Photovoltaics – Exercise 1
solar cell principles
The bandgap of GaAs is 1,52 eV at 0K and 1,42 eV at 300K.
a) Calculate the cutoff wavelength at each temperature.
𝝀𝝀𝑮𝑮,𝟎𝟎𝟎𝟎 = 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝝀𝝀𝑮𝑮,𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑 = 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
b) The internal quantum efficiency of the cell is 96% at 600nm and 89% at
800nm. The following also applies: R = 5% (600nm) and R = 3% (800nm).
Calculate the external quantum efficiency assuming T = 0 (λ <cut-off
wavelength) for both wavelengths.
𝜼𝜼𝒊𝒊𝒊𝒊𝒊𝒊 (𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔) = 𝟗𝟗𝟗𝟗, 𝟐𝟐%
𝜼𝜼𝒊𝒊𝒊𝒊𝒊𝒊 (𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖) = 𝟖𝟖𝟖𝟖, 𝟑𝟑%
c) Calculate the spectral sensitivity at 600, 800 and 900nm
𝑺𝑺(𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔) = 𝟎𝟎, 𝟒𝟒𝟒𝟒
𝑺𝑺(𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖) = 𝟎𝟎, 𝟓𝟓𝟓𝟓
⇒ 𝐒𝐒(𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗) = 𝟎𝟎
𝑨𝑨
𝑾𝑾
𝑨𝑨
𝑾𝑾
𝑨𝑨
𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 𝟗𝟗𝟗𝟗𝟗𝟗 𝒏𝒏𝒏𝒏 > 𝝀𝝀𝑮𝑮
𝑾𝑾
d) Consider an ideal GaAs solar cell. Calculate the open circuit voltage under the
following assumptions: Photo current density: 26mA/cm² and reverse
saturation current density Js,GaAs= 2*10-16 mA/cm², the temperature voltage
VT = 25,9mV at 300K.
𝑼𝑼𝑳𝑳 = 𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎
e) What is the efficiency of the GaAs solar cell (assumption: FF=82%)
𝜼𝜼 = 𝟐𝟐𝟐𝟐, 𝟖𝟖%
f) Calculate the diffusion voltage of the GaAs solar cell at room temperature. The
following charge carrier densities can be assumed:
ni,GaAs=1,8*106 cm-3, nD,GaAs = 5*1015 cm-3, nA,GaAs = 5*1017 cm-3
𝑼𝑼𝑫𝑫 = 𝟏𝟏, 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
g) Calculate the minority carrier densities in the p- and the n-regions.
𝒏𝒏𝒑𝒑 = 𝟔𝟔, 𝟒𝟒𝟒𝟒 ∙ 𝟏𝟏𝟏𝟏−𝟔𝟔
𝒑𝒑𝒏𝒏 = 𝟔𝟔, 𝟒𝟒𝟒𝟒 ∙ 𝟏𝟏𝟏𝟏−𝟒𝟒
Photovoltaics – Exercise 2
Current, voltage, efficiency
We consider an amorphous silicon solar cell with the following properties:
open circuit voltage
MPP-voltage
Short circuit current density
fill factor
VOC
VMPP
JSC
FF
STC
0,9 V
0,7 V
12,3 mA/cm²
63%
60 °C und 850 W/m²
0,86 V
0,65 V
10,6 mA/cm²
61%
The irradiance correction factor for the open circuit voltage is 0.03.
