how a module gets its stripes: thin film... · laser scribe glass m. a. alam, pv lecture notes. 21...
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How a Module gets its Stripes: Thin Film
M. A. Alam
Electrical and Computer Engineering
Purdue University
West Lafayette, IN USA
Theory and Practice of Solar Cells: A Cell to System Perspective
M. A. Alam, PV Lecture Notes
1
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33
Outline
1) Motivation: Power-loss vs. area loss
2) Theory of power loss
3) Power loss in series-connected cells
4) Cell geometry and power-loss
5) Conclusions
M. A. Alam, PV Lecture Notes
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Photoelectric effect and solar cells
UVelectron
metal
Sunlightelectron
p-n junction
load
4
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Soland
The Puzzle of Striping
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66
𝛿
Ala-Si:H
TCO
Laser Scribe
Glass
M. A. Alam, PV Lecture Notes
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The Puzzle of Striping
7
𝑃𝑎𝑟𝑒𝑎 = 𝑁 − 1 𝛿 × 𝐿 ∼ 𝐶2𝑁
𝑃 =𝐼
𝑁× 𝑉 × 𝑁 = 𝐶0
7
𝐴𝑇 = 𝑁 × 𝐴
𝛿
We do striping to reduce series resistance …M. A. Alam, PV Lecture Notes
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8
Modules have cells connected in series
Si module Thin-film module
M. A. Alam, PV Lecture Notes
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9
c-Silicon
*Google Images
a-silicon
M. A. Alam, PV Lecture Notes
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1010
Outline
1) Motivation: Power-loss vs. area loss
2) Theory of power loss
3) Power loss in series-connected cells
4) Cell geometry and power-loss
5) Conclusions
M. A. Alam, PV Lecture Notes
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Thin film solar cells: the case for rectangular cells
11*Google Images
𝑃 = න0
𝐿
𝑑𝑥 න0
𝑔 𝑥
𝑑𝑦 𝐽0 × 𝑦 2 𝜌
𝑃
𝐽02𝜌
≡ 𝑃0න0
𝐿
𝑑𝑥𝑔3 𝑥
3
𝐴 = න0
𝐿
𝑔 𝑥 𝑑𝑥
𝑔(𝑥)
𝑥
𝑦
𝐿𝐽0
Top view
M. A. Alam, PV Lecture Notes
𝜌 in ohms
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Lagrange optimization
12
𝐋 = 𝐽02𝜌න
0
𝐿
𝑑𝑥𝑔3 𝑥
3− 𝜆2{න
0
𝐿
𝑔 𝑥 𝑑𝑥 − 𝐴}
𝜆2 is the Lagrange multiplier …
𝑑𝐋
𝑑𝑔= 0 𝑔 𝑥 = 2𝜆/(𝐽0
2𝜌)
𝐴 = 0𝐿𝑔 𝑥 𝑑𝑥 =
2𝜆𝐿
𝐽02𝜌
→ 𝜆 =𝐴 𝐽0
2 𝜌
2𝐿= 𝑐𝑜𝑛𝑠𝑡
→ 𝐴/𝐿 = 𝑐𝑜𝑛𝑠𝑡. Rectangles perform the best …
M. A. Alam, PV Lecture Notes
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Power Loss in Rectangular Cells
13
𝑃0 ≡ 𝑃/𝐽02𝜌 = න
0
𝐿
𝑑𝑥𝑔3 𝑥
3
Let 𝑔(𝑥) = 𝑎𝑥𝑛 define the shape,
where 𝑎 = (𝑛 + 1)𝐴/𝐿𝑛+1
ensures that 𝐴 = 0𝐿𝑔(𝑥) 𝑑𝑥 =const
𝑃𝑜(𝑛) =𝑛+1 3
3 3𝑛+1×
𝐴3
𝐿2
𝑃𝑜 𝑛 × (𝐿2/𝐴3) =𝑛+1 3
3 3𝑛+1
𝑛 = 0
𝑛 = 1
𝑛 > 1
𝑛 < 1
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Shape and Loss
14
𝑃1 ≡ 𝑃𝑜 𝑛 × (𝐿2/𝐴3) =𝑛+1 3
3 3𝑛+1
𝑛 = 0 𝑛 = 1 𝑛 > 1𝑛 < 1
𝑃1
1/3 (n=0) 2/3 (n=1)
27/21 (n=2)
𝑃 𝑜(𝑛)=
𝑛+1
3
33𝑛+1×
𝐴3
𝐿2
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Observation 0:
15
𝑃1 ≡ 𝑃𝑜 𝑛 × (𝐿2/𝐴3) =𝑛+1 3
3 3𝑛+1
Power dissipation reduces with the cube of the area, so
segmenting a module in many cells is a good idea.
Making the cells longer is also helpful. Given the constraint
of the total module area.
The scribe area penalty will be discussed later.
