Chapter 5_PV Systems_April 11- 2011.pdf

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A material or device that is capable of converting theenergy contained in photons of light into an electricalvoltage and current is said to be photovoltaic.

A photon with short enough wavelength and high enoughenergy can cause an electron in a photovoltaic material to

break free of the atom that holds it.

If a nearby electric field is provided, those electrons can beswept toward a metallic contact where they can emerge asan electric current.

The driving force to power photovoltaics comes from thesun, and it is interesting to note that the surface of theearth receives something like 6000 times as much solarenergy as our total energy demand.



Spurred on by the emerging energy crises of the 1970s, thedevelopment work supported by the space program began topay off back on the ground.

By the late 1980s, higher efficiencies (Fig. 8.1) and lowercosts (Fig. 8.2) brought PVs closer to reality, and they beganto find application in many offgrid terrestrial applicationssuch as pocket calculators, off-shore buoys, highway lights,signs and emergency call boxes, rural water pumping, andsmall home systems.

• While the amortized cost ofphotovoltaic power did dropdramatically in the 1990s, adecade later it is still aboutdouble what it needs to be tocompete without subsidies inmore general situations.

Figure 8.1 Best laboratory PV cell efficiencies for various technologies.

(From National Center for Photovoltaics, www.nrel.gov/ncpv 2003



By 2002, worldwideproduction of

photovoltaics hadapproached 600 MWper year and wasincreasing by over 40%per year (bycomparison, global

wind power sales were10 times greater).

Figure 8.3 World production of photovoltaics is growing rapidly, but the U.S. share of the market is

decreasing. Based on data from Maycock (2004).

Critics of this decline point to thegovernment’s lack of enthusiasm to fund PVR&D. By comparison, Japan’s R&D budget is

almost an order of magnitude greater.

Figure 8.2 Possible evolution of turn-key PV system

prices



Before we can talk about solar power, we needto talk about the sun

Need to know how much sunlight is available

Can predict where the sun is at any time

Insolation : in cident so l ar radiation

Want to determine the average daily insolationat a site

Want to be able to chose effective locations andpanel tilts of solar panels



The sun

◦ 1.4 million km in diameter

◦ 3.8 x 1020 MW of radiated electromagnetic energy

Blackbodies

◦ Both a perfect emitter and a perfect absorber

◦ Perfect emitter – radiates more energy per unit ofsurface area than a real object of the same

temperature◦ Perfect absorber – absorbs all radiation, none is

reflected



Plank’s law – wavelengths emitted by ablackbody depend on temperature

8

5

3.74 10 (7.1)

14400exp 1

E

T

•λ = wavelength (μm)

• E λ = emissive power per unit area ofblackbody (W/m2-μm)

•T = absolute temperature (K)



Source: en.wikipedia.org/wiki/Electromagnetic_radiation

Visible light has a wavelength of between 0.4 and 0.7 μm, withultraviolet values immediately shorter, and infrared immediately longer

http://upload.wikimedia.org/wikipedia/commons/f/f1/EM_spectrum.svg



The earth as a blackbody

Figure 7.1Area under curve is the total radiant power emitted



Total radiant power emitted is given by theStefan –Boltzman law of radiation

4

(7.2) E A T s

• E = total blackbody emission rate (W)

• σ = Stefan-Boltzmann constant = 5.67x10-8 W/m2-K4


•A = surface area of blackbody (m2)



The wavelength at which the emissive powerper unit area reaches its maximum point

max

2898 (7.3)T


• λ = wavelength (μm)

• λmax =0.5 μm for the sun , T = 5800 K

• λmax = 10.1 μm for the earth (as a blackbody), T = 288 K





• h 1 = path length through atmosphere with sundirectly overhead

• h 2 = path length through atmosphere to spot onsurface

• β = altitude angle of the sun

Figure 7.3

As sunlight passes through theatmosphere, less energy arrives at the

earth’s surface



Air mass ratio of 1 (―AM1‖) means sun is directlyoverhead

AM0 means no atmosphere

AM1.5 is assumed average at the earth’s surface

2

1

1air mass ratio = (7.4)

sin

hm

h

Figure 7.3





• m increases as the sun

appearslower inthe sky.

• Notice there is a large

loss towards the blue endfor higher m, which is why

the sun appears reddish at

sun rise and sun set



One revolution every 365.25 days Distance of the earth from the sun

n = day number (Jan. 1 is day 1) d (km) varies from 147x106 km on Jan. 2 to

152x106 km on July 3 (closer in winter, further

in summer) Note that the angles in this chapter are in

degrees

8 360( 93)1.5 10 1 0.017sin km (7.5)

365

nd



In one day, the earth rotates 360.99˚ The earth sweeps out what is called the ecliptic

plane

Earth’s spin axis is currently 23.45˚

Equinox – equal day and night, on March 21and September 21

Winter solstice – North Pole is tilted furthest

from the sun Summer solstice – North Pole is tilted closest to

the sun





Figure 7.5For solar energy applications, we’ll consider the characteristics of

the earth’s orbit to be unchanging



Solar declination δ – the angle formed betweenthe plane of the equator and the line from thecenter of the sun to the center of the earth

δ varies between +/- 23.45˚

Assuming a sinusoidal relationship, a 365 dayyear, and n =81 is the spring equinox, theapproximation of δ for any day n can be foundfrom

360

23.45sin 81 (7.6)365

n



Predict where the sun will be in the sky at any time

Pick the best tilt angles for photovoltaic (PV) panels

Figure 7.6

•Another perspective-

Solar declination



Solar noon – sun is

directly over thelocal line oflongitude

Rule of thumb forthe NorthernHemisphere - asouth facingcollector tilted at anangle equal to thelocal latitude

• During solar noon, the sun’s rays are perpendicular tothe collector face

Figure 7.8



Find the optimum tilt angle for a south-facingPV module located at in Tucson (latitude 32.1˚)at solar noon on March 1

From Table 7.1, March 1 is day n = 60



The solar declination δ is

The altitude angle is

To make the sun’s rays perpendicular to thepanel, we need to tilt the panel by

360 360

23.45sin 81 = 23.45sin 60 81 = -8.3365 365

n

90 = 90 32.1 8.3 49.6 N L

90 = 40.4 N tilt



Altitude angle at solar noon β N – angle betweenthe sun and the local horizon

Zenith – perpendicular axis at a site

90 (7.7) N L

Figure 7.9



Find the optimum tilt angle for a south-facingPV module located at in Tucson (latitude 32.1˚)at solar noon on March 1

From Table 7.1, March 1 is day n = 60



Described in terms of altitude angle β andazimuth angle of the sun ϕ S

β and ϕ S depend on latitude, day number, andtime of day

Azimuth angle (ϕ S ) convention◦ positive in the morning when sun is in the east

◦ negative in the evening when sun is in the west

◦ reference in the Northern Hemisphere (for us) is true

south

Hours are referenced to solar noon



Figure 7.10

Azimuth Angle

Altitude Angle



Hour angle H - the number of degrees the earthmust rotate before sun will be over your line oflongitude

If we consider the earth to rotate at 15˚/hr,

then

At 11 AM solar time, H = +15˚ (the earthneeds to rotate 1 more hour)

At 2 PM solar time, H = -30˚

15

hour angle hours before solar noon (7.10)hour

H



sin cos cos cos sin sin (7.8) L H L cos sin

sin (7.9)cos

S

H

•H = hour angle

• L = latitude (degrees)

• Test to determine if the angle magnitude is less thanor greater than 90˚ with respect to true south-

tanif cos , then 90 , else 90 (7.11)

tan S S H

L



Find altitude angle β and azimuth angle ϕ S at 3 PMsolar time in Boulder, CO (L = 40˚) on the summersolstice

At the solstice, we know the solar declination δ ˚ =23.45

Hour angle H is found from (7.10)

The altitude angle is found from (7.8)

15

-3 h 45

h

H

sin cos 40cos 23.45cos 45 sin 40sin 23.45 0.7527

1sin 0.7527 48.8



The sin of the azimuth angle is found from (7.9)

Two possible azimuth angles exist

Apply the test (7.11)

cos 23.45 sin 45sin = -0.9848

cos 48.8S

1 = sin -0.9848 80S

1 = 180 -sin -0.9848 260 or 100S

cos cos 45 0.707 H tan tan 23.45

0.517tan tan 40 L

= 80 (80 west of south)S



Now we know how to locate the sun in the skyat any time

This can also help determine what sites will bein the shade at any time

Sketch the azimuth and altitude angles of trees,buildings, and other obstructions

Sections of the sun path diagram that are

covered indicate times when the site will be inthe shade



Trees to the southeast, small building to the

southwest Can estimate the amount of energy lost to

shading

Figure 7.15



The shading of solar collectors has been an area oflegal and legislative concern (e.g., a neighbor’s treeis blocking a solar panel)

California has the Solar Shade Control Act (1979) toaddress this issue◦ No new trees and shrubs can be placed on neighboring

property that would cast a shadow greater than 10 percent ofa collector absorption area between the hours of 10 am and 2pm.

◦ Exceptions are made if the tree is on designated timberland,

or the tree provides passive cooling with net energy savingsexceeding that of the shaded collector

◦ First people were convicted in 2008 because of theirredwoods



Source: NYTimes, 4/7/08



Most solar work deals only in solar time (ST) Solar time is measured relative to solar noon Two adjustments –

◦ For a longitudinal adjustment related to time zones◦

For the uneven movement of the earth around the sun Problem with solar time –two places can only

have the same solar time is if they are directlynorth-south of each other

Solar time differs 4 minutes for 1˚ of longitude Clock time has 24 1-hour time zones, each

spanning 15˚ of longitude



Source: http://aa.usno.navy.mil/graphics/TimeZoneMap0802.pdf



Time Zone Local Time Meridian

Eastern 75˚

Central 90˚

Mountain 105˚

Pacific 120˚

Eastern Alaska 135˚

Alaska andHawaii

150˚



The earth’s elliptical orbit causes the length ofa solar day to vary throughout the year

Difference between a 24-h day and a solar dayis given by the Equation of Time E

n is the day number

9.87sin 2 7.53 1.5sin minutes (7.12) E B B B

360

-81 (degrees) (7.13)

364

B n



Combining longitude correction and theEquation of Time we get the following:

CT – clock time ST – solar time LT Meridian – Local Time Meridian During Daylight Savings, add one hour to the

local time

Solar Time (ST) Clock Time (CT) +

4 min LT Meridian Local Longitude + (min)degree

E

(7.14)



Find Eastern Daylight Time for solar noon inBoston (longitude 71.1˚ W) on July 1

July 1 corresponds to n = 182

From the Equation of Time (7.12) and (7.13) weobtain

360 360 = ( 81) (182 81) 99.89

364 364

B n

= 9.87sin 2 7.53cos 1.5sin = 3.5 min E B B B



The local time meridian for Boston is 75˚, sothe difference is 75 ˚-71.7 ˚, and we know thateach degree corresponds to 4 minutes

Using (7.14)

But we need to adjust it for Daylight Savings, soadd 1 hour

= 4 min/ 75 71.1 ( 3.5min)CT ST = 12:00 12.1min 11:49.9 AM ESTCT

= 12: 49.9 AM EDTCT



Can approximate the sunrise and sunset times Solve (7.8) for where the altitude angle is zero

+ sign on HSR indicates sunrise, - indicatessunset

sin cos cos cos sin sin (7.8) L H L

sin cos cos cos sin sin 0 (7.15) L H L sin sin

cos = tan tan (7.16)cos cos

L H L

L

1

cos ( tan tan ) (7.17)SR H L

Hour angle of sunrise

Sunrise (geometric) 12 :00 (7.18)

15 /

SR H

h





Direct beam radiation I BC – passes in a straightline through the atmosphere to the receiver

Diffuse radiation I DC – scattered by molecules inthe atmosphere

•Reflectedradiation I RC – bounced off asurface near thereflector

Figure 7.18



Starting point for clear sky radiation calculations I 0 passes perpendicularly through an imaginary

surface outside of the earth’s atmosphere

I 0 depends on distance between earth and sun and

on intensity of the sun which is fairly predictable Ignoring sunspots, I 0 can be written as

SC = solar constant = 1.377 kW/m2

n = day number

2

0

360SC 1 0.034cos (W/m ) (7.20)

365

n I

These changes are due to the variation in

earth’s distance from the sun



In one year, less than half of I 0 reaches earth’ssurface as a direct beam

On a sunny, clear day, beam radiation mayexceed 70% of I 0

Figure 7.19



Can treat attenuation as an exponential decayfunction

(7.21)km

B I Ae

• I B = beam portion of the radiation thatreaches the earth’s surface

•A = apparent extraterrestrial flux• k = optical depth

• m = air mass ratio from (7.4)



(7.21)km

B I AeFrom curve fits of the table data, A and k are

approximately 23601160 75sin 275 (W/m ) (7.22)365

A n

360

0.174 0.035sin 100 (7.23)365

k n



Direct-beam radiation is just a function of the anglebetween the sun and the collecting surface (i.e., theincident angle q:

Diffuse radiation is assumed to be coming fromessentially all directions to the angle doesn’t matter; itis typically between 6% and 14% of the direct value.

Reflected radiation comes from a nearby surface, anddepends on the surface reflectance, r, ranging down

from 0.8 for clean snow to 0.1 for a shingle roof.

cos BC B I I q



1 cos

2 RC BH DH I I I r



Most residential solar systems have a fixedmount, but sometimes tracking systems arecost effective

Tracking systems are either single axis (usually

with a rotating polar mount [parallel to earth’saxis of rotation), or two axis (horizontal[altitude, up-down] and vertical [azimuth, east-west]

Ballpark figures for tracking system benefitsare about 20% more for a single axis, and 25 to30% more for a two axis



For a fixed system the total annual output issomewhat insensitive to the tilt angle, but there is asubstantial variation in when the most energy isgenerated





In 2007 worldwide PV peak was about 7800 MW, with almost half

(3860 MW) in Germany, 1919 MW in Japan, 830 in USA and

655 in Spain



Photovoltaic definition- a material or device that is capable

of converting the energy contained in photons of light intoan electrical voltage and current

Rooftop PVmodules on a

village health

center in West

Bengal, India

http://www1.eere.energy.gov/solar/pv_use.html

"Sojourner"

exploring Mars,1997

University of Illinois 2009

Solar Decathalon House – 2nd

place overall

http://www.solardecathlon.uiuc.edu/gallery.html#



Edmund Becquerel(1839)

Adams and Day (1876)

Albert Einstein (1904)

Czochralski (1940s) Vanguard I satellite(1958)

Today…

http://www.nrel.gov/pv/pv_manufacturing/cost_capacity.htm

Cost/Capacity Analysis(Wp is peak Watt)













Shadows

• Solar cell is a diode• Photopower coverted to DC

• Shadows & defects convert

generating areas to loads• DC is converted to AC by an

inverter

• Loads are unpredictable• Storage helps match

generation to load



When Prof. Chapman built a new house inUrbana in 2007 he added some solar PV.

