Solar cells
Parasuraman Swaminathan
Dept. of Metallurgical and Materials Engineering
IIT – Madras
MM5017
Electronic materials, devices, and fabrication
July-Nov 2021
1
Photovoltaics (PVs)
PVs: convert light into electricity using semiconducting materials
Solar cells: convert solar spectrum into electricity using semiconductors
2
Solar spectrum
• Broad wavelength range: 0.2 – 3 𝜇𝑚 (UV to IR region)
• Peak lies in the visible region
• Total areal intensity ~ 1.35 kWm-2 (area under the curve)
• Equivalent curve – black body at 5250 ℃
• Use air mass (AM) number to
classify spectra
• AM0 – spectrum outside
atmosphere
• AM1 – sun is at zenith
• AM2 – sun is at 60º
• Intensity and distribution are
3 altered
𝑝𝑛 junction solar cell
4
Open circuit voltage
𝑝𝑛 junction solar cell
• 𝑝𝑛 junction has a built-in electric field (at equilibrium)
• This is used for carrier separation (depletion region)
• Illumination is from left – penetration depth depends on wavelength
• Wide depletion region for carrier generation
• 𝐿! + 𝑤 + 𝐿" is the region from where useful charge carriers are
generated
• These can be controlled by dopant concentrations
• Low doping is preferred – since 𝑤 is large
• Patterned metal electrode at top and continuous metal electrode
at bottom for carrier collection
5
𝑝𝑛 junction solar cell
• Central stem or ‘bus’
electrode
• Fingers extending from bus
• Balance between charge
collection and transparency
EHP - Electron hole pair
• EHPs decrease
exponentially with depth
• Current generated is called
photo current or short circuit
current
6
Solar cell characteristics
• Solar cell – 𝑝𝑛 junction connected to a load (resistor)
• Short circuit current (no load): 𝐼#$ = −𝐼%! = −𝑘𝐼&%
• When an external load (𝑅) is connected – voltage drop (𝐼𝑅)
• Opposes the built-in potential – forward bias current - 𝐼'
• Net current, 𝐼 = 𝐼' − 𝐼%!
7
Equivalent circuit
𝑒𝑉
𝐼 = 𝐼' − 𝐼%! 𝐼' = 𝐼#( 𝑒𝑥𝑝 −1
𝑘) 𝑇
𝑒𝑉
𝐼 = −𝐼%! + 𝐼#( 𝑒𝑥𝑝 −1
𝑘) 𝑇
Two parameters:
Two components: a constant current
• 𝐼#$ or 𝐼%!
source + forward biased 𝑝𝑛 junction
• 𝑉&$
8
Solar cell parameters
• Under dark conditions 𝑝𝑛
junction 𝐼 − 𝑉 curve
• Illumination – shifting of curve
below – photo current
• Y-intercept - 𝐼#$
• X-intercept - 𝑉&$
𝑒𝑉&$
𝐼%! = 𝐼#( 𝑒𝑥𝑝 −1
𝑘) 𝑇
𝑘) 𝑇 𝐼%!
𝑉&$ ≈ ln
𝑒 𝐼#(
• To increase 𝑉&$ decrease 𝐼#( Voc - open circuit voltage
• Higher band gap – preferred – but light absorption decreases
9
Solar cell power
𝑒𝑉
𝑃 = 𝐼𝑉 = 𝐼#(𝑉 𝑒𝑥𝑝 − 1 − 𝐼%! 𝑉
𝑘) 𝑇
𝑑𝑃
For maximum power =0 Obtain 𝐼* and 𝑉*
𝑑𝑉
𝑘) 𝑇 𝑘) 𝑇 𝑒𝑉*
𝐼* ≈ 𝐼%! 1− 𝑉* ≈ 𝑉&$ − ln 1 +
𝑒𝑉* 𝑒 𝑘) 𝑇
𝑘) 𝑇 𝑒𝑉* 𝑘) 𝑇
𝑃* = 𝑉* 𝐼* ≈ 𝐼%! 𝑉&$ − ln 1 + −
𝑒 𝑘) 𝑇 𝑒
𝑉* 𝐼* • Fill factor of a solar cell
𝐹𝐹 =
𝑉&$ 𝐼#$ • Measure of solar cell quality
10
Si 𝑝𝑛 junction solar cell
Consider a Si 𝑝𝑛 junction solar cell operating at 300 K
𝑛+ = 10,( 𝑐𝑚-. (intrinsic carrier concentration)
𝜀/ = 11.9 (relative dielectric constant). 𝜀( = 8.854 × 10-,0 𝐹𝑐𝑚-,
𝑝-side 𝑛-side
𝑁1 = 5 × 10,2 𝑐𝑚-. 𝑁3 = 10,4 𝑐𝑚-.
