Mark Lundstrom (2011), "Solar Cells Lecture 2: Physics of Crystalline Solar Cells," http://nanohub.org/resources/11890.

1. NCN Summer School: July 2011
Solar Cell Physics:
recombination and generation
Prof. Mark Lundstrom
lundstro@purdue.edu
Electrical and Computer Engineering
Purdue University
West Lafayette, Indiana USA

2. copyright 2011
This material is copyrighted by Mark Lundstrom
under the following Creative Commons license.
Conditions for using these materials is described at
http://creativecommons.org/licenses/by-nc-sa/2.5/
Lundstrom 2011
2

3. acknowledgement
Dionisis Berdebes, Jim Moore, and Xufeng Wang
played key roles in putting together this tutorial.
Their assistance is much appreciated.
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3

4. solar cell physics
A solar cell is a simple device – just a pn junction with
light shining on it.
To maximize efficiency, we must maximize the
generation of e-h pairs and minimize the recombination
of e-h pairs.
This lecture is a short introduction to the physics of
crystalline solar cells – specifically Si.
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4

5. outline
1) Introduction
2) Recombination at short circuit
3) Recombination at open circuit
4) Discussion
5) Summary
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6. dark current and recombination
- +
N P
ID
s.s. excess s.s. excess
holes electrons
electron-injecting hole-injecting
contact contact
− VA +
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6

7. recombination in the N-type QNR
N - - P
+ ID
electron-injecting hole-injecting
contact contact
− VA +
Anytime an electron and hole recombine anywhere within the diode, one
electron flows in the external circuit. 2011
Lundstrom
7

8. Shockley-Read-Hall recombination
minority carriers injected across junction
ET
Fn qVA FP
SRH
recombination
ID
− VA +
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9. recombination at a contact
minority carriers injected across junction
Fn qVA FP
ID
− VA +
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10. light-current and generation
Vbi − VA “base”
EF (absorbing layer)
“emitter”
− VA + ID < 0
Every time a minority electron is generated and collected, one
10 electron flows in the external current.
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11. light-current and recombination
3 e-h pairs generated
“emitter”
1 e in external circuit
Every time a minority electron is generated and recombines before being
collected, the solar cell current suffers. 2011
Lundstrom 11

12. solar cells and recombination
• Carrier recombination lowers the short-circuit current and
reduces the open-circuit voltage.
• To optimize solar cell performance, we need a clear
understanding of how many carriers are recombining and
where they are recombining.
• Then we need to establish a quantitative relation between
recombination and solar cell performance.
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13. solar cells and recombination
= q ( RTOT (VA ) − GTOT )
J D (VA )
J p ( 0) Jn ( L)
ID L
J p ( 0) Jn ( L)
RTOT= ∫ R ( x )dx −
0
q
−
q
N P L
GTOT = ∫ Gop ( x )dx
0
0 L x
For a formal derivation of this result, see the appendix.
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14. outline
1) Introduction
2) Recombination at short circuit
3) Recombination at open circuit
4) Discussion
5) Summary
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14

15. generic crystalline Si solar cell
SF = 1000 cm/s
key device
n+ “emitter” (0.3 μm)
parameters
p-type “base” base doping: NA = 1016 /cm3
emitter doping ND = 6 x 1019 /cm3
200 um
(198.9 μm) minority carrier lifetime τn = 34 μs
(base)
p+ “Back Surface Field” (BSF) base thickness W = 198.9 μm
(0.8 μm)
front junction depth xjf = 0.3 μm
back junction depth xjb = 0.8 μm
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15

16. light-generated current
SF = 1000 cm/s = q ( RTOT ( 0 ) − GTOT )
J D ( 0)
n+ “emitter” (0.3 μm)
1) What is GTOT?
p-type “base”
2) How is GTOT spatially distributed?
200 um
3) What is RTOT?
(198.9 μm)
4) How is RTOT spatially distributed?
p+ “Back Surface Field” (BSF) 5) How do things change if we
(0.8 μm) remove the BSF?
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16

17. light-generated current: numbers
J SC J D (V= 0= q ( RTOT − GTOT )
= A )
n+ “emitter” (0.3 μm)
∞
WD ≈ 0.3 µ m =
GMAX ∫ Gop ( x = 2.97 ×1017 cm -2s -1
)dx
0
200 um
2L
G=
TOT ∫ Gop ( x )dx 2.79 ×1017 cm -2s -1
=
p-type “base” 0
(198.9 μm) Ln ≈ 320 µ m
J SC 39.4 mA/cm 2
p+ “Back Surface Field” (BSF) = = 2.46 ×1017 cm -2s -1
q q
(0.8 μm)
RTOT ( 0 ) 3.31×1016 cm -2s -1
=
17
CE = 0.88
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18. light-generated current: understanding
entire device near surface
xj x j + WD
18 Lundstrom 2011

