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 VI R == On the other hand, from KVL, we know that 1 S VV V 2 = + ( 1 ) 21 808 S VVV V =− = = 20 1.7 D V >= The diode does conduct. This contracts the above assumption. The actual value for is: 2 V 1.7 D V = = Again, from Eq.(1), 12 8 1.7 6.3 S V =−= = 1 6.3 0.0252 25.2mA 250 V IA R = = ( a ) (absorb power) (b) 2 1.7 25.2 42.84mW LED PV I × = (absorb power) (c) 1 6.3 25.2 158.76mW R I ==× = (supply power) (d) 8 25.2 201.6mW Source S I =− × 42.84 100% 100% 21.25% 201.6 LED LED LED R P PP η = × = + ( e ) Solution by Hongwei Liao
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Homework 8 Part 2 “LED shines” implies that the diode conducts and 20 1.7 D VV V = = . Thus, from above, 12 81 . 7 6 . 3 S VVV V =−= = . We want: 1 40mA V I R =≤ 1 6.3 0.1575k 157.5 40mA 40mA R ≥== = Part 3 Let us assume . Obviously, min max S V ≤≤ min 0 1.7 D V = = ; otherwise, the LED will not shine. In addition, satisfies: max V max 2 40mA 250 = Ω where, . 1.7 D V == max 11.7 = Therefore, 1.7 11.7 S 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
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Homework 8 Part 4 First, we assume that the diode does not conduct. Then, 0 I = . In this case (note the polarity of the voltage source), 2 6 VV = − . Since , there is no voltage drop across both resistors. Therefore, 0 I = 02 6 V =− =− If we change to 12V, the diode still does not conduct; and we also have . Similarly, S V V 0 I = 12 =− Source voltage, V 6 V 12 V Current I 0A 0A Output voltage V 0 -6V -12V Part 5 First, we assume that the diode does not conduct. Then, 0 I = . Assume to be the voltage across 300 Ω resistor, and to be the voltage across 400 3 V 4 V Ω resistor; both and have the positive terminal at the lower end of its corresponding resistor. From KVL, 3 V 4 V 34 12 VVVV = ++ By assumption, 0 I = and 3 0 V = 3 4 V = 0 12 0.7 D V => = V The diode does conduct. 0.7 = Solution by Hongwei Liao
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Homework 8 11.3 16.14mA 300 400 700 S VV V I == = Ω+ Ω Ω 4 400 6.46 VI V =× Ω = 04 6.46 =− =− V Now, we change the source voltage to 24V. Still, the diode does conduct. 0.7 = 23.3 33.29mA 300 400 700 S V I = Ω Ω 4 400 13.31 V = 13.31 V Source voltage, V 12 V 24 V Current I 16.14mA 33.29mA Output voltage V 0 -6.46V -13.31V Solution by Hongwei Liao
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Homework 8 Problem 2 Part 1 Lamp is turned on MOSFET conducts 2.7 GS T VV V ≥= (By KVL, also note that the diode is connected in parallel with terminals GS) 2.7 SR V −≥ 2.7 SP D VI R V 2.7 10 2.7 4.867 A 1.5M S PD VVVV I R μ ≤= = Ω Part 2 Following the same reasoning as above, Lamp is turned on 2.7 10 2.7 7.3 S PD I V R RR = Given 50nA PD I = 146M R Solution by Hongwei Liao
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Homework 8 Part 3 First, let us recall the fact (called as Fact* in the following discussion): For a reverse-biased photodiode, PD I increases when the ambient light gets brighter .
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This note was uploaded on 09/23/2009 for the course EECS EECS314 taught by Professor Ganago during the Winter '09 term at University of Michigan.

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

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