Homework Problem Solutions for MSE156/256:

Solar Cells, Fuel Cells and Batteries:Materials for the Energy Solution

Bruce M. ClemensStanford University

Autumn 2014

Copyright c2014 Bruce M. Clemens

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Contents

1 Energy and Solar Spectrum 1

2 Semiconductor Physics 13

3 Photovoltaic Devices 27

4 Solar Cells 47

5 Electrochemistry and Batteries 77

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iv

Chapter 1

Energy and Solar Spectrum

1. At noon on a sunny day, there is a solar power density 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 = 732 nm.

(a) Determine the number of photons per time incident on the panel.

(b) Determine the current generated if the panel has 100% efficiency (every inci-dent photon generates one electron that contributes to current).

Solution:

The easiest way to solve this problem is to think about it conceptually.

(a) You are given the Solar Irradiance, which is in units of Power/Area. This canbe converted to # photons/time by determining the area of the panel and theenergy of one photon of 732 nm light

power

area=

energy

time areaenergy

time area Area1

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.The photon energy can be found from

1

E =hc

=

6.626 1034 J s 3 108 m/s732 109 m = 2.716 10

19 J/photon

If we plug everything back into our original equation (Eq. 1.1) we get

1000 W/m2 1.6 m2 12.716 1019 J/photon = 5.89 10

21 photons/s

(b) Current can be thought of as the quantity of charge that passes a single pointin a wire per unit time. From part (a) we know the # of photons per secondhitting the panel. The problem states that one electron is generated perphoton. Thus, we can determine the # of electrons generated in the materialper second. Since we know the charge of an electron we can determine thecharge per second generated in the material, which is equal to the current.

photons

s=

electrons

s

electrons

s charge

electron=

charge

s= current

5.89 1021 electronss

1.6 1019 C

electron= 944 C/s = 944 Amps

2. (a) What is the power density of the sun shining on a solar cell lying flat onthe ground at the National Renewable Energy Laboratory in Golden, COon September 23, 2007 (the autumnal equinox) at noon? Assume the solarradiation to be parallel rays with a power density perpendicular to the rays of1.35kW/m2 and neglect losses due to atmospheric effects.

(b) What is the total power incident on a 1 m wide solar cell running from thenorth pole to the south pole on the same day?

Solution:

(a) 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

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P0 = 1.35 kW/m2. We must take into account the angle of the sun on the

cell, which lies flat on the ground. The latitude of Golden, CO is 39.74 N.According to Figure 1.1 we can see that the power density is given by thequantity

a/c P0 = P0 cos

Incident Sunlight

Golden, CO: 39.74 deg N

N

S

Solar Cell

ca

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

(b) To find the total power generated by a 1 m wide cell stretching from the N toS poles of the earth at the same time as the main question, we consult Figure1.2 below. To set up the problem, we consider an infinitesimal area elementof the solar cell, given by da in the illustration. This area element is 1 m wideand length dl along the long direction of the cell. We can write the length dl interms of the radius of the earth, Re, and the latitude angle, , as dl = Red sothat da = Rewd. Then we can write the power collected by the infinitesimalarea as the power density at that times the area of the infinitesimal area as

dP = P ()da = P0 cosRewd

where w is the width of the cell, here 1 m.

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Incident Sunlight

Golden, CO: 39.74 deg N

N

S

Solar Cell

ca

N

S

dl

Re

Figure 1.2: Schematic showing the setup for the pole-to-pole cell calculation

Next, to find the total power being collected we integrate up all the dP s frompole to pole as

Power =

dP =

pipiP0Rew cosd = P0Rew [sin pi sin(pi)]

= 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

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

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3. A 3.1 kW peak system that was recently installed on a campus house cost a total of$30,000 after rebates from the government. Graph the cost per KWh of the energygenerated by this system as a function of the lifetime of the solar panels used (from5 years to 50 years). Note: State any assumptions you made for this calculationand please use a computer for all graphs

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.

Figure 1.3: Cost/kWh vs. Lifetime

4. Cadmium Telluride (CdTe) is considered a promising 2nd generation solar celltechnology, due to its high efficiency, low cost, and long-term stability. However,

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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.

(a) If we use all the Te produced in 2007 to make 2m thick CdTe solar cellmodules with 10.5% efficiency, calculate the total power that can be generatedunder one sun (1 kW/m2) by these panels.

(b) Now calculate the total power that can be generated under one sun if weconsume all the Te in the worlds reserves to make these panels. (The densityof CdTe is 5.85 g/cm3)

(c) Suppose that tellurium costs $215/kg. What is the cost of tellurium used in1kW of CdTe solar cells?

(d) CdTe cell manufacturer First Solar claims that its production cost is $0.76/W.Compare this to the cost of the tellurium it uses.

Solution:

(a) To calculate the total power it is possible to generate with one years worthof Te production we must first know how much Te is in a solar cell per unitarea.

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=

MTeMTe +MCd

=127.6 g/mol

127.6 g/mol + 112.411 g/mol= 0.53

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)

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Annual power =(135 1000 1000 g) (1 kW/m2) (10.5%)

(5.85 106 g/m3) (0.53) (2 106m) = 2.28 GW

(b) 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 theearths crust:

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 106m) = 0.81 TW

(c) To calculate the cost of Te per kW peak solar production we just take thevolume of Te needed to make one kW times the density times the weightfraction times the price:

$Te

kW solar power=

($215/kg) (2 106 m) (5.85 103 kg/m3)fTe1 kW/m2 10.5%

= $12.74/kW

(d) The cost of the Te works out to about $0.013 per watt, and so is a prettysmall part of the total cost of the cell

5. For this problem you will need to use a spreadsheet program such as Excel orNumbers. Download the file SolarSpectrum from the class Coursework web-page (it is in the HW folder). This file comes from NREL and contains the ASTMG173-03 AM1.5 solar spectrum downloaded from url rredc.nrel.gov/solar/spectra/am1.5/.The spreadsheet will have four columns. The first is the list of wavelengths, thesecond is the extraterrestrial (AM0) irradiance, the third is the normal-incidence ir-radiance for AM1.5 and the third is the total (including scattered radiation) AM1.5irradiance.

(a) Make a plot of the total irradiance as a function of wavelength. Be sure tolabel the axes (including units) of the plot.

(b) From this spectra, calculate and plot the photon flux density (the number ofphotons per area per time per energy) as a function of photon energy.

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(c) Using numerical integration, calculate the total photon flux (number of pho-tons per m2 per s) which have an energy above the bandgap of Si (Eg = 1.12eV).

Solution:

There is a good explanation of this problem in the notes (Lecture 2, Slides 8-10).However, here it is again.

(a) See Fig. 1.4

Figure 1.4: The AM1.5 spectrum

(b) The data you downloaded from the website is in the form of irradiance (I())versus wavelength. We would like to find the photon flux density (n(E)) versusenergy. To calculate n(E), we will need to first calculate P (E), the irradiancein terms of energy, instead of wavelength. From P (E) we can then get n(E)for n(E) = P (E)/E. To get P (E) we recognize that

I()d = P (E)dE

We can then solve for P (E) as

P (E) = I() ddE

We know that wavelength and energy are related by E = hc/, so we canwrite

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=hc

E d

dE= hc

E2

Then

P (E) = I()hc

E2

and

n(E) =P (E)

E=I() hc

E2

E= I()

hc

E3

where h and c are Plancks constant and the speed of light, respectively. Toplot the data, we need to convert the ordinate into energy units as well throughE = hc/. The converted plot is shown in Fig. 1.5.

Figure 1.5: The AM1.5 spectrum in terms of photon flux per energy with integratednumber of photons shown in blue and indicated by the shaded region of the curve

(c) You are trying to calculate the area under the n(E) curve to determine thetotal photon flux over the different photon energies (or wavelengths) thatare above the bandgap energy of Si (1.12 eV). This can be accomplished byintegrating the n(E) curve over the appropriate limits. However, since ourdata is discrete, we have to replace the integral with a sum and numericallyintegrate the data. We need to be careful, though, because the n(E) data is

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not regularly spaced in energy so we will need to compute E for each datapoint. The number of photons can be found by

Total photons =

1.12 eV

n(E)dE

=E

E=1.12 eV

n(E)E(E) = 2.73 1021 photons

6. (From the Midterm Exam in 2007) Despite conservation efforts, an on-campusfaculty home has a $1053 annual heating bill for natural gas.

(a) Assuming the price of gas (including taxes) is $0.0125/cubic foot, how muchenergy was used in the year? Give your answer in both J/year and kWh/year.(Note: 100 cubic feet of gas contain approximately one therm of heat energy.There are 100,000 BTU in one therm, and there are about 1055 Joules in oneBTU.)

(b) What is the yearly averaged power (in kW) for heating?

(c) Assuming the gas heater is 80% efficient, and that an electric heater would be100% efficient, how many square meters of 12% efficient solar panels would beneeded to satisfy the heating need? Assume the average solar flux density is239 W/m2.

