Page 316 - Fiber Optic Communications Fund

Optical Amplifiers 297

10log 10 G max = 20,
2
G max = 10 = 100,
G = 10 5∕10 = 3.16,
s
√
(1 − 3.16R)× 10 =(1 − R) 3.16,
R = 0.276.

Example 6.14

A Fabry–Perot amplifier has to be designed such that its full 3-dB bandwidth is greater than 25 GHz. Calculate
the upper bound on the single-pass gain G . Assume R = R = 0.3, refractive index n = 3.5, amplifier length
s 1 2
L = 200 μm.
Solution:
From Eq. (6.128), we have
c 3 × 10 8
FSR = = = 214 GHz.
2nL 2 × 3.5 × 200 × 10 −6
From Eq. (6.132), the 3-dB bandwidth is
{ }
2 FSR −1 1 − RG s
f 3dB = sin ,
( 4RG ) 1∕2
s
or ( )
f 3dB 1 − RG s
sin = . (6.296)
2FSR (4RG ) 1∕2
s
If f = 25 GHz, the left-hand side of Eq. (6.296) becomes 0.1805:
3dB
2 2
(1 − 0.3G ) = 1.2G ×(0.1805) .
s s
Solving the above quadratic equation, we find

G = 4.79 or 2.31.
s
If G = 4.79, (1 − RG ) < 0, which corresponds to biasing above the threshold and violates the condition
s
s
given by Eq. (6.118) and, therefore, it has to be rejected. To have f 3dB > 25 GHz, G has to be less than 2.31.
s