaAs pin photodetector has an absorption coefficient
of 1.0 μm
−1 at 1550 nm. Show that the penetration depth at which 50% of the
photons are absorbed is 0.69 μm.
When the energy of an incident photon is greater than or equal to the bandgap
energy E g of the semiconductor material, the photon can give up its energy and excite
an electron from the valence band to the conduction band. This absorption process
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the optical power level as this photon flux passes through an incremental distance dx
in the semiconductor is given by dP(x) = −α s (λ)P(x)dx, where α s (λ) is the photon
absorption coefficient at a wavelength λ. Integrating this relationship gives the power
level at a distance x into the material as
P(x) = P in exp(−α s x)
(6.1)
Figure 6.1 gives an example of the power level as a function of the penetration
depth into the intrinsic region, which has a width w. The width of the p region
typically is very thin so that little radiation is absorbed there.
Example 6.1 If the absorption coefficient of In 0.53 Ga 0.47 As is 0.8 μm
–1 at 1550 nm,
what is the penetration depth at which P(x)/P in = l/e = 0.368?
Solution From Eq. (6.1)
P(x)
P in
= exp(−α s x) = exp(−0.8x) = 0.368
Therefore
−0.8x = ln 0.368 = −0.9997
which yields x = 1.25 μm.
Example 6.2 A high-speed In 0.53 Ga 0.47 As pin photodetector is made with a depletion
layer thickness of 0.15 μm. What percent of incident photons are absorbed in this
photodetector at 1310 nm if the absorption coefficient is 1.5 μm
–1 at this wavelength?
Solution From Eq. (6.1), the optical power level at x = 0.15 μm relative to the
incident power level is
P(x)
P in
= exp(−α s x) = exp[(−1.50)0.15] = 0.80
Therefore only 20% of the incident photons are absorbed.
Drill Problem 6.1 An InGaAs pin photodetector has an absorption coefficient
of 1.0 μm
−1 at 1550 nm. Show that the penetration depth at which 50% of the
photons are absorbed is 0.69 μm.
When the energy of an incident photon is greater than or equal to the bandgap
energy E g of the semiconductor material, the photon can give up its energy and excite
an electron from the valence band to the conduction band. This absorption process