PV Performance Limits by Shockley-Queisser (SQ) Triangle/uploads/Lecture13... · 3. What is the maximum short-circuit current achievable from AM1.5 illumination? Ans. 70 mA/cm^2

PV Performance Limits by Shockley-Queisser (SQ) Triangle

M. A. Alam

[email protected]

Electrical and Computer Engineering

Purdue University

West Lafayette, IN USA

Theory and Practice of Solar Cells: A Cell to System Perspective

1

M. A. Alam, PV Lecture Notes

Outline

3

1) Motivation

2) Derivation of the SQ-Triangle

3) Applications of SQ-Triangle

4) Conclusions

5) Appendix: Self-test questions


An Inefficient Machine!

4

𝜂𝐴 ∼ 1/2

𝜂𝑁 = 2/𝜋

𝜂𝑆𝑄 = 1/3

𝜂 = 𝜂𝑁 × 𝜂𝑆𝑄 × 𝜂𝑀 × 𝜂𝐴=2

𝜋×1

3×5

6×1

2∼

1

10

Tandem Bifacial

𝜂𝑀 = 4/5


Power and efficiency

5

𝐸1

𝐸2

Incre

asing

ban

dgap

Incre

asing b

andgap

𝑞𝑉𝑚𝑝 = 0.95𝐸𝑔 − 0.3

𝐽𝑚𝑝 = 82(1 − 0.428𝐸𝑔)

𝐽𝑆𝑈𝑁∗

𝐽𝑚𝑝

𝑉𝑚𝑝 (𝐸𝑔)

𝐽𝑚𝑝,1

𝑉𝑚𝑝,1

𝜂

𝐸𝑔,1

Counting the photons – a roadmap

6

Shockley-QueisserYablonovitch

Partition,

Cell-Module

Radiative

Angle Entropy

Carnot

Below bandgap

Above bandgap

LED

CPV

Tandem

MEG

>4n2

Striping

c

c c


Outline

7

1) Motivation

2) Derivation of SQ-Triangle

3) Application of SQ-Triangles

4) Conclusions


Defining the SQ-Triangle

8

𝐽𝑚𝑝 = 82(1 − 0.428𝐸𝑔)

𝑞𝑉𝑚𝑝 = 0.95𝐸𝑔 − 0.3

𝐽𝑚𝑝 = 72 1 − 0.52 𝑉𝑚𝑝

1. Calculate 𝑉𝑚𝑝 and 𝐼𝑚𝑝

2. Calculate efficiency:

𝜂 = 𝑉𝑚𝑝 𝐼𝑚𝑝 /𝑃𝑖𝑛

3. Locate in the triangle


Step 1: Calculate 𝐽𝑚𝑝

9

𝐽𝑚𝑝 = 83.75(1 − 0.428𝐸𝑔)


Step 2: Derive 𝑉𝑚𝑝 by a 2-Level Atom

E1

E2

m1

m2

nph(Ts)

TDSome deep sea Plackton live

in a similar diffused light environment

Single frequency filtered light

Isotropic illumination

Atom with only two levels

10


Detailed balance between Photons and Electrons

E1

E2

m1

m2

nph(TS)

TD

1 2(1 )( 1) phf nf

2 2(1 ) phf f nUpward transition

Downward transition

1 2 2 2(1 )( 1) (1 ) ph phf n f ff n

At steady state up and down transitions are equal …

1

2

1 /

2 /

1

1

1

1

D

D

T

T

f

f

e

e

1 1 1 2 2 2 m m B Bk E k E

11M. A. Alam, PV Lecture Notes

Detailed balance of photons and electrons

1 2 2 2(1 )( 1) (1 )ph phf f fn nf

2 1

1 2 2 1

/ /

/ / / /

1 1( 1) (

1)

1 1 1

D D

D D D Dph

T T

T T pT T hne

e en

e

e e

E1

E2

m1m2

TD

/

/ /

11

1 1

S

S S

B

B Bph p Th

k T

k T kn n

e

e e

Bose Einstein

distribution

12M. A. Alam, PV Lecture Notes

PV Efficiency of 2-level System

nph(Ts)

1 2 1 2 1( ) ( )m m

D

S

ET

ET

Single molecule-single frequency gives Carnot efficiency!

