Download Chapter 6 - Photovoltaic Devices - UAH Department of Electrical and

Optoelectronics
EE/OPE 451, OPT 444
Fall 2009 Section 1: T/Th 9:30‐ 10:55 PM
John D. Williams, Ph.D.
Department of Electrical and Computer Engineering
406 Optics Building ‐ UAHuntsville, Huntsville, AL 35899
Ph. (256) 824‐2898 email: [email protected]
Office Hours: Tues/Thurs 2‐3PM
JDW, ECE Fall 2009
PHOTOVOLTAIC DEVICES •
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6.1 Solar Energy Spectrum 6.2 Photovoltaic Device Principles
6.3 pn Junction Photovoltaic I‐V Characteristics 6.4 Series Resistance and Equivalent Circuit
6.5 Temperature Effects
6.6 Solar Cells Materials, Devices and Efficiencies Prentice‐Hall Inc.
© 2001 S.O. Kasap
ISBN: 0‐201‐61087‐6
http://photonics.usask.ca/
JDW, ECE Fall 2009
Blackbody Radiation
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Definition of a black body: An object that absorbs all radiation incident upon it
The energy absorbed by a blackbody is then emitted in a temperature dependent spectrum of light, called thermal radiation which is primarily in the infrared spectrum
Blackbody radiation is characterized by three fundamental equations
– Plank’s Law – blackbody absorption curve
8πhν 3
U (ν , T ) =
c3
–
e
hν
kT
λmax = 10.2μm
T = 200 K
T = 300 K
T = 400 K
T = 500 K
λmax = 12.8μm
λmax = 17.0μm
−1
5.1⋅10 m ⋅ K
T
Stefan‐Boltzmann Law ‐ energy radiated from the blackbody as a function of temperature j* = σT 4
σ = 5.67 ⋅10 −8
JDW, ECE Fall 2009
T = 100 K
Wien’s Law ‐ characteristic wavelength of the blackbody
−3
λmax =
–
1
Blackbody Radiation
Intensity (arbitrary)
•
W
m2 K 4
λmax = 25.5μm
0
0
5
10
15
20
Wavelength (um)
J(100K)=5.67 W/m2
J(200K)=90.7 W/m2
J(300K)=1450 W/m2
J(400K)=3540 W/m2
J(500K)=17900 W/m2
25
30
35
Photovoltaic Basics
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Photovoltaic cells convert incident solar radiation energy into electrical energy
Thus these devices are used as energy harvesters and converters
The goal is to convert as much incident photon energy as possible into electrical energy
To do this, one must have a measure of what incident energy is available
The figure presented shows three curves
–
–
–
–
JDW, ECE Fall 2009
First, is the ideal blackbody spectrum of a 6000K heat source that such as the sun
The red curve represents the actual available solar energy above the earth’s atmosphere
The green curve represents the actual available solar energy at the earth’s surface
Absorption of radiation in the atmosphere constitutes the difference between AM0 and AM1.5
JDW, ECE Fall 2009
Solar Energy
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Light intensity variation with wavelength is generally represented by Spectral Intensity which is Iλ = I/λ
(intensity/wavelength)
λ2
The total intensity over a spectral range is given by I = Idλ
∫λ
1
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The integrated intensity above the earths atmosphere, gives the total power flow through a unit area perpendicular to the direction of the sun. This quantity is called the solar constant (air‐mass zero) or AM0 radiation and is approximately 1.353 kW/m2
Actual intensity on the earth’s surface depends on absorption and scattering in the atmosphere. The effects of water vapor (clouds) in the air provide a general value of solar intensity at the surface that is approximately 70% of AM0 with specific bands that are reduced significantly due to water and nitrogen absorption. The received spectrum is called Air Mass 1 ( AM1)
Generically, Air Mass m (AMm) is defined as the ratio of actual radiation path to shortest possible path through an atmosphere. So for light AM0
entering an atmosphere at angle θ
m = h / ho = (h0 sec θ ) / h0
m = sec θ
AMm = AM sec θ
•
However, diffusive scattering plays a significant role as well, and the ratio of scattered light and scattering angles should also be taken into account
h0 Atmosphere
h
θ
α
Earth
AM1 AM(secθ)
α
Direct
Diffuse
Tilted PV device
(a)
(b)
(a) Illustration of the effect of the angle of incidence θ on the ray path length and the
definitions of AM0, AM1 and AM(secθ). The angle α between the sun beam and the horizon
is the solar latitude (b) Scattering reduces the intensity and gives rise to a diffused radiation
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Example: Solar Energy Conversion
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Suppose that a particular household consumes 500 W of electrical power daily
If the average solar intensity incident per day is 6kW h/m2, a photovoltaic device that converts solar energy into electrical energy has a 15% efficiency, then what is the required device area to meet the electrical needs of the home.
