Tut 3 solutions

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Page 1 of 2 SOLA2540 & SOLA9001 Applied Photovoltaics Tutorial 3 Solutions Question 1 (a) 1 0 2 1 1.5 3.9 2.42 600 62 4 4 2.42 n n n nm d nm n λ = = × = = = = × (b) 0 3 0 2 1 3 0 1 2 2 3 1 1.5 3.9 1.5 2.6 3.9 glass Si n n n n n n n n n n n n n n = = = = = = = = They may be any values that give this ratio. If we pick n 2 = 3.2, n 1 = 1.23. 0 2 2 0 1 1 600 47 4 4 3.2 600 122 4 4 1.23 nm d nm n nm d nm n λ λ = = = × = = = × Question 2 (a) The fractional power loss is given by: (b) 2 2 2 1 350 0.233% loss mp mp mp loss gen mp mp mp cm t m t P I R I A t I J t P A P P V I V ρ μ ρ ρ ρ = Ω = = = = = = = MP MP s gen loss frac V S J P P P 12 2 ρ = = ( ) ( ) % 75 . 1 42 . 0 12 10 3 10 28 35 loss power fractional the Hence mm 3 and mV, 420 , mA/cm 28 mV, 420 , / 35 2 3 3 2 = × × × × × = = = = = Ω = frac MP MP MP s P S V J V ρ Si (n 2 = 3.9) ARC (n 1 = ?) Glass (n 0 = ?)
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Page 2 of 2 Question 3 (a) (b) ρ =500m Ω cm (c) R b = ( ρ b x e) / A = 100m Ω ρ b = (R b x A) /e = 100m Ω x 0.5cm 2 = 500m Ω cm (d) ρ = ρ /f , ρ = t x dx 0 )
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