Solar PV Technology From Atoms to Arrays

Solar Photovoltaic Technology from atoms to arrays
Ahmed Ennaoui
Helmholtz-Zentrum Berlin für Materialien und Energie
Science Advisory Board Member of IRESEN - Morocco
E-mail: ennaoui@helmholtz-berlin.de
https://www.helmholtz-berlin.de
The International Renewable and Sustainable Energy Conference(IRSEC'13)
March 7-9 2013, Ouarzazate, Morocco
http://www.iresen.org/index.phpLecture 4 on Friday 09h30 10h15‐

Introduction: PV from atom to array
Arrays
 Absorbed photon creates 1 electron-hole pair.
 The electric field separates the electron-hole pair.
 The electrons are collected in the external load.
 Generation-Recombination.
Enery levels
Atom
Module
Solar cell

Prof. Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
 Photo absorption and photo generation
 Direct and indirect band gap.
 External and Internal Quantum Efficiency (EQE and IQE).
 Absorption coefficient, absorption length, excess minority carrier.
 Recombination: Non Radiative, Radiative, Auger.
 Shockley-Read Hall Recombination.
 Continuity equation and Transport process.
 Basic J-V equation.
 Equivalent Circuit model.
 Silicon Technology versus. Thin Film technology.
 Basic building block for PV: cells in series, cells in parallel.
 Change in short circuit current and open-circuit with solar radiation.
 Change in short circuit current and open-circuit with the temperature
 Performance measurement standard conditions
What we have to learn

Task of Photovoltacis: Photo absorption and photo generation
Light = wave λ, and particle with energy E = hν
Albert Einstein
1879 - 1955
Max Planck
1858 - 1947
)(
1239
)(
hc
hE
nm
eVE
λλ
ν =⇒==
)rkexp()rk,()rk,( ⋅= iunnψ
Function with
the periodicity
of the crystal
lattice
Plane wave
)rkexp()rk,()rk,( ⋅= iunnψ
Function with
the periodicity
of the crystal
lattice
Plane wave
 Use of Bloch functions
Band structure of Si E(k)
1000 nm  1.239 eV≅ 1.4 eV
 Solving Schrödinger
equation
ψψψ ErV
m
=+∇− )(
2
2
0
2

Particle in a box: wave functions and energies
n ; the quantum number (n= 1, 2, 3,....)
L ; the length one dimensional) molecular box
m ; the mass of the particle (electron)
h ; Planck's constant

Device fabrication
1. Surface etch, Texturing
2. Doping: p-n junction formation
3. Edge etch: removes the junction at the edge
4. Oxide Etch: removes oxides formed during diffusion
5. Antireflection coating: Silicon nitride layer reduces reflection
Cells
Purifying the silicon:
STEP 1: Metallurgical Grade Silicon (MG-Silicon is produced from SiO2 melted
and taken through a complex series of reactions in a furnace at T = 1500 to
2000°C.
STEP 2: Trichlorosilane (TCS) is created by heating powdered MG-Si at around
300°C in the reactor, Impurities such as Fe, Al and B are removed.
Si + 3HCl SiHCl3 + H2
STEP 3: TCS is distilled to obtain hyper-pure TCS (<1ppba) and then vaporized,
diluted with high-purity hydrogen, and introduced into a deposition reactor to form
polysilicon: SiHCl3 + H2→Si + 3HCl Electronic grade (EG-Si), 1 ppb Impurities
STEP 1
STEPE 2 and 3
Electronic
Grade Chunks
Source: Wacker Chemie AG, Energieverbrauch: etwa 250kWh/kg im TCS-Process, Herstellungspreis von etwa 40-60 €/kg Reinstsilizium
Ingot sliced
to create wafers
Making single
crystal silicon
Czochralski (CZ) process
crucible
Seed crystal slowly grows
Microelectronic
1G: Crystalline Si PV technology

P-N Junction
Si
14
Ge
32
Ga
31
As
33
Cd
48
Te
52
P
15
In
49
Al
13
Sb
51
Cu
29
Se
34
In
49
31
IIB IIIB IVB VB VIBIB
C
6
B
5
Zn
30
Sn
50
S
16
O
8
N
7
Periodic Table
Doping Technology of Silicon: pn junction of Silicon
Silicon (IV) Diamond Structure
Boron doping Phosphorus doping
Martin Green, UNSW’s cell concepts PIP 2009; 17:183–189 / http://www.unsw.edu.au/

