Workshop problems solving

Photovoltaic Solar Energy Conversion (PVSEC)الشمسية الطاقة من الكهرباء إنتاج الكهرباء من الطاقة الشمسيةإنتاج

Courses on photovoltaic for Moroccan academic staff; 23-27 April, ENIM / Rabat

W k h d bl l i ti iti

Courses on photovoltaic for Moroccan academic staff; 23 27 April, ENIM / Rabat

Ahmed EnnaouiHelmholtz-Zentrum Berlin für Materialien und Energie

Workshop and problem solving activities

This material is intended for use in lectures, presentations and as handouts to students, it canbe provided in Powerpoint format to allow customization for the individual needs of course

[email protected]

instructors. Permission of the author and publisher is required for any other usage.

Sources: Stanford University/JOHN WILEY & SONS, INC., PUBLICATIONSources: Arizona State University

To clarify and illustrate the concepts of the theoretical courses

Aims of Classroom Workshop

To clarify and illustrate the concepts of the theoretical courses

To sharpen the communicative skills of the students, this includes

communication in other languages than the native tongue .

To develop a global quality reflex and to accomplish a reality sense:

through critical observation, in learning the applicability and limits of

the theoretical concepts and experimental techniques

To gain experience in working in groups and communicate towards

group members.group members.

To develop the analytical mind and the sense for synthesis

To be able to report orally and by writing To be able to report orally and by writing

Basic Laws of Radiation/ Black Body Radiation

- The following issues should be discussed:- A photon is determined by the magnitude and the direction of its momentum- direction of the associated electric field: p = ħ k- Distinguish between: wave vector k with the Boltzmann constant kB.- 2 states for photons corresponding to the two possibilities of polarization of the electric field perpendicular to the direction of propagation.

N b f ibl t t f th h t i l V i i b

Tk1β =

- Number of possible states of the photons in a volume V is given by:

1n βs = kdV2 33)( TkB

Average number of photons per unit volume having their wave vector between k and k + dk the n mber of states probabilit of each state

1e Sβεs − 3π)2(

k and k + dk = the number of states x probability of each state


⎥⎤

⎢⎡

−⎟⎟⎞

⎜⎜⎛

=

1hcexp

1λ

2hc5

2

T),E(λ

⎥⎦

⎢⎣

⎟⎟⎠

⎜⎜⎝

1Tλk

expB

3 0 108 1 h 6 63 10 34 J k 1 38 10 23 JK 1 c = 3.0 × 108 ms-1 ; h = 6.63 × 10-34 J.s ; k =1.38 × 10-23 JK-1


Total area under theTotal area under the curve is the total

radiant power emitted.

This area in (W/m2)is the power emitted between λ and λbetween λ1 and λ2


Total area under theTotal area under the curve is the total

radiant power emitted.


[Watt] = [m] x [W/m2 K4] x [Kelvin][Watt] = [m] x [W/m2 . K4] x [Kelvin]

Consider the earth to be a blackbody with average surface temperature


Consider the earth to be a blackbody with average surface temperature15°C and area equal to 5.1 x 1014 m2. Find the rate at which energy isradiated by the earth and the wavelength at which maximum power isradiated. Compare this peak wavelength with that for a 5800 Kblackbody (the sun).The earth radiates:

W102 0273)K][(15)10(5 1)KW10(5 64(W )E 174214428

σ A 4T

The wavelength at which the maximum power is emitted

Watt10x 2.0273)K][(15 )m10x (5.1)KWm10x (5.64(Watt) E 17421442-8 =+= −−

28982898 2898 0.5μ.58002898(sun) μm 10.1

2882898

T(K) 2898(earth)max ===== λλ

Copyrighted Material, from internet


Sun (visible)λMAX = 0.5 μm

F = 64 million W m-2

Earth (infrared)λMAX = 10 μmF = 390 W m 2

FT = 64 million W m-2

FT = 390 W m-2

Copyrighted Material, from internet


Replicate this experience in your lab.

Black Body Radiationis the amount of incoming solar radiation per unit area that would be incident on a plane

Solar constant

is the amount of incoming solar radiation per unit area that would be incident on a plane perpendicular to the rays, at a the mean distance from the Sun to the Earth.