Since we know that the avg. light intensity incident ,
Total energy available per day = incident solar energy in 1 day per unit area X area X efficiency
Which must be equal to the total energy consumed in the household per day
Area = energy per house / (incident solar energy per unity area X efficiency)
Area =
500W ⋅ 3600s / hr ⋅ 24hr
= 13.3m 2
⎛
W ⋅ hr ⎞
⎜⎜ 6000 2
⎟⎟ ⋅ 3600s / hr ⋅ 0.15
m
day
⎝
⎠
the required panel size is 3.6 m X 3.6 m
NOTE: This solution does not account for heavy clouds, rain, or excess consumption!!!!
JDW, ECE Fall 2009
Photovoltaic Device Principles
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Consider a pn junction with a very narrow and heavily doped n‐
region at the top surface
Light passes through the n‐region and is absorbed in the depletion region which can take up a vast majority of the thickness of the device.
The built‐in Field, Eo in the depletion region drives EHP to their respective electrode
Finger electrodes are often used to optimize surface area while still providing uniform contact across large areas of the photovoltaic
Thin antireflection coatings on the surface are also used to prevent excess reflection and improve efficiency
The open circuit voltage in the system driven by the built‐in field and photoabsorbed EHP carriers provides a bias that one can draw off the device
Mobility in the device
– Long wavelength photons generate electrons in the p‐
type neutral region that must drift diffuse using the diffusion coefficient of the material to n‐type region
– Medium wavelength photons generate EHP in the depletion region yielding electron transit times based on carrier mobility in the electric field
– Short wavelength photons generate holes in the n‐region that must drift diffuse to the depletion region where they are driven by hole carrier mobility to the neutral p‐type region
Bus electrode
for current collection
Finger electrodes
n
p
Finger electrodes on the surface of a solar cell
reduce the series resistance
© 1999 S.O. Kasap,
Optoelectronics
(Prentice Hall)
Photovoltaic Device Principles
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In steady state, there can be no net current through an open circuit solar cell. Thus the number of carriers of each type must be balanced. (conservation of charge)
To insure against recombination losses, N is very short (0.2 um or less)
For λ = 1‐ 1.2 um in Si, the penetration depth is typically larger than 100 um and p‐side thicknesses are 200‐ 500 um. This gives a long minority carrier diffusion length and a wide depletion zone for mid wavelength absorption
Si bandgap = 1.2 EV which corresponds to a cutt‐off of 1.1 um, so energy at lambda greater than 1.1 um is wasted, which is about 25% of the thermal energy available at room temperature
Furthermore losses to do EHP recombinations at the surface can be as high as 40%. In addition the antireflective coating is never perfect, reducing the total number of collected photons down by about 9%
Now including limitations of the photovoltaic device itself that will be discussed throughout the chapter, the efficiency of a Si solar cell is no higher than 26%
EHPs
exp(−αx)
x
Lh
W
Le
Iph
Photogenerated carriers within the volume Lh + W + L e give rise to a photocurrent I ph. The
variation in the photegenerated EHP concentration with distance is also shown where α is th
absorption coefficient at the wavelength of interest.
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Example: Photocurrent
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Consider a photovoltaic device that is illuminated with light of such wavelength that photogeneration occurs over the device thickness. –
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Assume an EHP photogeneration rate of Gph which decays spatially as Goexp(‐αx), where Go is the surface photogeneration rate and α is the absorption coefficient.
Let us suppose that the device is shortened to allow carriers to flow around the external circuit. Also suppose that Lh is greater than the n‐layer thickness, ln so that all EHPS generated within the volume of the device contribute to photocurrent.
Silicon Example
Finally assume that recombination near the surface of the crystal is negligible.