𝜇" = 1350 𝑐𝑚5𝑉 -,𝑠 -, 𝜇" = 1350 𝑐𝑚5𝑉 -,𝑠 -,
𝜇! = 450 𝑐𝑚5𝑉 -,𝑠 -, 𝜇! = 450 𝑐𝑚5𝑉 -,𝑠 -,
𝜏" = 200 𝑛𝑠 𝜏" = 100 𝑛𝑠
𝜏! = 75 𝑛𝑠 𝜏! = 25 𝑛𝑠
Incident photocurrent is 30 mA cm-2 on a cell of area 1 cm2
Load resistance is 18 Ω. What is the current through the cell?
11 Calculate FF for this solar cell.
Si 𝑝𝑛 junction parameters
𝑘) 𝑇 𝑁1 𝑁3
Built-in potential of the junction 𝑉( = ln
𝑒 𝑛+5
𝑉( = 0.697 𝑉 ≈ 0.7 𝑉
2𝜖(𝜀/ 𝑉( 𝑁1 + 𝑁3 𝑤( = 5.25 × 10-6 𝑚 = 525 𝑛𝑚
𝑤( =
𝑒 𝑁1 𝑁3
𝑤7 = 175 𝑛𝑚
𝑤% 𝑁1 = 𝑤7 𝑁3 𝑤% = 2𝑤7
𝑤% = 350 𝑛𝑚
Next is to calculate the reverse saturation current density, 𝐽#(
𝐷! 𝐷"
𝐽#( = 𝑛+5𝑒 +
𝐿! 𝑁3 𝐿" 𝑁1
12
Diffusion lengths
𝐿! = 𝐷! 𝜏! 𝑘) 𝑇𝜇! 𝑘) 𝑇𝜇"
𝐷! = 𝐷" =
𝐿" = 𝐷" 𝜏" 𝑒 𝑒
• In a 𝑝𝑛 junction, current is due to minority carrier injection
• Electrons on the 𝑝-side and holes on the 𝑛-side
𝑘) 𝑇(450) 𝑘) 𝑇(1350)
𝐷! = = 11.6 𝑐𝑚5𝑠 -, 𝐷" = = 34.9 𝑐𝑚5𝑠 -,
𝑒 𝑒
𝐿! = 11.6 × 10-0 × 25 × 10-8 = 5.4 × 10-4 𝑚
𝐿" = 34.9 × 10-0 × 200 × 10-8 = 2.6 × 10-2 𝑚
13
Solar cell characteristics
𝐷! 𝐷"
𝐽#( = 𝑛+5𝑒 + 𝐽#( = 7.68 × 10-,, 𝐴𝑐𝑚-5
𝐿! 𝑁3 𝐿" 𝑁1
Incident photocurrent is 30 mA cm-2
𝐼%! = 30 𝑚𝐴 𝑐𝑚-5 Negative according to convention
𝑘) 𝑇 𝐼%!
𝑉&$ ≈ ln 𝑉&$ = 0.51 𝑉
𝑒 𝐼#(
Now, consider the situation when there is a load of 18 Ω
Current through the load is
𝑉 𝑒𝑉
𝐼 = − = −𝐼%! + 𝐼#( 𝑒𝑥𝑝 −1
𝑅 𝑘) 𝑇
Solve this graphically to obtain the voltage
14
Solar cell 𝐼 − 𝑉 curve
Voltage is 0.45 V. Current density is -0.025 A cm-2
15
Solar cell fill factor
• For a load of 18 Ω, 𝑉 = 0.45 𝑉 and 𝐽 = −0.025 𝐴 𝑐𝑚-5
• Now to calculate FF, need to calculate 𝑉* and 𝐼*
𝑘) 𝑇 𝑘) 𝑇 𝑒𝑉*
𝐼* ≈ 𝐼%! 1− 𝑉* ≈ 𝑉&$ − ln 1 +
𝑒𝑉* 𝑒 𝑘) 𝑇
Solve for 𝑉* recursively. This gives 𝑉* = 0.44 𝑉
From 𝑉* , calculate 𝐼* = −0.028 𝐴 𝑐𝑚-5
This represents the maximum
𝑉* 𝐼*
𝐹𝐹 = = 0.804 power that can be extracted
𝑉&$ 𝐼#$
from this solar cell
16
Solar cells (research)
17 https://www.nrel.gov/pv/cell-efficiency.html
Solar cell materials
18
Solar cell efficiencies
• Highest efficiency solar cells have band gaps ~ 1-2 eV
• Goal is to maximize absorption of the solar spectrum
19
• Tandem cells – two or more solar cells
Tandem cells
Two cells stacked on
top of each other
20 Data courtesy Dr. M. Sukeerthi, IIIT, Sri City
Multijunction solar cells
Data courtesy Dr. M. Sukeerthi, IIIT, Sri City
21
Perovskite solar cell
Tandem perovskite-Si solar cell
https://en.wikipedia.org/wiki/Perovskite_solar_cell
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