24

DR T σ = E ⎟⎠⎞

⎜⎝⎛

Solar constant

D ⎠⎝σ = 5,67 . 10-8 W/(m2.K4), T = 5800 K

R = 6 96 108 m D = 1 49 1011 mR = 6,96.108 m, D = 1,49.1011 m

E = 1402 W/m2

282 ⎞⎛⎞⎛

For more detailSee PVSEC1:Solar flux intercepted by the Earth

2

11

8442 8-

24

m 1,49.10m6,96.10x K) (5800x ).KW/(m10 . 5,67

DR T σ = E ⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛

The atmosphere will transmit a fraction (75%) of solar radiation

τ E = 0,75 E = 1052 W/m2

Altitude angle at solar noon

L

LE iβNoon = 90 + L- δ

δL

P Example 1: Tilt Angle of a PV Module. Find theoptimum tilt angle for a south-facingphotovoltaic module in Rabat (latitude 34° at

l M h 1LEquationLocal

horizontalsolar noon on March 1st.

March 1st. is the 60th. day of the year so the solar declination is:

°−=⎥⎦⎤

⎢⎣⎡ −=⎥⎦

⎤⎢⎣⎡ −= 8.381)(60

365360sin23.4581)(n

365360sin23.45δ

The tilt angle that would make the sun’s rays perpendicularto the module at noon would therefore be

⎥⎦⎢⎣⎥⎦⎢⎣)(

365)(

365

Altit d lto the module at noon would therefore be

7.473.8349090 =−−=+−°= δβ Lnoon

Altitude angleβnoon=47.7

°=−=−= 42.347.790β90Tilt noonS

Tilt = 42.3°

Find altitude angle β and azimuth angle at 3 PM solar time in Boulder CO (L = 40˚)

Altitude angle and azimuth angle

Find altitude angle β and azimuth angle S at 3 PM solar time in Boulder, CO (L = 40 ) on the summer solstice:.At the solstice, we know the solar declination δ = 23.45°

°=⎟⎠⎞

⎜⎝⎛ °

=⎟⎠⎞

⎜⎝⎛ °

= -45(-3h) x15 solarnoon) before (hours x15HA ⎟⎠

⎜⎝

⎟⎠

⎜⎝

5( 3 )h

so a oo )be o e( ou sh

0.7527(-45)(23.45)coscos(40)cos(23.45)sin(40)sinsin =+=β

°80( 0 9848)i -1

°== 48.8(0.7527)sinβ -1

0 9848sin(-45) cos(23.45)sin(Az) == °−== 80(-0.9848)sinφ 1S

0.9848cos(48.8)

sin(Az) −==

A.M.P.M.

°= 48.8β

°−= 80φS

Solar anglesSunrise and sunset can be found from a simple use of:Sunrise and sunset can be found from a simple use of:

δ)cos(ω)cos(L)cos(δ)sin(L)sin(sin(h) += 0=

Find the time at which sunrise (geometric and conventional) will occur in Boston (latitude 42.3°) on July 1 (n = 182). Also find conventional sunset.

Find Eastern Daylight Time for solar noon in Boston (longitude 71.1° W)

Example 1: Solar Time vs. Clock TimeFind Eastern Daylight Time for solar noon in Boston (longitude 71.1 W) on July 1st. Answer: July 1st. is day number n = 182. to adjust for local time,we obtain:

99.89 81)(182364360 81)(n

364360B °=−=−=

[ ] -3.589)1.5sin(99.-.89)7.53cos(99(99.89) 29.87sin2 1.5sinB-7.53cosB9.87sin2BE =−=−= [ ] )()(( )

For Boston at longitude 71.7° W in the Eastern Time Zone with local time meridian 75°

E(min) Longitude) Local - Meridian Time (Localdegree4min (CT) TimeClock (ST) Time Solar +°+=

degree

East 47.9A.M.:11 12.1min00:123.5)( 71.1) - 4(75 12 CT =−=−−°−=To adjust for Daylight Savings Time add 1 h, so solar noon will be at about 12:48 P.M.

