EELE408 Photovoltaics Wave – Particle Duality Lecture 03: Characteristics of Sunlight • Light behaves as both a wave and a particle Dr. Todd J. Kaiser – Wave Properties [email protected] • Refraction • Diffraction Department of Electrical and Computer Engineering • Interference c   f Montana State University - Bozeman – Particle Properties c: speed of light n: frequency • Emission and Absorption of Light l: wavelength • Blackbody Radiation E: Energy • Photoelectric effect h: Planck’s Constant E  h

c  3108 m / s h  6.6361034 J  s  4.1361015 eV  s 

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Electromagnetic Spectrum (wavelike) Photon Energy (Particle Like) 107 Wavelength 10 Radio Waves Frequency 8 (meters) 10 (Hz) hc 1.24 1 E  h   EeV  109 10‐1   10 ‐2 10 ROY G. BIV 10 Microwaves 1011 10‐3 Higher 0.77 microns Energy Red 1012 High energy photon for blue light 10‐4 0.6 microns Orange 1013 10‐5 Infrared Yellow 1014 10‐6 Visible Green 0.5 microns 1015 10‐7 Low energy photon for red light Blue Ultraviolet 1016 10‐8 Indigo 1017 10‐9 0.4 microns X Rays 18 Violet ‐10 10 0.38 microns 10 Very low energy photon for infrared 1019 (invisible) 10‐11 Gamma Rays 1020 10‐12 3 4

Photon Flux & Energy Radiant Power Density Density Number of Photons Photon Flux    second Area  W   hc   W  # photons EJ hc Energy Density : Photon Flux Energy per Photon : H    J  H 2   2    EJ    m2      m  sec m photon 

For the same intensity of light shorter wavelengths require fewer photons, since the energy content of each individual photon is greater

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1 Radiant Power Density Spectral Irradiance

• The total power density emitted from a light source Visible

 N H  E d  E i i 0  i0 • The spectral irradiance is multiplied by the wavelength range for which is was measured and summed over the measurement range F(B) F(R) F(O) F(Y) F(G) F(I) F(V) Xenon – (left axis) Sun - (right axis)

Halogen – (left axis) Power Density(log scale) at a 1 2 3 4 5 6 7 Mercury – (left axis) particular wavelength

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Blackbody Radiation

• Ideal absorber and emitters of electromagnetic radiation Solar Radiation – The hotter the body the more radiation emitted – The hotter the body the higher the energy of the spectrum peak • Classical physics unable to explain blackbody radiation – 1900-Max Planck: Quantization of Energy Radiation – 1905-Albert Einstein: Photoelectric Effect

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Planck’s Formula

• A body in thermal equilibrium emits and absorbs radiation at the same rate 8h 3  J  s  E  d   • A body that absorbs all the radiation incident on it is an ideal blackbody 3 • The power per area radiated is given by the Stefan-Boltzmann Law 3   h    m  c exp  1 kT 4 2 8  W      H T W / m    5.670410    m2K 4  Integrate over all energies to get the intensity emitted into a hemisphere

 •This function has a maximum given by Wien’s Displacement Law c 4 H  E d  T 2900 0    4 peak TK •The spectral irradiance of a blackbody radiator is given by Planck’s Law H: intensity of radiation (W/m2) 2hc2 1 Stefan‐Boltzmann Constant s = 5.67x10‐8 (W/(m2K4) E ,T  BB 5 hc e kT 1

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2 Plot of Planck’s Law Sun’s maximum at visible spectrum

Higher Energy

15 10 2900 peak   13 30 T K Temperature of 10 Solar Peak incandescent 11 in visible 25 'T=3500' 10 'T=4000' light bulb spectrum 20 UV Visible IR 'T=4500' 9

'T=5000' 10 Visible

(eyes 12 'T=5500' 15 7 adapted to x10 'T=6000' Temperature of 10 sunlight) the sun 10 5 10 5 3 10 0 1 2 0.5 1.0 1.5 2.0 10 2hc 1 2 3 4 5 6 7 8 9 E   Wavelength (microns) 0.1 1 BB 5 hc Light bulbs are e kT 1 inefficient Wavelength (microns)

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Sun Sunlight in Space

The energy is emitted from the surface of the sun as a sphere of light

The size of the sphere grows as the light moves farther from the sun causing the power Fusion reaction at density to reduce 20,000,000 degrees

Converts H to He 2 2 4Rsun Rsun 2 Photosphere (sun’s Convection heat H D  H 0  H 0  H EarthOrbit 1367W / m 4 D2 D2 surface) at 5760±50 K transfer

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Sunlight at Earth’s Orbit Sunlight at Earth’s Surface

• Atmospheric Effects – Absorption – Scattering – Local Variations • Clouds • Pollution • Weather • Latitude • Season of Year   360(n  2)  2 H EarthEllipticalOrbit 13671 0.033cos W / m   365  • Time of Day n is the day of the year 17 18 17 18

3 Air Mass Clear Sky Absorption100 % input & Scattering  3% Scattered 2% Absorbed Ozone To Space 20-40 km

0.5% Scattered to Space Upper 1% Absorbed Dust Layer 15-35 km 1% Scattered to Earth

18% Absorbed Air 1% Scattered to Space Molecules 4% Scattered to Earth Amount of 8% Absorbed 0-30 km atmosphere the

Water 0.5% Scattered to Space sunlight has to go 6% Absorbed Vapor through to reach 0-3 km 1% Scattered to Earth earth surface

0.5% Scattered to Space varies with Lower latitude, season 1% Absorbed Dust 0-3 km 1% Scattered to Earth and time of day

