Page 327 - Electrical Properties of Materials

Semiconductor lasers 309

of electrons. The rate of change of the number of electrons, due to spontaneous
recombination, is N e lwd/t rec , and this loss should be replenished by injection Active
of electrons, that is the number required is (I i /e)η, where I i is the injected I region
current and η, the quantum efficiency, is the fraction of injected electrons which
recombine radiatively, leading to the formula
d
N e lwd I i η
= . (12.35)
t rec e
w
Recognizing now that the recombination time in the above equation cor- l
responds to t spont discussed before, we may now use eqn (12.31) to find the
Fig. 12.9
amplification of the optical wave. For simplicity, we may take the population
Schematic representation of a laser
of the lower level as zero, and obtain
diode.
2
c g(ν)η
γ (ν)= I i . (12.36)
2 2
8πn ν elwd
For laser oscillations we need the loop gain to be unity. When both mirrors
have the same reflectivity, the condition of oscillation is

R exp(γ – α)l = 1, (12.37)

whence the threshold current density is
2 2
I i 8πn ν ed 1
= α – 1n R . (12.38)
2
lw c ηg(ν) l
As we have said before α represents the losses in the material. But are there
any losses at all? The optical wave propagating in the active region will surely
grow and not decay. True, but there is no reason why the optical wave should
be confined to the active region. A well calculated plunge (one we shall not
take here) into the mysteries of electromagnetic theory would show that a not
inconsiderable portion of the electric field propagates outside the active region,
where there is no population inversion. The losses there are mainly caused by
the so-called free-carrier absorption, which comes about by electrons and holes
excited to higher energies within their own bands.
What is the value of R? In the simplest laser diode the mirror consists of
the cleaved end of the semiconductor crystal, that is one relies on the differ-
ence in refractive index between semiconductor and air. A typical refractive
index is 3.35, which yields for the reflection coefficient, R = 0.54. For a prac-
tical case (see Exercise 12.10) the threshold current comes to a value of about
–2
820 A cm . This is quite a large value. Can we reduce it by some clever trick?
Yes, we can, and the trick is to use a heterojunction instead of a homojunction.
A schematic drawing of the device is shown in Fig. 12.10.
What is the role of the various layers? The insulating SiO 2 layers are there in
order to steer the current towards the middle of the device and thus increase the
current density. The heavily doped GaAs layers next to the metal electrodes are
there to provide ohmic contacts. The p-type and n-type AlGaAs layers serve
to provide the p–n junction, and then we come to the star of the show, the thin