Linear algebra with application manual ch6

Download Document
Showing pages : 1 - 4 of 38

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
MSE 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 p-n 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,
Background image of page 1
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
Background image of page 2
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 ).
Background image of page 3
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.