His system has 14 moduleswith 205 W each, for a

total of 2870W. He hasa 3300 W inverter.

Total cost was about $27,000,but tax credits reduced it

to $16,900. He should be getting about 3700 kWh per year

Source: www.patrickchapman.com/solar.htm

Solar Intensity: Atmospheric Effects



Solar Intensity: Atmospheric Effects

Sun photosphere

“AM” means “air mass”

I n t e n s i t y

Extraterestrialsunlight (AM0)

Sunlight at sea level

at 40° N Lattitude atnoon (AM1.5)



• The device •Efficiency, cost, manufacturability

automation, testing

• Encapsulation

•Cost, weight, strength,yellowing, etc.

• Accelerated lifetime testing •30 year outdoor test is difficult•Damp heat, light soak, etc.

• Inverter & system design •Micro-inverters, blocking diodes, reliability

Load



C u

r r e n t

Voltage

Open-circuit

voltage

Short-circuit

current

Maximum

Power Point

n - t y p e

p - t y p e

-+

Load

• Solar cells are diodes

• Light (photons) generatefree carriers (electronsand holes) which arecollected by the electricfield of the diode junction

• The output current is afraction of thisphotocurrent

•The output voltage is afraction of the diodebuilt-in voltage

Standard Equivalent Circuit Model



Standard Equivalent Circuit Model

P h o t o c u

r r e n t

s o u r c e D i o d

e

S h u n t

r e s i s t a

n c e

L o a d

Series

resistance

Where does the power go?

(minimize)

( m a x i m

i z e )



Electrons in solids fill states until you run out of them

Conduction band – top band, here electrons contributeto current flow, empty at absolute zero forsemiconductors

Valence band – highest energy band where electrons arenormally present at absolute zero

An electron must acquire the band gap energy to jumpacross to the conduction band, measured in electron-

volts eV

Silicon band gap energy is 1.12 eV



http://upload.wikimedia.org/wikipedia/commons/c/c7/Isolator-

metal.svg

•The probability of finding an electron in astate is the Fermi distribution

•The Fermi energy is the energy at which theprobability of finding an electron is 0.5





Photons are characterized by their wavelength(frequency) and their energy

(8.1)c v

(8.2)hc

E hvv

Quantity Si GaAs CdTe InP

Band gap (eV) 1.12 1.42 1.5 1.35

Cut-off wavelength(μm)

1.11 0.87 0.83 0.92

Table 8.2 Band Gap and Cut-off Wavelength Above Which Electron

Excitation Doesn’t Occur

•The gaps between allowable energy bands are called forbiddenb d th t i t t f hi h i th ti th



bands, the most important of which is the gap separating theconduction band from the highest filled band below it.

•The energy that an electron must acquire to jump across theforbidden band to the conduction band is called the band-gapenergy, designated Eg.

•The units for band-gap energy are usually electron-volts (eV),

where one electron-volt is the energy that an electron acquireswhen its voltage is increased by 1 V (1 eV = 1.6 × 10−19 J).

•The band-gap Eg for silicon is 1.12 eV, which means an electronneeds to acquire that much energy to free itself from the

electrostatic force that ties it to its own nucleus—that is, to jumpinto the conduction band.

When a photon with more than 1.12 eV of energy is absorbed by a solar cell, a singleelectron may jump to the conduction band



electron may jump to the conduction band.

When it does so, it leaves behind a nucleus with a +4 charge that now

has only three electrons attached to it.

That is, there is a net positive charge, called a hole, associated with that nucleus asshown in Fig. 8.7a. Unless there is some way to sweep the electrons away from theholes, they will eventually recombine, obliterating both the hole and electron as inFig. 8.7b.

When recombination occurs, the energy that had been associated with the electron

in the conduction band is released as a photon, which is the basis for light-emittingdiodes (LEDs).

Figure 8.7 A photon with sufficient energy can create

a hole – electron pair as in (a).The electron can

recombine with the hole, releasing a photon of energy

(b).

Figure 8.8 When a hole is filled by a nearby valence

electron, the hole appears to move

Example 8.1 Photons to Create Hole – Electron Pairs in Silicon . What maximum

l h h h h l l i i ili h



wavelength can a photon have to create hole – electron pairs in silicon? What

minimum frequency is that? Silicon has a band gap of 1.12 eV and 1 eV = 1.6 ×

10−19

J.

Solution. From (8.2) the wavelength must be less thanλ ≤ hc/ E= = [6.626 × 10−34 J · s × 3 × 108 m/s]/ [1.12 eV × 1.6 × 10−19J/eV]

= 1.11 × 10−6 m = 1.11 μm

and from (8.1) the frequency must be at least ν ≥ c λ = 3 × 108 m/s /1.11 × 10−

6 m

= 2.7 × 1014

Hz

For a silicon photovoltaic cell, photons with wavelength greater than 1.11 μm

have energy hν less than the 1.12-eV band-gap energy needed to excite an



have energy hν less than the 1.12 eV band gap energy needed to excite an

electron. None of those photons create hole–electron pairs capable of carrying

current, so all of their energy is wasted. It just heats the cell. On the other

hand, photons with wavelengths shorter than 1.11 μm have more than enough

energy to excite an electron. Since one photon can excite only one electron, anyextra energy above the 1.12 eV needed is also dissipated as waste heat in the

cell. Figure 8.9 uses a plot of (8.2) to illustrate this important concept. The band

gaps for other photovoltaic materials—gallium arsenide (GaAs), cadmium telluride

(CdTe), and indium phosphide (InP), in addition to silicon—are shown in

Table 8.2.

Figure 8.9 Photons with

wavelengths above 1.11 μm

don’t have the 1.12 eV needed

to excite an electron, and this

energy is lost. Photons with

shorter wavelengths have morethan enough energy, but any

energy above 1.12 eV is wasted

as well



Upper bound on the efficiency of a silicon solarcell:

Band gap: 1.12 eV, Wavelength: 1.11 μm

This means that photons with wavelengthslonger than 1.11 μm cannot send an electron tothe conduction band.

Photons with a shorter wavelength but moreenergy than 1.12 eV dissipate the extra energyas heat



For an Air MassRatio of 1.5, 49.6%is the maximumpossible fractionof the sun’senergy that can becollected with asilicon solar cell

Air mass ratio of 1 (designated ―AM1‖) means that the sun is directly overhead. By convention, AM0

means no atmosphere; that is it is the extraterrestrial solar spectrum For most photovoltaic work an air



means no atmosphere; that is, it is the extraterrestrial solar spectrum. For most photovoltaic work, an air

mass ratio of 1.5, corresponding to the sun being 42 degrees above the horizon, is assumed to be the

standard.

The solar spectrum at AM 1.5 is shown in Fig. 8.10. For an AM 1.5 spectrum, 2% of the incoming solar

energy is in the UV portion of the spectrum, 54% is in the visible, and 44% is in the infrared.

8.2.3 Band-Gap Impact on Photovoltaic

Efficiency Solar spectrum at AM 1.5. Photons with wavelengths

longer than 1.11 μm don’t have

enough energy to excite

electrons (20.2% of the

incoming solar energy); those

with shorter wavelengths can’t

use all of their energy, which

accounts for another 30.2%

unavailable to a silicon

photovoltaic cell. Spectrum is

based on ERDA/NASA (1977).

If we know the solar spectrum, we can calculate the energy loss due



p , gy

to these two fundamental constraints. Figure 8.10 shows the results

of this analysis, assuming a standard air mass ratio AM 1.5.

As is presented there, 20.2% of the energy in the spectrum is lost

due to photons having less energy than the band gap of silicon (hν

< Eg), and another 30.2% is lost due to photons with hν > Eg.

The remaining 49.6% represents the maximum possible fraction of

the sun’s energy that could be collected with a silicon solar cell.

That is, the constraints imposed by silicon’s band gap limit theefficiency of silicon to just under 50%.

•Even this simple discussion gives some insight into the trade-off between choosing a photovoltaic

material that has a small band gap versus one with a large band gap



material that has a small band gap versus one with a large band gap.

•With a smaller band gap, more solar photons have the energy needed to excite electrons, which is

good since it creates the charges that will enable current to flow.

•However, a small band gap means that more photons have surplus energy above the threshold needed

to create hole – electron pairs, which wastes their potential.

•High band-gap materials have the opposite combination. A high band gap means that fewer photons

have enough energy to create the current carrying electrons and holes, which limits the current that can

be generated.

•On the other hand, a high band gap gives those charges a higher voltage with less leftover surplus

energy.

•In other words, low band gap gives more current with less voltage while high band gap results in less

current and higher voltage.

•Since power is the product of current and voltage, there must be some middle-ground band gap,

usually estimated to be between 1.2 eV and 1.8 eV, which will result in the highest power and

efficiency.

Notice that the efficiencies in Fig. 8.11 are roughly in the 20 – 25% range — well

below the 49 6% we found when we considered only the losses caused by



below the 49.6% we found when we considered only the losses caused by

(a) photons with insufficient energy to push electrons into the conduction band

and(b) photons with energy in excess of what is needed to do so.

(c) Other factors that contribute to the drop in theoretical efficiency include:

1. Only about half to two-thirds of the full band-gap voltage across the terminals

of the solar cell.

2. Recombination of holes and electrons before they can contribute to current

flow.

3. Photons that are not absorbed in the cell either because they are reflected offthe face of the cell, or because they pass right through the cell, or because they are

blocked by the metal conductors that collect current from the top of the cell.

4. Internal resistance within the cell, which dissipates power.

Limitations to Solar Cell Performance



I n

t e n s i t y ( m W / m 2 - m m )

Photons used

Maximum energycollected = Egap

Usablepower 24%

UnusedPhotons 19%

31% Loss forEnergyabove Egap

Voc < Egap 16%

Fill Factor 5%Other Losses 5%

Other losses:

Absorption

CollectionReflection

Series R

Shunts

Analysis for a 24%-efficient Si solar cell

All photon energy above Voc is lost.Energy (eV)



Two regions: ―n-type‖ whichdonate electrons and ―p-type‖which accept electrons

p-n junction- diffusion of

electrons and holes, currentwill flow readily in onedirection (forward biased) butnot in the other (reverse

biased), this is the diode



Making a connection from an n-type semiconductor(doped with impurities with extra electrons) to a p-

type material (extra holes) induces an electric field This field is what separates charges generated by light

The depletion width is the region where carriers havediffused

http://en.wikipedia.org/wiki/File:Pn-junction-equilibrium.png



Voltage-Current (VI) characteristics for a diode/

0 ( -1) (8.3)d qV kT

d I I e38.9

0 ( -1) (at 25 C)d V

d I I e

Figure 8.15



where :

Id is the diode current in the direction of the arrow (A),

Vd is the voltage across the diode terminals from the p-side to

the n-side (V),

Io is the reverse saturation current (A),

q is the electron charge (1.602 × 10−19C),

k is Boltzmann’s constant (1.381 × 10−23 J/K), and

T is the junction temperature (K).

Example 8.2 A p – n Junction Diode. Consider a p – n junction diode at 25◦C

ith t ti t f 10 9 A Fi d th lt d th



with a reverse saturation current of 10−9 A. Find the voltage drop across the

diode when it is carrying the following:

a. no current (open-circuit voltage)

b. 1 A

c. 10 A

Solution





Current in the Device:



J0: Reverse saturation current

V: Junction voltage

k: Boltzmann constantT: Temperature (K)

a: Ideality factor

1: ideal

2: non-ideal

Photocurrent:Diode (dark) current:

Voltage (V)

Calculating Cell Parameters:



Open circuit voltage from current equation atzero current:

Solving for Voc gives:

Notes:This is for an ideal diode!!gop is proportional to light fluxVoc increases logarithmically with light flux.

Lp, Ln: Minority carrier diffusion lengths

pn, np: Minority carrier concentrations



The previous equations in terms of the book’snotation are (just use the book equations andnotation when working your problems)

Setting I to zero, the open circuit voltage is

/

0 ( -1) (8.8)qV kT

SC I I I e

0

ln 1 (8.9)SC OC

I kT V

q I

+-

IV+

-

LoadPV



Figure 8.19 Two important parameters for photovoltaics are the short-circuit current

ISC and the open-circuit voltage VOC

Two important parameters for photovoltaics are the short-

circuit current ISC and the open-circuit voltage VOC



Add impact of parallel leakage resistance RP (want

RP to be high)

Add impact of series resistance RS (want RS to besmall) due to contact between cell and wires andsome from resistance of semiconductor

( ) (8.12)SC d

P

V I I I

R

(8.14)d S V V I R

Figure 8.22. PV Cell with parallel resistance

Vd



Considering both RS and RP

0 exp 1 - (8.17)S S

SC

P

q V I R V I R I I I

kT R

Figure 8.26. Equivalent circuit for a PV Cell including series and parallel resistance

Series and Shunt Resistance Effects:Al b i ll



Algebraically:

J JL J0 e qVJRseries / akT 1 VRshunt

Series resistance dropssome voltage (reducesoutput voltage)

Shunt resistance drops

some current (reducesoutput current)

Voltage & Current are coupled

P h o t o c

u r r e n t

s o u r c e D i

o d e

S h u n

t

r e s i s t a n c e

L o a d

Seriesresistance

(minimize)

( m a x i m i z e )

Equivalent Circuit



Parallel (RP) –

current drops byΔI=V/RP

Series (RS) – voltage drops by

ΔV=IRS

Figure 8.23

Figure 8.25

Fill Factor andC ll Effi i



Cell Efficiency

Fill Factor(FF) =

Jmax•V

max

Pabs = Jsc•Voc

CellEfficiency

(h) =

Jsc•Voc•FF

IncidentPower

Jmax•Vmax

Incident

Power

=

“AM 1.5” IncidentSolar Power~100 mW/cm2

JSC



Absorption coefficient:

Thicker is better.You need at least 2

absorption lengths evenwith a back surfacereflector.

Carrier diffusion length: Thinner is better.

Need to be able to diffuseto the contacts.

Optimal performance:

10 nm for organics

1-2 microns for CdTe,CIS, a-Si:H

2-10 microns for GaAs

20-100 microns for Si, Ge

Material Lifetime (usec) Mobility (cm2/V-sec) Ln Lp (um)

x-Si ~ 100 1350 480 590 340

CdTe ~ 0.001 3 500 0.12 1.6

GaAs ~ 0.1 8500 400 50 10

CuInSe2 ~ 0.01 800 200 3 1

a-Si ~ 0.001 1 0.05

organics ~ 0.001 10-3 0.002

Absorption of Light in the Solar Cell



Light trapping can be used to extend thepath length of the light in the absorber,allowing a thinner layer to be used.

