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07S-final-sol

07S-final-sol - Solution of ECE65 Final(Spring 2007 Notes 1...

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Unformatted text preview: Solution of ECE65 Final (Spring 2007) 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: OpAmps have a unity-gain bandwidth of 10 6 Hz, a maximum output current limit of 100 mA, and a slew rate of 1 V/ μ s. OpAmps are powered by ± 15 V power supplies (power supplies not shown), NPN Si transistors have β = 200, β min = 100, r π = 3 kΩ, and r o = 100 kΩ. NMOS transistors have K = 0 . 25 mA/V 2 and V t = 2 V. 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 the voltage transfer function and the input resistance of the circuit below (Assume OpAmp is ideal). (7 pts) R 1 R 2 v o v i v n i i v p R +- Part A: Ideal OpAmp: i p = i n = 0. Negative feedback: v p = v n = v i . Then, by node-voltage method: Node v n v n − R 1 + v n − v o R 2 = 0 → parenleftbigg 1 + R 2 R 1 parenrightbigg v n − v o = 0 A v = v o v i = v o v n = 1 + R 2 R 1 Part B: i i = v p − v o R = v i − v o R = v i 1 − v o /v i R = v i 1 − (1 + R 2 /R 1 ) R = − v i R 2 RR 1 R i = v i i i = − RR 1 R 2 Solution of ECE65 Final (Spring 2007) 1 Problem 2. Consider circuit below with a Si diode and R 2 = 9 R 1 . Find v o for v i ranging from − 10 to +10 V. (10 pt)- + R 1 R 2 v i v o- + R 1 R 2 v i v o i 2 i D i 1 v D- + Case 1: Diode OFF, i D = 0, v D < v γ = 0 . 7 V. From the circuit: i 1 = i 2 = v i R 1 + R 2 = v i R 1 + 9 R 1 = 0 . 1 v i R 1 v = R 2 i 2 = 9 R 1 × . 1 v i R 1 = 0 . 9 v i For diode being off, v D < v γ = 0 . 7 V: v D = − i 1 R 1 = − . 1 v i < . 7 V → v i > − 7 V Thus, for for 10 > v i > − 7 V, diode will be OFF and v o = 0 . 9 v i .- + R 1 R 2 v i v o i 2 i D i 1- + 0.7 Case 2: Diode ON, v D = v γ = 0 . 7 V, i D > 0. From the circuit: KVL: v o = v i + 0 . 7 i 1 = − . 7 R 1 and i 2 = v o R 2 = v i + 0 . 7 9 R 1 For diode being ON, i D > 0. Then, KCL: i D = i 1 − i 2 = − . 7 R 1 − v i + 0 . 7 9 R 1 = − v i + 7 9 R 1 > − ( v i + 7) > → v i < − 7 V Thus, for for − 7 > v i > − 10 V, diode will be ON and v o = v i + 0 . 7....
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07S-final-sol - Solution of ECE65 Final(Spring 2007 Notes 1...

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