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EECS314 Hmwk8 Soln

# EECS314 Hmwk8 Soln - Homework 8 Problem 1 Part 1 Define V1...

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Homework 8 Problem 1 Part 1 Define to be the voltage across the load resistor; to be the voltage across the diode; 1 V 2 V I to be the current flowing through the resistor and diode (since they are in series). First, we assume that the diode does not conduct. Thus, there is no current in the circuit, namely, 0 I = 1 0 V IR = = On the other hand, from KVL, we know that 1 S V V V 2 = + (1) 2 1 8 0 8 S V V V V = = = 2 0 1.7 D V V V > = The diode does conduct. This contracts the above assumption. The actual value for is: 2 V 2 0 1.7 D V V V = = Again, from Eq.(1), 1 2 8 1.7 6.3 S V V V V = = = 1 6.3 0.0252 25.2mA 250 V I A R = = = = (a) (absorb power) (b) 2 1.7 25.2 42.84mW LED P V I = = × = (absorb power) (c) 1 6.3 25.2 158.76mW R P V I = = × = (supply power) (d) 8 25.2 201.6mW Source S P V I = − = − × = − 42.84 100% 100% 21.25% 201.6 LED LED LED R P P P η = × = × = + (e) Solution by Hongwei Liao

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Homework 8 Part 2 “LED shines” implies that the diode conducts and 2 0 1.7 D V V V = = . Thus, from above, 1 2 8 1.7 6.3 S V V V V = = = . We want: 1 40mA V I R = 1 6.3 0.1575k 157.5 40mA 40mA V V R = = = Part 3 Let us assume . Obviously, min max S V V V min 0 1.7 D V V V = = ; otherwise, the LED will not shine. In addition, satisfies: max V max 2 40mA 250 V V = Ω where, . 2 0 1.7 D V V V = = max 11.7 V V = Therefore, 1.7 11.7 S V V V Max power of LED in Part 2: ,2 1.7 40 68mW LED P = × = Max power of LED in Part 3: ,3 1.7 40 68mW LED P = × = Solution by Hongwei Liao