No light trappingabsorption orreflection atback surface

Back surfacepatterned toreflect & scatterlight

Front & backsurface patternedto refract &scatter light



Figure 8.20 Photovoltaic current – voltage relationship for

“dark” (no sunlight) and “light” (an illuminated cell). The

dark curve is just the diode curve turned upside-down. The

light curve is the dark curve plus ISC



Example 8.3 The I V Curve for a Photovoltaic

Cell. Consider a 100-cm2 photovoltaic cellwith reverse saturation current Io = 10

−12

A/cm2. In full sun, it produces a short-circuit

current of 40 mA/cm2 at 25◦C. Find theopen-circuit voltage at full sun and again for50% sunlight. Plot the results.





Figure 8.21 The simple equivalent circuit of a

string of cells in series suggests no current can

flow to the load if any cell is in the dark

(shaded).

Figure 8.22 The simple PV equivalent circuit

with an added parallel resistance

PV Shading Phenomenon



Figure 8.23 Modifying the idealized PV equivalent

circuit by adding parallel resistance causes the current

at any given voltage to drop by V/RP

Figure 8.24 A PV equivalent circuit with series resistance

For a large cell, ISC

might be around 7 A and

VOC might be about 0.6

V, which says its parallel

resistance should be

greater than about 9

ohms.

For a cell to have less than 1% losses due to the



For a cell to have less than 1% losses due to theseries resistance, RS will need to be less than about

RS

< 0.01 VOC

/ISC

which, for a large cell with ISC = 7 A and VOC = 0.6 V,would be less than 0.0009 ohms.

Equation for current and voltage:

Figure 8.25 Adding series resistance to

the PV equivalent circuit causes thevoltage at any given current to shift to

the left by ΔV = IRS

Under the standard assumption of a 25◦C cell



Figure 8.26 A more complex equivalent circuit for a

PV cell includes both parallel and series resistances.

The shaded diode reminds us that this is a “real”

diode rather than an ideal one.

temperature, (8.17) becomes



Figure 8.27 Series and parallel resistances in the PV equivalent circuit

decrease both voltage and current delivered. To improve cell performance, high

RP and low RS are needed.



Since an individual cell produces only about 0.5 V, it is a rareapplication for which just a single cell is of any use. Instead, the basicbuilding block for PV applications is a module consisting of a number

of pre-wired cells in series, all encased in tough, weather-resistantpackages.

A typical module has 36 cells in series and is often designated as a―12-V module‖ even though it is capable of delivering much higher

voltages than that. Some 12-V modules have only 33 cells, which, aswill be seen later may, be desirable in certain very simple batterycharging systems.

Large 72-cell modules are now quite common, some of which have allof the cells wired in series, in which case they are referred to as 24-V

modules.

Some 72-cell modules can be field-wired to act either as 24-Vmodules with all 72 cells in series or as 12-V modules with twoparallel strings having 36 series cells in each.



Figure 8.28 Photovoltaic cells,

modules, and arrays.

Figure 8.29 For cells wired in series,

their voltages at any given current add.

A typical module will have 36 cells



Example 8.4 Voltage and Current from a PV Module

. A PVmodule is made up of 36 identical cells, all wired in series.With 1-sun insolation (1 kW/m2), each cell has short-circuitcurrent ISC = 3.4 A and at 25◦C its reverse saturation

current is Io = 6 × 10−10 A. Parallel resistance RP = 6.6 Ω andseries resistance RS = 0.005 Ω.

a. Find the voltage, current, and power delivered when the junction voltage of each cell is 0.50 V.

b. Set up a spreadsheet for I and V and present a few lines ofoutput to show how it works.





Figure 8 32 Two ways to wire an



Figure 8.32 Two ways to wire an

array with three modules in series

and two modules in parallel.

Although the I – V curves forarrays are the same, two strings of

three modules each (a) is

preferred. The total I – V curve of

the array is shown in (c).

The maximum power point (MPP) is that spot near the knee of the I –V curve at



p p ( ) pwhich the product of current and voltage reaches its maximum.

The voltage and current at the MPP are sometimes designated as Vm and Im forthe general case and designated VR and IR (for rated voltage and rated current)under the special circumstances that correspond to idealized test conditions.

Figure 8.34 The I – V curve and power output

for a PV module. At the maximum power point

(MPP) the module delivers the most power that

it can under the conditions of sunlight and

temperature for which the I – V curve has been

drawn

Figure 8.35 The maximum power point

(MPP) corresponds to the biggest rectangle

that can fit beneath the I – V curve. The fill

factor (FF) is the ratio of the area (power) at

MPP to the area formed by a rectangle with

sides VOC and ISC.

Fill factors around 70–75% for crystalline silicon solar modules



Fill factors around 70 75% for crystalline silicon solar modulesare typical, while for multijunction amorphous-Si modules, it iscloser to 50–60%.

Fill factor (FF) = Power at the maximum power point/VOC ISC =V R I R / V OC I SC



Figure 8.36 Current-voltage

characteristic curves under various cell

temperatures and irradiance levels for

the Kyocera KC120-1 PV module

As can be seen in Fig. 8.36, as cell temperature

increases, the open-circuit voltage decreasessubstantially while the short-circuit current increases

only slightly.

Photovoltaics, perhaps surprisingly, therefore

perform better on cold, clear days than hot ones.

For crystalline silicon cells, VOC drops by about

0.37% for each degree Celsius increase intemperature and ISC increases by approximately

0.05%.

The net result when cells heat up is the MPP slides

slightly upward and toward the left with a decrease in

maximum power available of about 0.5%/◦C. Given

this significant shift in performance as celltemperature changes, it should be quite apparent that

temperature needs to be included in any estimate of

module performance.

Cells vary in temperature not only because ambient

temperatures change, but also because insolation on

the cells changes.



To help system designers account for changes in cell performancewith temperature, manufacturers often provide an indicatorcalled the NOCT, which stands for nominal operating celltemperature.

The NOCT is cell temperature in a module when ambient is 20◦C,

solar irradiation is 0.8 kW/m2, and windspeed is 1 m/s.

To account for other ambient conditions, the following expressionmay be used:

where Tcell is cell temperature (◦C), Tamb is ambient

temperature, and S is solar insolation (kW/m2).



Example 8.5 Impact of Cell Temperature on Power

for a PV Module.

Estimate cell temperature, open-circuit voltage,and maximum power output for the 150-W

BP2150S module under conditions of 1-suninsolation and ambient temperature 30◦C. Themodule has a NOCT of 47◦C.





When the NOCT is not given, another approach to

estimating cell temperature is based on thefollowing:

Tcell = Tamb + γ ( Insolation/1 kW/m2)

where γ is a proportionality factor that dependssomewhat on windspeed and how well ventilatedthe modules are when installed.

Typical values of γ range between 25◦C and 35◦C;that is, in 1 sun of insolation, cells tend to be 25–35◦C hotter than their environment.



Figure 8.37 A module with n cells in which the

top cell is in the sun (a) or in the shade (b).



Consider the case when the bottom n − 1 cells still have full sun and still somehow 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 VSH with onecell shaded will drop to

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

Combining (8.26) and (8.27) gives

The drop in voltage V at any given current I , caused by the shaded cell, isgiven by

Since the parallel resistance RP is so much greater than the series resistance RS, (8.30) simplifies to

At any given current, the I –V curve for the module with one shaded cell dropsby ΔV.



Figure 8.38 Effect of shading one cell in an n-cell module. Atany given current, module voltage drops from V to V − ΔV



Example 8.6 Impacts of Shading on a PV Module. The 36-cell PV

module described in Example 8.4 had a parallel resistance percell of RP = 6.6 Ω. In full sun and at current I = 2.14 A theoutput voltage was found there to be V = 19.41 V. If one cellis shaded and this current somehow 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?

Solution







Figure 8.40 In full sun a cell may

contribute around 0.5 V to the

module output; but when a cell is shaded, it can have a large voltage

drop across it.

Figure 8.41 Mitigating the shade

problem with a bypass diode. In the

sun (a), the bypass diode is cut offand all the normal current goes

through the solar cell. In shade (b),

the bypass diode conducts current

around the shaded cell, allowing just

the diode drop of about 0.6 V to

occur.

To see how bypass diodes wired in parallel with modules can help mitigateshading problems, consider Fig. 8.42, which shows I –V curves for a string offive modules (the same modules that were used to derive Fig. 8.39).



The graph shows the modules in full sun as well as the I –V curve that resultswhen one module has two cells completely shaded. Imagine the PVs delivering

charging current at about 65 V to a 60-V battery bank. As can be seen, in fullsun about 3.3 A are delivered to the batteries. However, when just two cells inone module are shaded, the current drops by one-third to about 2.2 A. With abypass diode across the shaded module, however, the I –V curve is improvedconsiderably as shown in the figure.

Figure 8.42 Impact of bypassdiodes. Drawn for five modules in

series delivering 65 V to a battery

bank. With one module having two

shaded cells, charging current

drops byalmost one-third when

there are no bypass diodes. With

the module bypass diodes there

is very little drop.



Figure 8.43 helps explain how the bypass diodes do their job.

Imagine five modules, wired in series, connected to a battery thatforces the modules to operate at 65 V. In full sun the modulesdeliver 3.3 A at 65 V.

When any of the cells are shaded, they cease to produce voltage and

instead begin to act like resistors (6.6 Ω per cell in this example)that cause voltage to drop as the other modules continue to try topush current through the string.

Without a bypass diode to divert the current, the shaded moduleloses voltage and the other modules try to compensate by increasing

voltage, but the net effect is that current in the whole string drops.If, however, bypass diodes are provided, as shown in Fig. 8.43c,then current will go around the shaded module and the chargingcurrent bounces back to nearly the same level that it was beforeshading occurred.



8.7.3 Blocking Diodes



Figure 8.44 Blocking

diodes prevent reverse

current from flowing

down malfunctioningor shaded strings.





On Monday US Environmental Protection

Agency (EPA) said that greenhouse gases are adanger to public health and welfare

This is a necessary first step to allow the EPAto regulate greenhouse gas emissions◦

Some industries are concerned these regulationsmay be more restrictive than a legislative approach

Some in Congress have called on EPA towithdraw its proposal because of recent emailreleases that question the underlying science



Grid-connected systems



Stand-alone systems which charge batteries



Stand-alone systems with directly-connectedloads



PV panels have I-V curves and so do loads Use a combination of the two curves to tell

where the system is actually operating

Operating point – the intersection point at

which the PV and the load I-V curves aresatisfied



Straight line with slope 1/R

As R increases, operating point moves to the right

V IR1

(9.1) I V R

• Can use a potentiometerto plot the PV module’sIV curve

• Resistance value that

results in maximumpower

(9.2)mm

m

V R

I

Figure 9.5



Maximumpower point(MPP) shouldoccur whenthe load

resistance R =VR/IR under 1-sun 25˚C, AM1.5 conditions

•A MPP tracker maintains PV system’s highestefficiency as the amount of insolationchanges





DC motors have an I-V curve similar to aresistor

e = k ω is back emf, R a is armature resistance

(9.3)aV IR k



Energy is stored in batteries for most off-gridapplications

An ideal battery is a voltage source V B

A real battery has internal resistance R i

(9.4) B iV V R I



Charging– I-V line tilts right with a slope of1/R i , applied voltage must be greater than VB

Discharging battery- I-V line tilts to the leftwith slope 1/R i , terminal voltage is less than VB

Figure 9.12





A buck-boost converter used as a the heart of amaximum power tracker

The duty cycle D is the fraction of the time the switch is closed

(a). Examples: (b) 50% duty cycle; (c) D < 0.5; (d) D > 0.5.

While the switch is closed, from time t = 0 to t = DT, the voltage across the inductor

is a constant Vi . The average power put into the magnetic field of the inductor

during one complete cycle is given by



Under the assumption that inductor current is constant, the average power into the

inductor is

When the switch opens, the inductor’s magnetic field begins to collapse, returning

the energy it just acquired. The diode conducts, which means that the voltage

across the inductor VL is the same as the voltage across the load V0. The average

power delivered by the inductor is therefore



With good design, both V0 and IL are essentially constant, soaverage power from the inductor is

Over a complete cycle, average power into the inductor equals

average powerout of the inductor.

(*)



Equation (*) is pretty interesting. It tells us we can bump

dc voltages up or down (there is a sign change) just byvarying the duty cycle of the buck-boost converter.

•Longer duty cycles allow more time for the capacitor to

charge up and less time for it to discharge, so the output

voltage increases as D increases.

•For a duty cycle of 1/2, the output voltage is the same as

the input voltage.

•A duty cycle of 2/3 results in a doubling of voltage,while D = 1/3 cuts voltage in half.



A PV module has its maximum power point atVm = 17 V and Im = 6A.

What duty cycle should its MPPT have if themodule is delivering power to a 10Ω

resistance?

Max power delivered by the PVs is 17V*6A =102W

2

31.9 R RV P V V

R

The converter must boost the 17 V PV voltage to thedesired 31 9 V V



desired 31.9 V

Solving gives

0 (9.9)

1i

V D

V D

31.9

1.8817 1

D

D

0.65 D

Redrawing the PV I – V curves with

an MPPT



Current at any

voltage isproportionalto insolation

V OC drops asinsolationdecreases

Can justadjust the 1-sun I -V curveby shifting itup or down

the dc motor has been well matched to the 1-sun I – V curve, but does

poorly in the early morning and late afternoon.



Hour-by-hour PV I – V curves with examples of three different load

types: dc motor, 12-V battery, MPPT

the 12-V battery is consistently somewhat below the maximum

power point.



Table 9.1 provides a compilation of the hourly performance of each of these

loads. The dc motor loses about 15% of the available daily energy because

it doesn’t operate at the maximum power point while the 12-V battery loses

17%.



Can have a combiner box and a singleinverter or small inverters for each panel

Individual inverters make the system modular Inverter sends AC power to utility service

panel Power conditioning unit (PCU) may include

◦ MPPT◦ Ground-fault circuit interrupter (GFCI)◦ Circuitry to disconnect from grid if utility loses

power◦ Battery bank to provide back-up power





Easily allow expansion

Connections to house distribution panel aresimple

Less need for expensive DC cabling



Net metering – customer only pays for the amount

of energy that the PV system is unable to supply

In the event of an outage, the PV system mustquickly and automatically disconnect from the grid

• A battery backupsystem can helpprovide power to thesystem’s owners during

an outage• Good grid-connectinverters haveefficiencies above 90% http://www.pasolar.ncat.org/lesson05.php

The power conditioning unit absolutely must be designed to quickly and automaticallydrop the PV system from the grid in the event of a utility power outage. When there is anoutage, breakers automatically isolate a section of the utility lines in which the fault has

d ti h t i f d t ―i l d ‖



occurred, creating what is referred to as an ―island.‖

A number of very serious problems may occur if, during such an outage, a self-generator, such as a grid-connected PV system, supplies power to that island.

Most faults are transient in nature, such as a tree branch brushing against the lines, andso utilities have automatic procedures that are designed to limit the amount of time theoutage lasts.

