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Homework Problem Solutions for :Solar Cells, Fuel Cells and Batteries:Materials for the Energy Solution
Bruce M. ClemensStanford University
Autumn 2013
Copyright c©2013 Bruce M. Clemens
i
ii
Contents
1 Problem Set 1 1
2 Problem Set 2 9
3 Problem Set 3 17
4 Problem Set 4 23
5 Problem Set 5 29
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iv
Chapter 1
Problem Set 1
1. At noon on a sunny day, there is a solar irradiance of 1 kW/m2 impinging on asolar panel that is 1 meter by 1.6 meters. To simplify the problem we assume herethat the light is all the same wavelength λ = 732 nm.
Determine the number of photons per second incident on the panel.
Solution:
The easiest way to solve this problem is to think about it conceptually.
You are given the Solar Irradiance, which is in units of Power/Area. This can beconverted to # photons/time by determining the area of the panel and the energyof one photon of 732 nm light
power
area=
energy
time× area
energy
time× area× Area× 1
energy/photon=
# photons
time(1.1)
Since it is a rectangular panel, the area is L ×W = 1 m × 1.6 m = 1.6 m2. Thephoton energy can be found from
E =hc
λ=
6.626× 10−34 J s× 3× 108 m/s
732× 10−9 m= 2.716× 10−19 J/photon
1
If we plug everything back into our original equation (Eq. 1.1) we get
1000 W/m2 × 1.6 m2 × 1
2.716× 10−19 J/photon= 5.89× 1021 photons/s
2. At noon on a sunny day, there is a solar irradiance of 1 kW/m2 impinging on asolar panel that is 1 meter by 1.6 meters. To simplify the problem we assume herethat the light is all the same wavelength λ = 732 nm.
Determine the current generated, in amperes, if the panel has 100% efficiency (everyincident photon generates one electron that contributes to current).
Solution:
Current can be thought of as the quantity of charge that passes a single point ina wire per unit time. From part (a) we know the # of photons per second hittingthe panel. The problem states that one electron is generated per photon. Thus, wecan determine the # of electrons generated in the material per second. Since weknow the charge of an electron we can determine the charge per second generatedin the material, which is equal to the current.
photons
s=
electrons
s
electrons
s× charge
electron=
charge
s= current
5.89× 1021 electrons
s× 1.6× 10−19 C
electron= 944 C/s = 944 Amps
3. What is the power density of the sun shining on a solar cell lying flat on the groundat the National Renewable Energy Laboratory in Golden, CO on September 22,2012 (the autumnal equinox, when the center of the sun and the earth’s equatorare in the same plane) at noon? Assume the solar radiation to be parallel rays witha power density perpendicular to the rays of 1.35kW/m2 and neglect losses due toatmospheric effects.
Solution:
Here we are neglecting all atmospheric contributions, so all we are concernedwith is the effective power density on the cell due to the incident radiation P0 =
2
1.35 kW/m2. We must take into account the angle of the sun on the cell, which liesflat on the ground. The latitude of Golden, CO is 39.74 N. According to Figure1.1 we can see that the power density is given by the quantity
a/c× P0 = P0 cosα
Incident Sunlight
Golden, CO: 39.74 deg N
N
S
α
Solar Cell
ca
α
Figure 1.1: Schematic showing the incident sunlight and orientation of the solar panel
So for our latitude, the power density is given by
1.35 kW/m2 × cos 39.74 = 1.04 kW/m2
4. Here we apply concepts used in the previous problem for a thought experiment.What is the total power incident on a 1 m wide solar cell running from the northpole to the south pole on September 22, 2012 (the autumnal equinox, when thecenter of the sun and the earth’s equator are in the same plane) at noon? Assumethe solar radiation to be parallel rays with a power density perpendicular to therays of 1.35kW/m2 and neglect losses due to atmospheric effects.
Solution:
To find the total power generated by a 1 m wide cell stretching from the N to S polesof the earth at the same time as the main question, we consult Figure 1.2 below.To set up the problem, we consider an infinitesimal area element of the solar cell,given by da in the illustration. This area element is 1 m wide and length dl alongthe long direction of the cell. We can write the length dl in terms of the radius ofthe earth, Re, and the latitude angle, α, as dl = Redα so that da = Rewdα. Then
3
we can write the power collected by the infinitesimal area as the power density atthat α times the area of the infinitesimal area as
Incident Sunlight
Golden, CO: 39.74 deg N
N
S
α
Solar Cell
ca
α
N
S
α
dl
Re
Figure 1.2: Schematic showing the setup for the pole-to-pole cell calculation
dP = P (α)da = P0 cosαRewdα
where w is the width of the cell, here 1 m.
