6.2 Noise Effects in Photodetectors
257
Example 6.11 Let the responsivity R = 0.90 A/W for an InGaAs photodetector
operating at 1550 nm. What is the NEP in the thermal-noise-limited case if the load
resistor R L = 1000 and T = 300 K?
Solution From Eq. (6.17) the value for NEP is
NEP = [4(1.38 × 10
−23 J/K)(300K)/1000 ]
1/2
/(0.90 A/W)
= 4.52 × 10
−12 W
√
Hz
Drill Problem 6.5 A particular silicon pin photodiode has an NEP = 1 ×
10
−13 W/Hz
1/2 . If the receiver operating bandwidth is 1 GHz, show that the
required optical signal power is 3.16 nW for a signal-to-noise ratio equal to 1.
The parameter detectivity, or D*, is a figure of merit for a photodetector used
to characterize its performance. The detectivity is equal to the reciprocal of NEP
normalized per unit area A.
D∗ = A
1/2
/NEP
(6.21)
Its units commonly are expressed in cm:
√
H z/W.
6.3 Response Times of Photodiodes
6.3.1 Photocurrent in the Depletion Layer
To understand the frequency response of photodiodes, first consider the schematic
representation of a reverse-biased pin photodiode shown in Fig. 6.7. Light enters
the device through the p layer and produces electron–hole pairs as it is absorbed
in the semiconductor material. Those electron–hole pairs that are generated in the
depletion region or within a diffusion length of it will be separated by the reversebias-voltage-induced electric field, thereby leading to a current flow in the external
circuit as the carriers drift across the depletion layer.
Under steady-state conditions, the total current density J tot flowing through the
reverse-biased depletion layer is
J tot = J dr + J diff
(6.22)
Here, J dr is the drift current density resulting from carriers generated inside the
depletion region, and J diff is the diffusion current density arising from the carriers
257
Example 6.11 Let the responsivity R = 0.90 A/W for an InGaAs photodetector
operating at 1550 nm. What is the NEP in the thermal-noise-limited case if the load
resistor R L = 1000 and T = 300 K?
Solution From Eq. (6.17) the value for NEP is
NEP = [4(1.38 × 10
−23 J/K)(300K)/1000 ]
1/2
/(0.90 A/W)
= 4.52 × 10
−12 W
√
Hz
Drill Problem 6.5 A particular silicon pin photodiode has an NEP = 1 ×
10
−13 W/Hz
1/2 . If the receiver operating bandwidth is 1 GHz, show that the
required optical signal power is 3.16 nW for a signal-to-noise ratio equal to 1.
The parameter detectivity, or D*, is a figure of merit for a photodetector used
to characterize its performance. The detectivity is equal to the reciprocal of NEP
normalized per unit area A.
D∗ = A
1/2
/NEP
(6.21)
Its units commonly are expressed in cm:
√
H z/W.
6.3 Response Times of Photodiodes
6.3.1 Photocurrent in the Depletion Layer
To understand the frequency response of photodiodes, first consider the schematic
representation of a reverse-biased pin photodiode shown in Fig. 6.7. Light enters
the device through the p layer and produces electron–hole pairs as it is absorbed
in the semiconductor material. Those electron–hole pairs that are generated in the
depletion region or within a diffusion length of it will be separated by the reversebias-voltage-induced electric field, thereby leading to a current flow in the external
circuit as the carriers drift across the depletion layer.
Under steady-state conditions, the total current density J tot flowing through the
reverse-biased depletion layer is
J tot = J dr + J diff
(6.22)
Here, J dr is the drift current density resulting from carriers generated inside the
depletion region, and J diff is the diffusion current density arising from the carriers