When there is a fault, breakers trip to isolate the affected lines, and then they areautomatically reclosed a few tenths of a second later. It is hoped that in the interim thefault clears and customers are without power for just a brief moment.

If that doesn’t work, the procedure is repeated with somewhat longer intervals until

finally, if the fault doesn’t clear, workers are dispatched to the site to take care of theproblem. If a self-generator is still on the line during such an incident, even for less thanone second, it may interfere with the automatic reclosing procedure, leading to a longer-than necessary outage.

And if a worker attempts to fix a line that has supposedly been disconnected from allenergy sources, but it is not, then a serious hazard has been created.



Estimate the AC output power under varying

conditions

P dc,STC - DC power of array from adding module

ratings under standard test conditions (STC) (1-sun,AM 1.5, 25˚C)

Conversion efficiency – includes losses from inverter,dirty collectors, mismatched modules, and differences

in ambient conditions These losses can derate power output by 20-40%,

even in full sun

, (Conversion Efficiency) (9.10)ac dc STC P P



Illustrates the impact of slight variations inmodule I -V curves

Only 330 W is possible instead of 360 W



As temperature increases, power decreases PVUSA test conditions (PTC) – 1-sun

insolation in plane of array, 20˚C ambienttemperature, wind-speed of 1 m/s

P ac(PTC) AC output of an array under PTC test

conditions is a better indicator of actualpower delivered in full sun than the morecommonly used P dc(STC)

Describing a system based on P dc(STC) without

correcting for temperature and the inverter ismisleading



VOC decreases by

~0.37% per ˚C forcrystalline silicon cells

ISC increases by about

0.05% per ˚C NOCT – Normal

OperatingTemperature

20S (8.24)

0.8cell amb

NOCT C T T

Figure 8.36

S= insolation

A PV array has rating of 1 kW under standard test



condtions (STC). Nominal operating temperature

(NOCT) from Chapter 8 is 47˚C DC power output drops by 0.5%/ ˚C above the

STC temperature of 25˚C Mismatched module loss= 3%

Dirt loss = 4% Inverter efficiency = 90%

Estimate P ac(PTC) , the AC output power under

PVUSA test conditions (PTC)Monitoring program called PVUSA: the PVUSA test conditions

(PTC) are defined as 1-sun irradiance in the plane of the array, 20◦C

ambient temperature, and a wind-speed of 1 m/s.



The estimated cell temperature is

With DC losses at 0.5%/ ˚C above 25˚C,

Including inefficiencies, estimated AC rated powerat PTC is

20S (8.24)

0.8cell amb

NOCT C T T

47 20

20 1 = 53.80.8cell

C

T C

,( ) 1 kW 1 0.005(53.8 25) = 0.856 kWdc PTC P

,( ) 8.56 kW 0.97 0.96 0.90 = 0.72 kWac PTC P



1-sun is 1 kW/m2

We can say that 5.6 kWh/(m2-day) is 5.6hours of ―peak sun‖

If we know Pac, computed for 1-sun, just

multiply by hours of peak sun to get kWh If we assume the average PV system efficiency

over a day is the same as the efficiency at 1-sun, then

Energy (kWh/day) kW h/day of "peak sun" (9.14)ac P



Energy kWh/yr kW CF 8760 h/yr (9.15)ac P

h/day of "peak sun"CF (9.16)

24 h/day

Figure 9.28

PV Capacity

Factors for

US cities



When the grid isn’t nearby, the extra cost andcomplexity of a stand-alone power systemcan be worth the benefits

System may include batteries and a backup

generator

S d b h f h l d h



PV System design begins with an estimate of the loads that

need to be served by the PV system

Tradeoffs between more expensive, efficient appliancesand size of PVs and battery system needed

Should you use more DC loads to avoid inverterinefficiencies or use more AC loads for convenience?

What fraction of the full load should the backup generatorsupply?

Power consumed while devices are off

Inrush current used to start major appliances



Table 9.10 – Power Requirements of some typical loads

Note that these tables are useful for getting an idea of the average

values, but the best data comes from actual measurements!



Consider the power when the device isactively used

Also consider the power consumed whendevice is in standby

Table 9.10 – Power requirements of some consumer electronics





+ 2

2 4 4 2Positive Plate: PbO + 4H + SO + 2 PbSO 2H O (9.21)e

2

2 4 4 Negative Plate: PbO + SO PbSO 2e (9.22)



• During discharge, voltage drops and specific gravity

drops• Sulfate adheres to the plates during discharge and

comes back off when charging, but some of it becomespermanently attached



Battery capacity has tended to be specified in

179



Battery capacity has tended to be specified in

amp-hours (Ah) as opposed to an energy value;multiply by average voltage to get watt-hours◦ Value tells how many amps battery can deliver over a

specified period of time.

◦ Amount of Ah a battery can delivery depends on its

discharge rate; slower is better

Figure shows

how capacity

degrades with

temperature

and rate

T D it C t C l Ch P



Type Density,

Wh/kG

Cost

$/kWh

Cycles Charge

time,hours

Power

W/kg

Lead-acid,

deep cycle

35 50-100 1000 12 180

Nickel-metal

hydride

50 350 800 3 625

Lithium Ion 170 500-100 2000 2 2500

The above values are just approximate; battery technology is rapidly changing,and there are many different types within each category. For stationary

applications lead-acid is hard to beat because of its low cost. It has about a 75%

efficiency. For electric cars lithium ion batteries appear to be the current front

runner



l f l d



Analysis of load◦ Determine daily demands for power and energy

◦ What fraction of the worst month ―design month‖ shouldyou cover with the PV system? How much should youcover with a backup generator?

◦

What PV system voltage should you have?◦ Convert total DC load to amp hours @ system voltage

PV sizing◦ Pick a PV module based on insolation data for the site

for the design month

◦ Determine how many parallel strings of modules andhow many modules in each string





As the name implies a microgrid can be thought of as a

185



p g gsmall electric grid with several generation sources◦ The microgrid can be configured to operate either connected

to the main grid or standalone

The military is a key proponent of microgrids, sincethey would like the ability to operate bases independentof any grid system for long periods of time

Renewable generation by be quite attractive because itdecreases the need to store large amounts of fossil fuel◦ Time magazine reported in Nov 2009 that average US solider

in Afghanistan requires 22 gallons of fuel per day at anaverage costs of $45 per gallon



Dr. William J.Makofske

August 2004

S lid t t d i th t t i id t l



Solid state device that converts incident solar

energy directly into electrical energy

Efficiencies from a few percent up to 20-30%

No moving parts

No noise Lifetimes of 20-30 years or more



Th j ti f di i il t i l ( d



The junction of dissimilar materials (n and p

type silicon) creates a voltage

Energy from sunlight knocks out electrons,creating a electron and a hole in the junction

Connecting both sides to an external circuitcauses current to flow

In essence, sunlight on a solar cell creates asmall battery with voltages typically 0.5 v. DC

S l ll b l t i ll t d i



Solar cells can be electrically connected in

series (voltages add) or in parallel (currentsadd) to give any desired voltage and current(or power) output since P = I x V

Photovoltaic cells are typically sold inmodules (or panels) of 12 volts with poweroutputs of 50 to 100+ watts. These are thencombined into arrays to give the desired

power or watts.



While a major component and cost of a PV



While a major component and cost of a PV

system is the array, several other componentsare typically needed. These include:

The inverter – DC to AC electricity

DC and AC safety switches Batteries (optional depending on design)

Monitor – (optional but a good idea)

Ordinary electrical meters work as net meters









Solar CalculatorsREMOTE POWER Lighting

Buoys Communications Signs Water Pumping

Mountain Cabins













As prices dropped PV began to be used for



As prices dropped, PV began to be used for

stand-alone home power. If you didn’t havean existing electrical line close to yourproperty, it was cheaper to have a PV system(including batteries and a backup generator)than to connect to the grid. As technologyadvanced, grid-connected PV with netmetering became possible.

In net metering when the PV system



In net metering, when the PV system

produces excess electricity, it is sent to thegrid system, turning the meter backwards. Ifyou are using more power than is beingproduced, or it is at night, the electricity isreceived from the grid system and the meterturns forwards. Depending on PV size andelectrical consumption, you may producemore or less than you actually use. Individualhouses may become power producers.



Batteries can be used to provide long-term or



Batteries can be used to provide long term or

short-term electrical supply in case of gridfailure. Many grid-connected houses chooseto have a small electrical battery system toprovide loads with power for half a day incase of outage. Larger number of batteriesare typically used for remote grid-independent systems.

If your load is 10 kw-hr per day, and you want



If your load is 10 kw hr per day, and you want

to battery to provide 2.5 days of storage,then it needs to store 25 kw-hr of extractableelectrical energy. Since deep cycle batteriescan be discharged up to 80% of capacity

without harm, you need a battery with astorage of 25/0.8 = 31.25 kw-hr. A typicalbattery at 12 volts and 200 amp-hourcapacity stores 2.4 kw-hr of electrical energy.

The relationship between energy in kw-hr and



The relationship between energy in kw hr and

battery capacity isE(kw-hr) =capacity(amp-hr) x voltage/1000

E = 200 amp-hr x 12 volts/1000= 2.4 kw-hr

So for 31.25 kw-hr of storage we need31.25 kw-hr/2.4 kw-hr/battery = 13

batteries

If we are happy with one half day, we need only

2 or 3 batteries



PV can be added to existing roofs. While



PV can be added to existing roofs. While

south tilted exposure is best, flat roofs dovery well. Even east or west facing roofs thatdo not have steep slopes can work fairly wellif you are doing net metering since the

summer sun is so much higher and moreintense than the winter sun. The exactperformance of any PV system in anyorientation is easily predictable.

















If it is impossible or you don’t want to put a



p y p

PV system on your existing roof, it is possibleto pole mount the arrays somewhere near thehouse as long as the solar exposure is good.Pole mounted solar arrays also have the

potential to rotate to follow the sun over theday. This provides a 30% or more boost to theperformance.







If you are doing new construction or a



reroofing job, it is possible to make the roofitself a solar PV collector. This saves the costof the roof itself, and offers a more aestheticdesign. The new roof can be shingled or look

like metal roofing. A few examples follow.















Solar modules are typically sold by the peak watt.



That means that when the sun is at its peakintensity (clear day around midday) of 1000 wattsper m2, a solar module rates at say 100 Wp (peakwatts) would put out 100 watts of power. The

climate data at a given site summarizes the solarintensity data in terms of peak sun hours, theeffective number of hours that the sun is at thatpeak intensity on an average day. If the averagepeak sun hours is 4.1, it also means that a kw of PV

panels will provide 4.1 kw-hr a day.

When the sky is clear and it is around midday,



the solar intensity is about 1000 watts perm2 or 1 kw/m2

In one hour, 1 square meter of the earth’ssurface facing the sun will intercept about 1

kw-hr of solar energy.

What you collect depends upon surfaceorientation and collector efficiency

A PV system can be sized to provide part or



all of your electrical consumption. If youwanted to produce 3600 kw-hr a year at asite that had an average of 4.1 peak sunhours per day,

PV Size in KWp = 3600 kw-hr

4.1 kw-hr/day x 365 days/yr x 0.9 x0.98

= 2.7 KWp

Note: the 0.9 is the inverter efficiency and the0.98 represents the loss in the wiring.

1 kW = 1000 watts = 1.34 hp (presumably



the maximum sustained output of a horse)1 kW-hr = 3413 Btu is the consumption of a1 kW device operated for an hour (E=Pxt)

Now think about a Sherpa mountain guidecarrying a 90 lb pack up Mount Everest, about29,000 ft or 8,839 meters high, over a week,the typical time for such a trip

Since we know that the energy in lifting is given by



mgh or 40.8 kg x 9.8 m/s2 x 8839m = 3,539,000 joules or about 1 kw-hr, we can say that roughly1 kw-hr = 1 Sherpa-week. Typical U.S.household consumption is 600 kw-hr per month or

20 kw-hr per day, or every day it is like hiring 20Sherpa to carry the 90 lb packs up Mt. Everest. Atthe end of the week, we have 140 Sherpa climbingthe slopes so the equivalent power that weconsume is like having 140 Sherpa climbing Mt

Everest continually. We might want to considerreducing this number before adding PV to our roof.

The actual area that you need depends on



the efficiency of the solar cells that you use.Typical polycrystalline silicon with around12% efficiency will require about 100 ft2 ofarea to provide a peak kilowatt. Less efficient

amorphous silicon may need 200 ft2 toprovide the same output. Modules are sold interms of peak wattage and their areas aregiven so you can easily determine the total

roof area that is needed for a given size array.

Assume it is a 100 Wp module and its area is



0.8 m2. Remember that the peak powerrating is based on an intensity of 1000watts/m2. So the maximum output with 100%efficiency is P = I x A = 1000 w/m2 x 0.8 m2

= 800 wattsThe actual efficiency = Pactual peak/Pmaximum

peak

= 100 watts/800 watts = 0.125 or 12.5%

Of course you don’t have to stop with home



based PV systems. They make equally goodsense for businesses and corporations whowant to reduce their cost of electricity byreducing their peak power consumption, or

who want to emphasize their greenness aspart of their image, or who need to operate ina grid failure.























PV systems as an alternative energy resource

C l E i h b id



Complementary Energy-resource in hybridsystems

Necessary:

high reliability

reasonable cost

user-friendly design

The standards

EN61000 3 2 IEEE1547



EN61000-3-2, IEEE1547, U.S. National Electrical Code (NEC) 690

IEC61727

power quality, detection of islandingoperation, grounding

structure and the features of the present and

future PV modules.





PV Generator Converter AC-DC Local Loads

Grid

Solar-electric-energy growth consistently

20% 25% th t 20



20%–25% per annum over the past 20 years

1) an increasing efficiency of solar cells

2) manufacturing-technology improvements

3) economies of scale

2001, 350 MW of solar equipment was sold

2003 574 MW f PV i t ll d



2003, 574 MW of PV was installed. In 2004 increased to 927 MW

Significant financial incentives in Japan,Germany, Italy and France

triggered a huge growth in demand

In 2008, Spain installed 45% of allphotovoltaics, 2500 MW in 2008 to an drop

to 375 MW in 2009

World solar photovoltaic (PV) installations

ere 2 826 giga atts peak (GWp) in 2007



were 2.826 gigawatts peak (GWp) in 2007,and 5.95 gigawatts in 2008

The three leading countries (Germany, Japanand the US) represent nearly 89% of the totalworldwide PV installed capacity.

2012 are and 12.3GW- 18.8GW expected



Market leader in solar panel efficiency

(measured by energy conversion ratio) is



(measured by energy conversion ratio) isSunPower, (San Jose USA) - 23.4

market average of 12-18%.

Efficiency of 42 achieved at the University of

Delaware in conjunction with DuPont(concentration) in 2007.