OrigineFor Easternzone

Time Belts of the U.S : Eastern Standard Time - E.S.T. is calculated to the 75th meridian west longitude. Central Standard Time - C.S.T. is calculated to the 90th meridian west. Mountain Standard Time - M.S.T. is calculated to the 105th meridian west. Pacific Standard Time - P.S.T. is calculated to the 120th meridian west. Alaska was standardized in 1918 on 150th Meridian west, but in actual practice, other zones are and have been in use: 120° 150°, 165°

zone

For the “Green community village” near Dubai (latitude angle = 25° N, local longitude angle = 55°12’ E, t d d ti UTC 4 d li ht i ti ) F b 3 t 14 00 D t i

Example 2: Solar Time vs. Clock Time

standard time zone = UTC +4, no daylight saving time) on February 3 at 14.00. Determine:a. the apparent solar time.b. solar declination and hour angle , solar altitude and solar azimuth angles.

UTC = Universal Time and GMT = Greenwich Mean Time. Atlantic Standard Time (AST) is 4 hours behind of Coordinated Universal Time(UTC)Atlantic Standard Time (AST) is 4 hours behind of Coordinated Universal Time(UTC)

http://www.timeanddate.com/worldclock/search.htmlhttp://www.timeanddate.com/worldclock/results.html?query=Morocco

46,48- 81)(34364360 81)(n

364360B 34 n 3th February °=−=−==⇒a) The apparent solar time ,)(

364)(

364y

min -13.95)sin(-46.48 1.5 - 8)7.53(-46.48)sin2(-46.4 9.87sinB 1.5 - 7.53Bsin2B 9.87ET =−=−=

Local longitude angle = 55°12’ E = -55.2° (conversion in °, and negative since location is in EastStandard time zone UTC+4 Local Time meridian (LTM)= -60° (negative since location is in East)( ) ( g )No daylight saving time February LST = 14:00

b) H l d l d li ti δ

( ) [ ] 27:13degree4min.)55.2(60)13.95(00:14angle alLongitudin Local - M LT

degree4minE(min) LST AST =°−−°−+°−+=++=

16.97365

34)(284360sin23.45365

n)(284360sin 23.45δ −=+°

°=+°

°=b) Hour angle ω and solar declination δ.AST = 13:27 = 13.45 h (conversion of time in hours) ω =15° (hours from local solar noon) = 15° (ST-12)ω = 15°.( 13.45-12) = 21.75°

[ ] °°° 42 9816 97)5) () (21 7(2516 97)) i (i (25ihδ) ( )(L) (δ)i (L) i (i (h) -1[ ] °=−°+−°=⇒+= 42.9816.97)5).cos().cos(21.7cos(2516.97)).sin(sin(25sinhδ)cos(ω)cos(L)cos(δ)sin(L)sin(sin(h) -1

°=⎥⎦

⎤⎢⎣

⎡ −=⇒= − 28.98

cos(42.98)(21.75)16.97).sincos(sinAz

cos(h)sinω . cosδsin(Az) 1

Di t B R di ti t th S f f th E th

Attenuation and air mass

Direct Beam Radiation at the Surface of the Earth –Find the direct beam solar radiation normal to the sun’s rays at solar noon on a clear day in Atlanta (latitude 33.7°C) on May 21.y ( ) y

360 ⎤⎡

May 21 is day number 141

22 W/m1104)(W/m 275)(n365360 . 75sin1160A =⎥⎦

⎤⎢⎣⎡ −+=

0.197 100)(n3600.035sin0.175k =⎥⎦⎤

⎢⎣⎡ −+= )(365 ⎥⎦⎢⎣

Optical Depth k and the apparent Extraterrestrial Flux A. The Sky Diffuse Factor C can be used later for diffuse radiation

Measurements for the 21st Day of Each Month after ; Source: ASHRAE (1993).

Insolation on a Collector

Use the the following equation to evaluate δ

⎥⎤

⎢⎡ −= )81(360sin4523 nδ ⎥⎦⎢⎣

−= )81(365

sin45.23 nδ

The altitude angle of the sun at solar noon is

The air mass ratio:

The value of clear sky beam radiation at the earth’s surface:

Direct-Beam Radiation, IBC

Insolation on a CollectorBC

The translation of direct-beam radiation IB (normal to the rays) into beam insolation striking a collector face IBC is a simple function of the angle of incidence

θ = incidence angle between a normal to the collector face and the incoming to the collector face and the incoming beam.

nPanel Tilt

n

S i l f b i l ti h i t l f ISpecial case of beam insolation on a horizontal surface IBH

