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