Next, to find the total power being collected we integrate up all the dP ’s from poleto pole as
Power =
∫dP =
∫ π
−πP0Rew cosαdα = P0Rew [sinπ − sin(−π)]
= P0Rew (1− (−1))
= 2P0Rew
and using values for all the parameters given above we get that
Power = 2× 1.35 kW/m2 × 6.37× 106 m× 1 m = 17.2 GW
Shortcut
4
Another way to solve the problem is to realize that the earth only collects fluxfrom the sun over a linear distance of the diameter of the earth (2Re) and so thecell running from pole to pole, since atmospheric effects are neglected, collects theequivalent power of a cell in space that is 1 m wide by 2Re long. Thus, the solutionto the problem is just
Power = 2P0Rew = 2× 1.35 kW/m2 × 6.37× 106 m = 17.2 GW
5. A 3.1 kW peak system that was recently installed on a campus house cost a totalof $30,000 after rebates from the government. For your reference, graph the costper KWh of the energy generated by this system as a function of the lifetime of thesolar panels used (from 5 years to 50 years). You will realize that the more robustand long-lasting the solar cell, the better its performance in terms of cost. Thisis generally referred to in energy systems as ’Levelized Cost of Energy/Electricity(LCOE)’
Most solar cell companies claim lifetimes of 30 years. What is the cost per kWh ofelectricity generated over this lifteime? You can assume that the solar cell producesat peak power for 6 hours in the day, all throughout the year.
Solution:
Since solar cells on a private/commercial rooftop require a significant initial invest-ment, the cost per KWh of the electricity the cells generate is directly dependenton the total electricity the cells generate over their lifetime:
Cost
kWh=
Initial Cost
Total Energy Generated=
30, 000
3.1 kWp× day
6 hours× year
365 days× Lifetime
x years
If we vary the number of years in a lifetime, the curve seen in figure 1.3 can begenerated.
Doing the calculation specifically for x=30 years, we get cost per kWh to be 14.73cents/kWh, which rounds to an answer of 15 cents/kWh
6. Cadmium Telluride (CdTe) is considered a promising 2nd generation solar celltechnology, due to its high efficiency, low cost, and long-term stability. However,one of the main issues with CdTe solar cells is the total amount of Tellurium (Te)available on the earth. According to the United States Geological Survey, the globalproduction of Te was estimated to be 135 metric tons in 2007, and the world reserveof Te is estimated to be about 48000 metric tons.
5
Figure 1.3: Cost/kWh vs. Lifetime
If we use all the Te produced in 2007 to make 2µm thick CdTe solar cell moduleswith 10.5% efficiency, calculate the total power that can be generated under onesun (1 kW/m2) by these panels.
Solution:
To calculate the total power it is possible to generate with one year’s worth of Teproduction we must first know how much Te is in a solar cell per unit area.
Te weight
unit area= (CdTe Density)× (Te weight fraction in CdTe)× (Thickness)
The weight fraction of Te in CdTe is:
Te weight fraction in CdTe ≡ fTe
=MTe
MTe +MCd
=127.6 g/mol
127.6 g/mol + 112.411 g/mol= 0.53
6
where M is the molecular mass.
The annual power available is then:
Annual power =(Annual global production)(Power density from the sun)(Efficiency)
(CdTe density)(Te weight fraction in CdTe)(Thickness)
Annual power =(135× 1000× 1000 g)× (1 kW/m2)× (10.5%)
(5.85× 106 g/m3)× (0.53)× (2× 10−6m)= 2.28 GW
7. Cadmium Telluride (CdTe) is considered a promising 2nd generation solar celltechnology, due to its high efficiency, low cost, and long-term stability. However,one of the main issues with CdTe solar cells is the total amount of Tellurium (Te)available on the earth. According to the United States Geological Survey, the globalproduction of Te was estimated to be 135 metric tons in 2007, and the world reserveof Te is estimated to be about 48000 metric tons.
Calculate the total power that can be generated under one sun if we consume all theTe in the world’s reserves to make these panels. (The density of CdTe is 5.85 g/cm3)
Solution:
The total power that could be generated if you use all the Te available can becalculated by replacing the annual global production by the total Te in the earth’scrust:
Total power =(Total production)(Power density from the sun)(Efficiency)
(CdTe density)(Te weight fraction in CdTe)(Thickness)
Total power =(48000× 1000× 1000 g)× (1 kW/m2)× (10.5%)
(5.85× 106 g/m3)× (0.53)× (2× 10−6m)= 0.81 TW
8. Cadmium Telluride (CdTe) is considered a promising 2nd generation solar celltechnology, due to its high efficiency, low cost, and long-term stability. However,one of the main issues with CdTe solar cells is the total amount of Tellurium (Te)available on the earth. According to the United States Geological Survey, the globalproduction of Te was estimated to be 135 metric tons in 2007, and the world reserveof Te is estimated to be about 48000 metric tons.
Suppose that tellurium costs $215/kg. What is the cost of tellurium used in 1kWof CdTe solar cells?
7
Solution:
To calculate the cost of Te per kW peak solar production we just take the volumeof Te needed to make one kW times the density times the weight fraction times theprice:
$Te
kW solar power=
($215/kg)× (2× 10−6 m)× (5.85× 103 kg/m3)×fTe1 kW/m2 × 10.5%
= $12.74/kW
The cost of the Te works out to about $0.013 per watt, and so is a pretty smallpart of the total cost of the cell
8
Chapter 2
Problem Set 2
1. The Fermi Dirac distribution gives the electron occupancy of a given state withenergy ε:
f(ε) =1
exp[(ε− µ)/kBT ] + 1
where µ is the electron chemical potential (known as the Fermi level), T is thetemperature and kB is Boltzmann constant, given by:
kB = 1.38× 10−23 J/K
= 8.62× 10−5 eV/K
(a) What is the value of f(µ)?
(b) Find the difference in energy between the points where f(ε) = 1/4 and f(ε) =3/4. Your answer should be in terms of kBT .
(c) Describe the behavior of the Fermi-Dirac distribution at extremely high tem-peratures.
Solution
See figure 2.1. Plotting this function is easy if you choose to use enough data pointsin Excel or a program like Mathematica that can plot functions.
(a) f(µ) is the value of the Fermi-Dirac distribution function when the energy isequal to the chemical potential.
f(ε) =1
exp[(µ− µ)/kBT ] + 1=
1
exp[0] + 1=
1
2
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Figure 2.1: Fermi-Dirac Distribution Function for T = 0.1 K, T = 300 K, and T = 2000K
(b) We begin by inverting the Fermi-Dirac distribution function to find:
ε− µ = kBT ln
[1
f(ε)− 1
]So the energy difference δε between the points where f(ε) = 1/4 and f(ε) =3/4 is then given by:
δε = kBT
[ln
(1
1/4− 1
)− ln
(1
3/4− 1
)]= kBT [ln(3)− ln(1/3)]
= 2kBT ln(3)
= 2.197 kBT
(c) At high temperature kBT ε− µ so
exp
(ε− µkBT
)≈ 1
So f(ε) ≈ 1/2.
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2. The Ge band gap is EG = 0.66 eV, the density of states electron effective mass isme = 0.55 m and the density of states hole effective mass is mh = 0.37 m, wherem is the mass of an electron.
Calculate the intrinsic carrier concentration in Ge at 300K (cm−3). Think aboutwhy this is higher than the number discussed for Si at comparable temperatures.
Solution
The intrinsic carrier density is given by:
ni = 2
(kBT
2π~2
)3/2
(mpmn)3/4 exp
(−Eg2kBT
)Plugging in the numbers given for Eg, me, and mh, Planck’s constant ~ = 1.055×10−34 Js, Boltzmann constant kB = 1.38 × 10−23 J/K = 8.62 × 10−5 eV/K, andvarying temperature we calculate for Germanium ni = 2.18× 1013 cm−3
The smaller band gap for Ge results in a much higher intrinsic carrier density thanSi, where ni = 1× 1010 cm−3
3. A piece of silicon is doped with 8×1010 atoms/cm3 of Boron and 3×1010 atoms/cm3
of Phosphorus. Assume all the dopants are ionized and the temperature is 300 K.
(a) Is this material n-type or p-type?
(b) Calculate the concentrations of holes in the valence band (cm−3)
(c) Calculate the concentrations of electrons in the conduction band (cm−3)
(d) Find the Fermi level (with respect to the valence band edge) of this material.Express your answer in terms of eV above valence band.
Solution
(a) The general neutrality condition is n+NA = p+ND, giving:
p− n = NA −ND
Since the concentration of boron (a group III acceptor) is greater than thatof phosphorous (a group V donor), NA > ND, so this material will be havep > n and so will be p-type.
(b) We modify the treatment in the notes which found the electron concentra-tion in n-type material to find the hole concentration in this p-type material.
11
Solving the neutrality condition for n we find:
n = p+ND −NA
Inserting this into np = n2i and solving for p we find:
p =NA −ND
2+
√(NA −ND
2
)2
+ n2i
Inserting numbers we find:
p = 5.19× 1010 cm−3
(c) The concentration of electrons can be found, knowing p = 5.19× 1010 cm−3 :
n =n2i
p
= 1.9× 109 cm−3
(d) To find the Fermi level we use the expression derived in class:
p = Nv exp
[−(µ− Ev)
kBT
]Solving for µ we find:
µ = Ev + kBT ln
(Nv
p
)Using the result above and Nv = 1.83× 1019 cm−3 we find:
µ = Ev + 0.51 eV
This is a little below the middle of the gap.
4. Suppose that you want to make a silicon temperature sensor, which works by detect-ing carrier concentration in a piece of silicon and correlates it to the temperature.Ideally, you would need a piece of intrinsic silicon, but it is expensive and hard tofind (Can you guess why?)
(a) Suppose that your prototype device has a doping concentration equal to theintrinsic carrier concentration at 400K. What doping concentration is this?Please use the material parameters for Si discussed in class.
12
(b) You use your prototype at 500K. What is the ratio of the carrier concentrationin your prototype to that of intrinsic silicon at 500K? Give four significantfigures.
Solution For Si, me = 1.18m, mh = 0.81m, where m is the mass of one electron= 9.11× 10−31 kg, Eg = 1.12 eV .
(a) To get intrinsic silicon, the silicon ingot from which the Si wafer is made needsto be extremely pure, which is very expensive and difficult to make, and hencehard to find.
ni(Ta) ≥ ND,A
where Ta is the desired temperature and ND,A is the doping level (either donoror acceptor). Using the expression for ni in the notes we find:
ni = 2
(kBT
2π~2
)3/2
(memh)3/4 exp
(−Eg2kBT
)≈ 4.67× 1015 T 3/2 exp
(−6499 K
T
)cm−3
where we have used:
me = 1.18m , mh = 0.81m , Eg = 1.12 eV
Inserting T = 400 K we find that the maximum doping level is:
ND,A ≤ 3.28× 1012 cm−3
(b) By performing the same calculation at 500K and comparing it to the one at400K, you get the ratio 1.014
(a) The intrinsic temperature TI is the temperature at which the intrinsic carrierconcentration is equal to the dopant-induced carrier concentration (the num-ber of carriers introduced by the dopant). This is an important temperatureto keep track of, since solar cells actually rely on semiconductors being dopedfor functioning. Given a donor dopant concentration of ND, find the intrinsictemperature in terms of the doping concentration, effective densities of statesand the band gap.
13
Evaluate this for the case of Si doped with a phosphorous concentration of1016 cm−3 Full credit will be given for assuming the effective density of statesare constant and equal to their room temperature values, but if you wish youcan include their temperature dependence by iteration. (For iteration, youwould essentially separate the two temperature dependences on two sides ofthe equation, guess a value on one side, calculate the other and then plug thecalculated value back.)
Solution:
The intrinsic carrier concentration is given by:
n = 2(kBT/2π~2)1/2e−Eg/2kBT
where Nc and Nv are the effective density of states for the conduction andvalence band respectively. Setting this equal to the donor density we andsolving for temperature we find:
TI =Eg2kB
[ln
(√NcNv
ND
)]−1This is not a closed form equation, since the effective densities of states dependon temperature also.
Inserting numbers using the 300 K effective densities of states we find:
TI =1.12 eV
2 · 8.62× 10−5 eV/K
[ln
(√3.22× 1019 · 1.83× 1019
1016
)]−1= 833 K
If we include temperature dependence of the effective densities of states, weget
ni = 2
(kBT
2π~2
)3/2
(memh)3/4 exp
(−Eg2kBT
)
TI =Eg2kB
[ln(2(kBT2π~2 )3/2(memh
)3/4ni
)]−1
14
We can insert the initial guess for temperature T = 833K on the right to finda value for TI . This can be re-substituted on the right side of the equation tofind a new TI . Iterating a few times we find:
TI = 714 K
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16
Chapter 3
Problem Set 3
1. The absorption coefficients for Si and GaAs for 700 nm light are 2345.48 cm−1 and27424.07 cm−1 respectively. Compare the depths where the photon flux is reducedto %10 of the value just as the light enters the material for 700 nm light. The depthyou will calculate is the minimum thickness your solar cell needs to be to absorba significant fraction of the light. The big difference between Si and GaAs depthsarises from the fact that Si is an indirect band gap semiconductor, while GaAs isdirect band gap, which can absorb more strongly.
Solution
We want the distance where the photon flux is 10% of that at the surface, so weuse the formula from class.
I
I0= e−αz = 0.1⇒ z =
− ln(0.1)
α
Plugging in the values of the absorption coefficients given, we get the values below.
λ Si GaAs
700 nm 9.8 µm 0.84 µm
2. We consider a Si p+n junction with p+ region doping of NA = 1018 cm−3 and nregion doping of ND = 1016 cm−3 at a temperature of T = 350 K. Assume all thedopants are ionized and take the top of the valence band to be at E = 0 eV .
17
(a) Find the fermi level in the p+ region (relative to EV ).
(b) Find the fermi level in the n region (relative to EV ).
(c) What is the built-in voltage for this junction?
(d) Find the width in microns of the depletion region. The relative permittivity ofSilicon, εr is 11.8 and permittivity of free space ε0 is 8.85×10−12 m−3kg−1s4A2
and εs = εr × ε0.
Solution:
(a) The fermi level can be found from our expression for the hole concentration:
p = Nv exp
[−(µ− Ev)
kBT
]Solving for µ and inserting numbers we find:
µp+ = Ev − kBT ln
(p
Nv
)= 0 + 0.088 eV
= 0.088 eV
where in the last steps we used the parameters for Si.
(b) We can find the fermi level in the n region using the expression for electronconcentration:
n = Nc exp
[−(Ec − µ)
kBT
]Solving for µ and inserting numbers we find:
µn = Ec + kBT ln
(n
Nc
)= 1.12 eV − 0.24 eV
= 0.88 eV
where in the last step we have again used the parameters for Si.
(c) The built in voltage will be the difference between the chemical potentials in
18
the two regions:
φbi = µn − µp+
= Ec + kBT ln
(n
Nc
)− Ev − kBT ln
(p
Nv
)= Eg + kBT ln
(np
NcNv
)= 1.12 eV − 0.33 eV
= 0.79 eV
(d) The widths Xn, Xp and W are all only dependent on the doping concentrationsand the built in voltage of the junction.
Xp =
√2εsVbiND
eNA(NA +ND)
Xn =
√2εsVbiNA
eND(NA +ND)
W = Xp +Xn
Inserting numbers, we find W = 0.335 µm
3. The file Pset3-3.csv represents the AM 1.5 solar spectrum, and gives the energy ofthe photon (ε ), and the number of electrons at that energy in the two columns (n(E) ). The file format is a csv, and you can use any spreadsheet software or evenGoogle Docs to solve this problem.
(a) From the data, you can calculate the number of photons above a certainband gap energy, by adding up all the values above a certain photon energy.Calculate this value for the band gap energy of Silicon (band gap 1.1 eV)
(b) Assume that each photon creates an electron-hole pair, generating an equiv-alent current JL and each electron is extracted at a voltage equivalent to theband gap energy Vbg . The power produced per area would then be equal toJLVbg.
19
Calculate JL and determine the power extracted per area as a function of bandgap energy. What is the optimum band gap for the solar cell, where you getthe maximum power?
Note that your answer will be different from the value of 1.4 eV that we havediscussed in class, because JL is not the only current contribution we considerwhen we deal with solar cells, but this calculation is still illustrative.
Solution
(a) To determine the total photon flux above the bandgap of Si, the n(E) curveneeds to be integrated over the appropriate limits. However, since our data isdiscrete, we have to replace the integral with a sum and numerically integratethe data. We need to be careful, though, because the n(E) data is not regularlyspaced in energy so we will need to compute ∆E for each data point. Thenumber of photons can be found by:
Total photon flux =
∫ ∞Eg eV
n(E)dE
=E→∞∑
E=Eg eV
n(E)∆E(E)
By performing this discrete sum using the spreadsheet software, the numberof photons above the band gap comes out to 2.75× 1021 photons
(b) The total photon flux above the bandgap of a material is the photon flux thatis absorbed and generates electrons. Thus, if we use this method to calculatethe photon flux above each bandgap energy we would get the green dashedcurve in figure 3.1. This curve has a maximum at zero and decreases withenergy. This makes sense because if we make the bandgap infinitely small,all the incident photons will be absorbed by the material. As the bandgapincreases, the photon flux that is absorbed decreases as well because there is asmaller number of photons with energies large enough to excite electrons overthe bandgap.
If we assume that each photon above the bandgap generates an electron inthe material that contributes to the current, we can calculate the total currentgenerated by the light as a function of bandgap:
20
Current Density(Eg) =charge
time · area= e
∫ ∞Eg
n(E)dE
The question however, asks for efficiency which is the output power dividedby the output power. To convert current density to power density we needto know the voltage that electrons are extracted it. In a real solar cell, thisvoltage is the V oc calculated from an IV curve. We assume that this voltageis equal to the band gap and solve:
Power Density(Eg) = Voltage · Current Density(Eg) = e ∗ Eg∫ ∞Eg
n(E)dE
This curve is the blue curve plotted in figure 3.1. The maximum occurs at aband gap of 1.122ev and 490 W/m2.
(c) Since the input power is 1000 W/m2, the efficiency works out to be 0.49.
4
3
2
1
0
x1
02
1
4321
400
300
200
100
0
Figure 3.1: Photon flux density and power density vs. band gap energy
21
22
Chapter 4
Problem Set 4
1. Download the file HW4IVdata.txt from the website. The file contains actual currentvs voltage data taken for an organic solar cell. Make a plot of current vs. voltage.This is what is called an I-V curve and is an important characteristic of any solarcell. Also make a plot of power vs. voltage.
(a) Based on your plots determine the maximum power the cell can deliver inmicrowatts.
(b) What is the optimum load resistance to operate the cell at?
(c) Calculate the fill factor as a fraction
Solution
(a) The I-V curve and power vs. voltage is shown in figure 4.1.
The voltage which corresponds to the maximum power is approximately Vmax =247 mV and the corresponding current is Imax = 0.259 mA. This correspondsto a power of 64.12 µW
(b) The resistance which would lead to optimum power would be:
Rmax =VmaxImax
= 954 Ω
(c) From the data the open circuit voltage and short circuit current are given by:
Voc = 384 mV and Isc = 0.372 mA
23
The fill factor is then:
FF =ImaxVmaxIscVoc
= 0.45
Figure 4.1: Top: Current-voltage curve for organic solar cell, showing shaded regioncorresponding the the maximum power condition. Bottom: Power (P = IV ) as a functionof voltage for organic solar cell.
2. Due to the high cost of high-efficiency solar cells, many are used with opticalconcentrators to increase the amount of light power per solar cell area. The sunlightintensity is often measured in units of suns, so that the irradiance for Ns suns wouldbe Ns times the irradiance of an AM1.5 solar spectrum. We wish to see how thisincrease in light changes the operation of the solar cell. For simplicity we use anideal solar cell where we can ignore the effects of the series and shunt resistances,and the fill factor as a function of open circuit voltage is given in figure 4.2. Wealso take
Js = 10−12 A/cm2
24
and that an illumination of one sun (Ns = 1) with a total power density of900 W/m2 produces a light current density of:
JL(1) = 4× 10−2 A/cm2
Further we assume that the light current scales with the number of suns (JL(Ns) =NsJL(1)) and that the temperature of the cell is 25C independent of the amountof light.
Figure 4.2: Fill factor as a function of open circuit voltage for Si solar cells at roomtemperature.
Find the ratio of open circuit voltage, efficiencies and powers per unit area under500 suns to the one without using optical concentrators.
Solution
The open circuit voltage is given by:
Voc =kBT
eln
(1 +
ILIs
)25
Inserting the numbers given above we find:
Voc(Ns = 1) = 0.63 V and Voc(Ns = 500) = 0.79 V
Therefore the ratio is 0.79/0.63 = 1.25.
The efficiency is given by
η =PcPs
=FF VocJsc
Ps
where Pc and Ps are the output power density of the cell and the input powerdensity of the sun respectively, FF is the fill factor, and Jsc is the short circuitcurrent density, which is equal to the light current density.
Using the graph in figure 4.2 we find the following:
Ns Ps Jsc Voc FF η Pc1 0.090 W/cm2 4× 10−2 A/cm2 630 mV 0.83 23% 0.021 W/cm2
500 45 W/cm2 20 A/cm2 790 mV 0.86 30% 13.6 W/cm2
So the ratio of efficiencies is 30/23 = 1.30
The power per area is given by:
Pc = FF VocJsc
Using the numbers from the table above, we find ratio of powers to be 13.6/0/021= 647.61.
Note that we produce 647 times more power per area using only 500 times as muchlight energy input.
3. As we have seen in the previous problem, increasing the light intensity can resultin a saving of cell area and an increase in Voc and hence efficiency. However theincreased light can also heat up the cell, resulting in a decrease in Voc and efficiency.In this problem address this issue.
Assume that the p-n junction being used has the following properties:
NA ND Dn Dp Ln Lp Eg me mh
cm−3 cm−3 cm2/s cm2/s µm µm eV m m1018 1016 30 15 39 12 1.1 1.18 0.81
Where NA and ND are the acceptor and donor doping densities, Dp and Dn are thehole and electron diffusivities in the n and p regions respectively, Lp and Ln are
26
the hole and electron diffusion length in the n and p regions respectively, Eg is theband gap, mh and me are the hole and electron density of states effective masses,and m is the electron mass. We also assume that the solar cell behaves as an idealdiode with a fill factor of FF = 0.80.
For this problem we illuminate our solar cell with 500 suns. This gives a lightcurrent density of
JL = 500 · 0.04 A/cm2 = 20 A/cm2
and a solar power density of
Ps = 500 · 900 W/m2 = 4.5× 105 W/m2
We assume that this heats our solar cell to a temperature of T = 400 K.
(a) Find the intrinsic carrier density (ni in cm−3) for this cell under these con-ditions. Keep in mind that the effective densities of states will change withtemperature
(b) Find the reverse saturation current density (Js) for this junction in A/cm2
under these conditions.
(c) Determine the open circuit voltage (Voc in volts) for this junction under theseconditions.
(d) Calculate the efficiency of the cell under these conditions.
Hint: To save you time you might want to use the following shortcut. Given theparameters for this problem the effective density of states for the conduction andvalence bands are given by:
Nc =
(T
300 K
)3/2
3.22× 1019 cm−3 and Nv =
(T
300 K
)3/2
1.83× 1019 cm−3
Solution:
(a) The intrinsic density is given by:
ni =√NvNc e
−Eg/2kBT
Using the hint and plugging in numbers we find:
ni = 4.4× 1012 cm−3
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(b) The reverse saturation current is given by:
Js = en2i
(Dp
NDLp+
Dn
NALn
)Plugging in the numbers given for this cell we find:
Js = 3.8× 10−6 A/cm2
(c) The open circuit voltage is given by:
Voc =kBT
eln
(1 +
JLJs
)Plugging in numbers we find:
Voc = 0.53 V
(d) The efficiency is given by:
η = FFVocJLPs
Once again plugging in numbers we find
η = 19%
For comparison, under one sun we found:
ni = 1.4×1010 cm−3 , Js = 3.9×10−11 A/cm2 , Voc = 0.54 V and η = 19%
So the increase in efficiency from the increased light current is almost exactlycanceled out by the decrease in efficiency due to the temperature increase.Still, we make a lot more solar power per area of solar cell.