a) Calculate the current density at the MPP:
•
under STC
•
at 60° C cell temperature and 850W / m² irradiation
𝑱𝑱𝑴𝑴𝑴𝑴𝑴𝑴,𝑺𝑺𝑺𝑺𝑺𝑺 = 𝟏𝟏𝟏𝟏, 𝟎𝟎
𝒎𝒎𝒎𝒎
;
𝒄𝒄𝒎𝒎𝟐𝟐
𝑱𝑱𝑴𝑴𝑴𝑴𝑴𝑴,𝟔𝟔𝟔𝟔°𝑪𝑪 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖/𝒎𝒎² = 𝟖𝟖, 𝟔𝟔
b) Calculate the temperature coefficients for VOC and VMPP
𝜷𝜷𝑼𝑼𝑳𝑳 = −𝟏𝟏, 𝟐𝟐 ∙ 𝟏𝟏𝟏𝟏−𝟑𝟑 𝑲𝑲−𝟏𝟏 ;
𝒎𝒎𝒎𝒎
𝒄𝒄𝒎𝒎𝟐𝟐
𝜷𝜷𝑼𝑼𝑴𝑴𝑴𝑴𝑴𝑴 = −𝟐𝟐, 𝟏𝟏 ∙ 𝟏𝟏𝟏𝟏−𝟑𝟑 𝑲𝑲−𝟏𝟏
c) Calculate the temperature coefficients for ISC und IMPP.
⇒ 𝜶𝜶𝑰𝑰𝑲𝑲 = 𝟒𝟒, 𝟎𝟎 ∙ 𝟏𝟏𝟏𝟏−𝟒𝟒 𝑲𝑲−𝟏𝟏 ;
𝜶𝜶𝑰𝑰𝑴𝑴𝑴𝑴𝑴𝑴 = 𝟑𝟑, 𝟒𝟒 ∙ 𝟏𝟏𝟏𝟏−𝟒𝟒 𝑲𝑲−𝟏𝟏
d) How much does the efficiency decrease under the above conditions compared to
STC conditions?
⇒ 𝜼𝜼𝑺𝑺𝑺𝑺𝑺𝑺 = 𝟕𝟕, 𝟎𝟎% ;
𝜼𝜼
𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝒎𝒎𝟐𝟐
𝟔𝟔𝟔𝟔°𝑪𝑪
= 𝟔𝟔, 𝟓𝟓%
⇒ 𝚫𝚫𝜼𝜼 = 𝟎𝟎, 𝟓𝟓%
e) The cell is to be used as radiation sensor. A short-circuit current density of 7.8 mA /
cm² is measured at 40 °C cell temperature. Calculate the irradiance.
Assumption: 𝜶𝜶𝑰𝑰𝑲𝑲,𝟒𝟒𝟒𝟒°𝑪𝑪 = 𝜶𝜶𝑰𝑰𝑲𝑲 ,𝟐𝟐𝟐𝟐°𝑪𝑪 = 𝟒𝟒, 𝟎𝟎 ∙ 𝟏𝟏𝟏𝟏−𝟒𝟒 𝑲𝑲−𝟏𝟏
𝑬𝑬𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 = 𝟔𝟔𝟔𝟔𝟔𝟔
𝑾𝑾
𝒎𝒎²
Photovoltaics - Exercise 3
grid connected PV-system
Dimension a grid connected PV-system on a southeast roof with a dimension of 5 x 8 m²
with 35° inclination in Cologne with the goal of reaching minimum cable cross-sections.
You should use 16 solar modules with 375 W nominal output power as given below.
Select one of the inverters shown in the annex (table 2).
a) Dimension the inverter by considering the nominal power range, the MPP-voltage
range und the maximum input voltage of the inverter at the following operating
conditions: -10°C, 70°C and 1000W/m² irradiation.
Choose a suitable inverter from table 2. One string shall be used, if possible.
⇒ inv4
b) Calculate the cross section of the cable with 50m length (one-way!) to the inverter.
Electric conductivity κCu =56m/(Ω*mm²), common cross-sections: 2,5mm², 4mm²,
6mm², 10mm² oder 25mm².