Solution

(a)

Energy =$1053

year 1 ft

3

$0.0125 1 thm

100 ft3 100, 000 BTU

thm 1055 J

BTU

= 8.89 1010 Jyear

= 8.89 1010 Wsyear

h3600 s

kWh1000 Wh

= 24687kWh

year

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(b) The average power is given by:

Power =24687 kWh

year 1 year

356 days 1 day

24 h= 2.81 kW

(c) The area needed is given by:

area =2.81 kW

239 W/m2 0.80

0.12 1000 W

1 kW

= 78.6 m2

7. The term horsepower was coined by the engineer James Watt (1736 to 1819) in1782 while working on improving the performance of steam engines. This occurredwhile using a mine pony to lift coal out of a coal mine. He conceived the idea ofdefining the power exerted by these animals to accomplish this work. He foundthat, on the average, a mine horse could pull (lift by means of a pulley) 22,000foot-pounds per minute. Rather than call this pony power, he increased thesetest results by 50 percent, and called it horsepower i.e. 33,000 foot-pounds of workper minute.1 This works out to a power output of 745.7 Watts/horsepower.

If you have a solar cell with 15% efficiency and area of 3 m2, how many tons (2000pounds) of coal could you lift from a mine that is 1000 ft deep in one hour with anincident radiation or 1.35 kW/m2? Assume that the solar cell is connected to anelectric motor that substitutes directly into the machinery where the mine ponieswould have been working and that the motor converts from electric energy to me-chanical energy with 80% efficiency.

Solution:

The energy available to lift coal is the solar electricity times the motor efficiency.By equating this to the gravitational potential energy we find:

mg =A S t solar motor

h

where mg is the weight that can be lifted (mass times the acceleration due togravity), A is the solar panel area, S is the solar power density hitting the panels, t

1from http://en.wikipedia.org/wiki/Horsepower

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is the time, h is the height lifted, and solar and motor are the solar panel and motorefficiencies. Putting in numbers we find:

mg =3m2 1.35 kW/m2 60 min 0.15 0.80

1000ft 1 hp

0.7457 kW 33000 ft lb

hp min 1 ton

2000 lb= 0.645 ton

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Chapter 2

Semiconductor Physics

1. In class we saw that the electron occupancy of a given state with energy is givenby the Fermi-Dirac distribution function:

f() =1

exp[( )/kBT ] + 1where is the electron chemical potential (known as the Fermi level), T is thetemperature and kB is Boltzmann constant, given by:

kB = 1.38 1023 J/K= 8.62 105 eV/K

(a) Make a plot of this for temperatures T = 0.1 K, T = 300 K, and T = 2000 K.For your plot, use an energy range from -3 eV to 3 eV, and take = 0.

(b) What is the value of f()?

(c) Find the difference in energy between the points where f() = 1/4 and f() =3/4. Your answer should be in terms of kBT .

(d) Describe the behavior of the Fermi-Dirac distribution at extremely high tem-peratures.

Solution

(a) See figure 2.1. Plotting this function is easy if you choose to use enough datapoints in Excel or a program like Mathematica that can plot functions.

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Figure 2.1: Fermi-Dirac Distribution Function for T = 0.1 K, T = 300 K, and T = 2000K

(b) f() is the value of the Fermi-Dirac distribution function when the energy isequal to the chemical potential.

f() =1

exp[( )/kBT ] + 1 =1

exp[0] + 1=

1

2

(c) We begin by inverting the Fermi-Dirac distribution function to find:

= kBT ln[

1

f() 1]

So the energy difference between the points where f() = 1/4 and f() =3/4 is then given by:

= kBT

[ln

(1

1/4 1) ln

(1

3/4 1)]

= kBT [ln(3) ln(1/3)]= 2kBT ln(3)

= 2.197 kBT

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(d) At high temperature kBT so

exp

( kBT

) 1

So f() 1/2.2. The Ge band gap is EG = 0.66 eV, the density of states electron effective mass

is me = 0.55 m and the density of states hole effective mass is mh = 0.37 m.The Si band gap is EG = 1.1 eV, the density of states electron effective mass isme = 1.18 m and the density of states hole effective mass is mh = 0.81 m

(a) Plot the intrinsic carrier density as a function of temperature (from 300K -800K) for Ge and Si on the same graph. Why does Ge have a higher carrierdensity than Si at comparable temperatures.

(b) The Si and Ge are then doped with 1013 atoms/cm3 and 1015 atoms/cm3 ofphosphorus respectively. Plot the carrier density (only electrons) as a functionof temperature (from 300K - 800K) for both on the same graph. Assume allthe dopants are ionized.

Note: You will have to use a log scale for the y-axis to fit the data for Si and Geon the same graph.

Solution

(a) The intrinsic carrier density is given by:

ni = 2

(kBT

2pi~2

)3/2(mpmn)

3/4 exp

( Eg2kBT

)Plugging in the numbers given for Eg, me, and mh, Plancks constant ~ =1.055 1034 Js, Boltzmann constant kB = 1.38 1023 J/K = 8.62 105 eV/K, and varying temperature we can plot the curves shown in fig-ure 2.2.

So we see that the smaller band gap for Ge results in a much higher intrinsiccarrier density than Si.

(b) For doped semiconductors, the carrier density is given by:

n =ND2

+

[(ND2

)2+ n2i

]1/2

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Figure 2.2: Intrinsic carrier concentration in Si and Ge

We can plot n as a function of temperature by plugging in the intrinsic con-centration from the previous part of the question for different temperatures.This is shown in figure 2.3.

3. A piece of silicon is doped with 81010 atoms/cm3 of Boron and 31010 atoms/cm3of 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 electrons and holes in the conduction andvalence bands, respectively.

(c) Find the Fermi level (with respect to the conduction band edge) of this ma-terial.

Solution

(a) The general neutrality condition is n+NA = p+ND, giving:

p n = NA NDSince 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.

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Figure 2.3: Carrier concentration in doped Si and Ge

(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.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 cm3

The concentration of electrons can then be found from:

n =n2ip

= 1.9 109 cm3

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(c) To find the Fermi level we use the expression derived in class:

p = Nv exp

[( Ev)kBT

]Solving for we find:

= Ev + kBT ln

(Nvp

)Using the result above and Nv = 1.83 1019 cm3 we find:

= Ev + 0.51 eV

This is a little below the middle of the gap.

4. (a) A 4mV voltage is applied across the block of silicon shown in Figure 2.4, withlength L = 8 cm and cross section area A = 6 cm2. It generates a current of2A in the circuit. Determine the number of charge carriers in the block. (Themobility of Si is given by Si = 1400 cm

2/V s). Is this material doped?

A L

I

V

Figure 2.4: Schematic of Si block and circuitry.

(b) You are given another rectangular block of the same type of silicon, a volt-age control power supply (generates a specific voltage across two terminals),and an ammeter (measures current moving through any circuit). Set up anexperiment to determine the dimensions of this new block without the use ofa ruler. You may need to use some of the information from part (a). (Hint:You will have to take more than one measurement and then set up a systemof three equations and three unknowns.)

Solution

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(a) Given the applied voltage and the current induced due to that voltage we candetermine the resistance of the block of material using

V = I R

The resistance of the block is related to the resistivity of the block and theblock dimensions by the formula

R = LA

where L is the length of the block in the direction of the current, A is thecross-sectional area of the surface perpendicular to the length, and is theresistivity of the silicon. We know that the number of charge carriers in thematerial are related to the conductivity by

= n e

where is the conductivity, n is the concentration of charge carriers and is the mobility of Si. Since the conductivity is inversely proportional toresistivity,

= 1/

we can put this all together to get concentration of charge carriers in terms ofquantities that can be determined from what is given in the question

n =L I

V A e

n =8 cm 2 A

4 mV 6 cm2 1.6 1019 J 1400 cm2/V = 2.98 1018carriers/cm3

(b) With the two pieces of equipment that are given to you the only experimentyou can really do is apply a voltage and measure the current generated in theblock. From this, we can calculate the resistance of the block (see above).

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The key to this problem lies in the fact that resistance is dependent on thedimensions of the block and resistivity is an inherent material property andis constant (assuming constant temperature) no matter how the resistancemeasurement is taken. Lets say we apply a voltage and generate a current inthe x direction. We can now calculate the resistance in the x direction andthen relate it to the resistivity and the dimensions of the block. (see figure2.5)

Rx = xy z

Figure 2.5: Silicon block with unknown dimensions

where is the resistivity of the material (calculated in the main part of theproblem) and x,y, and z are the dimensions of the block in their respectivedirections. Unfortunately, we now have one equation and three unknowns. Tosolve for x,y, and z we will need two more equations that come from takingthe resistance measurement in the y and z directions.

Ry = yx z

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Rx = zy x

Now we have three equations and three unknowns and can solve for x,y, andz.

x =

2

Rz Ry

y =

2

Rz Rx

z =

2

Rx Ry

5. Semyon the Semiconductor Maker wants to make a silicon temperature sensor forhis oven. The sensor works by detecting the carrier concentration in a piece ofsilicon and correlates this concentration to the temperature. Ideally, hed want apiece of intrinsic silicon for this application, but intrinsic silicon is expensive andhard to find (as Frank the TA found out the hard way). Therefore, he must makedo with doped silicon.

