Page 132 - Photonics Essentials an introduction with experiments

Light-Emitting Diodes

126 Photonic Devices

2
Here we have assumed that N/n D 1. Note that N and n D N are
2
each much greater than n i . The next step is to substitute the result of
Eq. 6.18 in Eq. (6.17):
J N – n D N – n D
2
2
2
= B(N – n D N + n i – n i ) + = BN(N – n D ) + (6.19)
qd n–r n–r
By comparing the two terms in Eq. (6.19), you can see that BN looks
like a reciprocal relaxation time. Since this term is associated with
the radiative recombination rate, we can define
1
BN (6.20)
r–r
where r–r is the radiative recombination time. So

J 1 1
= + (N – n D )
qd r–r n–r
The total recombination time can be calculated by combining the
rates from the two recombination channels, radiative and nonradia-
tive:

1 1 1 1
= + = BN + (6.21)
r r–r n–r n–r
and
J N – n D
= (6.22)
qd r
The ratio of the radiative recombination rate to the total recombi-
nation rate:
1
1

r–r r–r r
= = = int (6.23)
1 1 1
+ r–r
r
r–r r
gives the fraction of photons created with respect to the total number
of electron–hole recombination events. This ratio is the internal quan-
tum efficiency, or int .
From Eq. (6.22), we have: N – n D = r · J/qd. Rearranging Eq. (6.15)
gives

1
N – n D J J
2
B(NP – n i ) = = · r · · int · (6.24)
r–r r–r qd qd

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