Insolation on a Collector

At any particular time θ will be a function of the collector orientation At any particular time, θ will be a function of the collector orientation, the altitude and azimuth angles of the sun at any particular time

The incidence angle is given byThe incidence angle is given by

β: altitudeφS solar azimuth

P l i hφC Panel azimuth

Panel Tilt

Beam Insolation on a Collector at solar noon

Insolation on a CollectorBeam Insolation on a Collector at solar noonin Atlanta (latitude 33.7°) on May 21 the altitude angle of the sun was found tobe 76.4° and the clear-sky beam insolation was found to be 902 W/m2. Find the beam insolation at that time on a collector that faces 20° toward the

th t if it i ti d t 52° lsoutheast if it is tipped up at a 52° angle.The cosine of the incidence angle

β: altitudeThe beam radiation on the collector β: altitudeφS solar azimuthφC Panel azimuth

The beam radiation on the collector

Panel Tilt

76.4°

collector that

Solar noon ΦS = 0

Panel Tilt52°

collector that faces 20◦ toward the southeast East

Insolation on a CollectorDiffuse Radiation on a Collector - find the diffuse radiation on the same panel. Diffuse Radiation on a Collector find the diffuse radiation on the same panel. Solar noon in Atlanta on May 21 (n = 141), aCollector faces 20° toward the southeast Tipped up at a 52° angle. I l ti f d t b 902 W/ 2Insolation was found to be 902 W/m2.Diffuse insolation on a horizontal surface IDH is proportional to the direct beam radiation IB no matter where in the sky the sun happens to be: IDH = C x IB where C is a sky diffuse factor.

The diffuse energy striking the collector

52°

Added to the total beam insolation of 697 W/m2, this gives a total beam of 785 W/m2.

Reflected Radiation Onto a Collector.

Insolation on a CollectorReflected Radiation Onto a Collector. Reflectance of the surfaces in front of the panel is 0.2 (20%)Solar noon in Atlanta on May 21, Altitude angle of the sun β =76.4°, Collector faces 20° toward the southeast Tipped up at a 52° angleDiffuse sky factor C is 0.121Cl k b i l i i 902 W/ 2Clear-sky beam insolation is 902 W/m2.

The clear-sky reflected insolation on the collector is

52°

y

The total insolation on the collector

Insolation on a CollectorYou can use this equation, combining the equations for the three q , g qcomponents of radiation:

Direct-beamDiffuse radiation

collector, C

Reflected radiation

Tilt angle

Insolation on a Collector using obstruction diagram

Figure below

The sun path diagram with superimposed obstructions makes it easy to estimate periods of shading at a site.



Continuity equations:Physics of semiconductor devices

y q• Derived from Maxwell’s equations:

( ) =⋅∇∂∂+⋅∇=×∇⋅∇ cond t

DJH 0

( ) ( )⎧∂

−+−=ρ=∂ρ∂++⋅∇

∂

ADpn

n

NNnpqt

t JJ

1

,0

⎪⎩

⎪⎨

⎧

−+⋅∇−=∂∂

−+⋅∇=∂∂

⇒ppp

nnn

RGtp

RGqt

n

J

J

1

1

• Low-level injection (SRH lifetime dominated by the minority carrier lifetime):

⎩∂pppqt

pp ppnn −− 00

p

nnp

n

ppn

ppRR

τ=

τ= 00 ,

Poisson’s equation:Physics of semiconductor devices

q• Derived from the Maxwell’s equations (electrostatics case):

ϕ−∇=→=×∇ 0 EE

( )ερ−=ϕ∇→ϕ∇ε−=ε⋅∇=⋅∇

ϕ22ED

Quasi-Fermi levels:• In non-equilibrium conditions, one needs to define separate

Fermi levels for n and p:Fermi levels for n and p:

iFni

EEnn ⎟⎟

⎞⎜⎜⎛ −

= exp

FpppFnnn

Fpii

Bi

EpEn

EEnp

Tknn

∇μ=∇μ=

⎟⎟⎞

⎜⎜⎛ −

=

⎟⎟⎠

⎜⎜⎝

=

JJ

a

exp

exp

Bi Tknp ⎟⎟

⎠⎜⎜⎝

exp

Source: Arizona State University

Sample problems:Physics of semiconductor devices

Sample problems:

• Decay of the photo-excited carriers• Steady-state injection from one side• Surface-recombination

