80
S. Leu and D. Sontag
Again, we apply (4.2); this time to the ARC layer in order to calculate the refractive
index n ARC needed for the latter. For complete absorption, the refractive index R in
(4.2) has to be zero. This is only possible if the phase shift and the amplitude of
the reflected waves are equal. Now, this will happen (see [2]) when the following
formula is fulfilled:
n ARC = (n 1 n 2 )
1/2
(4.5)
n ARC is the desired refractive index of the ARC layer.
n ARC = (1 × 3.9)
1/2
= 1.97
(4.6)
Now, it’s all about finding a material that has the desired value of refractive index
n ARC . Silicon nitride with a refractive index in the range 1.9–2.1 comes very close
to having the required value. In addition, depending on the deposition technology
chosen, silicon nitride may contain hydrogen. Due to the small size of hydrogen
atoms, hydrogen diffuses very rapidly into silicon and can passivate impurities and
recombination centres; thereby reducing recombination losses in silicon (see Chaps. 5
and 7). Previously, SiO x (n = 1.3) and TiO 2 (n = 2.52) were used, both of which have
a refractive index, which is quite far from the required “ideal” value. The coating
with silicon nitride fulfils, thus, two functions: it serves as an antireflection coating
(ARC) and as a passivation layer.
In the next step we have to determine—in a more precise way—the thickness of
the ARC layer. Most of the incoming light on Earth has a wavelength of 550–600 nm,
which is in the green part of the solar spectrum (see Chap. 2, Fig. 2.1 and Chap. 3,
Fig. 3.6). Silicon, in turn, is only transparent to light with a wavelength larger than
1127 nm, due to its bandgap of E g = 1.12 eV. So, we can optimize the layer thickness
for a light wavelength of 575 nm, because it is for a wavelength of 575 nm that we
expect the largest quantity of photons, which can be used with a silicon solar cell.
As illustrated in Fig. 4.6, the light waves are reflected at the interface A between
air and the antireflective coating—ARC (r 1 ) and at the interface B between the
antireflective coating (ARC) and the substrate (r 2 ). Thus, the light wave r 2 makes a
longer optical path, since it also has to pass twice through the ARC layer. In order to
achieve destructive interference and, thus, minimize reflection, the phase difference
in the ARC layer must correspond to half a wavelength. Since the ARC layer is
passed through twice, its thickness must be—in the simplest form—one quarter of
the wavelength.
7 Taking into account the refractive index n ARC of the antireflective
coating (ARC) we obtain
d = λ/(4n ARC )
(4.7)
d
geometric layer thickness of the ARC
7 It could also be an odd multiple of one quarter of the wavelength, i.e. (3λ/4) n ARC or (5λ/4) n ARC
etc.
S. Leu and D. Sontag
Again, we apply (4.2); this time to the ARC layer in order to calculate the refractive
index n ARC needed for the latter. For complete absorption, the refractive index R in
(4.2) has to be zero. This is only possible if the phase shift and the amplitude of
the reflected waves are equal. Now, this will happen (see [2]) when the following
formula is fulfilled:
n ARC = (n 1 n 2 )
1/2
(4.5)
n ARC is the desired refractive index of the ARC layer.
n ARC = (1 × 3.9)
1/2
= 1.97
(4.6)
Now, it’s all about finding a material that has the desired value of refractive index
n ARC . Silicon nitride with a refractive index in the range 1.9–2.1 comes very close
to having the required value. In addition, depending on the deposition technology
chosen, silicon nitride may contain hydrogen. Due to the small size of hydrogen
atoms, hydrogen diffuses very rapidly into silicon and can passivate impurities and
recombination centres; thereby reducing recombination losses in silicon (see Chaps. 5
and 7). Previously, SiO x (n = 1.3) and TiO 2 (n = 2.52) were used, both of which have
a refractive index, which is quite far from the required “ideal” value. The coating
with silicon nitride fulfils, thus, two functions: it serves as an antireflection coating
(ARC) and as a passivation layer.
In the next step we have to determine—in a more precise way—the thickness of
the ARC layer. Most of the incoming light on Earth has a wavelength of 550–600 nm,
which is in the green part of the solar spectrum (see Chap. 2, Fig. 2.1 and Chap. 3,
Fig. 3.6). Silicon, in turn, is only transparent to light with a wavelength larger than
1127 nm, due to its bandgap of E g = 1.12 eV. So, we can optimize the layer thickness
for a light wavelength of 575 nm, because it is for a wavelength of 575 nm that we
expect the largest quantity of photons, which can be used with a silicon solar cell.
As illustrated in Fig. 4.6, the light waves are reflected at the interface A between
air and the antireflective coating—ARC (r 1 ) and at the interface B between the
antireflective coating (ARC) and the substrate (r 2 ). Thus, the light wave r 2 makes a
longer optical path, since it also has to pass twice through the ARC layer. In order to
achieve destructive interference and, thus, minimize reflection, the phase difference
in the ARC layer must correspond to half a wavelength. Since the ARC layer is
passed through twice, its thickness must be—in the simplest form—one quarter of
the wavelength.
7 Taking into account the refractive index n ARC of the antireflective
coating (ARC) we obtain
d = λ/(4n ARC )
(4.7)
d
geometric layer thickness of the ARC
7 It could also be an odd multiple of one quarter of the wavelength, i.e. (3λ/4) n ARC or (5λ/4) n ARC
etc.