ADVANCED METALLISATION METHODS FOR MONOCRYSTALLINE SILICON WAFER SOLAR CELLS

... used for metallisation of HET cells [38] Metallisation technologies for Si wafer solar cells (including HET cells) are discussed further in Section 2.6 Screen printing for HET cells and plating for ... technologies for silicon wafer solar cell metallisation The metallisation technologies are applied to silicon homojunction and amorphous silicon/ crystalli...

Ngày tải lên: 09/09/2015, 08:11

... integration into heterojunction silicon wafer solar cells and subsequently the modelling of heterojunction and hybrid heterojunction silicon wafer solar cells that utilize these films Three key ... monocrystalline/multicrystalline solar cells, thin- film solar cells, organic solar cells, tandem cells, concentrator cells and varying choice of mat...

Ngày tải lên: 09/09/2015, 11:15

... In summary, an improved surface passivation scheme using inductively coupled plasma deposited amorphous silicon suboxide thin films for heterojunction silicon wafer solar cells is successfully ... analysis of surface passivating amorphous silicon alloys for heterojunction silicon wafer solar cell applications As the narrow process window of conventional intrins...

Ngày tải lên: 09/09/2015, 11:31

... spatial homogeneity of a silicon wafer solar cell 23 Identification of defects in solar cells 27 Advancing luminescence- based characterization of silicon wafer materials and ... non-destructive luminescence characterization of silicon wafers and silicon wafer solar cells, electroluminescence and photoluminescence are frequently used For characterization, fe...

Ngày tải lên: 09/09/2015, 17:58

... this thesis 132 vi Direct writing for silicon wafer solar cells This page is intentionally left blank vii Table of contents Direct writing for silicon wafer solar cells Abstract Abstract ... applications in silicon wafer solar cells are investigated in detail in this PhD thesis Direct writing for silicon wafer solar cells §1.2 Thesis outline C...

Ngày tải lên: 09/09/2015, 11:16

... 12 16 12 16 12 19 12 24 12 25 12 48 12 49 12 51 12 51 12 55 12 63 12 70 12 72 12 72 12 78 12 78 12 80 12 81 12 83 12 83 12 84 12 87 12 88 12 94 12 97 13 02 13 03 13 04 25 .14 Solutions ... 10 25 10 27 10 27 10 29 10 32 10 34 10 35 10 41 10 41 10 43 10 45 10 46 10 48 10 50 10 52 10 59 10 59 10 61 10 65 10 65 10 68 10 71 10 74 10 74 1...

Ngày tải lên: 06/08/2014, 01:21

... = 10 9 = x0 11 10 = x0 x0 = 10 9 At 7:00 pm the number of bacteria is 10 11 10 60 = 11 60 ≈ 3.04 × 10 11 51 10 At 3:00 pm the number of bacteria was 10 9 11 10 18 0 = 18 10 189 ≈ 35.4 11 180 Figure 1. 13: ... b2 (−i) = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k Next we evaluate the determinant i j k a a a a a a a1 a2 a3 = i − j + k b2 b3 b1 b3 b1 b2 b1 b2 b3 =...

Ngày tải lên: 06/08/2014, 01:21

... = ( 1) n 1 (n − 1) !, for n ≥ By Taylor’s theorem of the mean we have, ln x = (x − 1) − (x − 1) 2 (x − 1) 3 (x − 1) 4 (x − 1) n (x − 1) n +1 + − + · · · + ( 1) n 1 + ( 1) n n n + ξ n +1 72 -1 0.5 1. 5 2.5 ... 6! (2(n − 1) )! (2n)! Here are graphs of the one, two, three and four term approximations 0.5 0.5 -3 -2 -1 -0.5 -1 -3 -2 -1 -0.5 -1 0.5 -3 -2 -1 -0.5 -1 0.5...

Ngày tải lên: 06/08/2014, 01:21

... approximate sin (1) , 13 15 1 + ≈ 0.8 41 6 67 12 0 10 7 To see that this has the required accuracy, sin (1) ≈ 0.8 41 4 71 Solution 3 .19 Expanding the terms in the approximation in Taylor series, ∆x3 ∆x4 ∆x2 f ... Example 4. 4 .1 Consider the partial fraction expansion of + x + x2 (x − 1) 3 The expansion has the form a0 a1 a2 + + (x − 1) (x − 1) x 1 127 The coefficients are (1 + x...