19. light-generated current: summary
∞ 2L
= ∫ Gop ( x = 2.97 ×10 cm s
)dx G= ∫ Gop ( x )dx 2.79 ×1017 cm -2s -1
=
17 -2 -1
GMAX TOT
0 0
low lifetime (Auger recombination) good
minority carrier lifetime
surface recombination collection
BSF
19 Lundstrom 2011

20. recombination at short circuit
entire device near surface
xj x j + WD
20 Lundstrom 2011

21. recombination at short circuit: summary
J SC 39.4 mA/cm 2
= = 2.46 ×1017 cm -2s -1 RTOT ( 0 ) 3.31×1016 cm -2s -1
=
q q
(0.37)
(0.49)
(0.14)
low lifetime (Auger recombination) good minority carrier lifetime
surface recombination collection BSF
21 Lundstrom 2011

22. about recombination in the base
∆n ( x ) d 2 ∆n ∆n
expect: R ( x ) ≈ 2
− =
0 Ln = Dnτ n
τn dx Ln
We find the excess minority
electron profile by solving the
∆n
minority carrier diffusion
equation:
J n = q sback ∆n ( L′ )
( L′ )
d
( J n −q ) =R
−
dx 0′ x j + W
=
d ∆n J n ( 0′ ) q s j ∆n ( 0′ )
= L′ L − xBSF
=
J n ≈ qDn
dx x
xj +W L
22 Lundstrom 2011

23. Adept simulation results
∆n ( x )
R ( x) ≈
τn
∆n ( x )
23 Lundstrom 2011

24. the BSF
∆E = eV
0.13
EC
EI Sback ≈ υ th e− ∆E kBT
EF  ; 0.6 × 10 7 cm s
EV
What happens if we
remove the BSF?
EC
EI
Sback ≈ υ th
EF
 ; 1 × 10 7 cm s
EV
24 Lundstrom 2011

25. without the BSF
BSF
no BSF
With BSF Without BSF
J SC = 39.4 mA/cm 2 J SC = 38.2 mA/cm 2
qRTOT = 5.3 mA/cm 2 qRTOT = 6.5 mA/cm 2
25
CE = 0.88
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CE = 0.85

26. internal quantum efficiency
With BSF
No BSF
J D (V = 0, λ )
IQE =
Finc ( λ )
26 Lundstrom 2011

27. questions
1) Can you determine a way to find the actual back surface
recombination velocity from the Adept simulation results.
(Hint: Use plots of n(x) and Jn(x).)
2) How much could the performance improve if the back
surface recombination velocity could be reduced to zero?
3) With the original BSF, how much would the performance
increase if the minority carrier lifetime was 10 times longer?
4) In the original design, how would the short-circuit current
change if the base was twice as thick?
5) Since most of the recombination loss occurs in the emitter,
why not just make the emitter junction depth a lot smaller?
27 Lundstrom 2011

28. 2D effects
ID I ( x)
VD V ( x ) < VD
xj
dx
dR = ρ S
W
ρ 1
ρS
= =
x j N D qµn x j
distributed series resistance
28 Lundstrom 2011

29. outline
1) Introduction
2) Recombination at short circuit
3) Recombination at open circuit
4) Discussion
5) Summary
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29

30. dark I-V
= q ( RTOT (VA ) − GTOT )
J D (VA )
= q ( RTOT = VOC ) − GTOT )
0 (VA
Under open circuit conditions:
RTOT = VOC ) GTOT
(VA =
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30

31. superposition
JD
= q ( RTOT (VA ) − GTOT )
J D (VA ) dark IV
J SC
dark:
= J 0 ( e qVD
JD nk B T
− 1)
J dark
D (VA ) = q Rdark
TOT (VA )
VA
illuminated: VOC
= q ( RTOT (VA ) − GTOT )
J D (VA )
light light
− J SC
JL < 0
illuminated at VOC: superposition:
J D (VOC ) = J SC
dark
RTOT (VOC ) = GTOT
light ?
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RTOT (VOC ) = J SC q
dark
31

32. dark current characteristics (sketch)
= J 0 ( e qVA
J D (VA )
dark nk B T
− 1)
J D (VA ) J 01 ( e qVA
=
dark kBT
− 1) + J 02 ( e qVA 2 kBT
− 1)
series
resistance
or…
n=1
dark
log10 J D shunt
resistance
or…
n=2
VA
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32

33. dark current characteristics (Adept)
= J 0 ( e qVA
J D (VA )
dark nk B T
− 1)
=
dark
(
J D (VA ) J 01 e qVA kBT
) (
− 1 + J 02 e qVA 2 kBT
)
−1
n >1
n =1
n≈2
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33

