SOL 101 › Unit 7

Unit 7 — The Atmosphere: Absorption, Scattering, Air Mass, AM1.5G/AM1.5D

Big idea. On its way down, sunlight is absorbed by specific molecules (ozone, oxygen, water vapour, CO₂) and scattered by air molecules and aerosols. How much depends on the path length through the air — the air mass. The industry standard ground spectrum, AM1.5G, describes sunlight on a tilted surface after passing through 1.5 atmospheres' worth of air and integrates to 1000 W/m².

Learning objectives

Why this matters

Every efficiency on a terrestrial datasheet is measured under the AM1.5G spectrum (IEC 60904-3, equivalent to ASTM G173). Factory flash testers are graded on how well they match it (SOL 401). The water-vapour gaps in the spectrum are also why silicon "wastes" little near-IR and why the photon budget for tandems looks the way it does.

Absorption by molecules

Molecules absorb photons whose energy matches their internal transitions:

AM0 (space) vs AM1.5 Global and Direct from ASTM G173-03. Labels mark the main absorption dips. AM0 integrates to 1348 W/m², AM1.5G to 1000 W/m² and AM1.5D to 900 W/m² over 280–4000 nm (computed).
AM0 (space) vs AM1.5 Global and Direct from ASTM G173-03. Labels mark the main absorption dips. AM0 integrates to 1348 W/m², AM1.5G to 1000 W/m² and AM1.5D to 900 W/m² over 280–4000 nm (computed).

Scattering

Rayleigh scattering by air molecules much smaller than the wavelength scales as \(1/\lambda^4\):

\[\text{scattering} \propto \frac{1}{\lambda^4}\]

so 450 nm blue light is scattered \((700/450)^4 \approx 5.9\) times more than 700 nm red. Scattered blue light fills the sky (blue sky); the direct beam loses blue, so the Sun looks yellow-white at midday and red at sunset when the path is long. Aerosol (Mie) scattering by particles comparable to the wavelength (dust, smoke, haze) is less wavelength-selective and makes the sky whitish.

Scattering does not destroy light — it redirects it. Some reaches the ground as diffuse light from the whole sky.

Air mass

Air mass (AM) is the path length of sunlight through the atmosphere relative to the straight-down path. For a flat atmosphere,

\[AM = \frac{1}{\cos\theta_z}\]

where \(\theta_z\) is the solar zenith angle (angle from straight up). Sun overhead: AM1. At \(\theta_z = 60°\): AM2. AM1.5 corresponds to \(\theta_z = 48.19°\). Above the atmosphere: AM0.

Air-mass geometry. At 48.19° from vertical, the path through the atmosphere is 1.5× the vertical path. Schematic.
Air-mass geometry. At 48.19° from vertical, the path through the atmosphere is 1.5× the vertical path. Schematic.
Air mass versus zenith angle. The simple \(1/\cos\theta_z\) formula (blue) blows up near the horizon; the Kasten–Young (1989) formula accounts for Earth's curvature and stays finite (≈ 38 at the horizon).
Air mass versus zenith angle. The simple \(1/\cos\theta_z\) formula (blue) blows up near the horizon; the Kasten–Young (1989) formula accounts for Earth's curvature and stays finite (≈ 38 at the horizon).

Worked example 1 — air mass in Chicago

Chicago is at latitude 41.9° N. At solar noon on the equinox, \(\theta_z \approx 41.9°\) → \(AM = 1/\cos 41.9° = 1.34\). At noon on the December solstice, \(\theta_z \approx 41.9 + 23.4 = 65.3°\) → \(AM \approx 2.4\). On the June solstice, \(\theta_z \approx 18.5°\) → \(AM \approx 1.05\).

A rough empirical estimate of direct-beam intensity (Meinel & Meinel, as given on PVEducation) is \(I_D \approx 1.353 \times 0.7^{(AM^{0.678})}\) kW/m². At AM1.5: \(0.7^{1.316} = 0.625\), so \(I_D \approx 0.85\) kW/m² — consistent with AM1.5D's 900 W/m² (which includes a little circumsolar light).

