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HW7_solutions - ECE 333 Homework 7 Solutions Spring 2010 1...

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ECE 333 Homework 7 Solutions :: Spring 2010 1 For the simple equivalent circuit for a 0.005 m 2 photovoltaic cell shown in Figure P8.3, the reverse saturation current is I 0 = 10 -9 A and at an insolation of 1-sun the short-circuit current is I SC = 1 A. At 25 o C, find the following: a) The open-circuit voltage. V OC ! 0.0257ln ! I SC I 0 " 1 " (8.11) # 0.0257ln ! 1 19 $ 9 " 1 " # V OC ! 0.533 V b) The load current when the output voltage is V = 0.5 V. I L ! I SC $ I d (8.7) , where I d ! I 0 % e q V d AkT $ 1 & (8.6) , using q/AkT at 25 o C to be 38.9 # I L ! 1 $ 10 $ 9 % e 38.9 ' 0.5 $ 1 &# I L ! 0.72 A c) The power delivered to the load when the output voltage is 0.5 V. The output power is simply the output voltage multiplied by the load current from (b) P ! IV ! 0.72 A ! 0.5 V ! P ! 0.36 W d) The efficiency of the cell at V = 0.5 V. The efficiency can be found by comparing the output power to the power in the sun collected by the cell (using 1000 W/m^2 – 1-sun). P out ! 0.36 W P in ! 0.005 m 2 ' 1000 W ( m 2 ! 0.5 W )! P out P in ! 0.36 W 5 W # )! 7.2% 2 The equivalent circuit for a PV cell includes a parallel resistance of R P = 10 ! (Figure P8.4). The cell has area of 0.005 m 2
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