Semyon only bakes at temperatures above 250F (he gets his recipes from an Amer-ican website). What is the highest doping concentration that Semyon can tolerate(and still have a reliable sensor for the temperature range)?

Solution:

In order to have a appreciable temperature dependence Semyon will need to havea dopant level low enough so that his Si is intrinsic in the temperature range ofinterest. So we need

ni(Ta) ND,Awhere Ta is the annealing temperature and ND,A is the doping level (either donor

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or acceptor). Using the expression for ni in the notes we find:

ni = 2

(kBT

2pi~2

)3/2(memh)

3/4 exp

( Eg2kBT

) 4.67 1015 T 3/2 exp

(6499 KT

)cm3

where we have used:

me = 1.18m , mh = 0.81m , Eg = 1.12 eV

Inserting T = 250 F 394 K we find that the maximum doping level is:

ND,A 2.5 1012 cm3

6. Imagine that you engineered a solar cell where the composition varied with depth(a likely consequence of heavy doping) and, as a result, the absorption coefficientalso varied with depth. Assume that the linear absorption coefficient is now afunction of depth as (z) = 0z. With this new absorption coefficient, find anexpression for the photon flux of monochromatic light as a function of depth intothe semiconductor.

Solution: We must recognize how we arrived at our first expression for the lightintensity as a function of depth. We wrote out how the intensity changes across asmall distance element as

dI = Idz

Here we need to replace with (z) and continue as from before.

dI = (z)Idz = 0zIdz

We can then integrate to get that

I(z) = I0e0z2

2

Where we have recognized the derivative through the reverse chain rule.

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7. (From 2008 Midterm) The intrinsic temperature TI is the temperature at which theintrinsic carrier concentration is equal to the dopant-induced carrier concentration(the number of carriers introduced by the dopant).

(a) Given a donor dopant concentration of ND, find the intrinsic temperature interms of the doping concentration, effective densities of states and the bandgap.

(b) Evaluate this for the case of Si doped with a phosphorous concentration of1016 cm3 Full credit will be given for assuming the effective densities of statesare constant and equal to their room temperature values, but if you wish youcan include their temperature dependence by iteration.

Solution:

(a) The intrinsic carrier concentration is given by:

n = (NcNv)1/2EEg/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

(NcNvND

)]1This is not a closed form equation, since the effective densities of states dependon temperature also.

(b) Inserting numbers using the 300 K effective densities of states we find:

TI =1.12 eV

2 8.62 105 eV/K[ln

(3.22 1019 1.83 1019

1016

)]1= 833 K

Iterating the above equation a few times using the temperature dependenceof the effective densities of states we find:

TI = 714 K

8. (From 2013 Final Exam) We found that the concentration of electrons in the con-duction band of a semiconductor is given by:

n =

Ec

De()fe() d

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with the occupancy of electron states given by:

fe =1

exp [( )/kBT ] + 1and

D() = 12pi2

(2me~2

)3/2( Ec)1/2

In class we found the result as:

n = 2

(mekBT

2pi~2

)3/2exp [ (Ec ) /kBT ]

However, to get this result we cheated. To get this result we made the approxima-tion that the occupancy of electron states is given by:

fe () = exp [ ( ) /kBT ]

By examination we see that this approximation is valid only if:

Ec kBT

Thus when the doping concentration is low, and the fermi level is far away fromthe band edge, our approximation is ok. However, at high doping concentrationsit breaks down.

(a) What is the value of Ec that will give a 10% error in the occupancy ofelectron states at the conduction band edge? Your answer should be a numbertimes kBT .

(b) For Si at 300 K, what doping concentration does this correspond to? For thisproblem, use the regular (approximate) relationship between the doping con-centration and the fermi level, and assume only donor doping, all the dopantsare ionized and that the doping concentration is large compared to the intrinsiccarrier concentration. Your answer should be in cm3.

Solution:

(a) The error is given by:

e =fe (Ec) fe(Ec)

fe(Ec)

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Inserting our expressions and simplifying we find:

exp [ (Ec ) /kBT ] = 0.1

orEc = ln(0.1)kBT = 2.3 kBT

(b) For Si the relationship between the doping concentration and the fermi levelis given by:

ND = Nc exp [ (Ec ) /kBT ]So the doping concentration for this level of error is given by:

ND = 0.1 Nc = 3.22 1018 cm3

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Chapter 3

Photovoltaic Devices

1. Monochromatic light with photon flux n0 photons per unit area per second is nor-mally incident on a semiconductor surface. A fraction R is reflected.

(a) Find an expression for n(z), the photon flux as a function of the depth zinto the semiconductor. Assume that the this light has a linear absorptioncoefficient in this material.

An important quantity in modeling solar cell performance is the generation rateg(z), which is the rate per volume at which light-induced carriers (electron-holepairs) are generated. This quantity is defined by noting that g(z) dz is the numberof carriers per area per time generated in the region between z and z + dz.

(b) What are the units on g(z)?

(c) Assuming that each photon absorbed creates one electron-hole pair, and usingthe parameters mentioned above, calculate an expression for the generationrate g(z) of these pairs as a function of depth in the semiconductor.

Solution

(a) Since the light is monochromatic, we dont have to worry about variations inthe absorption with energy. The photon flux goes simply as the intensity aswe have seen from class. To account for the reflected light at the surface wejust change n0 to n0(1 R) and we get, for the photon flux as a function of

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depthn(z) = n0(1R)ez

(b) The units on g(z) are number per time per volume.

(c) Now the generation rate is just the amount by which the photon flux decreaseswith depth, which is the negative derivative of the photon flux as a functionof depth.

g(z) = dndz

= n0(1R)ez

2. Using the website http://www.ee.byu.edu/cleanroom/OpticalCalc.phtml to findthe absorption coefficients for Si and GaAs at each wavelength, compare the depthswhere the photon flux is reduced to %10 of the value just as the light enters thematerial for 700 nm and 400 nm light.

Note: The depth you will calculate is the minimum thickness your solar cell needsto be to absorb a significant fraction of the light.

Solution

From the website listed in the problem set, we get the data shown below.

Si GaAs

400 nm 121579 cm1 674185 cm1

700 nm 2345.5 cm1 27424 cm1

We want the distance where the photon flux is 10% of that at the surface, so weuse the formula from class.

I

I0= ez = 0.1 z = ln(0.1)

Plugging in the values of the absorption coefficients from above, we get the valuesbelow.

3. A p-n junction is created by doping the right side of a piece of silicon with 1014 atoms/cm3

of phosphorus and the left side with 1018 atoms/cm3 of Boron. Enough time is givenfor the junction to come to equilibrium (ie. drift current = diffusion current).

28

Si GaAs

400 nm 0.19 m 0.034 m700 nm 9.8 m 0.84 m

(a) Draw a band diagram of the p-n junction, mark Ec, Ev, f,intrinsic, and f .This plot does not have to be to scale, but the relative positions of the energylevels must be correct.

(b) Plot the charge density and electron concentration (in C/cm3) as a functionof position in the silicon piece. This plot does not have to be to scale, but therelative magnitudes of the charge density (on each side of the junction) andelectron concentration must be correct.

(c) Determine xn, xp (length of depletion region on the n side and the p side),and the total depletion width W .

(d) Plot the electric field as a function of position (take into account the differencesin xn and xp).

(e) Plot the voltage as a function of position (take into account the differences inxn and xp).

Notes:

Plots may be hand drawn for this question. Assume x=0 is at the center of the p-n junction

Solution

(a) Figure 3.1 shows the band diagram of the p-n junction that is formed. Noticethat the distance between the Fermi Level and the conduction band on the nside is larger than the distance between the Fermi level and the valence bandon the p side.

(b) Figure 3.2 shows the space charge density and the electron concentration inthe p-n junction. While the electron concentration is zero in the depletionwidth, it is non-zero on the p-side outside the depletion width because theelectrons are the minority carrier on that side.

(c) Xn, Xp and W are all only dependent on the doping concentrations and the

29

Figure 3.1: Energy Band Diagram for P-N Junction

built in voltage of the junction.

Xp =

2sVbiND

eNA(NA +ND)

Xn =

2sVbiNA

eND(NA +ND)

W = Xp +Xn

Vbi is dependent on the doping concentrations, the intrinsic carrier concentra-tion of Si and the temperature. If we assume room temperature.

Vbi =kBT

elnNANDn2i

=1.38 1023 295

1.6 1019 ln1018 1014

(1010)2= 0.703 V

Plugging into the equations for Xp and Xn

xp =

2 1.04 1012 0.703 1014

1.6 1019 1018(1018 + 1014) = 3.022 108 cm = 3.022 A

30

Figure 3.2: Space Charge Distribution and Electron Concentration

xn =

2 1.04 1012 0.703 1018

1.6 1019 1014(1018 + 1014) = .0003052 cm = 3.052 m

W = xn + xp = 3.052 m

(d) Figure 3.3 shows the E-field and Voltage in the PN junction. The Electricfield is related the charge concentration by

E =

ba

(x)dx

It is related to the potential by

V = ba

E(x)dx

31

(e) See part d.