Direct, diffuse and global

On a clear day diffuse light is perhaps 10–20 % of global; under overcast skies it is nearly 100 % (estimates). Flat-plate PV modules use both; concentrators use only the direct beam.

The ASTM G173 / IEC 60904-3 reference spectra

To compare cells fairly, the industry fixed one spectrum. ASTM G173-03 (generated with the SMARTS radiative-transfer model) defines:

Condition Value
Surface tilted 37° toward the Sun (≈ average latitude of the contiguous US)
Air mass 1.5 (zenith angle 48.19°)
Atmosphere 1976 US Standard Atmosphere, rural aerosol model
Precipitable water 1.42 cm
Ozone 0.34 atm·cm
Aerosol optical depth 0.084 at 500 nm
AM1.5G global on the tilted plane, integrates to ≈ 1000 W/m²
AM1.5D direct + circumsolar (2.5° half-angle), ≈ 900 W/m²

IEC 60904-3 uses the same global spectrum for terrestrial PV testing. Why 1.5? It represents typical mid-latitude conditions over a year better than AM1 (Sun overhead, rare outside the tropics); the choice dates from 1970s–80s US/European standardisation.

Worked example 2 — what the atmosphere removes

From the G173 integrals: AM0 → AM1.5G loses 1348 − 1000 = 348 W/m² (26 %) on the 37° tilted plane. In the UV below 400 nm the drop is larger in proportion: 7.6 % of AM0 is UV, but only 4.6 % of AM1.5G. Beyond 1107 nm (silicon's cutoff), AM0 has 24.8 % of its power while AM1.5G has 19.2 % — the water bands remove much near-IR.

Real spectra vary

Real outdoor spectra shift with air mass (redder in morning/evening), humidity (deeper water bands), haze and clouds (bluer diffuse light). These shifts change cells differently depending on band gap: wide-gap cells (a-Si, perovskite) respond better to bluer light; silicon is relatively tolerant. This "spectral effect" is part of energy-yield modelling (SOL 304).

Common misconceptions

Misconceptions

  • "AM1.5 means 1.5 kW/m²." No — AM is a path length; AM1.5G is normalised to 1000 W/m².
  • "Scattered light is lost." Much reaches the ground as diffuse light, which flat panels use.
  • "Clouds block all light." Overcast days still give 10–30 % of clear-sky irradiance (rough estimate) — mostly diffuse.
  • "The standard spectrum is what my panel sees." It is a reference; real spectra vary hour by hour.

What this means for manufacturing and the job

Key equations

Equation In words
\(AM = 1/\cos\theta_z\) path length through air relative to overhead
scattering \(\propto \lambda^{-4}\) blue scatters much more than red
\(GHI = DNI\cos\theta_z + DHI\) horizontal global = projected beam + diffuse

Review questions

1. What zenith angle gives AM2?

cos θz = 0.5 → θz = 60°.

2. Why is the sky blue and the setting Sun red?

Rayleigh scattering ∝ 1/λ⁴ sends blue light across the sky; at sunset the long path removes blue from the direct beam, leaving red.

3. Name the molecule responsible for the 940 nm and 1400 nm dips.

Water vapour.

4. List four conditions that define AM1.5G in ASTM G173.

37° tilt; AM1.5 (48.19° zenith); 1.42 cm precipitable water; 0.34 atm·cm ozone; AOD 0.084 at 500 nm; US Standard Atmosphere 1976 with rural aerosol. Integrated ≈ 1000 W/m².

5. Why does AM1.5D integrate to less than AM1.5G?

AM1.5D includes only the direct (plus small circumsolar) beam; AM1.5G adds sky-diffuse and ground-reflected light on the tilted plane.

Sources