HW4soln.pdf - ELEC 352 Electric Energy Systems Spring 2018...

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Problem 1 . Problem 5.10 in Masters. Solution: a. With one cell shaded, the PV current and voltage are related as V = 19 . 5 - 5 I I = 19 . 5 - V 5 . With two cells shaded, the PV current and voltage are related as V = 19 - 10 I I = 19 - V 10 . 0 2 4 6 8 10 12 14 16 18 20 22 24 0 2 4 40 cells, each 1 / 2 V in full sun 1 shaded 2 shaded 12-V battery Voltage [V] Current [A] b. In full sun, I = 4 A. With one cell shaded, I = (19 . 5 - 12) / 5 = 1 . 5 A. With two cells shaded, I = (19 - 12) / 10 = 0 . 7 A. 1 ELEC 352 Electric Energy Systems Spring 201 8 Homework 4
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Problem 2. Problem 5.12 Solution: 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 0 1 2 3 4 5 1 sun, 1 module With bypass diodes Without bypass diodes Voltage [V] Current [A] (a) MPP without bypass diodes is found as follows: I = 4 - 0 . 1 V, P = V I = 4 V - 0 . 1 V 2 . (1) Now differentiating the power equation in (1) and equating to zero in order to maximize the power, we get dP dV = 4 - 0 . 2 V = 0 , Hence, V = 20 V, I = 2 A, P max = (2)(20) = 40 W. We notice that the output power went from 160 W down to 40 W when only 1 cell is shaded.
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  • Winter '18
  • Chiniforoosh Sina
  • Volt, Input/output, bypass diodes

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