The highest efficiency achieved without

concentration is by Sharp Corporation at35.8 using a proprietary triple-junctionmanufacturing technology in 2009.

IGBT technology

Inverters must be able to detect an islanding



Inverters must be able to detect an islandingsituation and take appropriate measures inorder to protect persons and equipment

PV cells - connected to the grid

PV cells - isolated power supplies



Central inverters

Module oriented or module integrated



Module-oriented or module-integratedinverters

String inverters

Integration of PV strings of different

technologies and orientations



technologies and orientations









Decoupling is necessary

p instantaneous



p –instantaneous P - average



Voltage in the range from 23 to 38 V at a power generation of

approximate 160 W, and their open-circuit voltage is below 45

V.

New technolgies - voltage range around 0.5 -1.0 V at several

hundred amperes per square meter cell

EX i l l h ld



EX.: ripple voltage should

be below 8.5% of the MPP

voltage in order to reach a

utilization ratio of 98%







.

(a) Irradiation distribution for

a reference year.

(b) Solar energy distribution

for a reference year.

Total time of

irradiation equals 4686 h per

year.

Total potential energy is equal

to 1150 kWh=(m2 year) 130

W/m2



Centralized Inverters

String Inverters



String Inverters Multi-string Inverters

AC modules & AC cell technology



Reduced version of the centralized



Reduced version of the centralizedinverter

single string of PV modules isconnected to the inverter

no losses on string diodes

separate MPPTs

increases the overall efficiency



Multi-String Inverters

AC Modules



AC Modules AC Cells

…

Flexible



Flexible Every string can be controlled

individually.

One large PV cell

connected to a dc–i



connected to a dcac inverter

Very low voltage

New converter

concepts

Single-stage inverter



Dual stage inverter

Multi-string inverter

Capacitors



Component to avoid (line transformers= high

size, weight, price)h f f



size, weight, price) High-frequency transformers

Grounding,



standard full-bridge three-level VSI



Half-bridge diode-clamped three-level VSI



1. 100-W single-transistor flyback-type HF-

link inverter100 W t 230 V i 48 V 96% f 0 955



100 W, out 230 V, in 48 V, 96%, pf=0,955



3. Modified Shimizu Inverter (160W, 230, 28V,

87%)



)

4. 160-W buck–boost inverter

in 100V out 160V



5. 150-W flyback dc–dc converter with a line-

frequency dc–ac unfolding inverterin 44V o t 120V



q y g in 44V, out 120V

6. 100-W flyback dc–dc converter with a PWM

dc–ac inverter30V 210 V



30V – 210 V

110-W series-resonant dc–dc converter with

an HF inverter toward the grid 30 230V 87%



g 30-230V , 87%



Single-stage

Dual-stage



a transformerless half-bridge diode-clamped

three-level inverter









component ratings

relative cost lifetime



lifetime

efficiency

Dual-stage CSI = large electrolytic decoupling

capacitor



VSI = small decoupling electrolytic capacitor.

Low efficiency=87%

C=68 mF 160V



High efficiency=93%

C=2,2 mF 45V

The dual-grounded multilevel inverters p.82

– good solution but quite large capacitors2x640mF 810V -> half-period loading



2x640mF 810V > half period loading

bipolar PWM switching toward the grid p.83 &84 (no grounding possible, large ground

currents) – 2x1200 mF 375 V current-fed fullbridge dc–dc converters with

embedded HF transformers, for each PVstring – p.85 – 3x 310 mF 400V

Large centralized single-stage inverters shouldbe avoided

Preferable location for the capacitor is in the dclink where the voltage is high and a large



link where the voltage is high and a largefluctuation can be allowed without compromisingthe utilization factor

HFTs should be applied for voltage amplification

in the AC module and AC cell concepts Line-frequency CSI are suitable for low power,

e.g., for ac module applications. High-frequency VSI is also suitable for both low-

and high-power systems, like the ac module, thestring, and the multistring inverters

PV inverters with dc/dc converter (with or

without isolation) PV inverters without dc/dc converter (with or



PV inverters without dc/dc converter (with orwithout isolation)

Isolation is acquired using a transformer thatcan be placed on either the grid or lowfrequency (LF) side or on the HF side

full-bridge

single-inductor push–pull double-inductor push–pull



double-inductor push–pull

number of cascade power processing stages

-single-stage -- dual-stage



-- dual-stage

-----multi-stage

There is no any standard PV inverter topology

very efficient PV cells

roofing PV systems PV modules in high building structures



PV modules in high building structures

PV systems without transformers - minimize

the cost of the total system cost reduction per inverter watt -make PV-



cost reduction per inverter watt make PVgenerated power more attractive

AC modules implement MPPT for PV modules

improving the total system efficiency „ plug and play systems‖

MPPT control

THD improvements reduction of current or voltage ripple



reduction of current or voltage ripple

standards are becoming more and more strict





Basic Physics andMaterials Science of Solar Cells

Original Presentation by J. M. Pearce, 2006

• Photovoltaic (PV) systems convert lightenergy directly into electricity.



gy y y

• Commonly known as ―solar cells.‖

• The simplest systems power the small

calculators we use every day. Morecomplicated systems will provide a largeportion of the electricity in the near future.

• PV represents one of the most promising

means of maintaining our energy intensivestandard of living while not contributing toglobal warming and pollution.

• 1839 – Photovoltaic effect discovered byBecquerel.1870s Hertz developed solid selenium PV (2%)



• 1870s – Hertz developed solid selenium PV (2%).

• 1905 – Photoelectric effect explained by A.Einstein.

• 1930s – Light meters for photography commonlyemployed cells of copper oxide or selenium.

• 1954 – Bell Laboratories developed the firstcrystalline silicon cell (4%).

• 1958 – PV cells on the space satellite U.S.Vanguard (better than expected).







• Silicon

• Group 4 elementalsemiconductor



• Silicon crystal forms thediamond lattice

• Resulting in the use offour valence electronsof each silicon atom.

Crystalline

Silicon



Amorphous Silicon

• Advantages:

– High Efficiency (14-22%)

– Established technology



(The leader)

– Stable

• Disadvantages: – Expensive production

– Low absorption coefficient

– Large amount of highly

purified feedstock

Advantages:• High absorption (don’t need

a lot of material)• Established technology• Ease of integration into

b ld



gbuildings

• Excellent ecological balancesheet

• Cheaper than the glass,metal, or plastic you depositit on

Disadvantages:

• Only moderate stabilizedefficiency 7-10%

• Instability- It degrades when lighthits it – Now degraded steady state



The physics view

• There are 3 types of

materials in BandTheory, which arediff i d b h i

Ef



differentiated by theirelectronic structure: –

insulators, – conductors, and

– semiconductors.

Eg

Metal Insulator Semiconductor

EfEf

•Conduction

Band – Ec – empty



empty

• Valence

Band – Ev – full ofelectrons

1.

Intrinsic

2.

n-type

3.

p-type



• Types 2 and 3 are semiconductors thatconduct electricity - How?

– by alloying semiconductor with an impurity, alsoknown as doping

– carriers placed in conduction band or carriersremoved from valence band.

Note: Color Protocol

•Pure semiconductor

(intrinsic): contains the

right number ofelectrons to fill valence



band, therefore,conduction band is

empty.• Because electrons in full

valence cannot move,the pure semiconductor

acts like an insulator.

• n-type: current iscarried by negatively

charged electrons -How?



– group 5 impurity atomsadded to silicon melt

from which is crystal isgrown

– 4/5 of outer electronsused to fill valence band

– 1/5 left is then put into

conduction band. Theseimpurity atoms are calleddonors.

Within conductionband the electrons aremoving, therefore,

crystal becomes aconductor

• p-Type: current carried bymissing electron holes

which act as positivelycharged particles. How?



charged particles. How?

– group 3 added to siliconmelt

– need 4 out of 5 outerelectrons but doping createslack of electrons in valenceband.

– missing electrons, a.k.a

holes, are used to carrycurrent.

• Prevailing charges are called the majoritycarriers – prevailing charge carrier in n-type: electrons

prevailing charge carrier in p-type: holes



– prevailing charge carrier in p type: holes

• There are four main types of semiconductor junctions



– p-n – p-i-n – Schottcky barrier – Heterojunction

• Each has a built in potential

V bi V bi



Ef Ef



• All the junctions contain strong electricfield

• How does the electric field occur? – When two semiconductors come into contact,

l t i t f f t t f



electrons near interface from n-type, transferover to p-type, leaving a positively charged area

– Holes from p-type by interface transfer over to

n-type leaving a negatively charged area. – Because electrons and holes are swapped, a

middle potential barrier with no mobile charges,is formed.

– This potential barrier created does not let any

more electrons or holes flow through.• Electric field pulls electrons and holes in

opposite directions.

• Equilibrium means thereis no net current

• Reduced barrier height iscalled forward bias



called forward bias (positive voltage applied top-side) – Result- increases current

through diode

• Increased barrier height iscalled reverse bias. – Result- decreases current to

a very small amount..



Diode I-V Characteristics

• Output current = I = I

l

-I

o

[ exp(qV/kT)-1]

– Il=light generated current – q = electric charge – V = voltage – k = Boltzman’s constant = 1.3807 × 10-23 J/K

Wh i i it (I 0) ll li ht t d



• When in open circuit (I=0) all light generatedcurrent passes through diode

• When in short circuit (V=0) all current passes

through external load2 Important points:

1) During open circuit the voltage of opencircuit,

V

oc

= (kT/q) ln( I

l

/I

o

+1)

2) No power is generated under short and opencircuit - but Pmax = VmIm=FFVocIsc



Fourth quadrant (i.e., power quadrant) of theilluminated I-V characteristic defining fill factor (FF) andidentifying Jsc and Voc

• Photovoltaic energy relies on light.• Light → stream of photons → carries energy • Example: On a clear day 4.4x1017 photons

hit 1 m2 of Earth’s surface every second.



• Eph()=hc/ =hfh = plank’s constant = 6.625 x 10-34 J-s

= wavelengthc = speed of light =3 x 108 m/sf = frequency

• However, only photons with energy inexcess of bandgap can be converted into

electricity by solar cells.



The entire spectrum is

not available to single

junction solar cell

• Photon enters, isabsorbed, and letselectron from VB getsent up to CB

• Therefore a hole is left



behind in VB, creatingabsorption process:electron-hole pairs.

• Because of this, onlypart of solar spectrumcan be converted.

• The photon fluxconverted by a solarcell is about 2/3 oftotal flux.

• Generation Current = light induced electrons acrossbandgap as electron current

• Electron current:= Ip=qNA – N = # of photons in highlighted area of spectrum – A = surface area of semiconductor that’s exposed to

light



g• Because there is current from light, voltage can also

occur.• Electric power can occur by separating the electrons and

holes to the terminals of device.• Electrostatic energy of charges occurs after separation

only if its energy is less than the energy of the electron-hole pair in semiconductor

• Therefore Vmax=Eg/q• Vmax= bandgap of semiconductor is in EV’s, therefore

this equation shows that wide bandgap semiconductorsproduce higher voltage.

• Everything just talked about, where allenergy in excess of bandgap of photons are

absorbed, are called direct-bandgapsemiconductors.

l d b h



• More complicated absorption process is theindirect-gap series

– quantum of lattice vibrations, of crystallinesilicon, are used in the conversion of a photoninto electron-hole pair to conserve momentumthere hindering the process and decreasing theabsorption of light by semiconductor.

• Electric current generated in semiconductor isextracted by contacts to the front and rear of cell.

• Widely spaced thin strips (fingers) are created so that

light is allowed through. – these fingers supply current to the larger bus bar.



• Antireflectioncoating (ARC) is

used to cover thecell to minimizelight reflection fromtop surface.

• ARC is made with

thin layer ofdielectric material.



Diffusion

Drift

Excitonic

• n-type and p-typeare aligned by theFermi-level

• When a photoncomes in n-type, ittakes the place of a



takes the place of ahole, the hole actslike an air bubble and

―floats‖ up to the p-type• When the photon

comes to the p-type,it takes place of anelectron, the electronacts like a steel balland ―rolls‖ down tothe n-type



• There is anintrinsic gapwhere the photon

is absorbed inand causes theelectron hole pair

f



pto form.

• The electron risesup to the top anddrifts downwards(to n-type)

• The hole driftsupwards (to p-type)

• Dye molecule

–

electron hole pairsplits because ithits the dye



y

– the electron shiftsover to the

electric conductorand the holeshifts to the holeconductor



• Opposite of carrier generation,

where electron-hole pair isannihilated



• Most common at: –

impurities – defects of crystal structure

– surface of semiconductor• Reducing both voltage and current

• Losses of resistance caused bytransmission of electric current



produced by the solar cell.

• I-V characteristic of device:• I = I

l

-I

0

[exp(qV+IRs / mkT) 1]

•

m= nonideality factor

• Current losses- called collection efficiency,ratio b/w number of carriers generated by

light by number that reaches the junction.• Temperature dependence of voltage



p p g – V decreases as T increases

• Other losses – light reflection from top surface

– shading of cell by top contacts

– incomplete absorption of light



• Tandem cell- several cells, – Top cell has

large bandgap

Middl ll id

Indium Tin Oxide

p-a-Si:H

Blue Cell

i a Si:H

Silver Grid



– Middle cell mideV bandgap

– Bottom cellsmall bandgap.

i-a-Si:H

n-a-Si:H

p

Green Cell

i-a-SiGe:H (~15%)

n

p

Red Cell

i-a-SiGe:H (~50%)

n

Textured Zinc Oxide

Silver

Stainless Steel SubstrateSchematic diagram of state-of-

the-art a-Si:H based substrate n-

i-p triple junction cell structure.



• This is the first in a series of presentationscreated for the solar energy community to

assist in the dissemination of informationabout solar photovoltaics.

• This work was supported from a grant from



• This work was supported from a grant fromthe Pennsylvania State System of HigherEducation.

• The author would like to acknowledgeassistance in creation of this presentation fromHeather Zielonka, Scott Horengic and JenniferRockage.



Part I


Updated by J.M. Pearce and R. Andews, 2010

• What is a photovoltaic system – Cell, Module, and Array

– BOS• Structure

El t i



• Electronics

– PV System Design Basics

– Hybrid Systems



• The BOS typically contains; – Structures for mounting the

PV arrays or modules

– Power conditioningequipment that massagesand converts the do



and converts the doelectricity to the proper formand magnitude required by

an alternating current (ac)load.

– Sometimes also storagedevices, such as batteries,for storing PV generatedelectricity during cloudy days

and at night.

• Stand-alone systems – Systems which use photovoltaics technology

only, and are not connected to a utility grid.•

Hybrid systems S hi h h l i d



– Systems which use photovoltaics and someother form of energy, such as diesel generation

or wind.•

Grid-tied systems

– Systems which are connected to a utility grid.

• Water pumping



• PV panel on a water pump in Thailand



• PV powers stock waterpumps in remotelocations in Wyoming . Note that batterystorage is not used inthis case, as the solarenergy is stored in theform of pumped water.