External Load +-
Emitter Base Rear Contact
Front Contact
Antireflection coating
Absorption of photon creates an
electron hole pair. If they are
within a diffusion length of the
depletion region the electric field
separates them.
The electron after passing
through the load recombines
with the hole completing the
circuit
n pFront contact
Task of Photovoltacis: Photo absorption and photo generation
1. Light absorption: Generation of free excess
2. Charge separation:
a) Photocurrent, I [A] (Ampere)
b) Photovoltage, V [V] (Volt)
3. Recombintion (defect  recombination centers)
V[A] x I[V] = Power [Watt]
Light flux
Valence band
Conduction band

ZnO
,2500
Å
CdS700
Å
Mo
0.5-1
µm
Glass, Metal Foil, Plastics
Glass
Cd2
SnO 4
SnO 2
0.2-0.5 µm
CdS
600-2000
Å
CdTe
2-8
µm
CIGS1-2.5µm
C-Paste
with
Cu,
CdTe based device
Quelle: Noufi, NREL, Colorado, USA,
*CIGS based device
CdTe and CIGS Thin Film Solar cells (2G)

Glass
Moly rear
contact
CIGS
Buffer
ZnO Front
contact
Technology: monolithic" interconnect from three scribes P1 to P3
P1
Step 1: Deposition of Cu, In,Ga (Se)
(sputtering, codeposition, Electrodeposition)
Step 2: Rapid Thermal Processing (RTP)
Pulsed Picosecond Laser
Front ZnO of one cell is connected to back Mo contatc of the next.
dead-zone width can be up to 500 μm for mechanical scribing.
Se Cu
Ga In
Cu(In,Ga)Se2
P3
P2 P1
P1 periodic scribes to defines the width of the cells.
P2 scribe removes the CIGS
down to the Moly back contact.
P3 scribe can also remove the whole layer stack down
to the Moly
Si
Module
Vmodule= Vcell x Ncell  24 V for battery charging
Quelle: HZB / M. Lux-Steiner

Radiative
recombination
EV
EC Augerrecombination
Excessenergygiven
toanothercarrierin
thesameband
EC
EV
Electron thermalizes
to band edge
Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
 Recombination (r) is the opposite of generation, leading to voltage and current loss.
Non-radiative recombination  phonons, lattice vibrations.
Radiative recombination  photons (dominating in a direct bandgap materials )
Auger recombination  charge carrier may give its energy to the other carrier.
E(eV)
Non-radiative
recombination
EC
EV
Phonon
 Recombination processes are characterized by the minority carrier lifetime τ.
 Equilibrium: charge distributions np = ni
2
Out of equilibrium: The system tries to restore itself towards equilibrium through R-G
 Steady-state rates: deviation from equilibrium
( )npnBgrR
BnnB.pg
.pnBr 2
i2
i00
−=−=



==
=
/scm102B(Si) 315−
×=
Generation vs. recombination processes

Summary: Generation & Recombination
Shockley-Read Hall
recombination
Direct
recombination
direct band
Auger recombination
(dominant effect at high carrier concentration)
EV
EC
Ekin
Ekin= -qELsc
Generation
Impact ionization is a
generation mechanism.
When the electron hits an
atom, it may break a
covalent bond to generate
an electron-hole pair.
The process continues with the newly
generated electrons, leading to avalanche
generation of electrons and holes.
τ : average time it takes an excess minority carrier to recombine
(1 ns to 1 ms) in Si
τ : depends on the density of metallic impurities and the density
of crystalline defects.
t/teff
τ
( )2
DAugern,DTn .NcBNNcΔn ++=








++∆=++=
AugerDirectSRH
111
n
τττ
RRRR AugerDirectSRH
( ) 1
eff
−
++=⇒ 2
DAugern,DTn .NcBNNcτ
Loss to thermal
vibrations

Sun
Task of Photovoltacis
• 100 W light bulb is turning on for one hour
• Energy consumed is 100 W·h = 0.1 kW.h.
Production vs. consommation
100 W
light bulb

Controller, (charge regulator) regulates the voltage and current
coming from the solar panels Determines whether this power is
needed for home use or whether it will charge a deep-cycle solar
battery to be drawn upon later on.
All other current must pass
through a DC to AC inverter,
transforming it into electricity
usable by general household
appliances.
DC-current from the
controller can be used to run
electronic devices that don't
require an AC-current.
all surplus electricity not being
drawn by your home can be
sent to your utility company's
power grid.
Photovoltaic
P > C
Traditional System
Photovoltaic
P < C
Copyrighted Material, from internet
Task of Photovoltacis

Efficiencies beyond the Shockley-Queisser limit
(1) Lattice thermalization loss (> 50%)
(2) Transparency to hν < Band gap
(3) Recombination Loss
(4) Current flow
(5) Contact voltage loss
Not all the energy of absorbed photon can be captured
for productive use (Th. Maxi efficiency ~32% ).
R.R. King; Spectrolab Inc., AVS 54th International Symposium, Seattle 2007
Reflection loss
Recombination
loss
Resistive loss Top contact
loss
Back contact
„Loss“
 Good surface passivation.
 Antireflection coatings.
 Low metal coverage of the top surface.
 Light trapping or thick material
(but not thicker than diffusion length).
 High diffusion length in the material.
 Junction depth optimized for absorption
in emitter and base.
 Low reflection by texturing

