Download as PPS, PPTX
![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](https://image.slidesharecdn.com/lectureourzazateennaouimarch-130519070157-phpapp02/75/Lecture-Conference-Ourzazate-ennaoui-9-2048.jpg)
![Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
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](https://image.slidesharecdn.com/lectureourzazateennaouimarch-130519070157-phpapp02/75/Lecture-Conference-Ourzazate-ennaoui-18-2048.jpg)
![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
Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
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](https://image.slidesharecdn.com/lectureourzazateennaouimarch-130519070157-phpapp02/75/Lecture-Conference-Ourzazate-ennaoui-24-2048.jpg)
![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
=](https://image.slidesharecdn.com/lectureourzazateennaouimarch-130519070157-phpapp02/75/Lecture-Conference-Ourzazate-ennaoui-25-2048.jpg)
![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](https://image.slidesharecdn.com/lectureourzazateennaouimarch-130519070157-phpapp02/75/Lecture-Conference-Ourzazate-ennaoui-28-2048.jpg)
![A. Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
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.
Standard conditions of your PV module](https://image.slidesharecdn.com/lectureourzazateennaouimarch-130519070157-phpapp02/75/Lecture-Conference-Ourzazate-ennaoui-43-2048.jpg)
![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](https://image.slidesharecdn.com/lectureourzazateennaouimarch-130519070157-phpapp02/75/Lecture-Conference-Ourzazate-ennaoui-53-2048.jpg)
Uploaded byProf. Dr. Ahmed Ennaoui
Lecture Conference Ourzazate ennaoui
1) The document discusses the basics of solar photovoltaic technology from the atomic level to full solar arrays. It covers topics like photo absorption and generation, energy levels, band structure, doping, p-n junction formation, and device fabrication for both crystalline silicon and thin film solar cell technologies. 2) The document then discusses generation and recombination processes in solar cells like radiative, non-radiative, Auger and Shockley-Read-Hall recombination and how they impact efficiency. It also covers the continuity equation and transport processes. 3) Finally, the basic one diode equivalent circuit model for solar cells is presented along with the Shockley diode equation and how it relates to the current-
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Lecture Conference Ourzazate ennaoui
- 1. Solar Photovoltaic Technologyfrom 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‐
- 2. Introduction: PV fromatom 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
- 3. 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
- 6. 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
- 7. Device fabrication 1. Surfaceetch, 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
- 8. 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 IIIBIVB 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/
- 9. External Load +- EmitterBase 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
- 10. 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)
- 11. 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
- 12. Radiative recombination EV EC Augerrecombination Excessenergygiven toanothercarrierin thesameband EC EV Electron thermalizes toband 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
- 13. Summary: Generation &Recombination Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie 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
- 14. Ahmed Ennaoui /Helmholtz-Zentrum Berlin für Materialien und Energie 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
- 15. 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
- 16. Efficiencies beyond theShockley-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
- 17. Route to highefficiency 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
- 18. Ahmed Ennaoui /Helmholtz-Zentrum Berlin für Materialien und Energie 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
- 19. = (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 ∇+= +∇= ∂ ∂ Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie A little bit of Math: Continuity equation and Transport process t n 0= ∂ ∂
- 20. 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 −= Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie Basic: Continuity equation and Transport process
- 21. Basic Diode J-Vequation 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 ∆ = Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
- 22. 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
- 23. 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
- 24. Photocurrent analysis: Quantumefficiency 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 Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie 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
- 25. EQE and andabsorption 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 =
- 26. hν Band Gap -absorption coefficient - absorption length Temperature changes: EG ↑ as T ↓ Changing the absorption edge Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie 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 ++=
- 27. 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 = Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
- 28. Design to highefficiency 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
- 29. Power output characteristics JscVOC 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 = A. Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
- 30. Solar cell efficiencyunder 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 A. Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
- 31. Principle of asun simulator The unit of the photon flux A. Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie Reference cell solar cell Sources: Thomas FU-Berlin
- 32. Contact grid Total Area Including grid Iluminated Area (2) JSC israther 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 A. Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie (1) effective area or (2) total area
- 33. A. Ennaoui /Helmholtz-Zentrum Berlin für Materialien und Energie 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
- 34. From Cells toa 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
- 35. From Cells toa Module 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 . . . . . + - + - + -
- 36. From Cells toa Module 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
- 37. From Module toarray 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.
- 38. For modules inparallel, 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. From Module to array
- 39. Two ways towire 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. From Module to array
- 40. Two ways towire 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. From Module to array
- 41. Standard conditions ofyour 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
- 42. Standard conditions ofyour PV module
- 43. A. Ennaoui /Helmholtz-Zentrum Berlin für Materialien und Energie 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. Standard conditions of your PV module
- 44. • Module withPower 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 % Standard conditions of your PV module
- 45. • Module withPower 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 % Siliken modules were awarded the Number one test modules 2010 and Number two test modules 2011. Standard conditions of your PV module
- 46. • Module withPower 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 % • 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 % • 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. Standard conditions of your PV module
- 47. Exercice: Electronic Structureof 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 speed of light, c = 3.00 x 108 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. Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
- 48. Exercice: Electronic Structureof 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 Ahmed Ennaoui / Helmholtz-Zentrum Berlin für Materialien und Energie
- 49. 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.
- 50. Exercice: Charge transportand 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 speed of light, c = 3.00 x 108 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?
- 51. Exercise : Crystallinesilicon 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
- 52. Exercice: Charge transportand 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?
- 53. PV module madeup 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
- 54. A spreadsheet mightlook something like the following: From Cells to a Module
- 55. 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?
- 56. Exercise : TandemSolar 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?
Editor's Notes
- #8 Just hit return once to drive the whole graphic
- #12 There are three most used methods for producing thin films of CIGS. The first one is the coevaporation process. This method is named coevaporation because in the beginning all needed elements were evaporated in vacuum at the same time. The thin film is produced by evaporating Cu, In, Ga, Se from elemental sources. In order to achieve the favored film composition, a precise control of the particular evaporation rates is necessary. There an electron impact emission spectrometer and an atom absorption spectrometer or a mass spectrometer is used. But the process also requires a substrate temperature between 300 and 550°C for a certain time during film growth. There are several processes of coevaporation, but one of the most favored ones is the inverted three-stage process, which you can see on the right. At first In, Ga, and selenium are evaporated with different rates and deposited as (In,Ga) 2 Se 3 at 300°C on the substrate. Afterwards Cu and selenium are evaporated and deposited on the substrate at elevated temperatures. At last In, Ga, and selenium are evaporated again. The inverted three-stage process leads to smoother film morphology and to high efficiency solar cells.