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View Full DocumentMSE 302 Solar Cells  Midterm Exam Solutions
May 9, 2008
1.
[30 points]
Suppose you shine a laser at a solar cell in order to transport energy to a
location that was difficult to reach. The laser beam does not diverge with distance, so
a lot of power can be sent to a small area. All of the photons have essentially the same
energy. Cost in not important for this application, but performance is.
a)
How would you design the solar cell? Focus on the design issues that are unique
for this application. It goes without saying that the material should be doped
appropriately, have suitable thickness, have suitable electrodes and not have
defects that cause recombination. What should the photon energy be? What is the
approximate energy conversion efficiency the solar cell could have for this
specific application? Assume a laser is available for any photon energy you would
like.
The first thing to realize is that you are dealing with a single wavelength light
source. Therefore you would want to choose a material with a bandgap that is
exactly the same (or slightly smaller than) the energy of the incoming photons
to ensure absorption but not lose energy to thermalization. This is the key
concept we are looking for in this question and significant portion of points is
deducted if you did not mention this. You would also want a direct bandgap
material to ensure strong absorption just above the bandgap. In addition,
since you are not concerned about absorbing over a wide range of
wavelengths we would choose a larger bandgap to maximize the power out of
the device since we know that the voltage out is typically
V
q
E
g
3
.
−
.
For the possible energy conversion efficiency, we will assume that we absorb
all of the photons and that all of the minority carriers make it to the pn
junction. This is reasonable since we are using single wavelength excitation
with an optimized architecture. We also lose very little to thermalization since
we are using excitation energy just above the bandgap energy. Therefore,
ISM:
Linear Algebra
Section 6.1
14. det
4
0
0
3
k
0
2
1
0
= 0
,
so the matrix will never be invertible, no matter which
k
is chosen.
15. det
0
k
1
2
3
4
5
6
7
= 6
k

3
.
This matrix is invertible when
k
6
=
1
2
.
16. det
1
2
3
4
k
5
6
7
8
= 60 + 84 + 8
k

18
k

35

64 = 45

10
k.
So this matrix is invertible
when
k
6
= 4
.
5.
17. det
1
1
1
1
k

1
1
k
2
1
= 2
k
2

2 = 2(
k
2

1) = 2(
k

1)(
k
+ 1)
.
So
k
cannot be 1 or 1.
18. det
0
1
k
3
2
k
5
9
7
5
= 30 + 21
k

18
k
2
=

3(
k

2)(6
k
+ 5)
.
So
k
cannot be 2 or

5
6
.
19. det
1
1
k
1
k
k
k
k
k
=

k
3
+ 2
k
2

k
=

k
(
k

1)
2
.
So
k
cannot be 0 or 1.
20. det
1
k
1
1
k
+ 1
k
+ 2
1
k
+ 2
2
k
+ 4
= (
k
+ 1)(2
k
+ 4) +
k
(
k
+ 2) + (
k
+ 2)

(
k
+ 1)

k
(2
k
+ 4)

(
k
+ 2)(
k
+ 2) = (
k
+ 1)(3
k
+ 6)

(3
k
2
+ 9
k
+ 5) = 1
.
Thus,
A
will always be invertible,
no matter the value of
k
, meaning that
k
can have any value.
21. det
k
1
1
1
k
1
1
1
k
=
k
3

3
k
+ 2 = (
k

1)
2
(
k
+ 2)
.
So
k
cannot be 2 or 1.
22. det
cos
k
1

sin
k
0
2
0
sin
k
0
cos
k
= 2 cos
2
k
+ 2 sin
2
k
= 2
.
So
k
can have any value.
23. det(
A

λI
2
) = det
±
1

λ
2
0
4

λ
²
= (1

λ
)(4

λ
) = 0 if
λ
is 1 or 4.
24. det(
A

λI
2
) = det
±
2

λ
0
1
0

λ
²
= (2

λ
)(

λ
) = 0 if
λ
is 2 or 0.
291
Chapter 6
ISM:
Linear Algebra
25. det(
A

λI
2
) = det
±
4

λ
2
4
6

λ
²
= (4

λ
)(6

λ
)

8 = (
λ

8)(
λ

2) = 0 if
λ
is 2 or
8.
26. det(
A

λI
2
) = det
±
4

λ
2
2
7

λ
²
= (4

λ
)(7

λ
)

4 = (
λ

8)(
λ

3) = 0 if
λ
is 3 or
8.
27.
A

λI
3
is a lower triangular matrix with the diagonal entries (2

λ
)
,
(3

λ
) and (4

λ
).
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