4. In a previous homework problem, you were asked to calculate and plot the photonflux as a function of photon energy for AM1.5, and then calculate the total photonflux above the band gap of Si. We assume that each photon above the band gap isable to produce an electron, and the voltage at which they are extracted is equalto the band gap (Vbg). The power per area produced will then be JLVbg, where JLis the light current per area, which in this case is equal to electron charge times theflux of photons with energy above the band gap energy (eVbg).

(a) Plot the power per area as a function of band gap energy eVbg and determinethe optimum band gap for solar cell applications (band gap that produces themost power).

(b) What solar cell efficiency does this correspond to?

Solution

(a) In Problem #4 of Homework #1 you calculated the photon flux density fromthe Irradiance by the following method:

I()d = P (E)dE

We can then solve for P (E) as

P (E) = I() ddE

We know that wavelength and energy are related by E = hc/, so we canwrite

=hc

E d

dE= hc

E2

Then

P (E) = I()hc

E2

and

32

n(E) =P (E)

E=I() hc

E2

E= I()

hc

E3

where h and c are Plancks constant and the speed of light, respectively. Toplot the data, we need to convert the ordinate into energy units as well throughE = hc/. The converted plot is shown in figure 3.4.

You were also asked to determine the total photon flux above the bandgap ofSi. This can be accomplished by integrating the n(E) curve over the appropri-ate limits. However, since our data is discrete, we have to replace the integralwith a sum and numerically integrate the data. We need to be careful, though,because the n(E) data is not regularly spaced in energy so we will need tocompute E for each data point. The number of photons can be found by:

Total photon flux =

Eg eV

n(E)dE

=E

E=Eg eV

n(E)E(E)

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 3.4. 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:

Current Density(Eg) =charge

time area = e Eg

n(E)dE

33

The question however, asks for power density as a function of band gap energy.To convert current density to power density we need to know the voltage thatelectrons are extracted it. In a real solar cell, this voltage is the V oc calculatedfrom an IV curve. We assume that this voltage is equal to the band gap andsolve:

Power Density(Eg) = Voltage Current Density(Eg) = e Eg Eg

n(E)dE

This curve is the blue curve plotted in figure 3.4. The maximum occurs at aband gap of 1.11ev and 441 W/m2.

5. (From the Final Exam in 2007) In a linearly graded PN junction, the space chargedensity can be varied as a function of position by varying the doping concentration.The charge density is given by (x) = x where is some constant. Figure 3.5shows a plot of this function. X0 is the magnitude of the depletion width on eachside of the junction.

(a) Mark the p and n side of the junction.

(b) Derive an expression for the electric field as a function of position and plot it.

(c) Determine the maximum electric field magnitude in terms of X0 and .

(d) Derive an expression for the voltage as a function of position and plot it.

(e) Determine Vbi in terms of X0 and .

Notes:

Assume x=0 is at the center of the p-n junction. It is convention to integrate any position functions from the left to right.

Solution:

(a) Since the holes diffusing out of the p-type region will leave behind a negativecharge, the p-type region is to the left.

34

(b) Using Poissons equation we find:

E =

(x)

sdx

=

xx0

x

sdx

=

2s(x2 x20)

(c) The maximum electric field magnitude occurs at x = 0 and has value:

Emax =x202s

This is the magnitude. The actual field value at that point is negative.

(d) The voltage is found from by integrating again:

V = xx0

E(x) dx

=2s

xx0

(x2 x20) dx

=

2s

(2x30 x3

3+ xx20

)(e) The total bias voltage is found by evaluating the voltage at x = x0. This

gives:

Vbi =2x303s

6. A p-n junction is created by doping the left side of a piece of silicon with 1014 atoms/cm3

of phosphorus and the right side with 1018 atoms/cm3 of Boron. Enough time isgiven for the junction to come to equilibrium (ie. drift current = diffusion current).A voltage is then applied to Reverse Bias the junction.

(a) Draw two band diagrams, one of the standard p-n junction and the other ofthe junction after the voltage is applied.

Mark Ec, Ev, intrinsic, f , Vbi, Va, n, p, xn, xp, and W on both graphs. Youdo not have to draw the graph with exact values but the relative values ofthese parameters should be drawn accurately and clearly.

35

(b) Determine the direction and relative sizes of the electron drift and diffusioncurrents as well as the hole drift and diffusion currents for both the unbiasedand biased junction.

(c) Draw the charge density for both conditions on the same graph to show thedifference between the biased and unbiased case. Mark Na, Nd, xn, xp.

(d) Draw the E-field for both conditions on the same graph to show the differencebetween the biased and unbiased case.

(e) Draw the voltage for both conditions on the same graph to show the differencebetween the biased and unbiased case. Mark xn and xp.

(f) Finally, we shine some light on the unbiased junction in an attempt to useit as a solar cell. Explain what happens in the junction to produce current.Concentrate on the changes in the drift and diffusion currents and why theyoccur. You may use words, figures and equations to explain yourself.

Notes:

Make sure all your graphs are oriented such that the phosphorus is on the leftand the boron is on the right.

Assume x=0 is at the center of the p-n junction. It is convention to integrate any position functions from the left to right.

Solution:

(a) Figure 3.8 and figure 3.9 show the band diagrams for an unbiased and biasedjunction respectively.

(b) Figure 3.10 shows the drift and diffusion currents for electrons and holes. Thesize of the arrow indicates the magnitude of the current.

(c) Figure 3.11 shows the charge density, electric field and voltage of the biasedand unbiased junctions.

(d) See part c.

(e) See part c.

(f) When you shine light on the junction, photons from the light are absorbedby interacting with electrons in the semiconductor. In the absorption pro-cess, the energy from the absorbed photons is transferred to the electrons.

36

Electrons in the valence band can absorb photons with energy sufficient topromote them to the conduction band, leaving behind a hole in the valenceband. So absorption of light results in electron-hole pairs in the semiconduc-tor. Electron-hole pairs produced in p-type region of the p-n junction, produceexcess minority electrons. These minority electrons can lower their energy bymoving down the electron energy profile into the n-type region. This is anincrease in the drift current. Similarly, electron-hole pairs produced in then-type region produce excess minority holes, which move toward lower holeenergy in the p-type region, also increasing the drift current. This movementof minority electrons from the p-type to the n-type region or minority holesfrom the n-type to the p-type region through the junction separates the elec-tron and hole in the electron hole pair. The resulting increase in drift currentbreaks the balance between the drift and diffusion current in the junction.Power can be extracted from this current as work can be done as the minoritycarriers are driven through the junction by the built-in voltage.

7. (From 2008 Midterm) We consider a Si p+n junction with p+ region doping ofNA = 10

18 cm3 and n region doping of ND = 1016 cm3 at a temperature ofT = 350 K. Assume all the dopants are ionized and take the top of the valenceband to be at E = 0.

(a) Find the fermi level in the p+ region.

(b) Find the fermi level in the n region.

(c) What is the built-in voltage for this junction?

Solution:

(a) The fermi level can be found from our expression for the hole concentration:

p = Nv exp

[( Ev)kBT

]Solving for and inserting numbers we find:

p+ = Ev kBT ln(p

Nv

)= 0 + 0.088 eV

= 0.088 eV

where in the last steps we used the parameters for Si.

37

(b) We can find the fermi level in the n region using the expression for electronconcentration:

n = Nc exp

[(Ec )kBT

]Solving for and inserting numbers we find:

n = 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 inthe two regions:

bi = n p+= Ec + kBT ln

(n

Nc

) Ev kBT ln

(p

Nv

)= Eg + kBT ln

(np

NcNv

)= 1.12 eV 0.33 eV= 0.79 eV

8. (From 2011 Midterm) A heterojunction solar cell is two materials with differentband gaps. For this problem we take material A to have a band gap of 1 eV andmaterial B to have a band gap of 1.5 eV. We assume that the the temperatureis 300 K, and assume that the effective density of states is the same for both thevalence and conduction band for both materials and is given by its free electronvalue:

Nc = Nv Nc,v = 2(mkBT

2pi~2

)3/2= 2.51 1019 cm3

where m is the free electron mass.

(a) What are the intrinsic (un-doped) electron and hole carrier densities for eachmaterial?

(b) What is the position of the fermi level in each isolated material when they areboth intrinsic (un-doped). Measure the position of the fermi level relative tothe top of the valence band.

38

We now dope material A with n-type dopants to a density of 1 1012 cm3 and wedope material B with a p-type dopants to a density of 1 1017 cm3. We assumethat all the dopants are ionized.

(c) What is the carrier density in each isolated material? Find both n and p forboth materials.

(d) What is the position of the fermi level in isolated material A and material B.Again, measure the position of the fermi level relative to the top of the valenceband.

