6.32 dan 7.

7
6.32
A solar cell used in Eskimo Point The intensity of light arriving at a point on Earth, where
the solar latitude is α can be approximated by the Meinel and Meinel equation:

I =1.353 ( 0.7 )( cosecα ) 0.678 kW m-2

where cosec α = 1∕(sin α). The solar latitude α is the angle between the sun’s rays and the
horizon. Around September 23 and March 22, the sun’s rays arrive parallel to the plane of the
equator. What is the maximum power available for a photovoltaic device panel of area 1 m 2 if
its efficiency of conversion is 10 percent?
A manufacturer’s characterization tests on a particular Si pn junction solar cell at 27
°C specifies an open circuit output voltage of 0.45 V and a short circuit current of 400 mA
when illuminated directly with a light of intensity 1 kW m −2. The fill factor for the solar cell
is 0.73. This solar cell is to be used in a portable equipment application near Eskimo Point
(Canada) at a geographical latitude (ϕ) of 63°. Calculate the open circuit output voltage and
the maximum available power when the solar cell is used at noon on September 23 when the
temperature is around −10 °C. What is the maximum current this solar cell can supply to an
electronic equipment? What is your conclusion? (Note: α + ϕ = π∕2, and assume η = 1 and that
Io ∝ n2i .)

Jawab :
The definition of α is shown in figure 1 (a) and (b). We can plot the light intensity versus α as
shown in figure 2 which shows the maximum is at α = 90° as expected.
Notice how broad the cave is, implying that these is still substantial intensity when α is a low
angle.
The light intensity is maximum when α = 90° thus
(cosecα)
I max=1.353 ( 0.7 ) =0.95kW m-2

Maximum power per unit area = (efficiency) x (maximum intensity)

There fore, the maximum power per unit area

¿ ( 0.1 ) ( 0.95 kW m ) =0.095 kW m =95W m

−2 −2 −2

So, at the best spot on earth, the best one can do with this solar cell is 95 W m−2
 Figure 1
(a) Effect of the angle of incidence on the ray path length
(b) At noon, at a location of latitude (ϕ) on september 23 and march 22. The angle α is the solar latitude
and ϕ is the geographic latitude.

Figure 2. Light intensity versus α

Under test conditions, we have Voc = 0.45 V, Iph = 400 mA at I kW m-3, at 27 °C, that is Vγ =
0.0259 V

V oc =η V γ ln
[ ]
I ph
Io

[ ]
−3
400 x 10
0.45=1(0.0259) ln
Io
−8
I o=1.10 A
At Eskimo point α = 90° - 63° = 27° (see figure 2) which gives a light intensity of

I EP=1.353 ( 0.7 )(cosecα )=1.353 ( 0.7 )( cosec27 ° )=0.736 kW m−2

Which means that the Eskimo point photocurrent Iph is determined by

I ph (Eskimo point) 0.736 kW m−2
=
I ph (Test ) 1 kW m
−2

I ph=294 mA

This is the maximum current that can be supplied. It does not depend on the temperature
since it is generated by photogeneration in the solar cell.
The dark current depends on the temperature.
At -10 °C, VT = KT/e = 0.0227 V.
The change in the dark current Io can be found as fallout let V g = bandgap voltage = εg/e =
1.1 V for si. Then η = 1 means that

2
I o α ni α exp
[ ] −V g
Vγ

So that
I o ( Eskimo point )
I o (Test )
=exp
[
−V g
+
Vg
V γ ( eskimo point ) V γ ( test ) ]
I o ( Eskimo point )
1.1 x 10
−8
A
=exp
[ −1.1
+
1.1
0.0227 0.0259 ]
Solving gives
Io = 2.77 x 10-11 A
So that the Eskimo point the open circuit voltage is

[ ]
I ph
[ ]
−3
294 x 10
V oc =η V γ ln =1 ( 0.0227 ) ln =0.523 V
Io 2.77 x 10
−11

The maximum power is

Pmax = FF/SC Voc = (0.73)(244 mA)(0.523 V) = 93 mV
The maximum current is
Iph = 294 mA
7.7
Relative permittivity, bond strength, bandgap, and refractive index Diamond, silicon, and
germanium are covalent solids with the same crystal structure. Their relative permittivities
are shown in Table 7.11
a. Explain why εr increases from diamond to germanium.
b. Calculate the polarizability per atom in each crystal and then plot polarizability
against the elastic modulus Y (Young’s modulus). Should there be a correlation?
c. Plot the polarizability from part (b) against the bandgap energy Eg . Is there a
relationship?
d. Show that the refractive index n is √εr . When does this relationship hold and when
does it fail?
e. Would your conclusions apply to ionic crystals such as NaCl?

Jawab :
a. In diamond, Si, and Ge, the polarization mechanism is electronic (bond). There are
two factors that increase the polarization. First is the number of electrons available for
displacement and the ease with which the field can displace the electrons. The number
of electrons in the core shells increases from diamond to Ge. Secondly, and most
importantly, the bond strength per atom decreases from diamond to Ge, making it
easier for valence electrons in the bonds to be displaced.

b. For diamond, atomic concentration N is :

D N A ( 3.52 x 103 kg m−3 ) ( 6.022 x 1023 mol−1 ) 29 −3
N= = =1.766 x 10 m
M at ( 12 x 10−3 kg mo l−1 )
The polarizability can then be found from the Clausius-Mossotti equation:
ε r −1 N
= α
εr + 2 3 εo e

3 ε o ( ε r−1) 3(8.854 x 10−12 F m−1 )(5.8−1)

α e= =
N (ε r + 2) ( 1.766 x 1029 m−3 ) (5.8+ 2)
α e =9.256 x 10−41 F m2

The polarizability for Si and Ge can be found similarly, and are summarized in Table :
N(m-3) αe(F m )
2

Diamond 1.766 x 1029 9.256 x 10-41

Si 4.995 x 1028 4.170 x 10-40
Ge 4.412 x 1028 5.017 x 10-40

Figure 1. Plot of polarizability per atom versus Young’s modulus

As the polarization mechanism in these crystals is due to electronic bond polarization,

the displacement of electrons in the covalent bonds depends on the flexibility or
elasticity of these bonds and hence also depends on the elastic modulus.

c. Polarizability per atom versus Bandgap

Figure 2. Plot of polarizability versus bandgap energy

There indeed seems to be a linear relationship between polarizability and bandgap

energy.
d. To facilitate this proof, we can plot a graph of refractive index, n, versus relative
permittivity, ε r.

The log-log plot exhibits a straight line through the three points. The best fit line is n
= Aε xr (Correlation coefficient is 0.9987) where x = 0.513 ≈ ½ and A= exp(0.02070) ≈
1. Thus n is ❑√ ε r .

The refractive index n is an optical property that represents the speed of a light wave,
or an electromagnetic wave, through the material (v=c/n). The light wave is a high
frequency electromagnetic wave where the frequency is of the order of 10 14 to 1015 Hz
(foptical). n and polarizability (or ε r) will be related if the polarization can follow the
field oscillations at this frequency (foptical). This will be the case in electronic
polarization because electrons are light and rapidly respond to the fast oscillations of
the field. The relationship between n and ε r will not hold if we take ε r at a low
frequency (<<foptical) where other slow polarization contributions (such as ionic
polarization, dipolar polarization, interfacial polarization) also contribute to ε r.

e. n = ❑√ ε r would apply to ionic crystals if ε r is taken at the corresponding optical

frequency rather than at below foptical. Tabulated data for ionic crystals typically quote
ε r that includes ionic polarization and hence this data does NOT conform to n = ❑√ ε r .