Route to high efficiency solar cells
Traditional cell design PERLPERCIBCPESCMINP
(1) (2) (3)
(1) PERL developed at UNSW (EFF. 25%) Passivated Emitter and Rear Locally diffused1
(2) Localized Emitter Cell Using Semiconducting Fingers. (EFF. 18.6%, CZ n-type)
(3) Laser-grooved, buried front contact (LGBC; EFF. 21.1%)
1
Martin Green, PIP 2009; 17:183–189, University of New South Wales, Australia
Copyrighted Material, from internet

We need to use most of the solar spectrum: Tandem solar cells
Power [Watt/cm2
] = Voltage [Volt ] x Current density [A/cm2
]
Materials with small Band gap
But low voltage
Excess energy lost to heat
 Generating a large current (JSC)
Materials with large band gap
But low current
Sub-band gap light is lost
 Generating a large voltage (VOC)
Solar cell
versus
Solar spectrum

= (in flow – out flow) + Rain -
Evaporation
rain
In flow
Out flow
Evaporation
Rate of
increase of
water level
in lake r-g.J
q
1
nnn +∇=
dt
dn
nnnn
nnn
qDEqnμJ
r-g.J
q
1
t
n
∇+=
+∇=
∂
∂
A little bit of Math: Continuity equation and Transport process
t
n
0=
∂
∂

Voc
0 La= 1/αLp
W
Rec
ΕF,n=µe
ΕF,p=µh
WLn
La= 1/α
Rec
ΕF,p=µh
ΕF,n=µe
 Generated closer to the junction
 Generated within a diffusion length of the junction
 Key issues:
Minority carrier diffusion
Surface recombination
Collection near front surface and also rear
conditionsBondaryGτ
L
x
Bexp
L
x
AexpΔn(x) n
nn
←+
+
+
−
=
t
n
0=
∂
∂
Differential equation is simple only when G = constant.
n
2
n
2
2
D
x)G(λ(
L
Δn
dx
Δnd
−=
p
2
p
2
2
D
x)G(λ(
L
Δp
dx
Δpd
−=
Basic: Continuity equation and Transport process

Basic Diode J-V equation
NL
nD
NL
nD
qJ
Dp
2
ip
An
2
in
0








+=
+ JD
)( 1
Tk
qV
expp
L
D
qn
L
D
qJ
B
n,0
p
p
p,0
n
n
−








+=
0J
L
Jcurrent,Dark
Tn.k
qV
0 J1expJJ
D
B
−








−=
  
- JL
W)LqG(L pn ++−
LJntPhotocurre
Applying boundary conditions (ideal diode case)
Differentiating to find the current
Equating the currents on the n-type and p-type sides
J0 : saturation current
kB : Boltzmann`s constant, 1.381 10-23
J/Kelvin
n : ideality factor
ni: carrier concentration
NA,ND. Doping concentration
dx
pd
qDJ n
pp
∆
=
dx
nd
qDJ
p
nn
∆
=

One diode model / Equivalent Circuit
RLoad
J
VD
JD
 Ideal diode (dark current , ID)
(Shockley diode equation)






−= 1exp0
nkT
qV
JJ D
D
SD RJVV .+=
 add a serie resistance RS
jsh . Rsh
Current
loss
R
J.RV
J-
Sh
S
L
+
+





−
−
= 1
nkT
)R.JV(q
expJJ S
0
 Add a shunt resistance
Sshsh RJVRi .. +=
JL
L
S
J
nkT
RJVq
JJ −



−
−
= 1
).(
exp0
 Under illumination
VOC
JSC
- JL
4TH
Quadrant
J = I/A
VReverse
Forward
0
 Solar cell in the dark






−
−
= 1
).(
exp0
nkT
RJVq
JJ S
D








+=
Dp
ip
An
in
NL
nD
NL
nD
qJ
22
0
J. RS
(Voltage drop)
V
Dark characteristics being shifted down by
photocurrent which depend on light
intensity.
P
N
Slope -1/RLoad
Photogenerated carriers can also flow through the crystal
surfaces or grain boundaries in polycrystalline devices

Two diodes model / Equivalent Circuit
R
J.RV
Sh
S+
+





−
−
+





−
−
= 1
).(
exp1
).(
exp
2
02
1
01
kTn
RJVq
J
kTn
RJVq
JJ SS
RLoad
J
+
-
RS
V
J01
,n1
J02
,n2
Rsh
JL
LJ-
R
J.RV
Sh
S+
−





