6.5 Summary
265
6.3 If the absorption coefficient of silicon is 0.05 μm
–1 at 860 nm, show that the
penetration depth x at which P(x)/P in = l/e = 0.368 is equal to 20 μm.
6.4 A particular InGaAs pin photodiode has a bandgap energy of 0.74 eV. Show that
the cutoff wavelength of this device is 1678 nm. [Thus, this GaAs photodiode
will not respond to photons that have a wavelength greater than 1678 nm.]
6.5 An InGaAs pin photodiode has the following parameters at 1550 nm: i D = 1.0
nA, η = 0.95, and R L = 500 . The incident optical power is 500 nW (–33
dBm) and the receiver bandwidth is 150 MHz. (a) Show from Eq. (6.6) that
the primary photocurrent is 0.593 μA. (b) Show that the noise currents given
by Eqs. (6.12), (6.13), and (6.15) are as follows:
σ
2
shot = 2qi p B e = 2.84 × 10
−17 A
2
σ
2
dark = 2qi D B e = 4.81 × 10
−20 A
2
σ
2
th =
4k B T
R L
B e = 4.85 × 10
−15 A
2
6.6 Consider an avalanche photodiode receiver that has the following parameters:
dark current i D = 1 nA, quantum efficiency η = 0.85, gain M = 100, excess
noise factor F = M
1/2 , load resistor R L = 10
4
, and bandwidth B e = 10 kHz.
Suppose a sinusoidally varying 850-nm signal having a modulation index m
= 0.85 falls on the photodiode, which is at room temperature (T = 300 K).
Show that the responsivity is 0.58 A/W.
6.7 Consider an avalanche photodiode receiver that has the parameters listed in
Problem 6.6. To compare the contributions from the individual noise terms
to the SNR given by Eq. (6.9), examine each of the noises independently in
terms of the incident optical power P in . Show that for this particular set of
parameters, the relative contributions to the SNR are as follows:
S N R shot =
i
2
s (t)
i
2
shot (t)
= 6.565 × 10
12 P in
S N R dark =
i
2
s (t)
i
2
dark (t)
= 3.798 × 10
22 P
2
in
S N R th =
i
2
s (t)
i
2
th (t)
= 7.333 × 10
22 P
2
in
6.8 A given InGaAs avalanche photodiode has a quantum efficiency of 90% at
a wavelength of 1310 nm. Suppose 0.5 μW of optical power produces a
multiplied photocurrent of 8 μA. Show that the multiplication M = 16.
6.9 Suppose an avalanche photodiode has the following parameters: i D = 1 nA, η
= 0.85, F = M
1/2 , R L = 10
3
, and B e = 1 kHz. Consider a sinusoidal 850-nm
signal, which has a modulation index m = 0.85 and an average power level P in
= –50 dBm, to fall on the detector at room temperature. (a) Using Eq. (6.16)
show that
265
6.3 If the absorption coefficient of silicon is 0.05 μm
–1 at 860 nm, show that the
penetration depth x at which P(x)/P in = l/e = 0.368 is equal to 20 μm.
6.4 A particular InGaAs pin photodiode has a bandgap energy of 0.74 eV. Show that
the cutoff wavelength of this device is 1678 nm. [Thus, this GaAs photodiode
will not respond to photons that have a wavelength greater than 1678 nm.]
6.5 An InGaAs pin photodiode has the following parameters at 1550 nm: i D = 1.0
nA, η = 0.95, and R L = 500 . The incident optical power is 500 nW (–33
dBm) and the receiver bandwidth is 150 MHz. (a) Show from Eq. (6.6) that
the primary photocurrent is 0.593 μA. (b) Show that the noise currents given
by Eqs. (6.12), (6.13), and (6.15) are as follows:
σ
2
shot = 2qi p B e = 2.84 × 10
−17 A
2
σ
2
dark = 2qi D B e = 4.81 × 10
−20 A
2
σ
2
th =
4k B T
R L
B e = 4.85 × 10
−15 A
2
6.6 Consider an avalanche photodiode receiver that has the following parameters:
dark current i D = 1 nA, quantum efficiency η = 0.85, gain M = 100, excess
noise factor F = M
1/2 , load resistor R L = 10
4
, and bandwidth B e = 10 kHz.
Suppose a sinusoidally varying 850-nm signal having a modulation index m
= 0.85 falls on the photodiode, which is at room temperature (T = 300 K).
Show that the responsivity is 0.58 A/W.
6.7 Consider an avalanche photodiode receiver that has the parameters listed in
Problem 6.6. To compare the contributions from the individual noise terms
to the SNR given by Eq. (6.9), examine each of the noises independently in
terms of the incident optical power P in . Show that for this particular set of
parameters, the relative contributions to the SNR are as follows:
S N R shot =
i
2
s (t)
i
2
shot (t)
= 6.565 × 10
12 P in
S N R dark =
i
2
s (t)
i
2
dark (t)
= 3.798 × 10
22 P
2
in
S N R th =
i
2
s (t)
i
2
th (t)
= 7.333 × 10
22 P
2
in
6.8 A given InGaAs avalanche photodiode has a quantum efficiency of 90% at
a wavelength of 1310 nm. Suppose 0.5 μW of optical power produces a
multiplied photocurrent of 8 μA. Show that the multiplication M = 16.
6.9 Suppose an avalanche photodiode has the following parameters: i D = 1 nA, η
= 0.85, F = M
1/2 , R L = 10
3
, and B e = 1 kHz. Consider a sinusoidal 850-nm
signal, which has a modulation index m = 0.85 and an average power level P in
= –50 dBm, to fall on the detector at room temperature. (a) Using Eq. (6.16)
show that