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Optimum number of cells
16
𝑃𝑗 = 𝑃𝑜 × 𝑁 = 𝐽02𝜌 ×
𝑛 + 1 3
3 3𝑛 + 1×
𝐴3
𝐿2× 𝑁
= 𝐽02𝜌 ×
𝑛+1 3
3 3𝑛+1×
𝐴𝑇
𝑁
3(𝑁/𝐿2) = 𝑐1/𝑁
2
𝑃𝑠𝑐𝑟𝑖𝑏𝑒 = 𝑁 − 1 𝛿 × 𝐿 × 𝑃𝑖𝑑𝑒𝑎𝑙/𝐴𝑇 ∼ 𝐶2𝑁
𝑃𝑖𝑑𝑒𝑎𝑙 =𝐼𝑚𝑝
𝑁× 𝑉𝑚𝑝 × 𝑁 = 𝐶0
M. A. Alam, PV Lecture Notes
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Optimum number of cells
17
𝑃𝑗 = 𝑐1/𝑁2
𝑃𝑜𝑢𝑡 = 𝑃𝑖𝑑𝑒𝑎𝑙 − 𝑃𝑗– 𝑃𝑠𝑐𝑟𝑖𝑏𝑒
= 𝐶0 − Τ𝐶1 𝑁2 − 𝑐2𝑁
𝑃𝑠𝑐𝑟𝑖𝑏𝑒 ∼ 𝐶2𝑁𝑃𝑖𝑑𝑒𝑎𝑙 = 𝐶0
𝑁𝑜𝑝𝑡 = Τ2𝐶1 𝑐21
3
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Optimum number of cells
18
𝑁𝑜𝑝𝑡 = Τ2𝐶1 𝐶21
3
= 𝑛 + 1 ×𝐴𝑇𝐿
×2/3
(3𝑛 + 1)
13
×𝐽02𝜌
𝐶0𝛿/𝐴𝑇
13
𝐴𝑇 = 𝑁 × 𝐴
𝛿
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Optimum Cell-size: Compared
19
𝑁𝑜𝑝𝑡,1𝑁𝑜𝑝𝑡,2
=𝐽0,12 /𝐶0,1
𝐽0,22 /𝐶0,2
13
=𝐽0,12 /𝜂1
𝐽0,22 /𝜂2
13
=𝑊2
𝑊1
Technology 𝐽0 W (cm)
C-Si 40 0.52
a-Si 16 1.18
CIGS 35 0.75
CdTe 26 Homework
Same ITO, same n
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Series Connection in Thin Film Cells
20
Ala-Si:H
TCO
Laser Scribe
Glass
M. A. Alam, PV Lecture Notes
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2121
Outline
1) Motivation: Power-loss vs. area loss
2) Theory of power loss
3) Power loss in thin-film cells
4) Cell geometry and power-loss
5) Conclusions
M. A. Alam, PV Lecture Notes
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Striping and Shadow Degradation
22
1
2
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Optimum number of cells: Two shapes
23
𝑁𝑜𝑝𝑡,1
𝑁𝑜𝑝𝑡,2=
𝐽0,12 /𝜂1
𝐽0,22 /𝜂2
13
=𝑊2
𝑊1
𝑁𝑜𝑝𝑡,2
𝑁𝑜𝑝𝑡,1= (𝑛 + 1) ×
1
3𝑛 + 1
13 𝐿0
𝐿𝑛
= 1.66 ×1
3
13 2
𝜋= 0.735
Same shape:
Different shapes
(n=0 and n=2/3)
M. A. Alam, PV Lecture Notes
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Observation 1: Shorter triangles are dangerous
24
𝑃1 𝑛 = 1, 𝐿 = 𝑏
𝑃1 𝑛 = 0, 𝐿 = 𝑏=
2313
= 2
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Observation 2: Diagonal Triangle Reduces Power
25
𝑃𝑜 =𝑛+1 3
3 3𝑛+1× (
𝐴3
𝐿2)
𝑃𝑜 𝑛 = 1, 𝐿 → 2𝑏
𝑃0 𝑛 = 0, 𝐿 = 𝑏= 1!
M. A. Alam, PV Lecture Notes
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Design of spiral cells
https://www.flickr.com/photos/16111938@N07/97359514
M. A. Alam, PV Lecture Notes
26
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Average Length: 𝐿~𝜋/2
27
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Observation 3: Spirals are long triangles!
28
𝑃𝑜 =𝑛+1 3
3 3𝑛+1× (
𝐴3
𝐿2)
𝑃𝑜 𝑛~2/3, 𝐿 → (𝜋/2)𝑏
𝑃0 𝑛 = 0, 𝐿 = 𝑏< 1!
M. A. Alam, PV Lecture Notes
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Striping and Shadow Degradation
29
Each radial and spiral have slightly different dissipation
Spiral design improves the performance by reducing the
distance travelled
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M. A. Alam, PV Lecture Notes 30
Conclusions
30
1) Grids allows one to reduce power-loss due to series resistance.
2) An optimum grid balances shadowing loss and series-resistance losses.
3) Grid spacing also depends on absorber materials.
4) Mathematically, spiral designs are optimum. In practice, grids are rectangular.
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Self-assessment
1. The power loss of a thin-film solar cell increases as the (a) first power, (b) second
power, (c) third power, or (d) fourth power of the module area.
2. Compare the power dissipation of two types of cells: n=0 and n=2.
3. What is the ratio of the optimum number of cells for CdTe and Perovskite cells.
4. The power dissipation of a module scales inversely as the (a) first power, (b)
second power, (c) third power, and (d) fourth power of the width of the cell (L).
5. Spiral cells reduce power dissipation by increasing (L) by (a) 20% (b) 50%, (c)
100%, and (d) 200% compared to a traditional cells.
6. Mention three limitations of the spiral design compared to traditional design.
31M. A. Alam, PV Lecture Notes