−
−
−





−
−
−= 1
).(
exp1
).(
exp
2
02
1
01
kTn
RJVq
J
kTn
RJVq
JJJ SS
L
1st
Quadrant
4th
Quadrant
1st
Quadrant
4th
Quadrant
J
V
A. Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie

Photocurrent analysis: Quantum efficiency measurments
Acceptor
Voc
x = 0 La= 1/α
x = Ln
x = W
EJ σ=
dx
dp
Dp
Donor
Rec
µh
µeE
→
p∇
→
Load
• How much light converted?
• Limited information on the electronic properties
• Information on the optical properties of the device
)(R λ−
=
1
EQE
IQE
λ
hc
e
J
Φ
EQE
)(
)(
1 λ
λ
=
This ratio can be measured
Φ 0
x R λ
Φ
0
[ ]
[ ]Joulehν
Watt/cmΦ
N
2
photons
in =
[ ]
[ ]Coulombe
A/cmJ
N
2
electrons
out =
Electrons
collected
Photons
absorbed
Φ 0
x Rλ
h(c/λ) < EG
x
0 ).eR.(1ΦΦ α
λ
−
−=
Φ
0

EQE and and absorption coefficient
Photon absorption
direct band-gap
( ) GG
2
1
E)E(hνvs..hν →−α
2
G )E(h
hν
B
−= να
Direct Bandgap Eg
EC
EV
Photon
Conduction
Band
Valence
Band
E(k)
GaAse.g.
+k-k
Photon absorption
indirect band-gap
( ) GG
2
E)E(hvs..h →−ννα
2
1
G )E(h
h
A
−= ν
ν
α
Photon
+k-k
Eg
EC
EV
Conduction
Band
Valence
Band
Phonon
EG+Ep
Ep
E(k)
Sie.g.
Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie and IRESEN
Cut-off λ vs. EG
[eV]E
1.24
m][μλ
G
G =
∫Φ=
λ
λλλ dEQEqJsc )()(
λ
hc
e
J
Φ
EQE
)(
)(
1 λ
λ
=

hν
Band Gap - absorption coefficient - absorption length
Temperature changes:
EG ↑ as T ↓
Changing the absorption edge
A(T)(0)E(T)E gG −=
Si Ge GaAs
EG (eV) 1.12 0.66 1.42
λλλ ++= TAR100%
)R.(1Φ
Φ
ln.
d
1
α
λ0 −
−=λ
αx-
o R).α).-(E).(1Φ
dx
dΦ
x)G(E, =−=
Absorption
x
0 ).eR.(1ΦΦ α
λ
−
−=
Generation
Φ
ΦΦΦΦ TAR0 ++=

Quantum efficiency measurements
2 – Cell Measurement
2CELL
CELL
sc .Φq.EQEJ =
2MON
MON,2
sc .aΦq.EQEJ =
.a.EQE
J
J
.aEQE MONMON,2
sc
CELL
sc
CELL =
3 – Final Result
REFREF
sc
MON,1
sc
MON,2
sc
CELL
sc
CELL EQE
J
J
.
J
J
EQE =
Monochromator
equipped with more gratings*Chopper
Beam splitter
*Gratings should have line density as high as possible for achieving high resolution and high
power throughput. (600 – 3000 lines/mm).
EG
EQE vs. λ
1REF
REF
sc .Φq.EQEJ =
1MON
MON,1
sc .aΦq.EQEJ =
1
MON,1
sc
MON
qΦ
J
.aEQE =
1 - Reference measurement
photon1ofrgyphoton/eneofpowerTotal
electron1ofargecurrent/ch
EQE""EfficiencyQuantumExternal =

Design to high efficiency solar cells
Light trapping
Reflection Loss: ARC
Material Parameter absorption
Important cost factor €/kg








+
−−= −αW
p
e
αL1
1
1R)(1η
λ
hc
e
)J(
Φ(λ)
1 λ
η =
Decisive Material Parameter
The band gap
0.3 0.5 0.7 0.9 1.1
20
0
40
60
80
100
0
1
2
3
4
5
NumberofSunlightPhotons(m-2
s-1
micron-1
)E+19
RExternalQuantumEfficiency,%
µc-Si:H junctiona-Si:H junction
AM 1.5 global spectrum
Wavelength, microns
a-Si:H/µc-Si:H Cell Spectral Response
Textured TCO
a-Si
Top cell
Back Reflector
Glass substrate
Thin film mc-Si
Bottom cell
[ ]∫ λ−=
GE
λ0λsc dλ.dα-exp.)().ΦR(1.η(λ).qJ
Light from the sun
Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie and IRESEN