2 2 2 1 1 1m m

D S D

E E E E

T T T

E1

E2

m1

m2

TD

300

6000 1 0.95

1 2

1 2

1m m

D

S

R T

E E R T

1 2( ) 0

if

m m

SDT T

Can not extract

energy at same

temperature

13


Problem of Angular Anisotropy

The angle mismatch leads to Vmp loss

12 1 2 2 2(1 )( 1) (1 ) ph phf f n f f n

negative number

12

SUN

1 2 2 1 2 1 1

2

or, ln

B D B S B D

E E E E

k T k T k T

m

m

2

11 2 1 ln

m m

Dmp g B D

S

TV E k T

T

14

Step 1: Derivation of Vmp complete

5

1 06 1

2 4

2 2

Without mirror

With mirror

1 2 1

1 2 1 2 2

1 lnD B D

S

T k T

E E T E E

m m

Consistent with all experimental data !

(1.10

0.706)

(1.45,

1.111)

(1.01

0.736)Green,

Prog. In PV, 2011

1 2 2

1

2

0.95 0.31

,

Set, 6000 , 300 ,

2 :

1 ln

m

S D

g

Dg B D

S

g

T K T K

E E E

TE k TqV E

T

12

SUN

15

Concentrator cells fool the cells

, S C D

1 1ln

S

D

Dmp g B D

S

Dg

S

TE

T

TV E k T

T

Number of suns … 2 / 104720SN

16


Step 3: Creating the SQ Triangle

𝐸1

𝐸2

𝑞𝑉𝑚𝑝 = 𝐸𝑔 1 −𝑇𝐷𝑇𝑆

– 𝑘𝐵 𝑇𝐷 ln𝜃𝐷𝑐 𝜃𝑆

= 𝑐𝑓𝐸𝑔 – Δ

𝐽𝑚𝑝 𝐸𝑔 = 𝑐𝐼𝑠𝑢𝑛 1 − 𝛽′𝐸𝑔

= 𝐽0 1 − 𝛽 𝑉𝑚𝑝

𝑉𝑚𝑝

𝐼𝑚𝑝

Hirst, PIP, 2011

Khan & Alam, APL, 2015,

Also, see AJP, 2012

𝛽 ≡ 𝛽’ /𝑐𝑓

𝑉0 ≡ (1 − 𝛽Δ Τ) 𝛽

)𝐽0 ≡ 𝑐𝐼𝑠𝑢𝑛 (1 − 𝛽Δ

17

Outline

18

1) Motivation

2) Derivation of SQ-Triangle

3) Applications of SQ-Triangle

4) Conclusions


𝑆1 = 𝑥 1 − 𝑥

𝑑𝑆1𝑑𝑥

= 1 − 2𝑥 = 0 → 𝑥 =1

2

Maximum Single Junction Efficiency

𝐽𝑠𝑢𝑛∗

𝐽𝑚𝑝


𝐽𝑠𝑢𝑛∗ /2

𝑉𝑚𝑝,1

1.0

𝐽𝑚𝑝/𝐽𝑠𝑢𝑛∗

𝛽𝑉𝑚𝑝

1 − 𝑥

𝑥 1.0

0.52 ∗ 𝑉𝑚𝑝 =1

2, 𝑉𝑚𝑝 = 0.96, 0.95 × 𝐸𝑔 − 0.3 = 𝑉𝑚𝑝 = 0.96, 𝐸𝑔 = 1.33𝑉

𝐽𝑚𝑝 = 72 1 − 0.52 𝑉𝑚𝑝


19

Tandem Solar Cell

20

𝑬𝟎

𝑬𝟏+

𝑬𝒊+

𝑬𝑷

TCO

𝑬𝟏−

𝑬𝒊−

𝑬𝑸

TCO

1-Sun

R-Sun


N-Junction Tandem Cell

𝐸𝑔,1, 𝑉𝑚𝑝,1



𝑉𝑚𝑝𝑖= Τ𝑖𝑉0 𝑁 + 1

𝐼𝑚𝑝𝑖= Τ𝐼0 (𝑁 + 1)

𝑉𝑚𝑝 = 𝑐𝑓 𝐸𝑔 – 0.31 – 𝑘𝐵 𝑇𝐷 ln 𝑐


21

Efficiency of N-junction Tandem

𝜂𝑁 𝑐 = (𝐼0 𝑉0 Τ/4𝑐) × 2𝑁 (𝑁 + 1)

)𝐼0 ≡ 𝑐𝐼𝑠𝑢𝑛 (1 − 𝛽Δ

𝑉0 ≡ (1 − 𝛽Δ Τ) 𝛽𝛽 = 𝛽 Τ’ 𝑐𝑓


22

Tandem Efficiency Limits

𝐽𝑆𝑈𝑁∗

𝐽𝑚𝑝


𝐽𝑆𝑈𝑁∗

2

𝑉𝑚𝑝,1

𝐽𝑆𝑈𝑁∗

𝐽𝑚𝑝

𝑉𝑚𝑝(𝑋𝑔)