Ngày tải lên: 06/08/2014, 01:21

... 87 − 6 1/ 3 1/ 3 6−2/3 + √ 2/3 87 √ + 87 , 0, − 6 1/ 3   ≈ (0 .58 9 755 , 0, 0.347 81) 1/ 3 The closest point is shown graphically in Figure 5 .10 1- 1 -0 .5 0 .5 -1 -0 .5 0 0 .5 1. 5 0 .5 Figure 5 .10 : Paraboloid, ... dx2 = (1 + 2x) x= 1 = 1 x= 1 = x= 1 (2) x= 1 =1 Then we can the integration + x + x2 dx = (x + 1) 3 1 − + (x + 1) (x + 1) x +1 1 + + ln |x + 1|...

Ngày tải lên: 06/08/2014, 01:21

... values For instance, (12 ) 11 /2 = 1 and 11 /2 = ( 1) 2 = Example 6. 6.2 Consider 21/ 5 , (1 + ı )1/ 3 and (2 + ı)5 /6 √ 21/ 5 = eı2πk/5 , for k = 0, 1, 2, 3, 19 9 = √ (1 + ı )1/ 3 = = (2 + ı)5 /6 = √ √ ... 0, z and z + ζ Hint 6 .12 Hint 6 .13 Hint 6 .14 Hint 6 .15 Hint 6 . 16 Polar Form Hint 6 .17 Find the Taylor series of eıθ , cos θ and sin θ Note that ı2n...

Ngày tải lên: 06/08/2014, 01:21

... 6 .16 225 -1 -1 Figure 6 .15 : ( 1) −3/4 Solution 6 .13 ( 1) 1/ 4 = (( 1) 1 )1/ 4 = ( 1) 1/4 = (eıπ )1/ 4 = eıπ/4 11 /4 = eıπ/4 eıkπ/2 , k = 0, 1, 2, = eıπ/4 , eı3π/4 , eı5π/4 , e 7 /4 + ı 1 + ı 1 − ı − ... Cartesian form √ 1+ ı 1 = √ 1+ ı √ −2 + ı2 1 = √ −2 + ı2 √ −2 + ı2 √ −8 − ı8 1 = = √ −2 + ı2 √ 12 8 + 12 8 10 √ 1+ ı √ = − 512 − ı 512 1 1 √ 512 + ı √ 1 − ı √ √ = 512 + ı −...

Ngày tải lên: 06/08/2014, 01:21

... , and draw a 2 71 -1 -2 y -1 y -1 -1 x -1 -2 -1 x -2 -2 -1 -2 01 x -2 -1 y -1 -2 01 x -1 -2 12 10-2 -1 y -1 -1 y -1 1 210 -2 -1 y Figure 7.24: Plots of x -1 -2 -1 z 1/ 2 (left) and 272 x -1 -2 z 1/ 2 ... arguments: log( 1) = log 1 = log (1) − log( 1) = − log( 1) , therefore, log( 1) = = 11 /2 = (( 1) ( 1) )1/ 2 = ( 1) 1/2 ( 1) 1/2 = ıı = 1, therefore, = 1 Hint,...

Ngày tải lên: 06/08/2014, 01:21

... consider = 11 /2 = (( 1) ( 1) )1/ 2 = ( 1) 1/2 ( 1) 1/2 = ıı = 1 There are three multi-valued expressions above 11 /2 = 1 (( 1) ( 1) )1/ 2 = 1 ( 1) 1/2 ( 1) 1/2 = (±ı)(±ı) = 1 Thus we see that the first and ... Hint 7.24 Hint 7.25 1/ 2 (z + 1) = (z − ı )1/ 2 (z + ı )1/ 2 1/ 2 (z − z) = z 1/ 2 (z − 1) 1/2 (z + 1) 1/2 log (z − 1) = log(z − 1) + log(z + 1) log z +1 z 1 = log(z...

Ngày tải lên: 06/08/2014, 01:21