(1) Decay of photo-excited carriers:Consider a sample illuminated with light source until t≤0. The generation rate equals to G At t=0 the light source is turned offgeneration rate equals to G. At t=0 the light source is turned off. Calculate pn(t) for t>0 .

hf

n-type samplen type sample

xx=0Source: Arizona State University

SolutionPhysics of semiconductor devices

• Boundary conditions:

• Minority hole continuity equation:0,0 =

∂∂

=xp

E n Minority hole continuity equation:

p

nn

p

nnpp

n ppG

ppG

qtp

τ−

−=τ−

−+⋅∇−=∂∂ 001 J Have this set up in your lab,.All

what you need a diode,oscilloscope, )

• General form of the solution:

ppq

BAetp pt+=

τ− /)(

• Boundary conditions:BAetpn +)(

0)0( pnn Gptp τ+=≤Gp pn τ+0

)(tpn

0

0)()0(

nn

pnnpp

Gptp=∞

τ+≤

t / 0np

Light turned off

ptpnn Geptp

τ−τ+=

/0)(

0np

tSource: Arizona State University

(2) Steady-state injection from one side:

Physics of semiconductor devices( ) y j

Consider a sample under constant illumination by a light sour-ce. Calculate pn(x).

n-type samplehf

Solution• Minority carrier continuity equation:

xx=0

• Minority carrier continuity equation:

nnpn ppJRG

p −−

∂−⇒−+⋅∇−=

∂ 011 J

nnnn

pppp

pppdD

p

xqRG

qt

−=

∂

τ∂⇒+∇=

∂

02

J

p

nnnp

n

dxD

t τ−=

∂0

2


• Steady state situation: 0∂pn

Physics of semiconductor devices

• Steady-state situation: 0=∂tpn

pppnnnnn

p DLppdpppdD τ==Δ

−Δ

→=−

− ,00 22

20

2

2

pp LxLxn BeAexp

//)( +=Δ

−

ppppp

p Ldxdx τ,222

• Boundary conditions for a long sample:Bpn =→=∞Δ 00)(

Diffusion length

• Final solution:

Appp nnn =−=Δ 0)0()0(

)0(np

)(xpn

[ ]

[ ] p

p

Lxp

Lxnnnn

qDJ

epppxp/

/00

)0()(

)0()(−

−−+=

0p

n

[ ] pnn

p

pp epp

LxJ 0)0()( −= 0np

xLpSource: Arizona State University

(3) Surface recombination:Physics of semiconductor devices

Consider a sample under constant illumination by a light sour-ce. There is a finite surface recombination at x=0. Calculate pn(x).

Gp

n type sampleS f bi ti

p

n-type sample

xx=0 x=L

Surface recombination:

[ ]00

)0( nnrn

p ppSdxdp

D −=

Solution• Minority carrier continuity equation at steady state:

x 0 x L0xdx =

• Minority carrier continuity equation at steady-state:

p

p

p

nnp

p

nnnp D

G

Lp

dxpdGpp

dxpdD −=

Δ−

Δ⇒=+

τ−

− 22

20

2

20

ppp dxdxSource: Arizona State University

Surface Recombination Velocity from Photocurrent Measurements in Journal of The Electrochemical Society, 146 (12) 4640-4646 (1999)




CBeAexp pp LxLxn ++=Δ

− //)(

• Boundary conditions for a long sample:B=0 (asymptotic condition)

GCΔ )(

ppn

ppn

GACApGCp

τ+=+=Δ

τ==∞Δ

)0()(

• Use boundary condition at the surface to determine Δpn(0):

[ ] pppnnnr

np

GDpppSdpD

τ=Δ→−= )0()0( 0

• Final solution:

[ ]prp

nnnrx

p LSDpppS

dxD

+Δ→

=)0()0( 0

0

⎥⎤

⎢⎡ τ

Δ − pLxprSG /1)( ⎥⎦

⎢⎣ τ+−τ=Δ p

prp

pppn e

SLGxp 1)(



• Graphical representation of the solution:

)(Δ )(xpnΔ

0=rSppGτ

Sr increasing

S ∞→rS

x


Solar cell and Modules

1:



2



3


The voltage produced by the 36-cells module



4

The NOCT is cell temperature in a module when ambient is20°C, solar irradiation is 0.8 kW/m2, and windspeed is 1 m/s.The following expression may be used:g p y

