how a module gets its stripes: thin film... · laser scribe glass m. a. alam, pv lecture notes. 21...

How a Module gets its Stripes: Thin Film

M. A. Alam

alam@purdue.edu

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

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

Photoelectric effect and solar cells

UVelectron

metal

Sunlightelectron

p-n junction

load

4

M. A. Alam, PV Lecture Notes

Soland

The Puzzle of Striping

66

𝛿

Ala-Si:H

TCO

Laser Scribe

Glass

M. A. Alam, PV Lecture Notes

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

8

Modules have cells connected in series

Si module Thin-film module

M. A. Alam, PV Lecture Notes

9

c-Silicon

*Google Images

a-silicon

M. A. Alam, PV Lecture Notes

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

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

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

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

M. A. Alam, PV Lecture Notes

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

M. A. Alam, PV Lecture Notes

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.

M. A. Alam, PV Lecture Notes

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

Optimum number of cells

17

𝑃𝑗 = 𝑐1/𝑁2

𝑃𝑜𝑢𝑡 = 𝑃𝑖𝑑𝑒𝑎𝑙 − 𝑃𝑗– 𝑃𝑠𝑐𝑟𝑖𝑏𝑒

= 𝐶0 − Τ𝐶1 𝑁2 − 𝑐2𝑁

𝑃𝑠𝑐𝑟𝑖𝑏𝑒 ∼ 𝐶2𝑁𝑃𝑖𝑑𝑒𝑎𝑙 = 𝐶0

𝑁𝑜𝑝𝑡 = Τ2𝐶1 𝑐21

3

M. A. Alam, PV Lecture Notes

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

M. A. Alam, PV Lecture Notes

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

M. A. Alam, PV Lecture Notes

Series Connection in Thin Film Cells

20

Ala-Si:H

TCO

Laser Scribe

Glass

M. A. Alam, PV Lecture Notes

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

Striping and Shadow Degradation

22

1

2

M. A. Alam, PV Lecture Notes

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

Observation 1: Shorter triangles are dangerous

24

𝑃1 𝑛 = 1, 𝐿 = 𝑏

𝑃1 𝑛 = 0, 𝐿 = 𝑏=

2313

= 2

M. A. Alam, PV Lecture Notes

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

Design of spiral cells

https://www.flickr.com/photos/16111938@N07/97359514

M. A. Alam, PV Lecture Notes

26

Average Length: 𝐿~𝜋/2

27

M. A. Alam, PV Lecture Notes

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

Striping and Shadow Degradation

29

Each radial and spiral have slightly different dissipation

Spiral design improves the performance by reducing the

distance travelled

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.

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