•Communications facilities can be poweredby solar technologies: even in remote,

rugged terrain. Also, if a natural or human-caused disaster disables the utility grid,solar technologies can maintain power to



solar technologies can maintain power tocritical operations

• This exhibit, dubbed"Solar Independence",is a 4-kW system

used for mobileemergency power.• while the workhorse

b tt i th t



batteries that canstore up to 51 kW-hrs

of electricity arehoused in a portabletrailer behind theflag.

• The system is thelargest mobile power

unit ever built

• Smiling childstands infront ofTibetan homethat uses 20



that uses 20W PV panelfor electricity

• PV panel onrooftop ofruralresidence



• Ranching the

Sun project inHawaiigenerates



g175 kW of PVpower and 50kW of windpower fromthe fiveBergey 10 kW

wind turbines

• A fleet of smallturbines;



• PV panels in the

foreground

• PV / diesel hybrid

power system - 12kW PV array, 20 kWdiesel genset



• This system servesas the master sitefor the "top gun"Tactical Air CombatTraining System(TACTS) on the U.S.Navy's FallonRange.

• There are two types of Grid Tied PV systems

Two-MeterNet Meter





• At the moment, the boom in PV installations

is led by government subsidies, which willpay above market rates for energy generatedusing renewable energies



using renewable energies

• In areas where subsidies are available, it is

more economical to sell all the generatedelectricity to the grid, and buy backinexpensive energy to run the house

• National Centerfor Appropriate



pp pTechnology

Headquarters

•

Theworld'slargestresidenti



residential PV

project

• These systems are designed to generate revenue,currently subsidized by government incentivesand will have a one directional meter to sell theirelectricity onto the grid.



14.2 MW PV

Industrial Roof top solar array

• Grid-tied systems use the grid as storage,reducing the demands of fossil fuelled

generators when the sun is shining.



• Large scale storage can allow solar to act like adispatchable power source

• Pumped water storage is one of the mostproven methods of energy storage to date





• The appliances and devices (TV's, computers,lights, water pumps etc.) that consume electricalpower are called loads.

• Important : examine your power consumption

and reduce your power needs as much aspossible.

• Make a list of the appliances and/or loads you



pp / yare going to run from your solar electric system.

• Find out how much power each item consumeswhile operating. – Most appliances have a label on the back which lists the

Wattage.

– Specification sheets, local appliance dealers, and the

product manufacturers are other sources ofinformation.

• Calculate your AC loads (and DC ifnecessary)

• List all AC loads, wattage and hours of useper week (Hrs/Wk).

Multiply Watts by Hrs/Wk to get Watt



• Multiply Watts by Hrs/Wk to get Watt-hours per week (WH/Wk).

• Add all the watt hours per week todetermine AC Watt Hours Per Week.

• Divide by 1000 to get kW-hrs/week

• Decide how much storage you would like yourbattery bank to provide (you may need 0 if grid tied) – expressed as "days of autonomy" because it is based on the

number of days you expect your system to provide powerwithout receiving an input charge from the solar panels or thegrid.

• Also consider usage pattern and critical nature ofyour application.



your application.• If you are installing a system for a weekend home,

you might want to consider a larger battery bankbecause your system will have all week to charge andstore energy.

• Alternatively, if you are adding a solar panel array asa supplement to a generator based system, yourbattery bank can be slightly undersized since the

generator can be operated in needed for recharging.

• Once you have determined your storagecapacity, you are ready to consider thefollowing key parameters:

– Amp hours, temperature multiplier, batterysize and number

• To get Amp hours you need:



g p y1. daily Amp hours

2. number of days of storage capacity( typically 5 days no input )

– 1 x 2 = A-hrs needed

– Note: For grid tied – inverter losses

Temp

o

F

80 F

70 F

60 F

Temp oC

26.7 C21.2 C

15.6 C

Multiplier

1.001.04

1.11



50 F

40 F

30 F

20 F

10.0 C

4.4 C-1.1 C

-6.7 C

1.19

1.301.40

1.59

Select the closest multiplier for the average ambient winter temperature

your batteries will experience.

• Determine the discharge limit for the batteries( between 0.2 - 0.8 )

– Deep-cycle lead acid batteries should never becompletely discharged, an acceptable dischargeaverage is 50% or a discharge limit of 0.5

Di id A h / k b di h li it d



• Divide A-hrs/week by discharge limit andmultiply by ―temperature multiplier‖

• Then determine A-hrs of battery and # ofbatteries needed - Round off to the nexthighest number. – This is the number of batteries wired in parallel

needed.

• Divide system voltage ( typically 12, 24 or48, controlled by inverter selected ) by

battery voltage. – This is the number of batteries wired in series

needed.



needed.

• Multiply the number of batteries in parallelby the number in series –

• This is the total number of batteriesneeded.





• On any given day the solar radiation variescontinuously from sunup to sundown anddepends on cloud cover, sun position andcontent and turbidity of the atmosphere.

• The maximum irradiance is available at solar

noon which is defined as the midpoint, intime, between sunrise and sunset.



• Insolation (now commonly referred as

irradiation) differs from irradiance because ofthe inclusion of time. Insolation is theamount of solar energy received on a givenarea over time measured in kilowatt-hoursper square meter squared (kW-hrs/m2) - this

value is equivalent to "peak sun hours".

• Peak sun hours is defined asthe equivalent number ofhours per day, with solar

irradiance equaling 1,000W/m2, that gives the sameenergy received fromsunrise to sundown.

• Peak sun hours only make



• Peak sun hours only makesense because PV panel

power output is rated with aradiation level of1,000W/m2.

• Many tables of solar dataare often presented as anaverage daily value of peak

sun hours (kW-hrs/m2

) foreach month.

• Determine total A-hrs/day from load analysisand increase by 20% for battery losses thendivide by the area's peak sun hours to get

total Amps needed for array• Then divide your Amps by the Peak Amps

produced by your solar moduleYou can determine peak amperage if you divide the



– You can determine peak amperage if you divide themodule's wattage by the peak power point voltage

• Determine the number of modules in eachseries string needed to supply necessary DCbattery Voltage

• Then multiply the number (for A and for V)together to get the amount of power you

need – P=IV [W]=[A]x[V]

• Charge controllers are included in most PVsystems to protect the batteries from overchargeand/or excessive discharge.

• The minimum function of the controller is todisconnect the array when the battery is fully



y y ycharged and keep the battery fully charged

without damage.• The charging routine is not the same for all

batteries: a charge controller designed for lead-acid batteries should not be used to control NiCdbatteries.

• Size by determining total Amp max for yourarray

• Selecting the correct size and type ofwire will enhance the performance andreliability of your PV system.

• The size of the wire must be largeenough to carry the maximum currentexpected without undue voltage losses.

• All wire has a certain amount of



resistance to the flow of current.•

This resistance causes a drop in thevoltage from the source to the load.Voltage drops cause inefficiencies,especially in low voltage systems ( 12V orless ).

• See wire size charts here:

www.solarexpert.com/Photowiring.html

• For AC grid-tied systemsyou do not need a battery orcharge controller if you donot need back up power –

just the inverter.• The Inverter changes the DC



gcurrent stored in the

batteries or directly fromyour PV into usable ACcurrent. – To size increase the Watts

expected to be used by your

AC loads runningsimultaneously by 20%

• Steven J. Strong andWilliam G. Scheller, The

Solar Electric House:Energy for theEnvironmentally-Responsive, Energy-



Responsive, Energy Independent Home, by

Chelsea Green Pub Co;2nd edition, 1994.

• This book will help withthe initial design and

contacting a certifiedinstaller.

• If you want to doeverything yourself alsoconsider these resources: – Richard J. Komp, and John

Perlin, PracticalPhotovoltaics: Electricityfrom Solar Cells Aatec Pub



from Solar Cells, Aatec Pub.,3.1 edition, 2002. (A

layman’s treatment). – Roger Messenger and Jerry

Ventre, PhotovoltaicSystems Engineering, CRCPress, 1999.

(Comprehensive specializedengineering of PV systems).

• Photovoltaics: Design &Installation Manual by SEISolar Energy International, 2004

• A manual on how to design,install and maintain aphotovoltaic (PV) system.

• This manual offers an overview ofh l l d



photovoltaic electricity, and adetailed description of PV systemcomponents, including PVmodules, batteries, controllersand inverters. Electrical loads arealso addressed, including lightingsystems, refrigeration, waterpumping, tools and appliances.



• This is the second in a series of presentationscreated for the solar energy community to

assist in the dissemination of informationabout solar photovoltaics.

• This work was supported from a grant fromthe Pennsylvania State System of Higher



the Pennsylvania State System of HigherEducation.

• The author would like to acknowledgeassistance in creation of this presentation fromHeather Zielonka, Scott Horengic and JenniferRockage.



Application Flexibilityof Solar Electricity


Distributed Energy Source◦ Roof Retrofits

◦ BIPV◦ Tiles, Shingles

◦ Facades



◦ Parking and Sound Barriers

Centralized PV PV Systems for Transportation

◦ Cars, Boats, and Bikes

Located near the consumer in order to eliminatetransmission losses (which can be higher than50% on some antiquated grids).

Panels can be placed on roofs, built into roofs,



p , ,building facades, carports, highway sound

barriers, etc. Any surface which is exposed to sunlight is fair

game.

This house on right ismodular constructionand includes 6 kW ofPV modules.

Below retrofits toschools



The design and integration of PV technologyinto the building envelope, usually replacing

conventional building materials:

• vertical facades,

• replacing view glassSolar modules come in a

variety of shapes,



replacing view glass,

• spandrel glass,

• or other facade material;

• into semitransparent skylight systems;

• into roofing systems,

• replacing traditional roofing materials;

• into shading "eyebrows" over windows;

• or other building envelope systems.

variety of shapes,colors, sizes and canbe integrated into anytype of architecture.

• Advantages

– Aesthetics

– Reduce the cost of thesystem (e.g. roof,awning, etc.)

Specifically designedfor PV.

This energy efficienthome in Oxfordgenerates more



electricity than it

uses.◦ Surplus electricity is

sold to the local utilitycompany and to power

an electric car

The PV shingle shownhere won Popular ScienceMagazine's grand awardfor What's new inEnvironmental

Technology. PV shingles can replacecommon roofingshingles.

PV shingles look muchlike ordinary roofing



like ordinary roofingshingles, but they

generate electricity. They were laid out and

nailed to the roof usingthe same methods as areused to lay conventionalshingles.

Like their non-PVcounterparts, theseshingles overlapproviding for watershedding capability.

Bill Ball, president of the Stellar Sun Shopin Little Rock, Arkansas, is clearly

comfortable with PV shingles on therooftop of his shop.



The flexibleshingles, rated at 17watts each, aremanufactured by

United SolarSystems Corp.

Thin film amorphoussilicon solar cells

which can bedeposited on plasticor steel (right) are

l il



extremely versatile

The panels on theright can be stuckdirectly to standardstanding seam metal

roofs

Buildings with standing-seammetal roofs can use solarmodule material referred to as"thin films" that can be rolledout inside the standing seams.



Roof-integrated PV

panels were installed on

the south-facing roof of

the BigHorn HomeImprovement Center in

Silverthorne, Colorado.

These panels have amorphous silicondeposited on glass and are semi-transparent – they can be used both as

tinted glass but also as a project screenbackdrop.



The Boston Edison―Impact 2000‖ home

incorporated◦ a 4-kWp utility-

connectedphotovoltaic array inh i i l



the original

construction◦ all electric

appliances,◦ solar electric water

heating, and◦ both passive and

active space heating.

Large rooftopsare ideal

locations for PV



PowerGuard slopedtiles, 150 Wp per tile,

shown on customer'sroof

Lowe’s executivesdetermined that clean,reliable distributed solargeneration offers manybenefits both to its retailoperations and to thesurrounding community.

Lowe’s is extremelycommitted to reducinggreenhouse gasemissions while



emissions, whilepromoting and selling

energy efficient homeimprovement products. Lowe’s sought a cost-

effective solution toreduce the operatingcosts associated withproviding reliableelectricity supply at its

West Hills store.

Wal-Mart McKinney

Location: TXOperator: Wal-MartConfiguration: 59 kWpPVOperation: 2005System supplier: RWE



System supplier: RWESchott

This pilot installationfor Wal-Mart has5,500sqft of crystallineand thin film PV in five

separate locations.

Hawaii’s Mauna LaniBay Hotel

This sprawling hotelhad acres of roofspace, making it theperfect host for aphotovoltaicsystem.

They installed aPowerGuard (R)

f l



( )system of insulating

PV roofing tiles thatcovers 10,000 ft2 and generates75kWs.

The hotel will bespared hundredsand thousands of

dollars in utilitybills.

The building’s mostadvanced feature isthe photovoltaic skin,a system that usesthin-film PV panels toreplace traditionalglass claddingmaterial.

The PV curtain wall



The PV curtain wallextends from the35th to the 48th flooron the south and eastwalls of the building,making it a highlyvisible part of themidtown New York

skyline.



ScheideggerBuilding with

photovoltaicfacade nearBern inSwitzerland.



Courtesy ofAtlantis SolarSysteme AG

Austria Swtizerland

Building Facades



73 kW system and generates ~ 55 000 kW-hrs of electricity per year in Sunderland, UK.

This 3500 m2 solar office building at theDoxford International Business Park nearSunderland in the UK incorporates 646 m2 of photovoltaic modules.



Solar cells can be made indifferent colors

Glass façades on office

buildings, winter gardens orsunroofs on automobileswill become energysuppliers with thetransparent solar cell



transparent solar cell.

The use of the transparent,dark blue solar cells allowsa beautiful play of light andshadow.

The standard product lets atenth of the light passthrough and has a 10 %efficiency rate.

The 20kW cube



The 20kW cube

stands 135 feettall on top of theDiscovery ScienceCenter in SantaAna







Prescott Airport

Location: AZOperator: Arizona PublicService

Configuration: 1,450 kWp

SGS Solar

Location: AZOperator: Tucson ElectricPower Co

Configuration: 3,200 kWp

Advantages

◦ Economy of scale

◦ Single location for maintenance◦ You can put a fence around it

Disadvantages



◦ All eggs in one basket – natural/terrorist disaster

◦ Transmission losses

◦ Land cost

Hysperia PV Power Station.



These 20-kW SolarSystems dishes dwarf

visitors in Alice Springs,Australia. The concentrators use

an array of mirrors tofocus sunlight ontoh h ff l



ghigh-efficiency solar

cells. Four supports hold the

cells in front of themirrors

The supports also

supply cooling waterand electricalconnections

In hybrid

energy systems

more than a

single source

of energy



supplies the

electricity.

Wind and Solar

compliment

one another

Unique totheproject istheintroduction of anewgeneration ofhybridpower



powerprocessingequipment.

