Page 148 - Fiber Optic Communications Fund

Lasers 129

From Eq. (3.122), we have
G = 1. (3.125)
ph
Using Eq. (3.120), we obtain

G (N − N ) = 1, (3.126)
0 e e0 ph
1
N = N + . (3.127)
e
e0
G
0 ph
From Eq. (3.125), it follows that
Γg = cav , (3.128)
which is a restatement of the fact that gain should be equal to loss. If the current I is very small, there will not
be enough electrons in the conduction band to achieve population inversion. In this case, the gain coefficient
will be much smaller than the loss coefficient and photons will not build up. For a certain current I,the gain
coefficient Γg becomes equal to the loss coefficient and this current is known as the threshold current, I .
cav th
If I > I , stimulated emission could become the dominant effect and the photon density could be significant.
th
Under steady-state conditions, there are two possibilities.
Case (i) I = I . Stimulated emission is negligible and N ≅ 0,
th− ph
dN N
e e I
=− + = 0. (3.129)
dt e qV
Let N = N e,th . From Eq. (3.129), we have
e
N e,th qV
I = . (3.130)
th
e
From Eq. (3.127), we have
1
N e,th = N + . (3.131)
e0
G
0 ph
Case (ii) I > I . When the current exceeds the threshold current, we may expect the electron density N to be
th e
larger than N . However, the electron density will be clamped to N when I > I . This can be explained
e,th e,th th
as follows. The threshold current is the minimum current required to achieve population inversion. When
I > I , the excess electrons in the conduction band recombine with holes and, therefore, the photon density
th
increases while the electron density would maintain its value at threshold. Fig. 3.36(a) and 3.36(b) shows the
numerical solution of the laser rate equations for I = 50 mA and 100 mA, respectively. The threshold current
in this example is 9.9 mA. In Fig. 3.36(a) and 3.36(b), after t > 5 ns, we may consider it as steady state since
N and N do not change with time. Comparing Fig. 3.36(a) and 3.36(b), we find that the steady-state electron
e
ph
density is the same in both cases, although the bias currents are different. In fact, it is equal to N e,th as given
by Eq. (3.131). Using Eqs. (3.124) and (3.125) in Eq. (3.123), we obtain
N N
ph I e,th
= − . (3.132)
qV
ph e
Using Eq. (3.130), Eq. (3.132) can be written as
(I − I ) ph
th
N ph = . (3.133)
qV