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09F-final-sol

# 09F-final-sol - Solution of ECE65 Final(Fall 2009 Notes 1...

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Unformatted text preview: Solution of ECE65 Final (Fall 2009) Notes: 1. For each problem, 20% of points is for the “correct” final answer. 2. Messy, incoherent papers lose point! Explain what you are doing! 3. Use the following information only in designing circuits: NPN Si transistors have β = 200, β min = 100, r π = 3 kΩ, and r o = 100 kΩ. In circuit design, use 5% tolarence commercial resistor and capacitor values of 1, 1.1, 1.2, 1.3, 1.5, 1.6, 1.8, 2, 2.2, 2.4, 2.7, 3., 3.3, 3.6, 3.9, 4.3, 4.7, 5.1, 5.6, 6.2, 6.8, 7.5, 8.2, 9.1 ( × 10 n where n is an integer). You can also use 5 mH inductors. Problem 1. Find I in the circuit below with a Si diode. (8pt) V 2 1k 5V 9V 1k 1k 1k I V 1 V 2 v D 1k 5V 9V 1k 1k 1k +- I- + V 1 V 2 I 1 I 2 1k 5V 9V 1k 1k 1k 0.7 V I Diode OFF Diode ON Case 1: Assume the diode is OFF ( i D = 0 , v D < v γ = 0 . 7 V). Replacing the diode with its circuit model (open circuit), we arrive at the circuit above. By voltage divider formula, we get V 1 = 5 × (1 k ) / (1 k + 1 k ) = 2 . 5 V, and V 2 = 9 × (1 k ) / (1 k + 1 k ) = 4 . 5 V. Then v D = V 2 − V 1 = 4 . 5 − 2 . 5 = 2 > . 7 = V γ . Thus, the diode OFF assumption is incorrect. Case 2: Assume the diode is ON ( i D > , v D = V γ = 0 . 7 V). Replacing the diode with its circuit model (a voltage source), we arrive at the circuit above. Using the node-voltage method, we note that V 2 = V 1 +0 . 7 and the combination of nodes V 1 and V 2 is a super-node: Super-node V 1 & V 2 V 2 − 9 1 , 000 + V 2 − 1 , 000 + V 1 − 5 1 , 000 + V 1 − 1 , 000 = 0 V 2 − 9 + V 2 + V 1 − 5 + V 1 = 0 2( V 1 + 0 . 7) + 2 V 1 − 14 = 0 → V 1 = 3 . 15 V We use KCL to find I : I = I 2 − I 1 = V 1 − 1 , 000 − 5 − V 1 1 , 000 = 2 V 1 − 5 1 , 000 = 1 . 3 mA Note that sine i D = I > 0, the assumption of diode ON is justified. Solution of ECE65 Final (Fall 2009) 1 Problem 2. Design a BJT switch that turns an LED off when v i = 0 and turns the LED on and brightly lit when v i = 5 V. The LED we have has a v γ = 1 . 8 V and is brightly lit when its current is 9 mA. Use a 5 V power supply. (10 pt) V i R B R C 5 V Prototype of the circuit is shown. The equations govern- ing the circuit are: BE-KVL: v i = R B i B + v BE CE-KVL: 5 = R C i C + v LED + v CE For v i = 0, BE-KVL gives v BE = 0 and i B = 0 leading to BJT being in cut off and LED being OFF....
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09F-final-sol - Solution of ECE65 Final(Fall 2009 Notes 1...

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