Page 215 - Fiber Optic Communications Fund

196 Fiber Optic Communications

through the photodiode is
P =(1 − R )P exp (−W), (5.12)
I
p
tr
where is the absorption coefficient and W is the thickness of the depletion or active region of the photode-
tector. Therefore, the power absorbed in the photodiode is
P abs =(1 − R )P − P =(1 − R )P [1 − exp (−W)]. (5.13)
I
tr
I
p
p
From Eq. (5.7), we find that the mean number of photons absorbed per unit time, or photon absorption
rate,is P ∕(hf ).If a photon is absorbed, an electron–hole pair is generated. Therefore, the number of
abs 0
electron-hole pairs generated per unit time is
N P
PC abs
= . (5.14)
T hf 0
Using Eqs. (5.14) and (5.13) in Eq. (5.10), we find
P abs
=
P I
=(1 − R )[1 − exp (−W)]. (5.15)
p
The quantum efficiency is equal to a product of:

1. the power transmission coefficient at the air–semiconductor interface, 1 − R ;
p
2. the photons absorbed in the active region of thickness W, given by the term 1 − exp (−W); and
3. the fraction of photocarriers that reach the device terminal and contribute to the measured photocurrent.

The third term is usually the most difficult to determine as it depends on a number of factors, such as carrier
lifetimes, transit paths, surface properties, and the physical dimensions of the device.

Example 5.1
If the incident optical signal on a pn photodiode is at a wavelength of 550 nm, its absorption coefficient =
−1
4
10 cm , width of the active region W = 3 μm, and optical power 1 nW, calculate (a) the photon incidence
rate, (b) the photon absorption rate, and, (c) the quantum efficiency. Assume R = 0 and = 0.9.
p
Solution:
(a) The energy of a photon is
hc −19
E ph = = 3.6 × 10 J.
0
The photon incidence rate is given by Eq. (5.8),

P I 1 × 10 −9 W
R incident = = photons/s
E ph 3.6 × 10 −19
9
= 2.77 × 10 photons/s.