Module data sheet
⇒ 𝑨𝑨𝑴𝑴 = 𝟒𝟒𝟒𝟒𝟒𝟒²
Pmpp
375
Wp
η
21,4
%
Nominal voltage at MPP
Vmpp
38,7
V
Nominal current at MPP
Impp
9,71
A
Open circuit voltage
Voc
45,2
V
Short circuit current
Isc
10,23
A
Maximum reverse current
IR
25,00
A
1000,00
V
Rated power at MPP
Module efficiency
Maximum system voltage
Temperature coeeficient of Vmpp and
Voc
Dimensions
inverter data-sheet
β
-0,24
% /° C
1,72 m x 1 m
inv1
inv2
inv3
inv4
inv5
Max. DC-power [W]
4.000
5.500
6.200
6.000
7.100
Max. input current [A]
16,5
9,2
14
17,4
18
Max. input voltage [A]
800
800
800
850
740
MPP-voltage range [V]
200-700
200-700
200-600
350-750
300-750
MPP-voltage range [V]
1
1
1
1
1
Photovoltaics – Exercise 4
power supply grid
A single-phase grid-controlled inverter with a pulse number of 12 and cosϕ = 1 has an AC
nominal power of 3.8 kW and a nominal voltage of 230 V. There is no information regarding
harmonics. We consider the following connection point (AP) on land (overhead line):
nominal transformer power
Reactance coating
Specific resistance of copper
single cable length
Cable cross section
Pnenn
X‘L
ρ
l
A
100
0,33
0,02
130
75
kVA
mΩ/m
Ωmm²/m
m
mm²
a) Calculate the resistance and reactance of the lines.
𝑹𝑹𝟏𝟏𝟏𝟏 = 𝟔𝟔𝟔𝟔, 𝟑𝟑𝟑𝟑𝛀𝛀
𝑿𝑿𝟏𝟏𝟏𝟏 = 𝟖𝟖𝟖𝟖, 𝟖𝟖𝟖𝟖𝛀𝛀
b) Determine the resistance and reactance of the transformer (see lecture 5).
𝑹𝑹𝑻𝑻 = 𝟐𝟐𝟐𝟐, 𝟖𝟖𝟖𝟖𝛀𝛀
𝑿𝑿𝑻𝑻 = 𝟔𝟔𝟔𝟔𝟔𝟔𝛀𝛀
c) Calculate the line impedance.
𝒁𝒁𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐𝛀𝛀
d) Calculate the short-circuit power at the AP.
𝑺𝑺𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟐𝟐𝟐𝟐𝟐𝟐, 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
e) Form the quotient of the short-circuit power at the AP and the nominal inverter
power. What conclusion can be drawn from this finding?
𝑺𝑺𝑲𝑲𝑲𝑲
= 𝟕𝟕𝟕𝟕, 𝟖𝟖
𝑺𝑺𝑾𝑾𝑾𝑾
f) Determine the permissible connection power with max. 3% increase in voltage.
𝑷𝑷𝑨𝑨𝑨𝑨𝑨𝑨 = 𝟏𝟏𝟏𝟏, 𝟗𝟗𝟗𝟗𝟗𝟗
g) Can the WR be connected? Discuss the result!
yes
Photovoltaics – Exercise 5
island system
You have bought a 1-room summer cottage near Weihenstephan (see Table for factor
„Z2“ in the lecture) You would like to use the cottage during the whole weekend (up to
48h). There is no connection to the grid (island-mode). You want to use the reading light
until 22:30 (6h Light at 21.12. and 3h Light on 21.9.). You use a light bulb with 11W. Your
refrigerator (nominal power 50W) will only be used during summer time (AprilSeptember) during weekend only and uses 300Wh per day.
The following cable are used:
Simple distance generator - battery
4
m
Simple distance battery-consumer
7
m
Cross section
2,5
mm²
conductivity
50
m/(Ωmm²)
System voltage
12
V
Conversion-, mismatch- AND cable-losses can be calculated with 10%, each.
a) Calculate the daily energy consumption during weekends durch summer and
winter.
𝑾𝑾𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 = 𝟑𝟑𝟑𝟑𝟑𝟑
𝑾𝑾𝑾𝑾
;
𝒅𝒅
𝑾𝑾𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 = 𝟔𝟔𝟔𝟔
𝑾𝑾𝑾𝑾
𝒅𝒅
b) Determine the month which yields the value of the minimum size oft he PV
generator (see Tables for factors „Z2-Z4“ in the lecture).
september
c) Which is the optimum inclination angle of the solar module (der PV-generator will
be mounted on a fixed surface, see Table „Z3“ in the lecture)?
