Microsoft Word - HW8solutions162

Microsoft Word - HW8solutions162 - Solutions Manual for...

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Solutions Manual for Optoelectronics and Photonics: Principles and Practices S.O. Kasap 6.1 23 April 2001 6.3 Solar cell driving a load a A Si solar cell of area 4 cm 2 is connected to drive a load R as in Figure 6.8 (a). It has the I-V characteristics in Figure 6.8 (b) under an illumination of 600 W m -2 . Suppose that the load is 20 Ω and it is used under a light intensity of 1 kW m -2 . What are the current and voltage in the circuit? What is the power delivered to the load? What is the efficiency of the solar cell in this circuit? b What should the load be to obtain maximum power transfer from the solar cell to the load at 1 kW m -2 illumination. What is this load at 600 W m -2 ? c Consider using a number of such a solar cells to drive a calculator that needs a minimum of 3V and draws 3.0 mA at 3 - 4V. It is to be used indoors at a light intensity of about 400 W m -2 . How many solar cells would you need and how would you connect them? At what light intensity would the calculator stop working? Solution a The solar cell is used under an illumination of 1 kW m -2 . The short circuit current has to be scale up by 1000/600 = 1.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 1.67 along the current axis. The load line for R = 20 W and its intersection with the solar cell I-V characteristics at P which is the operating point P . Thus, 22.5 mA and 0.45 V The power delivered to the load is P out = I¢V¢ = (22.5 × 10 -3 )(0.45V) = 0.0101W, or 10.1 mW. This is not the maximum power available from the solar cell. The input sun-light power is P in = (Light Intensity)(Surface Area) = (1000 W m -2 )(4 cm 2 ´ 10 -4 m 2 /cm 2 ) = 0.4 W The efficiency is η = 100 P out P in = 100 0.010 0.4 = 2.5 0 0 which is poor. b Point M on Figure 6Q3-2 is probably close to the maximum efficiency point, 23.5 mA and 0.44 V. The load should be R = 18.7 W, close to the 20 W load. At 600 W m -2 illumination, the load has to be about 30 W as in Figure 6.8 (b). Thus, the maximum efficiency requires the load R to be decreased as the light intensity is increased. The fill factor is FF = I m V m I sc V oc = (23.5 mA)(0.44 V) (27 mA)(0.50 V) 0.78 c The solar cell is used under an illumination of 400 W m -2 . The short circuit current has to be scale up by 400/600 = 0.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 0.67 along the current axis. Suppose we have
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