We now bring the doped A and B together to form a p-n junction. Eventually thediffusion current is countered by the drift current of the built in potential, as in ap-n junction formed from the same materials.

(e) Sketch the electron energy level diagram for material A and B before and afterforming the p-n junction. Show on you sketch the fermi level, the intrinsicfermi levels, as well as the conduction and valence band edges.

Solution:

(a) The intrinsic electron and hole carrier density is given by:

ni = pi = Nc,veEg/2kBT =

{1.0 1011 cm3 material A6.3 106 cm3 material B

(b) Since the effective densities of states are the same for both the conduction andvalence band, the fermi level will be right in the middle of the gap for bothmaterials So for intrinsic material A, the fermi level will be 0.5 eV above thetop of the valence band and for intrinsic material B, the fermi level will be0.75 eV above the top of the valence band.

(c) Turning first to material A we first calculate the electron density:

n(A) =ND2

+

(ND2

)2+(n

(A)i

)2= 1.01 1012 cm3

We find the hole concentration by:

p(A) =

(n

(A)i

)2n(A)

= 9.90 109 cm3

39

Similarly for material B we find:

p(B) =NA2

+

(NA2

)2+(n

(B)i

)2= 1.00 1017 cm3

We find the electron concentration by:

n(B) =

(n

(B)i

)2p(B)

= 3.98 104 cm3

Wow! This is really small.

(d) The fermi level position can be found from the equation for the carrier density.For electrons in material A we have:

n = Nc,ve(Ec)/kBT

Rearranging we find:

Ev = EG (Ec )= EG + kBT ln

(n

Nc,v

)=

{0.56 eV material A0.14 eV material B

40

Figure 3.3: E-field and Voltage in the PN Junction41

43

2

1

0

x1

02

1

4321

400

300

200

100

0

Figure 3.4: Photon flux density and power density vs. band gap energy

Figure 3.5: Charge density for a linearly graded PN junction

42

Figure 3.6: Electric field versus position for graded composition p-n junction.

Figure 3.7: Potential versus position for graded composition p-n junction.

43

Figure 3.8: Band Diagram for Unbiased Junction

Figure 3.9: Band Diagram for Biased Junction

44

Figure 3.10: Drift and Diffusion Currents

45

Figure 3.11: Drift and Diffusion Currents

46

Chapter 4

Solar Cells

1. Download the file HW4IVdata.txt from the coursework website. The file containsactual current vs voltage data taken for an organic solar cell.

(a) Make a plot of current vs. voltage.

(b) Make a plot of power vs. voltage and determine the maximum power producedby the cell.

(c) What is the optimum load resistance to operate this cell at?

(d) Calculate the fill factor.

Solution

(a) See figure 4.1.

(b) See figure 4.1.

(c) The voltage which corresponds to the maximum power is Vmax = 247 mV andthe corresponding current is Imax = 0.259 mA. Hence the resistance whichwould lead to optimum power would be:

Rmax =VmaxImax

= 954

(d) From the data the open circuit voltage and short circuit current are given by:

Voc = 384 mV and Isc = 0.372 mA

47

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. A silicon diode has doping in the p and n regions of NA and ND respectively. Theminority electrons in the p region have diffusion length Ln and diffusivity Dn, andthe minority holes in the n region have diffusion length Lp and diffusivity Dp. Thevalues for these are given in the table below.

n Region p RegionND (cm

3) Lp (cm) Dp (cm2/s) NA (cm3) Ln (cm) Dn (cm2/s)2 1016 11.4 103 13 2 1017 4.1 102 34

Also take T = 300 K and ni = 1010 cm3

48

(a) Find Js.

(b) Find the current density when this diode is forward biased with 0.66 V.

(c) Make a plot of the current density as a function of voltage for this diode.

Now we illuminate this junction with the spectra n(E) shown below.

(d) Assuming each photon above the band gap produces a carrier, what is thelight current density (JL) through the cell?

(e) What is the total current density through the cell when it is biased with avoltage of 0.66 V?

(f) Make a plot of the current density as a function of voltage for this cell whenit is illuminated.

(g) What is the maximum power density (W/cm2) for this cell?

(h) What is the efficiency for this cell?

(i) What is the open circuit voltage Voc for this cell under illumination?

4

3

2

1

0

x1

02

1

43210

2.0

1.5

1.0

0.5

0.0

x1

02

1

AM1.5

5960 K Spectra

Figure 4.2: Solar photon spectra. Dotted line shows the total number of photons withenergy above the Si band gap as a function of energy. (The vertical axis for the dottedline is on the right.)

49

0.12

0.08

0.04

0.00

-0.04

J

F

(A/cm

2

)

-1.0 -0.5 0.0 0.5 1.0

Voltage (V)

Figure 4.3: Current density versus voltage for the p-n diode described in the problem.

Solution

(a) We can find Js by inserting values into the equation:

Js = en2i

(Dp

LpND+

DnLnNA

)= 9.8 1013 A/cm2

(b) The current density at a given voltage is given by:

J = Js(eV/Vth 1)

where Vth = kBT/e, which, at T = 300 K, has the value Vth = 0.0259 eV.Taking V = 0.66 eV, we find:

J = 0.11 A/cm2

(c) The plot is shown below:

(d) Since each photon with energy above the band gap produces a carrier in thesemiconductor, the light current density is just the electron charge times the

50

number density of the photons with energy above the band gap. From thefigure we find

JL = e

EG

n(E) dE

=(1.602 1019 C)(2.5 1021 1

m2 s)

= 4.0 102 A/cm2

(e) The current density for a solar cell is given by:

J = JL Js(eV/Vth 1)

Using the results from above for V = 0.66 V we find:

J = 4.0 102 A/cm2 0.11 A/cm2 = 0.07 A/cm2

so in this case, the current is running backward through the cell since JF > JL

(f) The current density versus voltage plot is shown below.

(g) The power density P/A = JV is shown plotted in figure 4.5. From this graphwe can see that the maximum power density is 21 mW/cm2. We could alsohave found the maximum power voltage Vmax by solving the equation

VtheVoc/Vth = (Vmax + Vth) e

Vmax/Vth

and found the maximum power by multiplying Vmax by the correspondingcurrent density. This gives

P

A

max

= Vmax[JL Js

(eVmax/Vth 1)]

= (0.55 V) (38 mA/cm2)= 21 mW/cm2

(h) The efficiency for this cell can be found by dividing the maximum powerfound above by the total input solar power. The total input solar power canbe found by integrating the AM1.5 spectra shown above. When I do so I get1000 W/m2 = 10.0 mW/cm2. Hence the efficiency is

=21 mW/cm2

100 mW/cm2

= 21 %

51

-0.12

-0.08

-0.04

0.00

0.04

J (A/cm

2

)

-1.0 -0.5 0.0 0.5 1.0

Voltage (V)

Figure 4.4: Current density versus voltage for the illuminated p-n diode solar cell de-scribed in the problem.

25

20

15

10

5

0

Power Density (mW/cm

2

)

-1.0 -0.5 0.0 0.5 1.0

Voltage (V)

Figure 4.5: Current density versus voltage for the p-n diode described in the problem.

52

(i) The open circuit voltage Voc can be found from the plot in figure 4.4 or throughthe expression from the notes

Voc = Vth ln

(1 +

JLJs

)= 0.63 V

3. (From the Midterm Exam in 2007) An ideal (Rsh = , Rs = 0) Si p-n junctionsolar cell with the following properties:

NA = 5 1018 cm3 ND = 1016 cm3Dn = 25 cm

2/s Dp = 10 cm2/s

n = 5 107 s p0 = 107 sis exposed to the standard AM1.5 solar spectrum with an intensity of 900 W/m2

corresponding to a solar flux of photons with energy above the bandgap of 2.5 1017 cm2s1. The temperature of the device is held at 300 K.

(a) Find the light-induced current density JL Assume that a carrier is excited foreach incident photon with energy greater than the bandgap.

(b) Find the minority carrier diffusion lengths Ln and Lp.

(c) Find the reverse saturation current density Js.

(d) Find the open circuit voltage Voc.

(e) Again assuming ideal diode behavior, find the efficiency of this solar cell.

(f) What is the cell efficiency if we increase the solar flux by a factor of 100?

