Page 140 - Photonics Essentials an introduction with experiments

Light-Emitting Diodes

134 Photonic Devices

where we have defined a response time for ac modulation around a
steady-state operating current as ac = 1/BN. Solving Eq. (6.37) for the
modulation rate,
1 BJ 1/2
= (6.38)
ac 3qd
Under these conditions, the modulation bandwidth will be proportion-
al to the square root of the dc drive current around which the ac mod-
ulation is taking place.

Example 6.5. Calculate the ac Modulation Bandwidth
of an LED in the High-Injection Limit
Using the same parameters as before, the ac modulation bandwidth
can be calculated:
50 × 10 –3
J = · 1 × 10 –4 = 5 × 10 A-cm –2
3
100 × 10 –4
B = 8 × 10 –11 cm sec –1
3
d = 10 –5 cm

1
= 2.9 × 10 sec –1
8
ac
ac bandwidth = 90 MHz

Example 6.6. Calculate the ac Modulation Bandwidth
of an LED in the Low-Injection Limit
In this case, the excess carrier density introduced by the current re-
mains much less than the doping density, therefore N = n D . As before,
we assume that the LED is being modulated around a steady-state
operating point:

d J N
N = – B(N + P + N) N – = 0
dt qd n–r
J N
= BN N +
qd n–r
1
= Bn D + N
n–r

1
= N (6.39)
ac

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