45°
d) Determine the size of the PV-generator
𝑷𝑷𝑷𝑷𝑷𝑷 = 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟒𝟒𝟒𝟒
e) Dertermine the cable losses in each section.
generator – battery:
battery – consumer:
𝑷𝑷𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = 𝟔𝟔, 𝟏𝟏𝟏𝟏
𝑷𝑷𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = 𝟐𝟐, 𝟗𝟗𝟗𝟗
f) Dertermine the capacity of the battery
𝑪𝑪𝒏𝒏 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
Photovoltaics - Exercise 6
solar-radiation
1) Calculate sun position and “air mass” in Cologne on April 16th at 10:45 AM. Please
consider the summer time.
a) What is the apparent local time and the value for the hour angle?
𝑾𝑾𝑾𝑾𝑾𝑾 = 𝟗𝟗: 𝟏𝟏𝟏𝟏h
°
𝝎𝝎 = 𝟒𝟒𝟒𝟒, 𝟕𝟕 𝒉𝒉
b) Calculate the solar altitude
𝜸𝜸𝑺𝑺 = 𝟑𝟑𝟑𝟑, 𝟓𝟓°
c) Calculate the “air mass”
𝑨𝑨𝑨𝑨 = 𝟏𝟏, 𝟕𝟕
2) Consider a solar module which is inclined 30° to the horizontal and whose surface is
oriented south-west in Cologne on April 16th at 10:45 AM
What is the value for the incidence angle on the module surface?
𝜽𝜽𝒈𝒈𝒈𝒈𝒈𝒈 = 𝟓𝟓𝟓𝟓, 𝟒𝟒°
Photovoltaics – Exercise 7
power losses in a solar cell
A rectangular Si solar cell has the following values under STC conditions and AM1.5:
MPP-voltage
Short circuit current density
Fill factor
reverse saturation current density
temperature voltage
Sheet resistance of the emitter
VMPP
JK
FF
JS
UT
RS
0,5
38
82
-1*10-13
25,9
60
V
mA/cm²
%
A/mm²
mV
Ω/sq
The contact fingers have the following dimensions and values:
width
height
length
distance oft he fingers
specific resistance of the metal
b
h
L
d
ρ
110
20
36
2
-6
3*10
µm
µm
mm
mm
Ωcm
a) Calculate the maximum electrical power density of the cell without ohmic losses.
𝑷𝑷̇𝑴𝑴𝑴𝑴𝑴𝑴 = 𝟏𝟏𝟏𝟏, 𝟖𝟖 𝒎𝒎𝒎𝒎/𝒄𝒄𝒎𝒎𝟐𝟐
b) Calculate the absolute and relative ohmic losses in a contact finger.
𝑷𝑷𝑴𝑴,𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍𝒍 = 𝟎𝟎, 𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎
relative losses in the contact finger: 𝟎𝟎, 𝟗𝟗%
c) Calculate the absolute and relative ohmic losses in the emitter.
𝑷𝑷𝑬𝑬,𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽 = 𝟎𝟎, 𝟎𝟎𝟎𝟎 𝒎𝒎𝒎𝒎
relative losses in the emitter:𝟎𝟎, 𝟕𝟕%
d) Determine the shading losses and compare them tot he ohmic losses. How can the
contact grid be optimized?
𝑷𝑷𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 = 𝟎𝟎, 𝟕𝟕 𝒎𝒎𝒎𝒎
e) Calculate the efficiency of the solar cell before and after deducting the from b) - d).
𝜼𝜼𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗𝒗 = 𝟏𝟏𝟏𝟏, 𝟖𝟖%
𝜼𝜼𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟏𝟏𝟏𝟏, 𝟔𝟔%
Photovoltaics – Exercise 8
refractive index
The refractive index of Si is n = 3.5. The wavelength with maximum radiation intensity in
the solar spectrum is 550 nm.
a) Calculate the optimal refractive index and the optimal layer thickness of a simple AR
layer on Si, so that a minimum reflection of the solar cell in air (n = 1) is achieved for
perpendicular incidence of light.