Solution:

(a) Assuming that each photon with energy above the band gap creates one car-rier, the light-induced carrier density is just the the carrier charge times theincident photon density:

JL = e

Eg

n(Eph) dEph

= 1.602 1019C 2.5 1017 cm2s1

= 0.040A

cm2

53

(b) The minority carrier diffusion lengths Ln and Lp are given by:

Ln =Dnn =

(25 cm2/s 5 107 s)1/2 = 3.5 103 cm

andLp =

Dpp =

(10 cm2/s 107 s)1/2 = 1.0 103 cm

(c) The reverse saturation current density is given by:

Js =eDnnp0Ln

+eDppn0Lp

= en2i

(DnLnNA

+Dp

LpND

)where we have used n2i = np and set n = ND in the n-region and p = NA inthe p-region. Inserting numbers we find:

Js = 1.602 1019 C (1010 cm3)2 (25 cm2/s

3.5 103 cm 5 1018 cm3 +10 cm2/s

1.0 103 cm 1016 cm3)

= 1.604 1011 A/cm2

(d) Since for the ideal diode Jsc = JL, the open circuit voltage is given by:

Voc =kBT

eln

(1 +

JLJs

)= 0.026 V ln

(1 +

0.040

1.604 1011)

= 0.559 V

(e) From the attached graph for an ideal diode solar cell fill factor, we find thatfor Voc = 559 mV the fill factor is given by FF = 0.818. Hence the efficiencyof this cell is given by:

=VocJscFF

Ps

=0.559 V 0.040 A/cm2 0.818

900 W/m2

= 20%

54

(f) Increasing the solar flux by a factor of 100 increases JL by a factor of 100.Thus the open circuit voltage becomes:

Voc =kBT

eln

(1 +

JLJs

)= 0.026 V ln

(1 +

4.0

1.604 1011)

= 0.678 V

From the attached graph we now find FF = 0.84 and thus the efficiency is:

=VocJscFF

Ps

=0.678 V 4.0 A/cm2 0.84

90000 W/m2

= 25%

4. (From the Midterm Exam in 2007) As mentioned in class, advanced solar cellshave a back contact field associated with a p+ layer that is below the p layer ofa solar cell. To examine the effect of this layer we consider the simple case of ap+p structure as shown schematically below, where the Si p+ layer has a doping of

N(p+)A = 10

18 cm3 and the Si p layer has a doping of N (p)A = 1016 cm3

(a) Calculate the position (in eV) of the fermi level in an isolated p+ layer andan isolated p layer with the above dopings. (Assume that the position of thevalence band edge is Ev = 0, and that the temperature is 300K.)

(b) Calculate the potential difference (in volts) across the junction between thesetwo layers.

(c) Sketch the fermi level, valence band edge and conduction band edge for thisp+p structure.

Solution:

(a) The relationship between hole concentration, fermi level and valence bandedge is:

p = Nv exp[( Ev)/kBT ]we can find:

= Ev + kBT lnNvp

55

0.90

0.85

0.80

0.75

0.70

0.65

0.60

Fill Factor

800700600500400300200

V

oc

(mV)

Figure 4.6: Ideal diode fill factor as a function of open circuit voltage.

Figure 4.7: Schematic of p+p structure.

56

Using Nv = 1.83 1019 cm3 from the notes, and p = NA for each layer wefind:

(p+) = 0.026 eV ln

(1.83 1019

1018

)= 0.0756 eV

and

(p) = 0.026 eV ln

(1.83 1019

1016

)= 0.195 eV

(b) The difference in potential across the layer will be:

=(p

+) (p)e

= 0.120 V

5. Figure 4.8 shows the IV curve and max power for this PN junction after it is madeinto a Solar Cell.

(a) What would be the optimal load resistance to operate this cell at?

(b) If I change the top contacts of the cell to a higher conductivity material howwould that change the IV curve? Draw the new curve on the graph.

(c) Calculate the Fill Factor and efficiency. (assume that the light incident on thesample has a power of 5 Watts)

Figure 4.8: IV Curve and Max Power for Solar Cell

Solution

57

(a) We know that the power coming from the cell is dependent on the current andvoltage that cell is operating at. The only way that we have of controllingthe current and voltage coming from the cell is by adjusting either the lightincident on it or the load resistor. Since in the operation of a real solar cellwe cannot adjust the light we are only left with the load resistor. The optimalload resistance will allow us to extract the maximum power from the cell.From the graph we can see that this point occurs at 195mV and 2.75mA. Theload resistance can then be found by

R =V

I=

195 mV

2.75 mA= 70.9

(b) Changing the top contacts to a higher conductivity material lower the seriesresistance of the cell. This increases the IV curve and the max power point.See figure 4.9.

Figure 4.9: IV Curve and Max Power for Solar Cell

58

(c) The fill factor is given by

FF =Imax VmaxIsc Vsc =

195 mV 2.75 mA305 mV 4.3 mA = 0.41

Efficiency is given by

=Isc Voc FF

Ps=

305 mV 4.3 mA 0.415W

= 0.00011 = 0.011%

6. (From 2008 Midterm) Due to the high cost of high-efficiency solar cells, many areused with optical concentrators to increase the amount of light power per solar cellarea. The sunlight intensity is often measured in units of suns, so that the irradiancefor Ns suns would be Ns times the irradiance of an AM1.5 solar spectrum. Wewish to see how this increase in light changes the operation of the solar cell. Forsimplicity we use an ideal solar cell where we can ignore the effects of the series andshunt resistances, and the fill factor as a function of open circuit voltage is givenin figure 4.10. We also take

Js = 1012 A/cm2

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) = 4 102 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 25

C independent of the amountof light.

(a) What is the open circuit voltage Voc for Ns = 1 and Ns = 500?

(b) What is the efficiency of the solar cell under these two conditions of solarirradiance?

(c) What is the power per area for the solar cell under these two conditions ofsolar irradiance?

Solution

(a) The open circuit voltage is given by:

Voc =kBT

eln

(1 +

ILIs

)59

Figure 4.10: Fill factor as a function of open circuit voltage for Si solar cells at roomtemperature.

60

Inserting the numbers given above we find:

Voc(Ns = 1) = 0.63 V and Voc(Ns = 500) = 0.79 V

(b) The efficiency is given by

=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 graphin figure 4.10 we find the following:

Ns Ps Jsc Voc FF Pc1 0.090 W/cm2 4 102 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

(c) The power per area is given by:

Pc = FF VocJsc

Using the numbers above we find the numbers in the table for part (b). Wenote that we produce 650 times more power per area using only 500 times asmuch light energy input.

7. (From 2009 Midterm) One approach for improving the efficiency of a thin filmsolar cell is to pattern the junction as shown in Figure 4.11 so that photo-excitedcarriers are always within a diffusion length to the junction, while still allowing fora long optical penetration depth so all the sunlight is absorbed. One disadvantageof this geometry is that the junction area is significantly increased, leading to anincreased dark current. For this problem we take the junction area to be AJ whilethe area exposed to the sunlight is AL. Since we have a different solar light areaand junction area, we now re-examine the formalism for the cell current voltagebehavior. Assume ideal diode behavior.

(a) Write down the equation for the solar cell current I as a function of the reversesaturation current density Js, the light-induced current density JL, the loadvoltage V , temperature T , junction area AJ and sunlight exposure area AL.

(b) Find the equation for the open circuit voltage in terms of the reverse satura-tion current density Js, the light-induced current density JL, temperature T ,junction area AJ and sunlight exposure area AL.

61

Figure 4.11: Schematic of structured solar cell with interdigitated n and p regions givinga long light absorption pathway and a short minority carrier distance to the junction.

(c) Taking T = 300 K, JL = 0.04 A/cm2, and Js = 10

12 A/cm2. Find the opencircuit voltage for the cases AJ/AL = 1 and AJ/AL = 10.

(d) If we assume our cell has thickness 3/, where is the optical absorptioncoefficient, which corresponds to absorption of 95% of the light. We takeour pattern feature width to be 2Lm, where Lm is the minority carrier diffusiondistance. This gives a junction area of:1

AJ = AL

(1 +

3

2Lm

)Taking = 3 106 m1 and Lm = 100 nm, calculate the ratio AJ/AL.

Solution

(a) The solar cell I V curve is given by:

I = ALJL AJJs[exp

(eV

kBT

) 1]

(b) To find Voc we set the current equal to zero and solve for V . This gives:

Voc =kBT

eln

(1 +

ALJLAJJs

)(c) Inserting numbers we find for AJ/AL = 1, Voc = 0.63 V and for AJ/AL = 10,

Voc = 0.57 V.

1Dont spend time deriving this.

62

(d) Using the given expression we find:

AJAL

= 1 +3 m

3 106 2 100 109 m= 6

8. (From 2009 Midterm) As we have seen in one of the example problems, increasingthe light intensity can result in a saving of cell area and an increase in Voc andhence efficiency. However the increased light can also heat up the cell, resulting ina 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 mhcm3 cm3 cm2/s cm2/s m m 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 arethe 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) for this cell under these conditions.

(b) Determine the reverse saturation current density (Js) for this junction underthese conditions.

(c) Determine the open circuit voltage (Voc) for this junction under these condi-tions.

(d) Calculate the efficiency of the cell under these conditions.

63

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

)3/23.22 1019 cm3 and Nv =

(T

300 K

)3/21.83 1019 cm3

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 cm3

(b) The reverse saturation current is given by:

Js = en2i

(Dp

NDLp+

DnNALn

)Plugging in the numbers given for this cell we find:

Js = 3.8 106 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:

= FFVocJLPs

Once again plugging in numbers we find

= 19%

64

For comparison, under one sun we found:

ni = 1.41010 cm3 , Js = 3.91011 A/cm2 , Voc = 0.54 V and = 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.

