光电子学与光子学的原理及应用S.O.Kasa chapter6课后答案
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Solutions Manual for Optoelectronics and Photonics: Principles and Practices S.O. Kasap 6.1
23 April 2001
6.3 Solar cell driving a load
a A Si solar cell of area 4 cm
2
is connected to drive a load R as in Figure 6.8 (a). It has the I-V
characteristics in Figure 6.8 (b) under an illumination of 600 W m
-2
. Suppose that the load is 20 Ω and it
is used under a light intensity of 1 kW m
-2
. What are the current and voltage in the circuit? What is the
power delivered to the load? What is the efficiency of the solar cell in this circuit?
b What should the load be to obtain maximum power transfer from the solar cell to the load at 1
kW m
-2
illumination. What is this load at 600 W m
-2
?
c Consider using a number of such a solar cells to drive a calculator that needs a minimum of 3V
and draws 3.0 mA at 3 - 4V. It is to be used indoors at a light intensity of about 400 W m
-2
. How many
solar cells would you need and how would you connect them? At what light intensity would the
calculator stop working?
Solution
a The solar cell is used under an illumination of 1 kW m
-2
. The short circuit current has to be scale
up by 1000/600 = 1.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 1.67 along
the current axis. The load line for R = 20 W and its intersection with the solar cell I-V characteristics at P
which is the operating point P. Thus,
I¢ ≈ 22.5 mA and V¢ ≈ 0.45 V
The power delivered to the load is
P
out
= I¢V¢ = (22.5×10
-3
)(0.45V) = 0.0101W, or 10.1 mW.
This is not the maximum power available from the solar cell. The input sun-light power is
P
in
= (Light Intensity)(Surface Area)
= (1000 W m
-2
)(4 cm
2
´ 10
-4
m
2
/cm
2
) = 0.4 W
The efficiency is
η
= 100
P
out
P
in
= 100
0
.
010
0.4
= 2.5
0
0
which is poor.
b Point M on Figure 6Q3-2 is probably close to the maximum efficiency point, I¢ ≈ 23.5 mA and
V¢ ≈ 0.44 V. The load should be R = 18.7 W, close to the 20 W load. At 600 W m
-2
illumination, the
load has to be about 30 W as in Figure 6.8 (b). Thus, the maximum efficiency requires the load R to be
decreased as the light intensity is increased. The fill factor is
FF =
I
m
V
m
I
sc
V
oc
=
(
23
.
5
mA
)(
0
.
44
V
)
(27 mA)(0.50 V)
≈ 0.78
c The solar cell is used under an illumination of 400 W m
-2
. The short circuit current has to be
scale up by 400/600 = 0.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 0.67
along the current axis. Suppose we have N identical cells in series, and the voltage across the calculator
is V
calculator
. The current taken by the calculator is 3 mA in the voltage range 3 to 4 V and the calculator
stops working when V
calculator
< 3 V. The cells are in series so each has the same current and equal to 3
mA, marked as I¢ in Figure 6Q-2. The voltage across one cell will be V¢ = V
calculator
/N. which is marked
in Figure 6Q3-2. V¢ = 0.46 V. Minimum number of solar cells in series = N = 3 / 0.46 = 6.5 or 7 cells ,
since you must choose the nearest higher integer.
If we want the calculator continue to work under low intensity levels, then we can connect more
cells in series until we reach about 4 V; N = 4 V / 0.46 V = 8.7 or 9 cells in series.
The easiest estimate for the minimum required light intensity is the following: The calculator
will stop working when the light intensity cannot provide energy for the solar cell to deliver the 3 mA
calculator current. The short circuit current at 400 W m
-2
is 11 mA in Figure 6Q3-2. Thus
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