𝒏𝒏𝟏𝟏 = 𝟏𝟏, 𝟖𝟖𝟖𝟖
𝒅𝒅𝟏𝟏 = 𝟕𝟕𝟕𝟕, 𝟓𝟓𝟓𝟓𝟓𝟓
b) What is the minimum reflection of the same solar cell in air without an AR layer (with
perpendicular incidence of light)?
𝑹𝑹𝒎𝒎𝒎𝒎𝒎𝒎 = 𝟑𝟑𝟑𝟑, 𝟗𝟗%
c) What is the minimum reflection from glass (n = 1.5) with an AR layer that has been
optimized for a solar cell?
𝑹𝑹𝒎𝒎𝒎𝒎𝒎𝒎 = 𝟏𝟏𝟏𝟏, 𝟎𝟎%
Photovoltaics – Exercise 9
energy pay back time
A solar system with modules from crystalline Si is installed in Cologne and produces an
specific energy yield of 950Wh/Wp/year.
Date of production:
2005
2015
Thickness of Si-Wafer
280 µm
180 µm
Sawing losses
150 µm
110 µm
Efficiency
12%
15%
Energy to produce the Si ingot
200 kWh/kg
200 kWh/kg
Cost poly-Si
100 US$/kg
22 US$/kg
Cost crystalline Si- solar module
3 US$/Wp
0,66 US$/Wp
a) Calculate the energy pay-back time for each year.
2005:
2014:
𝟏𝟏, 𝟕𝟕𝟕𝟕
𝟎𝟎, 𝟗𝟗𝟗𝟗𝟗𝟗 ⟶ 𝟒𝟒𝟒𝟒, 𝟕𝟕% 𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝒕𝒕𝒕𝒕 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
b) How much is the cost of poly-Si as a portion of the total solar module cost for each
year?
2005:
2014:
𝟐𝟐𝟐𝟐, 𝟓𝟓%
𝟏𝟏𝟏𝟏, 𝟗𝟗%
Photovoltaics – Exercise 10
economic paramters
The following parameters for financing a 10kWp-PV-system are known:
Cost PV-roof-system 2020
Cost PV-roof-system 2010
Specific energy yield
Electricity tarif 2020
KPV-system
KPV-system
Eyear
CElectricity
Feed in tarif in Germany for a PV-system ≤ 10 kWp:
June 2020
VEEG
June 2010
VEEG
Cost allocation due to
UmlageEEG
renewable energy law 2020
1250
2700
850
31
9,44
39,14
6,75
€/kWp
€/kWp
kWh/kWp
€Cent/kWh
€Cent/kWh
€Cent/kWh
€Cent/kWh
a) Calculate the annual cost of operation. The following costs should be included:
• The inverter of the PV-system must be change once during the operation
time (20 years). This will cost 1500 €.
• PV insurance:
150 €/year
• Meter rent: 40 €/year
𝟐𝟐𝟐𝟐𝟐𝟐€
b) Calculate the pay-back period for a system with a 100% feeding into the public
grid which has been installed in:
•
June 2010
•
June 2020
𝑻𝑻𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨,𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟖𝟖, 𝟖𝟖 𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚
𝑻𝑻𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨,𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟐𝟐𝟐𝟐, 𝟑𝟑 𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚
c) Calculate the pay-back period for a system which has been installed in 2020 and
for a self-consumption of 30% of the electricity produced. You can assume that no
cost allocation due to renewable energy law (UmlageEEG) has to be paid for selfconsumption.
𝑻𝑻𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 = 𝟏𝟏𝟏𝟏, 𝟓𝟓 𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚
d) How does the pay-back period change if you need to pay 40% of the cost
allocation due to renewable energy law (UmlageEEG) for self-consumption?
𝑻𝑻𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨 = 𝟏𝟏𝟏𝟏, 𝟐𝟐 𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚𝒚