Page 154 - Fiber Optic Communications Fund

Lasers 135

Since the electron rest mass, m = 9.109 × 10 −31 kg, m eff,1 = 0.07m = 6.37 × 10 −32 kg, using Eq. (3.145), we
obtain
1 1 1 1 1 −1
= − = − kg
m eff,2 m r m eff,1 5.62 × 10 −32 6.37 × 10 −32
−31
m = 4.78 × 10 kg.
eff,2

Example 3.11
−1
A laser diode has a 320-μm cavity length, the internal loss coefficient is 10 cm . The mirror reflectivities are
0.35 at each end. The refractive index of the active region is 3.3 under steady-state conditions. Calculate (a) the
optical gain coefficient Γg required to balance the cavity loss and (b) the threshold electron density N . Assume
e
−3
3
23
that the gain can be modeled as G = G (N − N ), G = 1.73 × 10 −12 m ∕s and N = 3.47 × 10 m .
0 e e0 0 e0
Solution:
The total cavity loss coefficient is given by Eq. (3.39),
= + ,
cav int mir
where [ ]
1 1 1 [ 1 ] 3 −1
= ln = ln = 3.28 × 10 m .
mir −6 2
2L R R 2 × 320 × 10 0.35
1 2
−1
3
The internal loss coefficient is = 10 m ,
int
3
3
3
−1
= 10 + 3.28 × 10 m −1 = 4.28 × 10 m .
cav
The optical gain coefficient Γg to balance the cavity loss is
−1
3
Γg = = 4.28 × 10 m .
cav
(b) The threshold electron density is given by Eq. (3.131),
1
N e,th = N e,0 + ,
G
0 ph
where
1
ph = ,
cav
c 3 × 10 8 7
= = = 9.09 × 10 m∕s,
n 3.3
1
= = 2.57 ps,
ph 7 3
9.09 × 10 × 4.28 × 10
23 1 −3
N = 3.47 × 10 + m
e,th −12 −12
1.73 × 10 × 2.57 × 10
23
−3
= 5.71 × 10 m .