Page 292 - Fiber Optic Communications Fund

Optical Amplifiers 273

be the small signal gain. Using Eqs. (6.150) and (6.153), Eq. (6.152) can be expressed as
[ ]
(G − 1)P(L)
s
G = G exp − . (6.154)
s
s 0
G P
s sat
When P(0)(= P(L)∕G ) ≪ P , the exponent in Eq. (6.154) is close to zero and G ≈ G .Fig.6.14shows
s sat s s 0
the gain G as a function of the input signal power P(0). When P(0) exceeds P ,the gain G decreases from
s sat s
its unsaturated value G .
s0
For a TWA, the overall gain G ≈ G . For a cavity-type SOA, the gain is given by Eq. (6.124). For both types
s
of amplifier, G decreases as the input power increases due to gain saturation. The saturation power for a TWA
is higher than that for a cavity-type SOA [7]. This is because the electron lifetime is lower at higher carrier
e
density and from Eq. (6.147), we see that P is inversely proportional to . The carrier density is higher for
sat e
a TWA since G can be much larger and still RG < 1.
s S
Example 6.7
A 1530-nm TWA has the following parameters:
Effective area of mode A = 5 μm 2
Active volume = 7.5 × 10 −16 m 3
Carrier lifetime = 1ns
Gain cross-section = 7.3 × 10 −20 m 2
g
N = 3.5 × 10 23 m −3
e,0
Overlap factor Γ= 0.3
Calculate (a) the saturation power and (b) the bias current I to have the small signal gain coefficient g =
0
3
−1
4.82 × 10 m .
Solution:
The saturation power is given by Eq. (6.148),
hf A
o
P = , (6.155)
sat
Γ
g e
c 3 × 10 8
f = = = 196.08 THz, (6.156)
0
1530 × 10 −9
0
12
6.626 × 10 −34 × 196.08 × 10 × 5 × 10 −12
P sat = W
0.3 × 7.3 × 10 −20 × 1 × 10 −9
= 29.7 mW. (6.157)
The relation between g and I is given by Eq. (6.144),
0
( )
I N e,0
g = − , (6.158)
0 g e
qV
e