34. what determines J0 and n?
J D (VA )
= J 0 e qVA
dark
( nk B T
)
−1
J A (VA ) = q RTOT (VA )
dark dark
Answer:
Electron-hole recombination determines I0.
The location of recombination within the solar cell
determines the ideality factor, n.
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34

35. recombination in the dark (VA = 0.7 V)
Emitter Base
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35

36. recombination summary: (VA = 0.7 V)
Short-circuit recombination VA = 0.7 V recombination
qRTOT ( 0 ) = 5.3 mA/cm
light 2 qRTOT ( 0.7 ) = 465 mA/cm 2
dark
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36

37. what happens if we remove the BSF? (VA = 0.7 V)
With BSF Without BSF
~70%
~85%
J D ( 0.7 ) = 644 mA/cm 2 J D ( 0.7 ) = 1372 mA/cm 2
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37

38. dark current physics (n = 1)
FB: minority carriers injected across junction I D (VA ) = qRTOT (VA )
1) Recombination in QNRs:
Fn qVA FP
2) Electrons and holes can also
recombine within the SCR of
ID > 0 Lundstrom 2011
the junction.
38

39. n = 1 device physics
I D (VA ) = qRTOT (VA )
nP ( 0′ ) ≈ n0 P e qVA kBT
Qn
qRTOT (VA ) =
tn
q (Vbi − VA )
ni2 qVA
Qn ∝
NA
e ( kBT
−1 )
Fn FP
tn : minority carier lifetime
n0P ≈ ni2 N A or base transit time
Recombination in quasi-neutral regions gives rise to n = 1 currents.
Lundstrom 2011 39

40. dark current characteristics (sketch)
= J 0 ( e qVA
J D (VA )
dark nk B T
− 1)
J D (VA ) J 01 ( e qVA
=
dark kBT
− 1) + J 02 ( e qVA 2 kBT
− 1)
series
resistance
or…
n=1
dark
log10 J D shunt
resistance
or…
n=2
VA
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40

41. recombination in the dark (VA = 0.2 V)
emitter region base region
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41

42. recombination summary: (VA = 0.2 V)
VA = 0.7 V recombination VA = 0.2 V recombination
qRTOT ( 0.7 ) = 465 mA/cm 2
dark
qRTOT ( 0.7 ) 8.4 ×10−6 mA/cm 2
dark
=
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42

43. dark current physics
FB: minority carriers injected across junction I D (VA ) = qRTOT (VA )
1) Recombination in QNRs:
Fn qVA FP
2) Electrons and holes can also
recombine within the SCR of
the junction.
ID > 0 Lundstrom 2011
43

44. recombination in SCRs
J D (VA ) = qRTOT (VA )
dark
q (Vbi − VA )
Maximum recombination
occurs when n(x) ≈ p(x)
n ( x ) p ( x ) = ni2 e qVA kBT
Fn FP
n ≈ p ∝ ni e qVA
ˆ ˆ 2 kBT
qni e qVA 2 kBT
np = ni2 e qVA kBT
qRTOT (VA ) ∝
dark
τ eff
Recombination in space-charge regions gives rise to n = 2 currents.
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44

45. recombination in SCR
J D (VA ) = qRTOT (VA )
n ≈ p ∝ ni e qVA
ˆ ˆ 2 kBT
ˆ
n ni e qVA / 2 kBT
R (VA )
ˆ = =
τ eff τ eff
J D (VA ) = q R Weff
ˆ
k BT q
Weff =
Eˆ
k BT q
E ˆ = 2.3 × 104 V cm =
Weff ≈ 11 nm
E ˆ
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45

46. dark IV
J D (VA )
= J 02 e qVA ( 2 kBT
) ( )
− 1 + J 01 e qVA = J 0 e qVA
1k B T
−1 ( nk B T
)
−1
Recombination in Recombination in
depletion regions neutral regions
J 02 ∝ ni ∝ e − EG / 2 kBT J 01 ∝ ni2 ∝ e − EG / kBT
large bandgaps and small bandgaps and
low temperatures high temperatures
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46

47. questions
1) What do you expect to happen if the BSF were removed?
Run an Adept simulation to confirm.
2) What do you expect to happen if the minority carrier lifetime
were reduced to 0.1 microseconds? Run an Adept
simulation.
3) Why is recombination in the emitter so important under short-
circuit conditions, but not under FB in the dark?
4) How much could VOC be increased if a BSF with near-zero
surface recombination velocity could be achieved?
5) Series resistance affects the dark current, but it has no effect
at open-circuit. What are the implications?
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47

48. outline
1) Introduction
2) Recombination at short circuit
3) Recombination at open circuit
4) Discussion
5) Summary
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48