BOR0606C 70% Direct to 7% Scattered Earth To Earth 20 19

Calculating Air Mass Measuring the Air Mass

• The path length that light takes through the atmosphere normalized to the shortest possible path length L cos   h q 2 1 h S 2  L2  S  x 1 AM       1 cos   cos L L  L  h h 1 h qObject of h  AM  height (L) cos Neglects the curvature of the earth’s atmosphere  x Shadow from object (S) h 1 Curvature AM  compensated cos  0.5057296.07995  1.36364

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Air Mass Values Sun Radiation

• The sun is approximately a blackbody radiator at 6000K • Earth is approximately a blackbody radiator at 300K • Total output of the sun is 4x1026 W Theta 1 Air Mass • Power reaching earth is 1.72x1011 W  ( ) cos • AM0 = 1353 W/m2 0⁰ 1AM1.0 • AM1.5 = 1000 W/m2 Used as Standard 48.2⁰ 1.5 AM1.5 60⁰ 2AM2.0 75⁰>Not Accurate

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4 Earth Temperature Calculation Earth Temperature Calculation 2

Assume the sun at 6000K is a blackbody radiator, then calculate the earth temperature if it radiates all the solar energy it captures as a blackbody. 8 8 rs = 6.955x10 (m) rs = 6.955x10 (m) 9 9 de = 150x10 (m) de = 150x10 (m) 3 3 re = 6.378x10 (m) re = 6.378x10 (m) s = 5.67x10‐8 (W/(m2K4)) s = 5.67x10‐8 (W/(m2K4) rs rs re re de de

Need to calculate the power density at Earths orbit

26 4 2 8 4 7 2 4.510 H  T W / m  5.6710 6000  7.3510 W / m P  4.51026 W  H 4 d 2  H  1590W / m2 e 9 2 2 7 8 2 26 4 15010 P  H 4rs  7.3510 4 6.95510  4.510 W

Power Radiated from the sun Power density at Earths orbit 25 26

Earth Temperature Calculation 3 Earth Temperature Calculation 4

8 8 rs = 6.955x10 (m) rs = 6.955x10 (m) 9 9 de = 150x10 (m) de = 150x10 (m) 3 3 re = 6.378x10 (m) re = 6.378x10 (m) s = 5.67x10‐8 (W/(m2K4) s = 5.67x10‐8 (W/(m2K4) rs rs re re de de

Need to calculate the power captured by the earth Assume the earth is radiating like a blackbody 1 11 11 4 2   2 2 3 11 210 W 4 210 H 1590W / m  P  H re 1590 6.37810  210 W H   T  T     288K 2  3 2 8  4 re  4 6.37810 5.6710 

Power captured by earth Temperature of earth 27 28

Planet Temperatures Green House Effect

• Venus atmosphere is 96% CO2 1/ 4 4 r 2  r 2  r 2  – The carbon dioxide captures the incoming radiation energy and prevents P   T 4 s e   T 4  T  s   T it from radiating S 2 2 e s  2  e 4 de 4 re  4de  • Pre-1860Pre-Industrial Revolution

–CO2 concentration = 280 ppm – Found by analyzing ice core samples from the Arctic and Antarctic Planet Distance to Calculated Average Sun (109) Temperature Temperature • Today

Mercury 57 469K 452K –CO2 concentration = 335 ppm Venus 108 340K 740K – Burning of Fossil Fuel: Trapped Organic Matter (Carbon) – Coal and Oil Earth 150 288K 288K – What took 1 million years to form is used in 1 year. Mars 227 235K 210K

Why the big discrepancy? 29 30

5 Green House Effect 2 Sun Light Attenuation (70%)

• Water absorbs in the band  ~ 4-7 • Rayleigh Scattering by molecules in the atmosphere (~-4 • Carbon Dioxide absorbs at  ~13-19 dependence) • Most energy escapes in the band ~ 7-13 – Significant at short wavelengths • Delicate balance of energy in = energy out • Scattering by aerosols and dust • Absorption by atmospheric gases • Disruption in balance create temperature changes  Climate Changes

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Solar Light Distribution Diffused Light

2500

80 Blackbody Radiation at 6000 K (yellow) Air Mass 0: Outside Earth's Atmosphere (orange) • 2 Components of sunlight •Overcast Day  100% diffuse 2000 Air Mass 1.5: Sun 48.2 degrees from horizon (blue) AMGlobal = AMdirect + AMdiffuse • 20% of maximum possible 60 • 100%=90%+10% at AM0 1500

x10 • Diffuse Increases with 1 2

40 increasing AM 1000 Asorption of energy by water vapor and carbon dioxide

20 500

0 0 0.5 1.0 1.5 2.0 2.5

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Direct and Diffuse Radiation Intensity Based on Air Mass

• The intensity of direct sunlight can be experimentally fit to the equation: AM 0.678 2 I D 1.3670.7 kW / m

•ID: Intensity of direct sunlight • 1.367 Solar Constant Diffused light peaks in the blue • 0.7: 70% of the light reaches the Earth’s surface spectrum (Sky is • 0.678 Experimental fit of the data which accounts for blue) atmospheric variations

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6 Intensity with Altitude Adjustment Estimate of Global Intensity

• The intensity of direct sunlight increases with altitude: • The diffuse radiation is approximately 10% of the direct radiation AM 0.678 2 I D 1.3671 ah 0.7  ahkW / m • The clear day global intensity is estimated as:

• a: empirical fit constant = 0.14 • h: height above sea level in kilometers 2 IG 1.1I D kW / m 

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