Find the photocurrent in the system
EHP _ per _ unit _ time _ in _ ln + W + Le = A∫
x =l n +W + Le
x =0
{
{
JDW, ECE Fall 2009
α= 2000/m (δ = 500 μm) λ= 1.1 μm
Iph = 20 mA
}
dN EHP Go A
=
1 − e[−α (ln +W + Le )]
dt
α
eG A
dN
I ph = e EHP = o 1 − e[−α (ln +W + Le )]
dt
α
Go e −αx dx
A = 5 cm X 5 cm
Ln = 0.5 μm
W =2 μm
Le =50 μm
Go = 1 x 1018 cm3/s
}
For _ long _ wavlengths, _ α _ is _ small
⇒ I ph = eGo A(ln + W + Le )
α= 105/m (δ = 10 μm) λ= 0.83 μm
Iph = 40 mA
λ < 0.45 μm
Then α= 1X106/m (δ = 0.5 μm)
Iph = 0.2 mA
Pn junction Photovoltaic I‐V Characteristics
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Consider an ideal pn photovoltaic connected to a resistive load
If the circuit is short circuited, then only current generated by incident light is present. We define this as photocurrent.
The greater the light intensity, the more EHPs generated, the higher the photocurrent
The short circuit current is found to be I sc = I ph = − KI
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The photocurrent does not depend on a bias across the junction, because there is always some internal field to drift EHPs across the diode.
Note: we exclude the secondary effect of voltage modulating the width of the device and assume a large spectral bandwidth Therefore photocurrent flows without a forced bias as long as light is present
IF R is not a short, then a positive bias appears across the junction resulting from the built in potential, Vo of the pn junction.
where K is a device constant
JDW, ECE Fall 2009
I‐V Characteristics
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In addition to the photocurrent in the device, there is a forward current developed across the resistor, Id
eV
⎛ nk T
⎞
B
⎜
− 1⎟
Id = Io e
⎜
⎟
⎝
⎠
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Where Io is the reverse saturation current and n is the ideality factor that depends on both the material and the fabrication characteristics (n=1‐2)
In an open current, the net current is equal to zero and Id=Iph
Thus the total current in a solar cell is ⎛ nkeVT
⎞
B
⎜
− 1⎟
I = − I ph + I o e
⎜
⎟
⎝
⎠
The IV characteristics in a solar cell correspond to a negative shift in amplitude of the dark current in the system as the optical intensity of the light is increased
For silicon the total current in the system is canceled between a light induced voltage value of 0.4 – 0.6V.
I (mA)
20
Dark
Voc
V
0
Iph
0.2
0.4
0.6
Light
–20
Twice the light
Typical I-V characteristics of a Si solar cell. The short circuit current is Iph
and the open circuit voltage is Voc. The I-V curves for positive current
requires an external bias voltage. Photovoltaic operation is always in the
negative current region.
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
The Load Line
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Consider again a solar cell connected to a resistive load, R.
– R maintains the same voltage as the solar cell
– R carries the same current as the solar cell
– However the current through R travels in the opposite direction of conventional current. The current in this circuit passes from low to high potential
I = ‐V/R
The actual current and voltage in the circuit must satisfy both the solar cell IV curve and that of the load. A graphical solution is presented here
– Easiest method of solution is to construct a “load line” with the slope = ‐1/R
– The load line intercepts the solar cell IV curve at point ‘P’ where the load and the solar cell have the same voltage and current
– Point ‘P’ satisfies both governing conditions and represents the operating point of the circuit.
– The power delivered to the load is Pout =I’V’
– The goal of any circuit design is to optimize R within Isc
and Voc such that when illuminated, one generates the maximum power output available from the load
– One figure of merit is the fill factor (FF) which gives the designer a quick measure of the closeness of the solar cell IV curve to a square shape and thus allows one to quickly identify the optimal resistive that need be applied
JDW, ECE Fall 2009
–
One figure of merit is the fill factor (FF) which gives the designer a quick measure of the closeness of the solar cell IV curve to a square shape and thus allows one to quickly identify the optimal resistive that need be applied
FF =
I mVm
I scVoc
Example: Open Circuit Voltage and Illumination
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A solar cell is illuminated with intensity 600 W/m2 and has a short circuit current, Isc = 16.1 mA and an open circuit voltage, Voc = 0.485 V. What are the short circuit current and open circuit voltages when the light intensity is doubled?