49. reducing recombination
higher material quality (longer lifetimes)
thinner base layer (but optically thick)
J D (VA ) q ( RTOT (VA ) − GTOT ) built-in fields
back-surface-fields / minority carrier mirrors
reducing contact areas
….
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49

50. high-efficiency Si solar cells
24.5% at 1 sun
Martin Green Group UNSW – Zhao, et al, 1998
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50

51. how good is superposition?
V = 0.62 V - Dark VOC = 0.62 V - Illuminated
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51

52. how good is superposition? (ii)
dark
JD
Jdark
light
JD
J D + J D (V = 0 )
dark light
superposition
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52

53. outline
1) Introduction
2) Recombination at short circuit
3) Recombination at open circuit
4) Discussion
5) Summary
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53

54. summary
1) Diode current = q times (total recombination – total
generation)
2) At VOC, recombination = optical generation
3) At V = 0, recombination lowers the collection efficiency
4) Dark current tells us much about the internal
recombination mechanisms
5) Solar cell design is all about maximizing total generation
and minimizing total recombination.
6) Simulations can be useful for understanding –especially
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if you look “inside” and not just at the IV. 54

55. questions
1) Introduction
2) Recombination at short circuit
3) Recombination at open circuit
4) Discussion
5) Summary
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55

56. Appendices
1) Formal derivation of the relation between current and
recombination/generation.
2) Mathematical justification of superposition
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57. Appendix 1: current and recombination
Formal derivation of the relation between current and
recombination/generation.
J D (V ) q ( RTOT − GTOT )
=
J p ( 0) Jn ( L)
ID L
J p ( 0) Jn ( L)
RTOT= ∫ R ( x )dx −
0
q
−
q
N P L
GTOT = ∫ Gop ( x )dx
0
0 L x
Lundstrom 2011 57

58. continuity equation for electrons
Wabash
River
Rate of increase of
water level in lake = (in flow - outflow) + rain - evaporation
∂n 
∂t
= (
−∇ • J n −q )+ G − R
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58
58

59. solar cell physics
“semiconductor equations”
Conservation Laws: Relations:
  
κε −κε
D = 0 E = 0 ∇V

∇• D =ρ ρ q( p − n + N D − N A )
= + −
  
 J= nq µn E + qDn∇n
(
∇ • J n −q
= ) (G op − R) 
n
 
J= pq µ p E − qD p ∇p
p
 R = f (n, p )
(
∇• Jp =
q ) (G op − R)
Gop = optical generation rate
(steady-state) etc.
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60. diode current and recombination

(
∇ • J n −q
= ) (G op − R)
d ID
( J n −q ) = Gop − R (1D) ID
dx
L L
N P
= q ∫  R ( x ) − Gop ( x )  dx
∫ dJ n  
0 0
0 L x
L
J n ( L ) − J n ( 0= q ∫  R ( x ) − Gop ( x )  dx
)  
0
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60

61. current and recombination-generation
L
J n ( L ) − J n ( 0= q ∫  R ( x ) − Gop ( x )  d + J p (x ) − J p ( 0 )
)   0
0
L
− { J n ( 0 ) + J p ( 0 )} = J D (V ) = q ∫  R ( x ) − Gop ( x )  dx − J n ( L ) − J p ( 0 )
 
0
J D (V ) q ( RTOT − GTOT )
=
ID
L
qRTOT q ∫ R ( x )dx − J n ( L ) − J p ( 0 )
=
0
L N P
GTOT = ∫ Gop ( x )dx
0
61
Lundstrom 2011 0 L x
61

62. current and generation-recombination
= q ( RTOT (VA ) − GTOT )
J D (VA )
The diode current is q times the total recombination minus the total
generation.
The total recombination is the integrated recombination rate within
the device plus the flux of minority carriers into each contact.
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62 62

63. Appendix 2: justifying superposition
= q ( RTOT (VA ) − GTOT )
J D (VA ) (valid in light or dark)
J D (VA ) = qRTOT (VA )
dark dark
(dark current)
= q ( RTOT ( 0 ) − GTOT )
J D ( 0)
light light
(short circuit current)
J D (= J D + J D ( 0 )
super
VA ) dark light
(principle of superposition)
J D (VA ) = qRTOT (VA ) + q ( RTOT ( 0 ) − GTOT )
super dark light (How does this compare to
the exact answer?)
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63

64. mathematical justification for superposition
( ) ( ( )
J D V A = q RTOT V A − GTOT ) (valid in light or dark)
( ) ( ( )
J D V A = q RTOT V A − GTOT
light light
)
( ) ( ) (
J D V A = qRTOT V A + q RTOT 0 − GTOT
super dark light
() ) (principle of superposition)
( )
light
( )
dark light
()
RTOT V A = RTOT V A + RTOT 0 ?? (criterion to justify superposition)
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64