9. (From 2010 Midterm) Thin-film silicon solar cells typically use a p-i-n structure,rather than a p-n structure. A p-i-n junction is similar to a p-n junction, exceptthat a layer of intrinsic material is sandwiched in between the p and n layers. He weconsider a p-i-n solar cell formed from p-doped, intrinsic, and n-doped Si. Assumethat the n-type silicon is doped with ND = 10

18 cm3 donors, and the p-type isdoped with NA = 10

15 cm3 acceptors. We consider a temperature of 300 K andassume all the dopants are ionized.

Figure 4.12: Schematic of p-i-n solar cell.

(a) Start with separate pieces of n-type, intrinsic, and p-type Si. Draw the bandstructures for each, labeling their Fermi levels.

(b) For the n and p materials, calculate the position of the fermi level relative tothe intrinsic fermi level.

(c) Now imagine placing these three segments together to make a p-i-n device.Draw a band diagram for the unbiased p-i-n device after it has reached asteady state (i.e. after carriers are done migrating). Be sure to label thevalence and conduction bands, the Fermi level, and the intrinsic Fermi level.Indicate on your graph for part a) the directions in which electrons and holeswould migrate after the layers are brought together.

(d) Draw a diagram of the space charge distribution and the electron concentrationin the device after the device has reached a steady state.

(e) Draw a graph of the electric field with respect to position.

(f) Draw a graph of the voltage with respect to position.

65

(g) What is the built-in voltage for this cell?

(h) p-i-n structures with wide intrinsic layers are used in thin-film silicon solar cellsbecause of the poor diffusivity of carriers in amorphous and microcrystallinesilicon. How might using a p-i-n structure improve the efficiency of these solarcells? Hint: See part (e).

Solution:

(a) The band diagram is shown in figure 4.13.

Figure 4.13: Band diagram of p-type, intrinsic and n-type regions, before assembly intop-i-n diode.

(b) We can find the differences in Fermi levels by using the relationship betweencarrier concentration and Fermi level. In the n-type region we have:

nn = Nce(Ecn)/kBT

where Nc is the conduction band effective density of states, Ec is the conduc-tion band edge, and n is the Fermi level in the n-type region. In the intrinsicregion we have:

ni = Nce(Eci)/kBT

66

Taking the ratio of these and solving for the Fermi level difference we find:

n i = kBT ln(nnni

)= 0.026 ln

(1018

1010

)eV

= 0.48 eV

where we have used ni = 1010 cm3 Considering the p-doped region we simi-

larly have:pp = Nve

(pEv)/kBT

andpi = Nve

(iEv)/kBT

Again taking the ratio and solving for the Fermi level difference we find:

p i = kBT ln(pipp

)= 0.026 ln

(1010

1015

)eV

= 0.30 eV

(c) As the device is assembled from the p-type, intrinsic, and n-type Si, electronsfrom the n-type region and holes from the p-type region diffuse into the in-trinsic region, setting up a built in potential that opposes further diffusion.Eventually the Fermi levels from the three regions align. The resulting banddiagram is shown in figure 4.14. The direction of diffusion of the electrons andholes is shown in figure 4.13.

(d) As electrons from the n-type and holes from the p-type region diffuse into theintrinsic region they leave behind fixed charges. The resulting space chargedistribution is shown in figure 4.15.

(e) The space charge results in a field as shown in figure 4.16.

(f) The field results in a voltage profile as shown in figure 4.17.

(g) The total built in voltage for this cell is given by:

Vbi = e(n p) = 0.78 V

67

Figure 4.14: Band diagram of p-i-n diode.

Figure 4.15: Space charge and electron distribution

68

Figure 4.16: Electric field vs. position

Figure 4.17: Voltage vs. position

69

(h) Minority carrier diffusion lengths in thin-film silicon materials are very short.In a p-n junction, many photogenerated carriers would be unable to diffuseto the junction to get collected. A p-i-n structure results in a field acrossthe entire intrinsic region, which is the majority of the material in a thin-filmsilicon cell. Hence most of the light-induced carriers will be formed in theintrinsic region, and will be moved by the field (rather than diffusion) towardthe appropriate region (electrons toward the n-type and holes toward the p-type regions). This field assisted transport can take place more rapidly thandiffusion so the likelihood of recombination is reduced, resulting in greaterefficiency.

10. (From 2011 Midterm) An ideal Si solar cell (low junction recombination so m = 1,no series resistance and infinite shunt resistance) operating at T = 300 K is exposedto a solar power density of Ps = 1000 W/m

2. This gives a light induced currentdensity of JL = 0.04 A/cm

2 and an open circuit voltage of 700 mV.

(a) What is the reverse saturation current density (Js) for this solar cell?

(b) What is the current density of the cell when the operating voltage is 616 mV?

(c) What is the efficiency for the solar cell under this condition?

We now consider a solar cell that has the same low junction recombination, som = 1, and has the same reverse saturation current density (Js) and is exposedto the same solar illumination, with the same light induced current density (JL).But this cell has a shunt resistance of Rsh = 50 for a solar cell with an area ofA = 1 cm2.

(d) What is the current density for this cell when the operating voltage is 616mV?

(e) What is the efficiency for the solar cell under this condition?

Solution:

(a) Using the expression for the current-voltage behavior of an ideal solar cell forthe open circuit condition we find:

Js = JLeeVoc/kBT = 7.0 1014 A/cm2

(b) Again using the expression for the current-voltage behavior of an ideal solar

70

cell we find:

J = JL Js[exp

(eV

kBT

) 1]

= 38.4 mA/cm2

(c) The efficiency can be found by dividing the product of the current density andvoltage by the input power density:

=JV

Ps= 23.7%

Pretty good solar cell!

(d) The current density for this non-ideal solar cell is given by:

J = JL Js[exp

(eV

kBT

) 1] VRshA

where A is the cell area. Plugging in numbers we find:

J = 26.1 mA/cm2

(e) The efficiency for this cell under this condition is:

=JV

Ps= 16.1%

Which is quite a bit less. Interestingly the open circuit voltage only decreaseda little from 700 mV to 689 mV.

11. (From 2012 Final) We derived the current voltage behavior of a single junction cellin class. Here we consider the current-voltage behavior of a two-junction, tandemcell. Here we have two different cells in series, so the current running through bothis the same, and each cell contributes their voltage. So the current densities follow:

J1 = J2 Jwhere J1 and J2 are the net current densities through the junction 1 and 2 respec-tively, and J is the total current density. We assume ideal diode behavior for eachcell, so the current densities for the two cells follow:

J1 = JL1 JS1(eeV1/kBT 1) and J2 = JL2 JS2 (eeV2/kBT 1)

71

where JL1 and JL2 are the light-induced currents for the two cells, JS1 and JS2 arethe reverse bias saturation current densities for both cells, and V1 and V2 are thevoltages across both cells. The total voltage across the tandem cell will be:

V = V1 + V2

Recalling that the reverse saturation current density is proportional to the squareof the intrinsic carrier concentration, we make the assumption:

JS1 = JS0eEg1/kBT and JS2 = JS0eEg2/kBT

where JS0 is a constant and Eg1 and Eg2 are the band gaps for the two cells. Lastly,we design the cell with the band gaps chose so that the light-induced current is thesame for both cells, and is equal to half the light-induced current for the lower bandgap cell operating as a single junction cell. Mathematically: JL1 = JL2 = JL/2,where JL is the light-induced current for the lower band gap cell operating as asingle junction cell

(a) Write down the equation for the total voltage for the tandem solar cell interms of JL, J , JS0, Vth = kBT/e, Eg1 and Eg2. Note, you will have to makean approximation to get the answer in the answer sheet.

(b) Invert your expression to find the current density J as a function of JL, JS0,Eg1, Eg2 and V .

(c) If we choose to use Si for the low band gap cell (recall the band gap for Si is1.12 eV), what is the approximate band gap for the upper cell that would resultin the same light-induced current for both cells when the device is exposedto AM 1.5. Hint: You can get this answer from the solar photon flux plot inUnit 1.

Solution:

(a) We start by inverting the expressions given above for the current densities tofind:

V1 = Vth ln

(JL1 + JS1 J

JS1

)and V2 = Vth ln

(JL2 + JS2 J

JS2

)We recognize that at appreciable operating voltages the numerator of the ln

72

term will have to be much larger than the denominator, so we approximate:

V = V1 + V2 = Vth

[ln

(JL/2 JJS1

)+ ln

(JL/2 JJS2

)]= Vth [2 ln (JL/2 J) ln (JS1JS2)]= Vth

[2 ln

(JL/2 JJS0

)+Eg1 + Eg2kBT

]=

Eg1 + Eg2e

+ 2Vth ln

(JL/2 JJS0

)(b) Inverting this we find:

J =JL2 JS0 exp

[V (Eg1 + Eg2) /e

2Vth

]Figure 4.18 shows plots of current and power are shown for single-junctionand tandem cells.

(c) The plot from Unit 1 is shown below in Figure 4.19 with the lines used to findthe value of Vg2 = 1.75 eV.