1

3𝐽𝑆𝑈𝑁∗

2

3𝐽𝑆𝑈𝑁∗

𝐽𝑆𝑈𝑁∗

𝐽𝑚𝑝

𝑉𝑚𝑝(𝑋𝑔)

1

𝑁𝐽𝑆𝑈𝑁∗

𝑁 − 1

𝑁𝐽𝑆𝑈𝑁∗

𝐽𝑚𝑝 = 70 1 − 0.52 𝑉𝑚𝑝 𝑃𝑜𝑢𝑡 = 𝐽𝑚𝑝 × 𝑉𝑚𝑝

𝜂 = 33% 44% 66%


23

Bifacial Solar Cell

24

𝜂 = 𝜂𝑁 × 𝜂𝑆𝑄 × 𝜂𝑀 × 𝜂𝐴=2

𝜋×1

3×5

6×1

2∼

1

10

𝜂𝐴 ∼ 1/2


𝐽𝑚𝑝

𝐽0= 1

𝑉𝑚𝑝 /𝑉01 = 1

𝑥 2𝑥

𝐸1+

𝐸2+

For inverted bifacial design

𝑆4 = 𝑥(1 − 𝑥) + 𝑥 (1 − 2𝑥) + 𝑥 1 − 2𝑥 − 𝛾𝑥 + 𝑥 1 −𝑥

𝑅

𝐽𝑚𝑝

𝐽0= 𝑅

2𝑥+𝛾𝑥

𝑥+(1

−𝛾)𝑥

𝑥

𝐸0𝐸0

𝐸2−1

𝛾 = 1/(1 + 𝑅)

1 − 2𝑥 − 𝛾𝑥

1 −𝑥

𝑅


25

SQ Triangle for Bifacial PV

𝑆𝑁𝑆1

=2 1 + 𝑅 𝑁2

𝑁 𝑁 + 1 𝑅 + 1 − 2𝑅

𝑆𝑁𝑆1

=8𝑅 1 + 𝑅 𝑁2

2𝑅 2𝑁2 + 4𝑁 − 1 – 𝑅2 − 1

𝑁𝑐𝑟𝑖𝑡≤

1+

𝑅−1


26

Advantages of Bifacial PV

𝑆𝑁𝑆1

=2 1 + 𝑅 𝑁2

𝑁 𝑁 + 1 𝑅 + 1 − 2𝑅

𝑆𝑁𝑆1

=8𝑅 1 + 𝑅 𝑁2

2𝑅 2𝑁2 + 4𝑁 − 1 – 𝑅2 − 1

𝑁𝑐𝑟𝑖𝑡 ≤ 1 + 𝑅−1


27

Plant-PV Tandem cells

𝜆

𝜆𝜆


28

Rescaled Tandem System

𝐽𝑆𝑈𝑁

𝐽𝑆𝐶

𝐸𝑔

For plants

𝐽𝑆𝑈𝑁𝑛𝑒𝑤

𝐽𝑆𝐶

𝑋𝑔

𝐽𝑆𝑈𝑁 83.75 mA/cm2

𝛽 0.428 eV−1

𝐽𝑆𝑈𝑁𝑛𝑒𝑤 63.2095 mA/cm2

𝛽𝑛𝑒𝑤 0.5666 eV−1

With the new parameters

we can now redesign any

N-J tandem

𝐽𝑆𝐶 𝑋𝑔 = 𝐽𝑆𝑈𝑁(1 − 𝛽𝑋𝑔)

𝐽𝑆𝐶𝑛𝑒𝑤 𝑋𝑔 = 𝐽𝑆𝑈𝑁 1 − 𝛽𝑋𝑔 − 𝐽𝑆𝐶

𝑐𝑢𝑡

= 𝐽𝑆𝑈𝑁𝑛𝑒𝑤 1 − 𝛽𝑛𝑒𝑤𝑋𝑔


29

Conclusions

• Solar cells are inefficient machines. A thermodynamic analysis

shows how we can improve the performance.

• A simple SQ Triangle anticipates thermodynamic limits for

tandem PV, bifacial PV, and concentrator PV cells.