This new115 kWPV/hybridenergy

system

• Located among the rolling hills of Northern California, these solarelectric sunflowers elegantly combine art and technology.E h fl d 1 kW f l l i i d i f f



• Each sunflower produces 1 kW of solar electricity and consists of four

240 Wp polycrystalline PV modules custom made with yellow back-skins and yellow frames.• The custom PV modules are attached to a two-axis-tracking

mechanism and mounted on custom-painted green poles producing afield of 36 kW solar electric sunflowers.

• The sunflowers wake up each day and follow the sun's path fromsunrise to sunset, increasing solar harvest over a fixed array by 25%.

• This system produces enough energy for approximately eight to ten

homes and was an ideal engineering solution for the steep hillsidesite.



PV will be most applicable totransportationif it can be stored

BatteriesElectrolysis of water to produce hydrogen

Car #195, CalStateUniversity/LongBeach, at the start

line of Sunrayce1995



Sunrayce is a biennial solar-powered car

race for colleges and universities inNorth America. Students design, buildand then race their car 1,250 km fromIndianapolis, IN, to Colo. Springs, CO.

From July 15th tothe 25th, 2001,2300 miles ofsolar ―raycing‖

challenged teamsfrom around theworld.

High tech andhigh efficiencysolar cars crossed



the Great Plains,climbed the RockyMountains, anddashed over theGreat AmericanDesert to thefinish line inSouthernCalifornia

This solarpowered race carwas built in

Philadelphia,Pennsylvania, bystudents atDrexel Universityduring the 1989-



during the 1989

1990 school year. Car #43,

University ofMissouri/Columbia, on the

road in Sunrayce1995

Electric or hybridelectric vehicles canrecharge theirbatteries at PV powerstations such as thisone at the Universityof S. Florida.



Covered parking inCalifornia providesshading and makeselectricity

Solar Panels on Sound

Barriers

This solar power, in

proposed system, could give

EVs essentially unlimited

range on freeways that

supply in-transit power.



Highway sound barrier

above planted slope, with

integrated PV panels, in

Austria.

An upcoming project in

Holland will use

PVs, installed on railway

right-of-way, to power an

entire electric rail system.

Team #2, from GreenMountain High Schoolin Lakewood, CO wonthe women's divisionof the solar bike race

• These golf carts arepowered by photovoltaicmodules from Shell Solar



300-watt system powerson-boardcommunications andlighting, an application

that has been in use for15 years



On board, the solar modulescreate a fascinating play of light,shadow, and transparency.

The largest solar boat in the world

began operating on the AlsterRiver in Hamburg on May 23,2000.

This boat can hold up to 120passengers for excursions and chartertours



tours

The boat can hold up to

120 passengers for

excursions and charter

tours. The "Alster Sun"reaches a speed of 5

kmph just from solarpower.If it has to go faster, therest of the energy comesfrom batteries.



Catamaran ―Sol 10‖:direct currentpowering from 550to 1600 watts.

No drivers license

needed, unsinkable,and easy to use.



This is the third in a series of presentationscreated for the solar energy community to

assist in the dissemination of informationabout solar photovoltaics. This work was supported from a grant from

the Pennsylvania State System of Higher



Education. The author would like to acknowledge

assistance in collecting information for thispresentation from Heather Zielonka.



PV Economics Basics• Solar Photovoltaic Cells convert sunlight directly into

electricity

•

They are sold on a $/Wp basis or $/power• Wp is the power in Watts for Peak sun hours -- the

equivalent number of hours per day, with solar irradianceequaling 1,000 W/m2, that gives the same energy receivedf i t d



from sunrise to sundown.

• To convert power to energy simply multiply by the amountof time that the cell is illuminated – W * hr = 1 W-hr

• Electricity (energy) is normally billed $/kW-hr

PV Economics Terms• kW = kilowatt = 1 000 Watts

• MW = Megawatt = 1 000 000 Watts

• kW-h/kW/ * – * year or month or day

– Amount of power predicted to be produced from a 1 kWsolar panel in the desired location



p

• Payback = minimum time it takes to recoverinvestment costs.

Economics of a Solar Electric Home

• A typical American uses ~11,000 kW-hrs/year

• A well-designed U.S. home needs 4kW-5kW of PVto provide for its energy needs averaged throughoutthe year



–

Depends on location (solar flux) – Energy use of home

– Because calculating on /Wp basis you do not need to worry about efficiency

How much for a

Solar Electric House?• The 2nd presentation discussed the components of a

grid-tied solar home system

• The price tag for the complete installed systemincluding all labor as of 2010 is between $5/Wp to$10/Wp



•

For a 4kW system: – 4000Wp x $5/Wp = $20,000

– 4000Wp x $10/Wp = $40,000

Financing PV• For new homes a PV system can be folded

into the mortgage – long term low interest

loan• For retrofits of existing homes PV can be

economic with:



–

Financial assistance through grants, subsidies, orother incentives

– High costs of electricity in your area

– Green power purchase agreements

– Off-grid Applications

PV Incentives• One stop shop for financial incentives is www.dsireusa.org/

• The Database of State Incentives for Renewable Energy (DSIRE) is acomprehensive source of information on state, local, utility, and federal

incentives that promote renewable energy.• Lists includes:

– Corporate Tax Incentives

– Direct Equipment Sales

G t P

http://www.dsireusa.org/





– Grant Programs

– Leasing/Lease Purchase Programs

– Loan Programs

– Personal Income Tax Incentives

– Production Incentives

– Property Tax Incentives

– Rebate Programs

– Sales Tax Incentives

Feed-In Tariff• Solar FIT rates for Ontario:

• Rooftop

– Less than 10 kW - 80.2 ¢/kWh

– 10 - 100 kW - 71.2 ¢/kWh

– 100-500 kW - 63.5 ¢/kWh



– Greater than 500 kW - 53.9 ¢/kWh

• Ground Systems

– Less than 10 MW - 44.2 ¢/kWh

• You are guaranteed these rates for 20 years.

Where PV makes

Economic Sense Now

• Remote sites that are too far from power

• Or where the power is too unreliable for a givenapplication (e.g. internet server)

– Costs for power lines range from $8000 to $75,000



per mile. – As a general rule, if you are more than ½ mile from a

line, solar is probably the best alternative.



Portable Radio Station



The Developing World



Stand Alone Systems



Coast Guard Stations

and Aircraft



Bus Stops and

Emergency Phones



Solar in



Space

Parking Lights



Running Trails and

Lighthouses



Solar powered

monasteries



!

When will PV make economicsense for me?



Economy of Scale$3.12/Wp to $3.56/Wp



0 subsidiesGrid-tied

Market

Module Costs



Component Costs



Industry-Developed PV Roadmap



World PV Module Production (MW) Increases

00

00

Rest of worldEuropeJapanU.S.

287.7

390.5

512.200

World PV installations in 2004 rose to 930MW --

growth of 62 % Consolidated world production of

PV now 1.15 GW+



198819891990199119921993199419951996199719981999200020010

00

00

33.640.246.555.4 57.960.169.477.688.6125.8

154.9201.3

2002Source: PV News, March 2003

World PV Module Production (MW) Increases



World PV Module Installation (MW) Increases



So Why Can’t We Do It?



PV System vs. Electricity Costs

20

24

28

32

36

40

44

r a t e d E l e c t r i c i t y ( c e n t s / k W h )

Japanese Retail Rate

German Retail RateCapactiy Factor = 0.25

(South West U.S)

Capacity Factor = 0.2

(U.S. Average)



0

4

8

12

16

$0.00$1.00$2.00$3.00$4.00$5.00$6.00$7.00$8.00$9.00

Installed PV System Cost ($/Wp)

C o s t o f G e n e

rPennsylvania Retail Rate

Additional Assumptions:

System Lifetime = 20 years

Real Interest Rate = 6%

O&M = 0.1 cent per kWh

California Retail Rate

New Technology

Could Play a Role• Heterojunction with Intrinsic

Thin-layer

• Sanyo

• 18.5%

SANYO, plans to continue to



grow its unique solar business,aiming to reach a productionscale of approximately 2GW forHIT solar cells by 2020.

New Technology

Could Play a Role



Built-in Incentives

$/W Value

Material avoided

by BIPV Installation

Material

Credit

$1/sq-ft

$5/sq-ft

$10/sq-ft

Asphalt Shingle roof,

monolithic glazing

Laminated glass w/coatings

metal roofing/cladding

Roofing slates, clay tile,

$0.10/W

$0.50/W

$1/W

Building Material Replacement Value



$10/sq ft

$20/sq-ft

Roofing slates, clay tile,

high performance coatings

Stainless steel,

photochromic glass

$1/W

$2/W

Utilizing Financial Incentives



Subsidies for Fossil Fuel• Fossil fuels and nuclear energy receive 90% of the

government money, (with PV receiving <3%).

• Hidden costs that we all subsidize for the energy industry which include:

– Air pollution leads to the death of 120,000 Americansevery year and costs $40 billion in health care annually. /



– Hidden Subsidies – pollution, global climate change, war•Military (U.S. military spends billions/yr just defending the oilsupplies in the Persian Gulf).

The Question of Energy

Unemployment

• If we switch to solar what about all the fossil fuel

jobs?• It is estimated that solar on average creates 26 total

jobs per MW in the US

– Coal only produces 8.7 total jobs per MW in the US.



y p j p

• Ontario predicts 31.6 jobs/MW for solar PV

! PV: Net Job Producer

Jobs created with every

million dollars spenton:

– oil and gas

exploration: 1.5

– on coal mining: 4.4



g

– on producing solar

water heaters: 14

– on photovoltaicpanels: 17

Jobs

Coal vs. Solar

– Coal only employs 80,000

– As of 2009 there are 46,000jobs. From 2009 to 2010,approximately 17, 000 newjobs were created as a result



of the increased demandthrough the installation ofsolar systems.

People Want Solar• The Program on International Policy Attitudes

found that the American public wants the federal

budget for renewable energy research like solar PVto increase by 1090 %.



Solar Photovoltaicsis the Future



Acknowledgements

• This is the fourth in a series of presentations createdfor the solar energy community to assist in thedissemination of information about solarphotovoltaic cells.

• This work was supported from a grant from thePennsylvania State System of Higher Education.

•



The author would like to acknowledge assistance incollecting information for this presentation fromHeather Zielonka and Michael Pathak.

The Environmental Impact

of Solar Photovoltaic Cells The Energy Challenge and How to Solve it with

Solar Cells



Original Presentation by J. M. Pearce,

2006

Updated by A. Shahed and J.M.

Pearce, 2010

Email: [email protected]

The Energy Challenge

• World population is expanding rapidly and will likelyreach over ~9 billion before stabilizing

• Energy use is directly proportional to standard ofliving

• Energy demand is skyrocketing

•



Standard methods of producing energy have alimited supply and have unacceptable impacts on theenvironment

The Population Explosion

• UN Projections to 2050 place

population level at ~9 billion



• Exponential population

rise with advancements in

energy production and

technology

Human Development Index



HDI: longevity (life expectancy), educational attainment, andstandard of living (gross domestic product per capita)

World Energy Consumption

By Fuel Type1980-2030



Future Energy Needs



How to Produce Electricity?

• Conversion of mechanical energy into

electricity (hydro, wind)

• Conversion of chemical energy into electricity

( l l )



(coal, natural gas, etc.)

• Conversion of nuclear energy into electricity

• Conversion of photon energy into electricity

Energy Sources – Global Primary

Energy Supplies



Peak Oil (U.S. Production Peak – 1970)



• Measured in 1000 barrels per day

• World energy reserves and peak under considerable debate…

U.S. and World Natural Gas Supplies



World and U.S. Coal Reserves



Even if we discover more fossilfuels….

W b h l d h



We can not burn what we already have.

Recent Global Atmospheric

Carbon Dioxide Concentration

From 1860 to 1990the industrial nations

released 185 billiontons of carbon into theatmosphere from

burning fossil fuels.

Toda 6 + billion tons



Today 6 + billion tonsof CO2 are releasedinto the air every year!

380ppm

v

Historical CO2 Concentration



Global Warming



This is the Consensus View

The Intergovernmental Panel on Climate Change

In 1995, over 2,500 scientists representing more than 80countries analyzed over 20,000 articles from the relevantliterature

It is now quite clear that certain gases such as carbondi id l i l l i d i i h h' li



It is now quite clear that certain gases, such as carbondioxide, play a crucial role in determining the earth's climate by preventing heat from escaping the atmosphere.

Temp. gains up to 14oF!!!

• In 50 years Earth will be ~3-5 degrees hotter

• Up to 14 ft rise in global sea levels. (3ft floods Florida & NYC)

• Weather patterns disrupted: droughts in some places andt ti l i i th di t bi i lt l d ti

Potential Consequences :

Manageable to Catastrophic



torrential rain in others – disturbing agricultural production.

• EPA estimates that by 2050 the southern boundary of forestecosystems could move northward by 600km.

• Massive forest death and species extinction.

Global Temperatures for Various Growth

Scenarios



Solar Photovoltaic Energy as aSolution to the Energy Challenge



Environmental Benefits

• During use PV produces no :atmospheric emissions

radioactive waste

•

During use PV produces no greenhouse gases so it will help offset CO2 emissions and global climatedestabilization

– PV does have an embodied energy andembodied CO2 emissions



• PV curtails air pollution, which produces acid rain,soil damage, and human respiratory ailments.

Environmental Benefit of a 4 kWp

PV system

This solar energy array would prevent:

• 2.4 tons of coal from being burned

• 6.2 tons of CO2 = decreasing the greenhouse effect

• over 3,600 gallons of water from being used

• ~34 pounds each of NOx and SO2 from polluting the

atmosphere



atmosphere• 1.8 pounds of particulates from causing a health hazard

(and no nuclear waste)

EACH YEAR - FOR 30+ YEARS!

Busting PV Related Myths



Myth 1:

PV use more energy to make than

they produce over their lifetime

For cells in production now the

energy payback is between 6



energy payback is between 6months and 5 years!

Net Energy for 1kW-hr invested in PV Roof in Detroit, MI

N

etEnergy(kW-hrs)

3

4

5

6

7

8

9

10

11

12

13

14

15 c-Si Low

c-Si High

p-Si Low

p-Si Higha-Si Low

a-Si High

14+breeds



Time (Years)

0 1 2 3 4 5 6 7 8 9 101112131415161718192021222324252627282930

-1

0

1

2

3

Built in

PV

Integrated into the roof



Integrated into the roof.

Integrated into an awning over a backporch California, generates electricity

while shading the family's outdoor

activities

Additional BIPV



Solar Shingles

• Various styles of PV

shingles are available



• Solar Shingles are rated at 17W• Lightweight and Flexible

• 20 year warranty on power output

Aesthetic Applications

• A Canadian team fashioned a

window with solar electric

cells and a motif of autumn

leaves

• Semi-Transparent and

Transparent PV cells



Transparent PV cells

• Building Integration

Myth 2: There is Not Enough Land

• Each day the sun casts moreenergy on the earth than all

people would consume in 27years

• The entire world’s energy needs



• The entire world s energy needscould be provided by ½ the areaof the Gobi desert covered withPV.

