Page 287 - Fiber Optic Communications Fund

268 Fiber Optic Communications

gain and bandwidth. As an example, consider R = R = 0.3, G = 2.5, and L = 300 μm. Using Eqs. (6.127)
s
2
1
and (6.132), we find G peak = 20 and f 3dB = 36 GHz. For optical communication systems, the desired amplifier
bandwidth is typically greater than 1 THz and, therefore, a cavity-type semiconductor amplifier is unsuitable
as an in-line amplifier for high-bit-rate optical communication systems. In addition, they are very sensitive to
fluctuations in bias current, temperature, and polarization of the incident field [5].

Example 6.6
In a cavity-type SOA, the cavity length is 500 μm, R = R = 0.32, and the peak gain is 15 dB. Find the
1 2
single-pass gain and the 3-dB bandwidth. Assume n = 3.2.
Solution:
G peak = 15 dB,

G = 10 G peak (dB)∕10 = 31.62.
peak
From Eq. (6.127), we have
2
(1 − RG ) G =(1 − R )(1 − R )G ,
s peak 1 2 s
2
2
1 + 0.32 G − 2 × 0.32G = 0.0146G ,
s
s
s
or
2
0.1024G − 0.6546G + 1 = 0.
s
s
This quadratic equation has solutions
G = 2.52 or 3.86.
s
Eq. (6.127) is derived under the assumption that |h| < 1or RG < 1. When G = 3.86, RG > 1, and therefore
s
s
s
it is not consistent with Eq. (6.127) and this solution is rejected.
The 3-dB bandwidth is given by Eq. (6.132),
( )
c −1 1 − RG s
f = sin
3dB √
Ln 2 RG s
( )
3 × 10 8 −1 1 − 0.32 × 2.52
= sin
× 500 × 10 −6 × 3.2 2(0.32 × 2.52) 1∕2
= 6.44 GHz.
6.6.2 Traveling-Wave Amplifiers

From Fig. 6.11, it can be seen that as R decreases, the bandwidth increases. In the limiting case of R = 0,
Eq. (6.124) becomes G = G , i.e., the overall gain is equal to the single-pass gain G . This should be expected
s
s
since there are no partial fields due to round trips when R = 0. Such an amplifier is known as a traveling-wave
amplifier (TWA).