• Tandem efficiency scales as: 𝜂𝑁 𝑐 = 𝐼0 𝑉0 Τ𝑁 2 𝑁 + 1 𝑐

• Bifacial PV has phase transition at 𝑁𝑐𝑟𝑖𝑡 ≤ 1 + 𝑅−1

• Thermodynamic limit serves as a beacon and a guardrail for

next generation PV.


30

31

Single (SJ) or multi-junction (MJ)SJ: J-V or Eg-sweepMJ: J-V or N-sweep

Input spectrum: AM1.5G or ideal BlackbodyDistance from sun (for Blackbody input)Spectral low-pass filter (for MJ calculations)Reflectance of the ground (for MJ calc.)

Simulation setup

Simulation specific input

Spectral input

These set of inputs change based on the choice of simulation setup

(a) (b) (c)

nanohub.org/resources/pvlimitshttp://arxiv.org/abs/1606.01176


Self-assessment

1. How does the average bandgap of tandem cell compare to that of a single

junction solar cell?

2. How does the efficiency of a triple junction solar cell compare to that of a

single junction solar cell? Hint. Efficiency ratio is given by 2N/(N+1).

3. What is the maximum short-circuit current achievable from AM1.5

illumination? Ans. 70 mA/cm^2.

4. What is the maximum short circuit current for a 4-junction solar cell? Ans.

70/(N+1) = 14 mA/cm^2

5. For R=0.3, what is efficiency of a 4-junction bifacial solar cell?

6. If R_s increases by a factor of 5, how much does the critical concentration for

a CPV decrease by? 32M. A. Alam, PV Lecture Notes

34

Wait, Wait don’t tell me …

Maximum Jsc (in mA/cm2) from AM1.5

spectrum is

a) 7000 b) 700 c) 70 d) 7

70 mA/cm2


35


Maximum Jsc for Eg=1.7 eV perovskite

under AM1.5 illumination is

a) 100 b) 50 c) 20 d) 5

Jsc=83.75(1-0.43Eg)

=23 mA/cm2


36


For a 3-junction tandem, maximum

Jsc in mA/cm2 is

a) 100 b) 70 c) 17 d) 4

Jsc=Jmax/(N+1)Jsc

Vmp


37


If Eg=1eV, maximum 𝑉𝑚𝑝 is

a) 1.0 b) 0.65 c) 0.45 d) 0.25

Voc=0.95Eg-0.31


38


In a N=3 tandem,

E1=1.9eV, E3=0.97,

what is E2?

a) 2.5 b) 1.6 c) 1.3 d) 1.1

Eg,ave=Eg,SJ=1.33


39


In a N=3 tandem, What is Vmp?

a) 4 b) 3 c) 2 d) 1

Vmp/N=0.95Eavg-0.31


40

A Little Formula Sheet

𝐸𝑔,𝑆𝐽𝑜𝑝𝑡

≅ 2.55 𝑘𝑇𝑠

𝑞𝑉𝑜𝑐,𝑆𝐽 = 0.95 × 𝐸𝑔 − 0.232

𝑞𝑉𝑚𝑝,𝑆𝐽 = 0.95 × 𝐸𝑔 − 0.31

𝐽𝑠𝑐,𝑆𝐽 = 𝐽0 1 − 𝛽𝐸𝑔 (AM1.5, 𝐽0 = 83.75, 𝛽 = 0.428).

𝐹𝐹 ~ Τ𝑣𝑜𝑐 𝑣𝑜𝑐 + 4.7

𝜂𝑇.𝑆𝐽 = −26.45𝐸𝑔2 + 70.77𝐸𝑔 − 14.42 (AM1.5,

empirical)

𝐽𝑠𝑐 𝑁 =2

𝑁+1𝐽𝑠𝑐( 𝐸𝑔 )

𝑞𝑉𝑚𝑝/𝑁 = 𝐸𝑔 1 −𝑇𝐷

𝐸𝑔

𝐸𝑔,𝑚𝑎𝑥

𝑇𝑠– 𝑘𝐵 𝑇𝐷 𝑙𝑛

𝜃𝐷

𝜃𝑆

𝐸𝑔,max =𝑁 − 1

𝛽 𝑁+𝛽 1 + 𝑅 𝐸0 − 𝑅

𝛽 × 𝑁

(𝑇 − 𝑇𝑎) = 𝑃/ℎ = 1000(1 − 𝜂 − 𝑅)/ℎModule

Tandem

SJ cell


Documents

PV Performance Limits by Shockley-Queisser (SQ) Triangle/uploads/Lecture13... · 3. What is the maximum short-circuit current achievable from AM1.5 illumination? Ans. 70 mA/cm^2