Power output characteristics
Jsc VOC
Pmax
Sun
OCSC
P
.FF.VJ
EFF. =
Vmpp
Pmpp= Impp x Vmpp
OCSC
mppmpp
.VJ
.VJ
Inverse of slope Vmpp/Impp
is characteristic resistance
Jmpp mmp
Rmpp
V
J
mpp = Maximum Power Point
P=I.V
Fill Factor
OCSC
mppmpp
.VJ
VxJ
Sun
mppmpp
P
V.J
EFF =

Solar cell efficiency under simulated sun light
Earth´ s Surface
AM1
AM0
AM1.5
d=1.5 atmos d=1 atmos
Challenges
To simulate a spectrum as similar as possible to the sun spectrum with excellent
homogeneity over relatively large areas

Principle of a sun simulator
The unit of the photon flux
Reference cell
solar cell
Sources: Thomas FU-Berlin

Contact
grid
Total
Area
Including
grid
Iluminated
Area (2)
JSC is rather accurately determined by EQE measurements
0.5 cm
1 cm
Iluminated
Area (1)
0.5 cm
1 cm
∫Φ=
λ
λλλ dEQEq )()(J 0sc
From
monochromator
Performance measurement standard conditions
(1) effective area or
(2) total area

Photons to electrons, solar power to electrical power
You need a computer for this exercise
Physical constants: elementary charge, e = 1.60 x 10-19
C
Planck’s constant, = 6.63 × 10-34
J s
speed of light, c = 3.00 x 108
m s-1
Exercice: An ideal solar cell has a band gap energy . The solar cell absorbs 100% of photons with energy
and 0% of photons with energy . All absorbed photons are converted to current with 100% quantum efficiency.
The solar cell has a fill factor of 70% and an active area of 1 cm2. The external quantum efficiency (EQE)
spectrum and current-voltage (I-V) curves are sketched below:
0
20
40
60
80
100
120
Eg
Current
Voltage V
OC
I
SC
V
mpp
EQE(%)
a) The international standard AM1.5 solar spectrum is provided in the text file “Solar spectral irradiance.txt”
(from NREL.gov). Use it to calculate the short circuit current, ISC
, for the ideal solar cell made from:
Crystalline silicon Si, EG = 1.1 eV; Germanium Ge, EG = 0.67 eV ; Gallium arsenide GaAs EG = 1.42 eV
Amorphous Si, EG = 1.75 eV.
b) If the open-circuit voltage is given by Voc = EG/e, what is the maximum power conversion efficiency of
each of the four cells? (The total terrestrial irradiance is 1000 W m-2
.).
c) What is the optimum band gap for an ideal solar cell?
0
0.5
1
1.5
2
2.5
0 500 1000 1500 2000
Spectralirradiance(Wm-2nm-1)
Solar spectral irradiance
Extraterrestria
lTerrestrial

From Cells to a Module
 The basic building block for PV applications is a module consisting of a number
of pre-wired cells in series.
 Typical module Silicon technolog/ 36 cells in series referred to as 12V.
 Large 72-cell modules are now quite common.
 Multiple modules can be wired in series to increase voltage
and in parallel to increase current.
Such combinations of modules are referred to as an array
Cells wired in series

0.6 V each cell
N°1
N° 36
4 cells
4 x 0.6V
36 x
36 x 0.6V = 21.6 V
Adding cells in series
Vmodule = n (Vd – I.RS)
Series resistance RS
Cell 1 Cell 2 Cell 36
. . . . .
+ - + - + -

A parallel association of n cells is possible and enhances the output current of the
generator created.
In a group of identical cells connected in parallel, the cells are subjected to the same
voltage and the the resulting group is obtained by adding currents
VSC,n
Cell n
n Cells
Cell 1
n Cells
in parallele
n x ISC
ISC,n

From Module to array
For modules in series, the I –V curves are simply added along the voltage axis at any given
current which flows through each of the modules), the total voltage is just the sum of the
individual module voltages.

For modules in parallel, the same voltage is across each module and the total
current is the sum of the currents at any given voltage, the I –V curve of the
parallel combination is just the sum of the individual module currents at that
voltage.

Two ways to wire an array with three modules in series and two modules in parallel.
The series modules may be wired as
strings, and the strings wired in parallel.
The parallel modules may be wired together
first and those units combined in series
V V
If an entire string is removed from service
for some reason, the array can still
deliver whatever voltage is needed by the
load, though the current is diminished,
which is not the case when a parallel
group of modules is removed.