What about in the USA?

• 100 miles by 100 miles in Nevada would provide

the equivalent of the entire US electrical demand

• Distributed (to sites with less sun) it would take

l th 25% f th d b US d



less than 25% of the area covered by US roads.

Solar Cell Land Area Requirements for

the World’s Energy with Solar PV



6 Boxes at 3.3TW Each

Solar Cells• PV cells can provide a

substantial portion of

energy needs.

• For the US, only 3% of

the land area would be

required using 1999

efficiency measures



• PV technology is improving

continuously

• Greater efficiencies

• Lower costs

Myth 3: We do not have Enough Raw Materials

• Si - 2nd most abundant element in

Earth’s crust

• The amorphous silicon cellsmanufactured from one ton of sandcould produce as much electricity as



could produce as much electricity as burning 500,000 tons of coal.

• Philadelphia home

Solar Photovoltaicsis the Future

• Photovoltaic technology provides an effective means ofsupplying clean energy

• The technology is ready to use and proven

• There is extensive research and development providing

opportunities for higher efficiencies and new market players



opportunities for higher efficiencies and new market players

• Abundant silicon resource

• Will play a major role in tackling our increasing energy

needs and curbing our GGH emissions

Acknowledgements

• This is the fifth in a series of presentations createdfor the solar energy community to assist in thedissemination of information about solarphotovoltaics.

• This work was supported from a grant from thePennsylvania State System of Higher Education.

•

The author would like to acknowledge assistance incollecting information for this presentation from



The author would like to acknowledge assistance incollecting information for this presentation fromHeather Zielonka.

The full series can be found here:

http://www.appropedia.org/Solar_Photovoltaic_Open _Lectures

Prof. Le Chi Hiep

Director Program on Rene able Energ &

http://www.appropedia.org/Solar_Photovoltaic_Open_Lectures







Director, Program on Renewable Energy &Energy Conservation (VNU-HCM)

Berlin, October 9, 2009

519

1. GENERAL INFORMATION

2. SOLAR ENERGY and ITS APPLICATIONSIN VIETNAM

3 SOLAR PHOTOVOLTAICS IN VIETNAM



3. SOLAR PHOTOVOLTAICS IN VIETNAM

4. DISCUSSION & CONCLUSION

520



521

VIETNAM

Population – 85.8



pmillions (2009)

Area – 332,000km 2

Capital – Hanoi

522

Location Latitude Longitude

Ha Giang 22o54’N

105oE

Hanoi 21o03’N 105o54’E

Hue 16o29’N 107o36’E

Da nang 16o03’N 108o12’E

Qui Nhon 13o47’N 109o15’E



Qui Nhon 13o47 N 109o15 E

HoChiMinh City 10o45’N 106o41’E

Phu Quoc 10o12’N 103o58’E

Ca Mau 9o11’N 105o09’E

523

Installed electric generating capacity (9GW ,2004).

In 2004, Vietnam generated 40.1 billion

kilowatthours (Bkwh) of total electricity, ofwhich 52 percent was supplied byconventional thermal sources and 48percent came from hydroelectric sources.

Electricity demand has increased steadily in



524

Electricity demand has increased steadily inVietnam during the last decade, but thecountry’s per capita energy consumptionremains one of the lowest in Asia.



525



526



527



528

North: 1800 – 2100 hours of sunshine a year,

on average. South: 2000 – 2600 hours of sunshine a year,

on average.



529

January 2.2

February 1.6

March 1.4

April 2.7

May 5.3

June 5.2July 5 9



July 5.9

August 5.3

September 5.4

October 5.3

November 4.2

December 3.5530

January 4.4

February 5.1

March 3.4

April 6.9

May 8.3

June 7.9July 8 3



July 8.3

August 6.7

September 5.8

October 4.7

November 4.0

December 3.6531

January 7.9

February 8.8

March 8.8

April 7.7

May 6.3

June 5.7July 5 8



July 5.8

August 5.6

September 5.4

October 5.9

November 6.7

December 7.2532

January 2.24

February 2.40

March 2.53

April 3.46

May 5.23

June 5.31July 5 59

Annual Mean Solar

Radiation:

3.93kWh/m2 /day



July 5.59

August 5.10

September 4.79

October 4.18

November 3.45

December 2.97533

January 3.5

February 4.3

March 5.2

April 5.8

May 6.4

June 5.9July 6.5

Annual Mean Solar

Radiation:

4.85kWh/m2 /day



July 6.5

August 5.7

September 5.2

October 4.2

November 3.1

December 2.5534

January 4.66

February 5.29

March 5.69

April 5.91

May 5.90

June 5.66July 5.66

Annual Mean Solar

Radiation:

5.15kWh/m2 /day



July 5.66

August 5.51

September 4.92

October 4.42

November 4.04

December 4.15535

January 5.1

February 6.3

March 6.6

April 5.7

May 5.0

June 4.9July 5.1

Annual Mean Solar

Radiation:

5.2kWh/m2 /day



July 5.1

August 5.0

September 4.8

October 4.5

November 4.3

December 4.6536

400

500

600

700

800

900

1000

"Average Solar Radia tion" in "10 Minute Summ ary" of "Current" data base a t "28-004-2005"

W

/ m

2



537

0

100

200

300

9A M28 Thu A pr 2005

12P M 3P M

40 0

50 0

60 0

70 0

80 0

90 0

"A verage S ola r R adia t ion" in "10 M inute Summ ary" of "C urrent" da tabas e a t " 28-06-2005"

W

/ m

2



538

0

10 0

20 0

30 0

9A M28 Tue Jun 2005

12P M 3P M

09:00 323 10:40 255 12:20 886

09:10 494 10:50 249 12:30 725

09:20 489 11:00 650 12:40 791

09:30 638 11:10 1064 12:50 934

09:40 658 11:20 464 13:00 797

09:50 650 11:30 871 13:10 785



10:00 508 11:40 1044 13:20 773

10:10 659 11:50 1031 13:30 838

10:20 503 12:00 497 13:40 453

10:30 629 12:10 957 13:50 474 539

14:00 685 15:40 387 17:20 139

14:10 611 15:50 344 17:30 98

14:20 344 16:00 385 17:40 66

14:30 392 16:10 381 17:50 17

14:40 557 16:20 306 18:00 7

14:50 627 16:30 135 18:10 7



15:00 649 16:40 139 18:20 0

15:10 643 16:50 140 18:30 0

15:20 583 17:00 182 18:40 0

15:30 469 17:10 168 18:50 0 540

Two main applications:

- Solar hot water- Solar photovoltaics



541

Solar hot water has been developed since1990. At the early period, it was very difficultto get the attention of the community.

But, since around 1998, the number ofinstalled solar hot water systems hasgradually increased, especially in Ho Chi Minh

city.



542

Currently, solar hot water has become one ofproducts at high competition. That leads tothe increase of the number of companies

doing business in this field.

But, there is so far no full industry referringto solar hot water. The main components of

the system (solar collector,…) are imported,mainly from China



mainly from China.

543



544

Researches referring to solar photovoltaicshave been done from 1975.



545

1. Monocrystalline silicon

2. Polycrystalline silicon

3. Amorphous silicon



546

Determination of the light- induced degradationrate of the solar cell sensitized N719 onnanocrystalline TiO2 particles

Thermal degradation kinetics of solar cell dyeN719 bound to nanocrystalline TiO2 particles

Fabrication of solar cells based on N719, D520-

dyed nano-crystalline titanium dioxide andinvestigation of their performances



g p

Improvement of the Dye-Sensitized Solar CellOpen-circuit Voltage by Electrolyte Additives and

Cell Treatment with 4-tert-butylpyridine

547

The Relationship between ElectrochemicalImpedance Spectra and Photovoltaic PerformanceCharacteristics during the Light and ThermalAgeing of Dye-Sensitized Solar Cells

Effects of Electrolyte Additives on the Open-circuitVoltage of Dye-Sensitized Solar Cells

Decomposition and degradation of dyes in solarcells under prolonged thermal and light ageing

Dye-sensitized solar cell based on nano-crystallinetitani m dio ide



titanium dioxide

548

Fabrication of solar cells based on titanium dioxideand organometallic dyes

Grid connected systems



549

But, due to many difficulties, the research on

solar photovoltaics is nearly activities ofscientists in laboratory only.

It has not exercised its influence on thesociety.



550

In 1990, the first 300Wp -unit was installedat Can Gio district, Ho Chi Minh city.

From 1994, solar photovoltaics has beenexpanded in the whole country.

Currently, total installed photovoltaics isd 2MW



around 2MWp .

551

Installed locations:

- Rural areas (PV units were installed at

more than 3000 houses, there have beenaround 8500 families who can reach PVindirectly by using batteries charged fromstations).

- Cultural centers.M di l t



- Medical centers.

- Telecommunication units.

- Traveling boats.552

- Post offices

- Ambulances.

- Public lighting systems.- Traffic lighting systems.

- Battery charging stations.

- Schools.

- Islands.



553

2000Wp



554

Reference:

http://www.soltechvn.com/vn/

810Wp



555

Reference:




556

Reference: http://www.selco-

vietnam.com.vn/index.php?lang=vn

Main features of photovoltaic developmentin Vietnam:

- Most budgets funded by internationalorganizations and several funded bynational agencies (there is so far nearly nopersonal budget invested to set up the

system).- It leads to unstable and unsustainable



development.

- All solar panels are so far imported.

557

Currently, there are only several companiesdoing business on photovoltaics in Vietnam.

SELCO-VIETNAM could be considered as the

biggest company in this field (Installedphotovoltaics – 262kWp )



558

Several related websites for your reference:

1. http://www.selco-vietnam.com.vn/index.php?lang=en

2. http://www.soltechvn.com/vn/

http://www.selco-vietnam.com.vn/index.php?lang=en









559

Recently, a factory producing solar panelshas been installed in Long An – near Ho Chi

Minh City.

By planning, its expected capacity is3MWp/year (first step) and 5MWp/year

(next step).



560



561



562

There is high potential to exploit solar energy

in Vietnam.

There is also high demand on solartechnologies such as solar photovoltaics andsolar hot water.

(Survey : 50,000 families in southern and

central parts of the country need PV systems)



563

Current situation on solar photovoltaics :

1. Solar hot water has been developed well,especially in the southern parts of thecountry. Currently, it is easy to see solar hot

water systems on roofs of many houses inHo Chi Minh city.

2. In the meantime, solar photovoltaics has

been almost done by projects and has beeninstalled dispersedly . There have been onlyf l h i h



a few people who want to invest thesystems by their own budget.

564

Referring to energy policy, currently there are

good enough macroscopic policies androadmap to promote renewable energyactivities in Vietnam.

In June 2009, the draft of the law on energy

conservation

was submitted to theconsideration of the national assembly.Hopefully, the similar draft of the law on

renewable energy could be submitted nextyear .



565

According to the current energy policy,renewable energy including solar energy isexpected to be one of main energy resources

of the country.



566

But, unfortunately, the current applications ofrenewable energy, especially of installed solarphotovoltaics, are still far from our expectedtargets.

Although there have been activities referring tophotovoltaics in Vietnam leading to around2MWp installed, but Vietnam is still nearly a

blank area in terms of photovoltaics.



567



568

YES, we need solar photovoltaics.

But, like everywhere, the first main reason isalways the price .



569

To set up a photovoltaic system, we oftenneed:- Solar panel (Kyocera, SolarWorld,…)

- Control system- Battery (Phoenix,…) - Converter- Others

Current mean price in Vietnam (including



Current mean price in Vietnam (includingsolar panel and main components):8USD/Wp - 10USD/Wp

570

We can see:

- Rural area : very high demand on solar

photovoltaics, but because of high pricepoor people can not dream of getting it.

- Urban area : good electricity supply fromthe national grid. But, if the price is good

enough, solar photovoltaics can also attractthe attention of the people living in theurban area to prevent unexpected power



urban area to prevent unexpected powershortage (It is quite often in Vietnam,especially in the dry season).

571

- Currently, there is general macroscopic energy

policy. But, there are so far no concrete policies

such as tax exemption and financial supports,…

- Solar panels and corresponding parts aremainly imported. There is nearly no industry

f i t l h t lt i i Vi t



572

referring to solar photovoltaics in Vietnam,

except one factory named RED SUN in Long An

– near Ho Chi Minh City.

Price reduction:1. In country fabrication by local andforeign investment.2 Detailed policies focusing on tax



2. Detailed policies focusing on taxexemption, or tax reduction, or financialsupports ….

573



The new projects should be invested bypeople who are intending to do business inthis field.

Under this point of view, the new projectsshould not be decentralized and should beconnected to the grid…..

In order to set up the projects, it must berepeated that we need good financialli i h f ll hi h ld b



policies, hopefully, which would bepublished in the first quarter of the next

year.575

2. Following the success of the first step, thesolar photovoltaics together with its low pricecan get the attention of the community.

After this step, hopefully personal budgetscan be expended to buy solar panels.



576

―SOLAR ROOF should be considered as one ofmain strategic approaches parallel with thenew projects (step 1) to promote photovoltaic

application in Vietnam, especially at NEWURBAN AREAS‖.

There is high potential to develop the notion

SOLAR CITY at NEW URBAN AREAS.



577



578

Nearly blank market and blank area in termsof solar photovoltaics are awaiting you.



579

―Vietnam Needs Investors To Promote Use

Of Renewable Energy Technologies‖.

―The Government of Vietnam has been

exploring more and more the possibilitiesfor investment in the country as they



o est e t t e cou t y as t eyrecognize the toll that electricity shortagesnationwide are causing‖.

580

―The climate of Vietnam is conducive topositive impacts from solar energy use, andthe sun is a priceless commodity in bothsummer and winter. Research information

shows that sunshine is available between1,800 hours and 2,700 hours annually‖.



, , y

581

―Experts agree that solar energy is a greatway to fill off-grid electricity needs,especially in areas of a higher concentrationof pollution where mini-grids are not a

good idea‖.

Sherry Irvin

on behalf of theBascoTec Internet LimitedTechnologie Park 13



Technologie Park 1333100 Paderborn

Germany582

The draft of the decree on the means supportingthe development of renewable energy has just beensubmitted to the government by the ministry ofindustry and commerce (Tuoi Tre daily newspaper,

Sept. 8, 2009).

Main contents of the draft of the decree:- The government commits to exempt those

who invest to develop the electric generation by

renewable energy from taxes.- The price of electricity produced by renewable

energy would be discussed to guarantee the



energy would be discussed to guarantee theinvestor’s rational benefit.

583

It’s time to invest and to do businessreferring to solar photovoltaics in Vietnam.



584

Documents

Chapter 5_PV Systems_April 11- 2011.pdf