set
⎛ eV
⎞
I = − I ph + I o ⎜ e nk BT − 1⎟ = 0
⎜
⎟
⎝
⎠
assume
Voc >> nk BT / e
then
nk T ⎛ I ph ⎞
⎟⎟
Voc = B ln⎜⎜
e
⎝ Io ⎠
nk BT ⎛⎜ I ph 2 ⎞⎟ nk BT ⎛ I 2 ⎞
ln
ln⎜⎜ ⎟⎟
=
⎜I ⎟
e
e
⎝ I1 ⎠
⎝ ph1 ⎠
⎛I ⎞
= I sc1 ⎜⎜ 2 ⎟⎟ = (16.1mA)(2 ) = 32.2mA
⎝ I1 ⎠
Voc 2 − Voc1 =
I sc 2
next _ assume _ for _ convienence _ that
n =1
Voc 2 = Voc1 +
JDW, ECE Fall 2009
nk BT ⎛ I 2 ⎞
ln⎜⎜ ⎟⎟ = 0.485 + 1(0.0259 ) ln(2) = 0.503V
e
⎝ I1 ⎠
I (mA)
20
Dark
Voc
V
0
Iph
0.2
0.4
0.6
Light
–20
Twice the light
Typical I-V characteristics of a Si solar cell. The short circuit current is Iph
and the open circuit voltage is Voc. The I-V curves for positive current
requires an external bias voltage. Photovoltaic operation is always in the
negative current region.
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Example: A Solar Cell Driving a Resistive Load
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Consider a solar cell driving a 30 Ω load. Suppose the cell has an area of 1 cm X 1 cm and is illuminated with light of intensity Io = 600 W/m2 and has the IV characteristics shown in the figure below. – What are the current and voltage in the circuit?
– What is the power delivered to the load?
– What is the efficiency of the solar cell in the circuit?
V
30Ω
V ' ≈ 0.425V
I =−
I'≈ −
0.425V
= 14.2mA
30Ω
Pout = I 'V ' = (14.2mA)(0.425V ) = 6.035mW
recall
(
)
Pin = 600W / m 2 (0.01m ) = 60mW
efficiency
η = 100
JDW, ECE Fall 2009
Pout
= 10.06%
Pin
2
Series Resistance and Equivalent Circuit
Neutral Depletion
n-region region
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Along with the load resistance, RL
One must also consider the effective series resistance of electrons traveling along the surface of the solar cell prior to reaching the finger electrode, RS
There are also a much smaller series resistances due to the neutral p‐type region that also contribute to Rs
We know the photocurrent process is represented by a constant current generator Iph
where current is proportional to light intensity
And we know that the normal diode current flows opposite of the photcurrent
There is also the possibility for carriers to flow through the crystal surfaces or grain boundaries (polysilicon devices) instead of through the external load. These effects are represented by an internal parallel resistance, Rp
– Typically Rp is less important than Rs unless the device is highly polycrystalline
Neutral
p-region
Back
electrode
Rs
Finger
electrode
Rp
RL
Series and shunt resistances and various fates of photegenerated EHPs.
© 1999 S.O. Kasap,Optoelectronics (Prentice Hall)
Rs
Iph
A
I
Id
Iph
Rp
RL
V
B
Solar cell
Load
The equivalent circuit of a solar cell
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
JDW, ECE Fall 2009
Series Resistance within the Solar Cell
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Rs can significantly deteriorate solar cell performance. In an ideal device FF is nearly square and the series resistance inside the cell is negligible
However as Rs increases, the FF value deteriorates significantly, reducing the power that can output from the cell
Note that when Rs becomes large enough, that it can decrease the saturation current of the solar cell
However it has no effect on the open circuit voltage.
Vo c
0
0
0.2
0.6
V
Rs = 50 Ω
Rs = 0
5
Rs = 20 Ω
Is c
Ip h
10
I (mA)
The series resistance broadens the I-V curve and reduces the maximum
available power and hence the overall efficiency of the solar cell. The example
is a Si solar cell with n ≈ 1.5 and Io ≈ 3 × 10-6 mA. Illumination is such that
the photocurrent Iph = 10 mA.
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
JDW, ECE Fall 2009
0.4
Example: Solar Cells in Parallel
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Consider two identical solar cells with Io = 25 nA, n = 1.5, Rs = 20 Ω. Both cells are subjected to the same illumination such that Iph = 10 mA.
Explain the characteristics of the two cells connected in parallel
Find the Maximum power that can be delivered by one cell and two cells in series and also the corresponding voltage and current at the maximum power point if Rp = ∞
I/2
I/2
A
Id
Rs
I
Iph
Id
Iph
RL
V
B
Two identical solar cells in parallel under the same illumination and
driving a load RL.