PV Module Performance Data Under Standard Test


Conditions (1 kW/m2, AM 1.5, 25°C Cell Temperature)


See table

Effect of Shading

To help understand shading phenomenon consider Figure (a) in which an n-cell moduleTo help understand shading phenomenon, consider Figure (a) in which an n-cell modulewith current I and output voltage V shows one cell separated from the others (shown asthe top cell, though it can be any cell in the string). The equivalent circuit of the top cellhas been drawn, while the other (n − 1) cells in the string are shown as just a module withhas been drawn, while the other (n 1) cells in the string are shown as just a module withcurrent I and output voltage Vn-1.

all cells are in the sun and since they are in series the same current I flows through series, the same current , I flows through each of them.

Effect of Shading

Now the top cell in figure (b) is shaded and its current source ISC has been reduced to Now , the top cell in figure (b) is shaded and its current source ISC has been reduced to zero. The voltage drop across RP as current flows through it causes the diode to be reverse biased, so the diode current is also (essentially) zero. That means the entire current flowing through the module must travel through both RP and RS in the shaded current flowing through the module must travel through both RP and RS in the shaded cell on its way to the load. That means the top cell, instead of adding to the output voltage, actually reduces it.

We consider the bottom n 1 cells still have full sun and We consider the bottom n − 1 cells still have full sun and still some how carry their original current I so they will still produce their original voltage Vn−1. This means that the output voltage of the entire module p gVSH with one cell shaded will drop to:

Effect of Shading

With all n cells in the sun and carrying I the output voltage was V so the voltageWith all n cells in the sun and carrying I , the output voltage was V so the voltageof the bottom n − 1 cells will be

The drop in voltage ΔV at any given current I , caused by the shaded cell, is given by

Since the parallel resistance RP is so much greater than the series resistance RS

Effect of Shading

At any given current the I V curve for the module with one shaded cell dropsAt any given current, the I –V curve for the module with one shaded cell dropsby V. The huge impact this can have is illustrated in the figure below:

Effect of Shading

Example 5 : The 36-cell PV module described in Example 3 had a parallel resistance per cell of Example 5 : The 36-cell PV module described in Example 3 had a parallel resistance per cell of of RP = 6.6 Ω. In full sun and at current I = 2.14 A the output voltage was found there to be V = 19.41 V. If one cell is shaded and this current some how stays the same, then:a. What would be the new module output voltage and power?b. What would be the voltage drop across the shaded cell?c. How much power would be dissipated in the shaded cell?

(a) The drop in module voltage will be

The new output voltage will be 19.41 − 14.66 = 4.75 V.The new output voltage will be 19.41 14.66 4.75 V.Power delivered by the module with one cell shaded would be

F i i f ll th d l d i 41 5 WFor comparison, in full sun the module was producing 41.5 W.(b) All of that 2.14 A of current goes through the parallel plus series resistance (0.005Ω ) of the shaded cell, so the drop across the shaded cell will be

(normally a cell in the sun will add about 0.5 V to the module; this shaded cell subtracts over 14 V from the module).

Effect of ShadingThe power dissipated in the shaded cell is voltage drop times current, which isp p g p ,

All of that power dissipated in the shaded cell is converted to heat, which can cause a local hotAll of that power dissipated in the shaded cell is converted to heat, which can cause a local hotspot that may permanently damage the plastic laminates enclosing the cell.

Final remarks:The proced res demonstrated in this e ample can be e tended to de elop I V c r es nderThe procedures demonstrated in this example can be extended to develop I –V curves undervarious conditions of shading. The following figures shows such curves for the examplemodule under full-sun conditions and with one cell 50% shaded, one cell completely shaded,and two cells completely shaded. Also shown on the graph is a dashed vertical line at 13 V,p y g pwhich is a typical operating voltage for a module charging a 12-V battery. The reduction incharging current for even modest amounts of shading is severe. With just one cell shaded outof 36 in the module, the power delivered to the battery is decreased by about two-thirds!

Effect of Shading

Effects of shading on the I –V curves for a PV module. The dashed lineshows a typical voltage that the module would operate at when charging a 12-V battery;

the impact on charging current is obviously severe.