Standard conditions of your PV module
Standard Test Conditions:
• 1 kW/m2
, AM 1.5, 25°C Cell Temperature
• Solar irradiance of 1 kW/m2
(1 sun)
• Air mass ratio of 1.5 (AM 1.5).
• Key parameter: rated power PDC,STC
• I –V curves at different insolation and cell temperature
• NOCT: Nominal Operating Cell Temperature
(T = 20°C,Solar Irradiation= 0.8 kW/m2
, winds speed 1 m/s.)
.S
0.8
C20NOCT
TT ambCell 




 °−
+=
cell temperature (°C)
ambient temperature (°C)
Insolation
(1 kW/m2
)
VMPP
MPP
VMPP
V
MPP

Impact of Cell Temperature on Power for a PV Module.
Estimate cell temperature, open-circuit voltage, and maximum power output for the
150-W BP2150S module under conditions of 1-sun insolation and ambient
temperature 30°C. The module has a NOCT of 47°C.
C64.1
0.8
C204
3.S
0.8
C20NOCT
TT ambCell °=




 °−
+=




 °−
+=
7
0
From The table for this module at the standard T = 25°C, VOC = 42.8V
VOC drops by about 0.37% per °C , the new VOC = 42.8[1 − 0.0037(64 − 25)] = 36.7 V
with decrease in maximum power available of about 0.5%/°C.
With maximum power expected to drop about 0.5%/°C, this 150-W module at
its maximum power point will deliver:
Pmax = 150 W· [1 − 0.005(64 − 25)] = 121 W
This is a significant drop of 19% from its rated power.

• Module with Power of 240 WC
• 240 Wc and efficiency 14.8%
• 1.64×0.99=1.6236 m².
• ηSTC=240/(1000×1.6236) = 14.78 %≈ 14.8 %

• 1.64×0.99=1.6236 m².
• ηSTC=240/(1000×1.6236) = 14.78 %≈ 14.8 %
Siliken modules were awarded the
Number one test modules 2010 and
Number two test modules 2011.

• 1.64×0.99=1.6236 m².
• ηSTC=240/(1000×1.6236) = 14.78 %≈ 14.8 %
• 1.64×0.99=1.6236 m².
• ηSTC=240/(1000×1.6236) = 14.78 %≈ 14.8 %
• KT(P) = -0.41 %/°C  Power decreases by (0.41% × 240W)
= 0.984 W /°C
• KT(Uco) = -0.356 %/°C Load voltage decreases by
(0.356 × 37V) = 0.13 V / °C.
• KT(Icc) = 0.062 %/°C  Isc enhanced by
(0.062% × 8.61 = 0.0053 A / °C
• NOCT = 49°C (±2°C).
).S(kW/m
0.8
C20C249
C)(TC)(T 2
ambCell 




 °−°±
+°=°
NOCT terms:
Level of illumination: 800 W / m²
Outdoor temperature: 20 ° C
Wind speed: 1 m / s
Air mass AM = 1.5
Siliken modules were awarded the
Number one test modules 2010 and
Number two test modules 2011.

Exercice: Electronic Structure of Semiconductors and Doping
Physical constants: Planck’s constant, h = 6.63 × 10-34
J s
Boltzmann’s constant, k = 1.38 × 10-23
J K-1
= 8.62 x 10-5
eV K-1
m s-1
Rest mass of an electron, m0 = 9.11 x 10−31
kg
Elementary charge, e = 1.60 x 10-19
C
1) Germanium has an effective density of states (DOS) NC = 1019
cm-3
for the conduction band and a band gap
EG = 0.66 eV. The intrinsic carrier density at 300 K is 1.8 x 1013
cm-3
.
i) What is the effective DOS for the valence band, NV ?
ii) If the material is n-doped to give an electron density of ne = 1018
cm-3
, what is the hole density?
iii) What is the intrinsic carrier density at 100 K? You can assume that the effective DOS do not change with temperature.
2) (Only attempt this question if you like calculus or use a program like Mathematica)
The conduction-band DOS in a direct band gap semiconductor is given by
where is the conduction band minimum and is the electron effective mass. Show that the conduction band electron
density can be approximated by:
where EF is the Fermi level and is the effective DOS.