12. (From 2013 Final Exam) A single-junction Si solar cell is exposed to AM1.5 atT = 300 K and a open circuit voltage of Voc = 700 mV is achieved. Calculate thereverse bias saturation current Js. Assume ideal diode behavior and take the Siband gap to be Eg = 1.12 eV. Your answer should be in A/cm

2.

Solution:

Using our expression for the open circuit voltage

Voc =kBT

eln

(1 +

JLLs

)we find:

Js =JL

exp (eVoc/kBT ) 1We have calculated that for a Si band gap of 1.12 eV, AM1.5 produces a lightcurrent of:

JL = 2.5 1017 e/(cm2s) = 0.040 A/cm2Inserting this we find:

Js = 7.13 1014 A/cm2

73

400

200

0

-200Curre

nt D

ensit

y (A

/m2 )

2.01.51.00.50.0 Cell Voltage (V)

tandem single

400

300

200

100

0Powe

r Den

sity

(W/m

2 )

2.01.51.00.50.0Cell Voltage (V)

tandem single

35%

23.5%

Figure 4.18: The current density and power density for tandem and single junction cells.The single junction cell has a band gap of 1.12 eV and the tandem cell has cells withband gaps of 1.12 eV and 1.75 eV. The percentages listed on the power density plot arethe solar cell efficiencies. For these plots JS0 was that for a single-junction solar cell witha 1.12 eV band gap and a open circuit voltage of 0.7 V.

74

43

2

1

0

x102

1

43210

2.0

1.5

1.0

0.5

0.0

x1021

AM1.5 5960 K Spectra

Figure 4.19: The photon flux as a function of energy. Shown also is the integrated photonflux above the Si band gap energy of 1.12 eV.

75

76

Chapter 5

Electrochemistry and Batteries

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.

(a) Draw the cell you could construct with the radio as the load. Mark thecathode, anode, electron flow and cation/anion flow. (assume you have anappropriate electrolyte available).

(b) Determine the half cell reactions and the full cell reaction for this cell. (ignorethe electrolyte)

(c) Determine the standard potential and the Gibbs free energy for the cell reac-tion.

Solution:

(a) The drawing is shown in figure 5.1.

(b) By examination of the half-cell potentials in the notes we see that the potentialfor oxidization of Fe is greater than that for oxidization of Ag. In fact, thepotential for oxidization of Ag is negative, indicating that reduction of Ag isfavorable. Hence we write the half-cell reactions with Fe as the anode, where itis oxidized and Ag as the cathode, where it is reduced. The half-cell reactionsare:

Fe Fe2+ + 2e E = 0.44 V2Ag+ + 2e 2Ag E = 0.80 V

77

!" #"

$%"&'()%"

&*"+,-.()%"

/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.

So the total cell reaction is

Fe + 2Ag+ 2Ag + Fe2+ E = 1.24 V

(c) From above we see that the standard potential is E = 1.24 V. The standardstate free energy can be found from:

G = nFE = 239 kJ/mol

2. I place 20 grams of NaCl in 56 grams of water. Calculate the atomic fraction,molarity and 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:

78

element H O Na ClM(g/mole) 1 16 23 35.45

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 so we have:

mNa =0.34 mol

0.02 kg + 0.056 kg= 4.47 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. You are making a battery out of Ti and Mn.

(a) Write out the balanced electrochemical reaction.

(b) What is the potential of the cell according to the half-cell reaction chart?

(c) Calculate the gram-equivalent for each material.

(d) What is the theoretical capacity of the cell, in Ah/g? Ignore the masses of theelectrolyte, case, etc.

(e) What is the specific energy available from the cell, in J/g?

79

Solution:

(a) The half cell reactions are:

Ti Ti2+ + 2e E = 1.75 VMn2+ + 2e Mn E = 1.05 V

So the total cell reaction is

Ti + Mn2+ Ti2+ + Mn E = 0.70 V

(b) From above see the standard potential is E = 0.70 V.

(c) The gram equivalent of each material is found from the atomic mass dividedby the number of electrons gained or lost in the reaction. For Ti we find:

geqTi =47.88 g/mol

2= 23.94 g/mol eq

and for Mn:

geqMn =54.94 g/mol

2= 27.47 g/mol eq

(d) To find the theoretical capacity we first find the number of Ah/g for each ofthe electrodes:

Ah

g

Ti

=Fn

MA=

26.8 Ah/mol

23.94 g/mol eq= 1.12 Ah/g

For Mn we find:

Ah

g

Mn

=Fn

MA=

26.8 Ah/mol

27.47 g/mol eq= 0.98 Ah/g

The theoretical capacity for the cell is then:

Ah

g

cell

=

(1

1.12 Ah/g+

1

0.98 Ah/g

)1= 0.52 Ah/g

(e) The specific energy available from the cell can be found by multiplying thevoltage by the theoretical capacity:

Energy

g= 0.70 V 0.52 Ah/g 3600 s/h = 1.32 kJ/g

80

4. The cell reactions in a Li ion battery are given by:

anode LixC6 xLi+ + xe + 6Ccathode Li1yCoO2 + yLi+ + ye LiCoO2

cell y(LixC6) + x(Li1yCoO2) 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.

(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 cAMLiInserting numbers we find

x = 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

81

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 power density is just the cell capacity times the voltage. We find

power density = 0.098 Ah/g 3.8 V = 0.37 Wh/g

5. (From the Final Exam in 2007)

(a) Using the table for standard electrode potential from Lecture 15, slide 7,calculate the standard potential E and the standard free energy G forthe reaction:

H2 +1

2O2 H2O (5.1)

(b) What are E and G for the reaction:

2H2 + O2 2H2O

We are interested in making a hydrogen-oxygen fuel cell from reaction 5.1. Howeverwe want a higher voltage. In a flash of brilliance we realize that cranking up thepressure of the hydrogen and oxygen will drive the reaction more to the right andthus should give a higher cell voltage.

(c) Calculate the theoretical cell potential we will get if we operate the cell usingpure H2 at a pressure of 10 atmospheres, and air at a pressure of 5 atmospheres.The cell is operated at a temperature below 100C so the water is liquid. Takeair to be 21% O2.

Solution:

(a) The reaction above can be found by summing the reactions:

12 (O2 + 4H+ + 4e 2H2O) E = 1.23 V2 (H+ + e 1

2H2) E

= 0.00

Thus we findE = 1.23 eV

82

since the standard potential is not dependent on the amount of material inthe reaction. Using the standard state Nernst equation we find the reactionfree energy:

G = nFE= 2 96500 C/mole 1.23 V= 237.4 kJ/mole

(b) The reaction above can be found by summing the reactions:

O2 + 4H+ + 4e 2H2O E = 1.23 V

4 (H+ + e 12H2) E

= 0.00

Thus we findE = 1.23 eV

same as before. Again, the standard potential is not dependent on the amountof material in the reaction. Using the standard state Nernst equation we findthe reaction free energy:

G = nFE= 4 96500 C/mole 1.23 V= 474.7 kJ/mole

(c) Here we use the Nernst equation for this reaction:

E = E RTnF

lnaH2O

aH2aO2

= E RTnF

ln1

PH2PO2

= 1.23 V 8.314 J/(mole K) 300 K2 96500 C/mole

ln1

10

0.21 5= 1.26 V

So not much change for all the work in increasing the pressure.

6. (From the Final Exam in 2007) The measured activation potential versus currentfor a battery is shown in Figure 5.2.

(a) From the plot in Figure 5.2 find the exchange current density j0.

83

Figure 5.2: Measured activation potential versus current density for hypothetical cell.Also shown is the fit to the Tafel equation (dashed line).

(b) Also from the plot, find the Tafel slope and calculate . (Assume there aretwo electrons transfered in the cell reaction and that the measurement is doneat T = 300 K.)

(c) What is the activation potential when we drive this cell at a current densityof j = 0.5 A/cm2?

Solution:

(a) The exchange current density is the forward current when the activation po-tential act = 0. This corresponds to the horizontal axis intercept of theextrapolation of the Tafel equation (act = a+ b ln j) fit to the high activationpotential behavior. In this case we find:

J0 = 0.05 mA/cm2

(b) We can estimate the slope of the Tafel equation fit by using the points:

j (mA/cm2) act (V)0.1 0.0381.0 0.158

84

We find:

b =1 2

ln(j1/j2)

=0.12 V

ln(10)

= 0.052 V

From the notes:

b =RT

nF

So

=RT

bnF= 0.25

(c) We can now use the Tafel equation to find the activation potential at a givencurrent, so long as it is in the high-bias regime. From the plot we see that thisis indeed the case: the Tafel does a good fit to the data well before we get tothe given current density. Rearranging the Tafel equation a bit we find:

act = b lnj

j0

= 0.051 V ln500

0.05= 0.48 V

7. (From the Final Exam in 2007) Your brand new Honda FCX Clarity is powered by ahydrogen fuel cell. The electric motor draws 100 kW at full power and requires 1000V to operate. The hydrogen fuel cell oper