Exercice: Spectral efficiency of photovoltaic (PV) cells

This exercise will enable you to determine the theoretical maximum spectral efficiency of a photovoltaic cell, and to determine the effect of concentration on cell efficiency.

Exercice: Spectral efficiency of photovoltaic (PV) cells

Table 2: Fractional function F0−λΤ

Solution: Spectral efficiency of photovoltaic (PV) cells




1) For the following materials determine the maximum wavelength of solar

PROBLEMS

1) For the following materials, determine the maximum wavelength of solarenergy capable of creating hole-electron pairs:a. Gallium arsenide, GaAs, band gap 1.42 eV.b. Copper indium diselenide, CuInSe2, band gap 1.01 eVpp 2 g pc. Cadmium sulfide, CdS, band gap 2.42 eV.

2) A p-n junction diode at 25°C carries a current of 100 mA when the diodevoltage is 0 5 V What is the reverse saturation current I ?voltage is 0.5 V. What is the reverse saturation current, I0?

3) For the simple equivalent circuit for a 0.005 m2 photovoltaic cell shownbelow, the reverse saturation current is I0 = 10−9 A and at an insolation of1-sun the short-circuit current is ISC = 1 A,. At 25°C, find the following:

PROBLEMSa The open-circuit voltagea. The open circuit voltage.b. The load current when the output voltage is V = 0.5 V.c. The power delivered to the load when the output voltage is 0.5 V.d. The efficiency of the cell at V = 0.5 V.

4) The equivalent circuit for a PV cell includes a parallel resistance of RP =10 . The cell has area 0.005 m2, reverse saturation current of I0 = 10−9 Aand at an insolation of 1-sun the short-circuit current is ISC = 1 A, At 25◦C,and at an insolation of 1 sun the short circuit current is ISC 1 A, At 25 C,with an output voltage of 0.5 V, find the following:

PROBLEMSa The load currenta. The load current.b. The power delivered to the load.c. The efficiency of the cell.

5) The following figure shows two I-V curves. One is for a PV cell with an equivalentcircuit having an infinite parallel resistance (and no series resistance).What is the parallel resistance in the equivalent circuit of the other cell?

PROBLEMS6) The following figure shows two I-V curves. One is for a PV cell with an6) The following figure shows two I V curves. One is for a PV cell with anequivalent circuit having no series resistance (and infinite parallel resistance).What is the series resistance in the equivalent circuit of the other cell?

PROBLEMS7) Estimate the cell temperature and power delivered by a 100-W PV module7) Estimate the cell temperature and power delivered by a 100-W PV modulewith the following conditions. Assume 0.5%/°C power loss.a. NOCT = 50°C, ambient temperature of 25°C, insolation of 1-sun.b. NOCT = 45°C, ambient temperature of 0°C, insolation of 500 W/m2.

2c. NOCT = 45°C, ambient temperature of 30°C, insolation of 800 W/m2.

8) A module with 40 cells has an idealized, rectangular I-V curve with ISC = 4 A and VOC = 20 V. If a single cell has a parallel resistance of 5 and negligible seriesVOC 20 V. If a single cell has a parallel resistance of 5 and negligible series resistance, draw the I-V curve if one cell is completely shaded. What current would it deliver to a 12-V battery (vertical I-V load at 12 V)?

PROBLEMS

9) Suppose a PV module has the 1 sun I V curve shown below Within the module9) Suppose a PV module has the 1-sun I-V curve shown below. Within the moduleitself, the manufacturer has provided a pair of bypass diodes to help the paneldeliver some power even when many of the cells are shaded. Each diode bypasseshalf of the cells, as shown. You may consider the diodes to be “ideal;” that is, theyhave no voltage drop across them when conducting.

Suppose there is enough shading on the bottom cells to cause the lower diodeto start conducting. Draw the new “shaded” I-V curve for the module.

Tracking system for photovoltaics

Tracker systems are either: 2 axis trackers which track the sun both in azimuth andTracker systems are either: 2-axis trackers, which track the sun both in azimuth andaltitude angles so the collectors are always pointing directly at the sun, or 1-axistrackers, which track only one angle or the other.