Exercice: Electronic Structure of Semiconductors and Doping
3) The effective electron mass in crystalline GaAs is .
The effective hole mass is = 0.47 m0 , and the band gap is EG = 1.42 eV.
i) Sketch the band structure (energy versus momentum) for GaAs.
ii) Using the expression for the effective DOS given in question 2, determine the intrinsic carrier density at 300K.
iii) A GaAs crystal is doped with 1016
cm-3
Si atoms, acting as electron donors by replacing Ga atoms in the lattice.
What is the electron and hole density, assuming that all dopants are ionised?
iv) What is the position of the Fermi-level relative to the conduction band onset? (Give your answer in electron volts.)
4) Crystalline silicon has an effective DOS of NC = 3 x 1019
cm-3
for the conduction band and NV = 2 x 1019
cm-3
for valence band, and a
band gap EG = 1.1 eV. A silicon crystal is doped with 1017
cm-3
boron (B) atoms. (Boron is a group III element.)
i) What is the position of the Fermi-level relative to the valence band maximum, EV, and conduction band maximum, EC, at 300 K?
ii) If the acceptor state energy, ED, is 0.05 eV above the valence band maximum (see diagram), use the Fermi-Dirac
distribution and the Fermi level calculated in (i) to calculate the fraction of dopant atoms that are ionized.
iii) Using the same approximations, calculate the Fermi-level and fraction of ionized dopants at 77 K. Is the assumption of
complete ionization still valid?
iVI Roughly sketch the variation of hole density with temperature over a wide temperature range.
energy
EV
ED
EC
0.05 eV

At wavelength λ = 1050 nm, the refractive index of silicon is n = 3.6 and the absorption
coefficient is α = 15 cm-1
.
i)A Si wafer of thickness d = 0.25 mm is illumined by light at wavelength λ = 1050 nm at normal
incidence. What fraction of light is reflected and what fraction of light is absorbed?
ii)A perfect reflective surface is added to the back side of the wafer. What is the absorptivity
now?
iii)An estimate for the absorptivity of a wafer with a light-trapping surface is given by Würfel as:
(P. Würfel, Physics of Solar Cells, p144) where R is the reflectivity of the surface. By what
factor would you expect the external quantum efficiency at 1050 nm of the corresponding solar
cell to be improved by a light-trapping surface?
Use the absorption coefficient and refractive index for silicon as a function of wavelength and
the solar irradiance spectrum to calculate the reflectivity and absorptivity spectra for a 0.25 mm-
thick Si solar cell with a reflective back side and no light trapping (assume normally incident
light). Plot these spectra as functions of wavelength. Assuming that each absorbed photon
generates one electron in the external circuit (external quantum efficiency = absorptivity),
calculate the short-circuit current for the cell under AM1.5 illumination.

Exercice: Charge transport and p-n diodes
Physical constants: Reduced Planck’s constant = 1.05 × 10-34
J s = 6.58 × 10-16
eV s
Boltzmann’s constant, k = 1.38 × 10-23
J K-1
= 8.62 x 10-5
eV K-1
m s-1
elementary charge, e = 1.60 x 10-19
C
1. A crystalline silicon wafer, has a band gap EG = 1.1 eV and an intrinsic carrier density of ni = 1.3 x 1010
cm-3
at 300 K.
The wafer is 200 µm thick and has an area of 1 cm-2
. The electron mobility is µe = 1000 cm2
V-1
s-1
, and the hole
mobility is µh = 100 cm2
V-1
s-1
.
i) What is the conductivity of the undoped wafer?
ii) The wafer is doped with a donor density ND = 1018
cm-3
. Is the doped wafer n-type or ptype? Which
carrier types (electron or hole) are the minority and majority carriers?
iii) A voltage of 1.0 V is applied across the wafer. Sketch the energy band diagram (energy vs depth),
indicating the direction of travel of holes and electrons. How large is the drift current?
iv) The minority carrier lifetime is 1 µs. On average how far does a minority carrier travel (under 1.0 V
applied bias) before recombining? How does this affect the photocurrent?
v) Why is the n-type region made thin relative to the p-type region in typical crystalline silicon solar cells?

Exercise : Crystalline silicon solar cells
2. A 250 micrometer-thick crystalline silicon wafer is doped with 5×1016
acceptors per cubic centimetre. A 1 micrometer-
thick emitter layer is formed at the surface of this wafer with a uniform concentration of 3×1019
cm-3
donors. Assume that
all doping atoms are ionized. The intrinsic carrier concentration in silicon at 300 K is ni = 1.3 x 1010
cm-3
. How large is (at
300 K and thermal equilibrium):
i)The electron and hole concentration in the p-type region and n-type region? Which charge carriers are the majority
carriers in the p-type region and what is their concentration?
ii)What is the position of the Fermi level (in eV) in respect to the conduction band in the ptype and n-type region,
respectively?
iii)The built-in voltage of the p-n junction? iv) Draw the corresponding band diagram of the p-n junction.
iv)The width of the depletion region of the p-n junction. Compare it with the total thickness of the Si wafer.
3. A 200 micrometer-thick multicrystalline silicon cell is doped with 5×1017
acceptors per cubic centimetre. A 1 micrometer-
thick n-type emitter layer is formed at the surface of this cell with a uniform concentration of 3×1019
cm-3
donors. Assume
that all doping atoms are ionized. The intrinsic carrier concentration in silicon at 300 K is ni = 1.3 x 1010
cm-3
, and the
dielectric constant is ε = 11.7. At 300 K and thermal equilibrium:
i) The electron mobility is µe
= 500 cm2
V-1
s-1
, and the hole mobility is µh
= 50 cm2
V-1
s-1
. The minority
carrier lifetime for electrons is τe
= 400 ns and τh
= 100 ns for holes. The diffusion constant is given
by the Einstein relation, D