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
For _ one _ cell :
⎛ nkeVdT
⎞
⎛ e (Vnk− IRT s ) ⎞
B
⎜
⎟
− 1 = − I ph + I o ⎜ e B − 1⎟
I = − I ph + I o e
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
Pmax ⇒ 2.2mW , I ≈ 8mA,V ≈ 0.27V
from
I = −V / R ⇒ R ≈ 34Ω
6
Rs
Iph
Power (mW)
4
2
Voltage (V)
0.2
0.4
0
5
⎛
I
= − I ph + I o ⎜ e
⎜
2
⎝
eVd
nk BT
⎛ e ⎛⎜⎝ V − IRs 2 ⎞⎟⎠ ⎞
⎜
⎟
⎞
− 1⎟ = − I ph + I o ⎜ e nk BT − 1⎟
⎟
⎜
⎟
⎠
⎝
⎠
Comparing _ the _ two _ one _ sees _ that _ the _ _ device
has _ 1 / 2 _ the _ series _ resis tan ce
has _ double _ the _ photocurrent
1 cell
has _ double _ the _ reverse _ saturation _ current
2 cells in parallel
20
Current (mA)
Current vs. Voltage and Power vs. Current characteristics of one cell and two
cells in parallel. The two parallel devices have Rs/2 and 2 Iph .
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
each _ devices _ has
0.6
10
15
In _ Series
PII = 4.4mW ⇐ I ≈ 16mA,V ≈ 0.27V
yielding
R ≈ 17Ω
JDW, ECE Fall 2009
Temperature Effects
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As one might expect, output voltage and efficiency in solar cells increase as the temperature is lowered.
substituting
assume
Voc >> nk BT / e
− Eg
n = N c N v e k BT
2
i
then
nk T ⎛ I ph ⎞
⎟⎟
Voc = B ln⎜⎜
e
I
⎝ o ⎠
next _ assume _ for _ convienence _ that
n =1
⎛n
⎛I ⎞
eVoc 2 eVoc1
−
== ln⎜⎜ o1 ⎟⎟ ≈ ln⎜⎜
k BT2 k BT1
⎝ Io2 ⎠
⎝n
2
i1
2
i 2
⎞
⎟
⎟
⎠
and _ to _ ease _ the _ solution,
neglect _ the _ temp _ dependence _ of _ N c _ and _ N v
we _ get
eVoc 2 eVoc1 E g ⎛ 1 1 ⎞
⎜ − ⎟
−
=
k BT2 k BT1 k B ⎜⎝ T2 T1 ⎟⎠
⎛ T ⎞ E g ⎛ T2 ⎞
⎜⎜1 − ⎟⎟
⇒ Voc 2 = Voc1 ⎜⎜ 2 ⎟⎟ +
T
e
⎝ 1⎠
⎝ T1 ⎠
For Silicon:
Voc1 = 0.55 V @ T1 = 293 K
Voc2 (T2 = 333 K) = 0.475 V JDW, ECE Fall 2009
Solar Cell Materials, Devices, And Efficiencies
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Most solar cells are silicon based due to the vast infrastructure and inexpensive fabrication schemes available
The figure presented shows how various aspects of a silicon solar cell effect its efficiency
Note: some 25% of all the incident energy is wasted because of photons having insufficient energy to generate EHPs
Of the 75% of light absorbed, only 60% of the is effectively utilized due to near surface absorption in the n‐region
etc. etc. etc. JDW, ECE Fall 2009
Surface Shaping of Solar Cells to Improve Light Collection
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Reduces reflections from surface
Increases path length for absorption and improves EHP and collection efficiencies
JDW, ECE Fall 2009
III‐V Materials for Thermal Photovoltaics
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Heterojunction devices used to overcome surface recombination limitations and improve efficiencies to as much as 24%
Wider bandgaps provide longer wavelength absorbtion
AlGaAs absorb hv > 2eV
GaAs absorb 1.4 eV <hv < 2eV
In more advanced cell designs, the composition of Al is graded slowly from the surface allowing the device to absorb the broad spectrum between 1.4 and 2.0 eV
more evenly.
Tandem cells designed to absorb short wavelength light at the top pn junction and longer wavelength light below it improve EHP efficiencies while maintaining high FF values for the device
p-AlGaAs window (< 0.02 μm
Passivated GaAs surface
p-GaAs
n-GaAs
AlGaAs window layer on GaAs passivates the surface states
and thereby increases the low wavelength photogeneration
efficiency
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)
Connecting region.
p
n
Cell 1 (Eg1)
p
n
Cell 2 (Eg2 < Eg1)
A tandem cell. Cell 1 has a wider bandgap and absorbs energetic
photons with hυ > Eg1. Cell 2 absorbs photons that pass cell 1 and
have hυ > Eg2.
JDW, ECE Fall 2009
© 1999 S.O. Kasap, Optoelectronics (Prentice Hall)