Two-axis tracking angular relationships


A single-axis tracking for photovoltaic consists of a mount having a manually g g p g yadjustable tilt angle along a north-south axis, and a trackingmechanism that rotates the collector array from east-to-west

When the tilt angle of the mount is set equal to the local latitude (called a polar

Tracking system for photovoltaicsWhen the tilt angle of the mount is set equal to the local latitude (called a polarmount), not only is that an optimum angle for annual collection, but the collectorgeometry and resulting insolation are fairly easy to evaluate as well.

If a polar mount rotates about its axis at the same rate as the earth turns, 15°/h, thenthe centerline of the collector will always face directly into the sun.

Under these conditions the incidence angle θ between a normal to the collector and the

Tracking system for photovoltaicsUnder these conditions, the incidence angle θ between a normal to the collector and thesun’s rays will be equal to the solar declination δ. That makes the direct-beam insolationon the collector just egal to IB cos δ. To evaluate diffuse and reflected radiation, we needto know the tilt angle of the collector.gThe axis of rotation has a fixed tilt Σ = L, unless it is solar noon, the collector itself is cocked at an odd angle with respect to the horizontal plane. The effective tilt, which is the angle between a normal to the collector and the horizontal plane, is given by

Th b diff d fl t d di ti l t i t k i The beam, diffuse, and reflected radiation on a polar mount, one-axis tracker are given by

One-Axis, Polar Mount:

Example: Compare the 40° latitude clear-sky insolation on a collector at solar noon

Tracking system for photovoltaicsExample: Compare the 40 latitude, clear-sky insolation on a collector at solar noonon the summer solstice for a two-axis tracking mount versus a single-axis polarmount. Ignore ground reflectance.1) Solution: For two-Axis Tracker, the beam insolation from IB = Aexp (−km), we need ) , p ( ),the air mass ratio m, the apparent extraterrestrial flux A, and the optical depth k. To find m, we need the altitude angle of the sun. Using:

with a solstice declination of 23.45°,


You can use the above relation to find A = 1088 W/m2, k = 0.205, and C = 0.134.The direct beam insolation on the collector is therefore

The diffuse radiation on the collector is

Tracking system for photovoltaicsThe diffuse radiation on the collector is

2) One-Axis Polar Tracker: The beam portion of insolation is given by

The diffuse portion

The total is

The two-axis tracker provides 994 W/m2, which is only 9% higher than the single-axis mount.

Physics of semiconductor devicesSources: Internet Materials: Arizona State University

Solar cells: operating principles, technology, and system applications by Martin Green (You have a sample of this book)

Practical photovoltaics: electricity from solar cells by Richard J. Komp.Solar cells: What they are and how they work ; How solar cells are made ; Solar cells and modules ; Using photovoltaics ; Batteries and other storage systems ; New developments in photovoltaic technology ; The future of photovoltaics ; Assembling your own modules.

The Physics of Solar Cells by Jenny Nelson.P id h i i t d ti t th h i f th h t lt i ll It i it bl f d d t d t t d t d Provides a comprehensive introduction to the physics of the photovoltaic cell. It is suitable for undergraduates, graduate students, and researchers new to the field. It covers: basic physics of semiconductors in photovoltaic devices; physical models of solar cell operation; characteristics and design of common types of solar cell; and approaches to increasing solar cell efficiency. The text explains the terms and concepts of solar cell device physics and shows the reader how to formulate and solve relevant physical problems. Exercises and worked solutions are included.

Physics of Semiconductor DevicesClassic book has set the standard for advanced study and reference in the semiconductor device field. Designed for graduate textbook adoptions and reference needs, this new edition includes: New devices such as three-dimensional MOSFETs, MODFETs, resonant-tunneling diodes, semiconductor sensors, quantum-cascade lasers, single-electron transistors, real-space transfer devices, Problem sets at the end of each chapterspace transfer devices, Problem sets at the end of each chapter

Photovoltaic Design & Installation For Dummies by Ryan Mayfield. Gives you a comprehensive overview of the history, physics, design, installation, and operation of home-scale solar-panel systems. You'll also get an introduction to the foundational mathematic and electrical concepts you need to understand and work with photovoltaic systems. Covers all aspects of home-scale solar-power systems , Viable resource for professionals, students, and technical laymen syste s Co e s a aspects o o e sca e so a po e syste s , ab e esou ce o p o ess o a s, stude ts, a d tec ca ay eCan be used to study for the NABCEP exam (the North American Board of Certified Energy Practitioners, NABCEP) http://www.nabcep.org/about-us

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