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Chapter 5
Problem Set 5
1. You are trapped on a desert island with a radio but no batteries to run it. Youremember that you could construct a basic battery to generate voltage and powerthe radio. However, you only have some iron nails and silver jewelry to work with.Draw the cell you could construct with the radio as the load and determine the halfcell reactions and the full cell reaction for this cell. (ignore the electrolyte)
Determine the standard potential E0 and the Gibbs free energy ∆G0 in kJ/mol forthe cell reaction. We assume that the constituents are in their standard states.
Solution:
The drawing is shown in figure 5.1.
By examination of the half-cell potentials in the notes we see that the potential foroxidization of Fe is greater than that for oxidization of Ag. In fact, the potential foroxidization of Ag is negative, indicating that reduction of Ag is favorable. Hencewe write the half-cell reactions with Fe as the anode, where it is oxidized and Agas the cathode, where it is reduced. The half-cell reactions are:
Fe Fe2+ + 2e− E = 0.44 V2Ag+ + 2e− 2Ag E = 0.80 V
So the total cell reaction is
Fe + 2Ag+ 2Ag + Fe2+ E = 1.24 V
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!" #"
$%"&'()%"
&*"+,-.()%"
/0%1-2('"3(4"
+,5('"3(4"
&'6('"3(4"
/0%1-2(07-%"
8,)6("9(,)"
Figure 5.1: Schematic of Fe-Ag electrolytic cell.
From above we see that the standard potential is E = 1.24 V. The standard statefree energy can be found from:
∆G = −nFE = −239 kJ/mol
2. For 20 grams of NaCl in 56 grams of water, calculate the atomic fraction, molarityand molality for the sodium.
Note: To calculate the molarity you will need the volume of the solution. We canfind this by using the partial molar volumes for H2O and NaCl in solution through:
V = nNaClVNaCl + nH2OVH2O
where ni and and Vi are s the number of moles the molar volume of i. Use:
VH2O = 18 ml/mol and VNaCl = 16.7 ml/mol
You might also want to know the atomic masses:
element H O Na ClM(g/mole) 1 16 23 35.45
30
Solution:
To start we calculate the number of moles of each constituent:
NNa = NCl =20 g
23 g/mol + 35.45 g/mol= 0.34 mol
and
NH2O =56 g
18 g/mol= 3.11 mol
The atomic fraction is:
xNa =0.34 mol
3.11 mol + 2 · 0.034 mol= 0.09
The molality is the number of moles per mass of solvent so we have:
mNa =0.34 mol
0.056 kg= 6.10 mol/kg
To find the molarity we need to calculated the volume of the solution. Using theequation and information given in the problem statement we find”
V = nNaClVNaCl + nH2OVH2O
= 0.34 mol · 16.7 ml/mol + 3.11 mol · 18 ml/mol = 61.7 ml
The molarity is then:
MNa =0.34 mol
0.062 L= 5.5 mol/L
3. The cell reactions in a Li ion battery are given by:
anode LixC6 → xLi+ + xe− + 6C
cathode Li1−yCoO2 + yLi+ + ye− → LiCoO2
cell y(LixC6) + x(Li1−yCoO2)→ 6yC + x(LiCoO2)
The anode has a capacity of 0.34 Ah/g and the cathode has a capacity of 0.137Ah/g.
(a) Calculate the cell capacity (Ah/g).
(b) Find x and y.
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(c) If the cell operates at 3.8 volts, calculate the energy mass density (Wh/g).
Solution:
(a) The cell capacity is found from:
Ah
g=
(1
0.34 Ah/g+
1
0.137 Ah/g
)−1= 0.098 Ah/g
(b) Beginning with the anode, we note that there are x unit charges for each moleof LixC6. So the anode capacity is given by:
cA =xF
MLixC6
=xF
xMLi + 6MC
where I have used a shorthand symbol cA for the anode capacity in Ah/g.Solving for x we find:
x =6cAMC
F − cAMLi
Inserting numbers we findx = 1
Turning to the cathode, we note that for there is y unit charges for each moleof LiCOO2. Thus the cathode capacity is given by:
cC =yF
MLiCoO2
where I have used a shorthand symbol cC for the cathode capacity in Ah/g.Solving for y we find:
y =MLiCoO2cC
F
Plugging in numbers we findy = 0.5
(c) The energy mass density is just the cell capacity times the voltage. We find
power density = 0.098 Ah/g × 3.8 V = 0.37 Wh/g
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