Exercice: Charge transport and p-n diodes
iVI) Estimate the saturation current density for the cell, neglecting recombination in the depletion zone. How does the
saturation current affect the open-circuit voltage of the cell?
iV) Minority carriers generated within one diffusion length of the depletion zone will be collected and will contribute to the
measured photocurrent. Those generated outside of this region will recombine and will not contribute to the current. The
absorption coefficient for silicon at 950 nm is α(950nm) = 104
m-1
. Using the Beer-Lambert law for absorption, estimate the
quantum efficiency for this cell at 950 nm. (The light has normal incidence and shines on the n-type side of the cell.)
Vi) Sketch the energy band diagram for the cell, labelling all relevant distances. Explain why
reducing the doping in the p-type region might increase the short-circuit current of the cell. How
might this affect the open-circuit voltage?

PV module made up of 36 identical cells, all wired in series. With 1-sun insolation
(1 kW/m2
), each cell has short-circuit current ISC = 3.4 A and at 25°C its reverse saturation
current is I0 = 6 × 10−10
A. Parallel resistance RP = 6.6 Ω and series resistance RS = 0.005Ω .
a) Find the voltage, current, and power delivered when the junction voltage of each cell is
0.50 V.
b) Set up a spreadsheet for I and V and present a few lines of output to show how it works.
Using Vd = 0.50 V along with the other data
The voltage produced by the 36-cell module:
Vmodule = n(Vd − I x RS ) = 36(0.50 − 3.16 x 0.005) = 17.43 V
Power dilevred:
P(watts) = Vmodule x I = 17.43 × 3.16 = 55.0 W
R
I.RV
1
n.k.T
)I . Rq(V
exp.-III
p
SS
0ph
+
−





−
−
=
[ ]
p
dV9.38
0ph
R
V
1e.-III d
−−=
[ ] A6.3
6.6
5.0
1e.10x6-4.3I 5.0x9.3810
=−−= −
Voltage and Current from a PV Module

A spreadsheet might look something like the following:

Gonçalves et al., Dye-sensitized solar cells, Energy Environ. Sci. 1, 655 (2008), is a very nice
summary of the current state of DSSCs. Use it as a reference to answer the following questions:
(only brief answers required)
What is the main reason for the lower efficiency of DSSCs compared to crystalline silicon
cells?
What is the main difference in the physical process of charge generation and transport
compared to silicon cells?
After excitation, what prevents the dye from returning to its ground state via
fluorescence?
What are the main requirements when choosing a dye?
What are the main requirements that the semiconductor (TiO2) layer must satisfy to in
order to make an efficient cell?
What causes the lack of stability of DSSCs? How can this potentially be solved?

Exercise : Tandem Solar Cells
A tandem cell is made from two sub-cells, A and B. The individual sub cells are ideal diodes, with current-voltage (J-V) characteristics
given by:
Where J0 is the reverse saturation current density, and Jph is the photocurrent density.
These have values of J0,A = 10-10
mA/cm2
, Jph,A= 25 mA/cm2
and J0,B = 10-18
mA/cm2
, Jph,B= 20 mA/cm2
for sub-cells A and B
respectively at temperature T = 300 K.
Calculate the open-circuit voltage for each sub-cell. Which sub cell do you suppose has the highest band gap?
The two sub-cells can be connected together in series or in parallel to make a tandem cell. Sketch the J-V characteristics of the
individual sub-cells as well as the two possible configurations of tandem cell.
Write an expression for the J-V characteristic of the parallel-connected tandem cell. Calculate the short-circuit current and the open-
circuit voltage.
Calculate the short-circuit current and open-circuit voltage for the series-connected tandem cell.
(optional) Using a computer or otherwise, calculate the fill-factor for each sub-cell and the two possible tandem-cell configurations. (Hint:
it is simpler to calculate power as a function of current for the series-connected cell.) Assuming the J-V curves were generated with
AM1.5 radiation (100 mW/cm2
), what are the corresponding power conversion efficiencies?
The series configuration is more efficient than the parallel configuration. Why?
Light passes through sub-cell B before reaching sub-cell A. The band gaps of each sub-cell can be adjusted to optimise the overall
efficiency. How are the J-V curves of sub-cells A and B affected by changing the band gaps of the two materials? What is an important
criterion for optimising the efficiency of a series-connected stacked tandem cell?

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