06W-final-sol

06W-final-sol - Solution of ECE65 Final(Winter 2006 Notes 1...

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Unformatted text preview: Solution of ECE65 Final (Winter 2006) 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 in solving or 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. In the circuit below, find v o in terms of v 1 and v 2 (Assume OpAmps are ideal). (10 pts) V 2 V 1 R 2 R 1 R 3 V o V 1 V 2 V A V B + _ +-- + Both OpAmps have negative feedback, so v p 1 = v n 1 = v 1 and v p 2 = v n 2 = v 2 . Then, by node-voltage method: v 1- v 2 R 2 + v 1- v A R 1 = 0 → v A = v 1 1 + R 1 R 2- v 2 R 1 R 2 v 2- v 1 R 2 + v 2- v B R 3 = 0 → v B =- v 1 R 3 R 2 + v 2 1 + R 3 R 2 Then: v o = v A- v B = v 1 1 + R 1 R 2 + R 3 R 2- v 2 1 + R 1 R 2 + R 3 R 2 v o = ( v 1- v 2 ) 1 + R 1 + R 3 R 2 Solution of ECE65 Final (Winter 2006) 1 Problem 2. An OpAmp circuit containing only one OpAmp is attached to a 500 Ω load. (A) Design the OpAmp circuit such that if an input voltage of V i sin( ωt ) is applied to the circuit, the output voltage would be 4 V i sin( ωt ) for a frequency range of DC to 100 kHz (Assume OpAmp is ideal for this part ONLY). (B) An input voltage of V i cos(10 5 t ) is applied to the circuit. What is the maximum value of V i for the circuit to operate per its design specifications of part A. (10 pt) V o V i R L R 1 R 2 +- Part A: First, we find the transfer function of the circuit: H = v o v i = 4 Since H is independent of frequency, this circuit is an am- plifier. Because there is no phase shift between input and output, it should be a non-inverting amplifier. Prototype of the circuit is shown with A = 1 + R 2 R 1 Setting A = 4 and choosing R 1 = 10 kΩ, we find R 2 = 30 kΩ (both commercial values)....
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This note was uploaded on 05/28/2009 for the course ECE ece 65 taught by Professor Najamanverde during the Winter '08 term at UCSD.

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06W-final-sol - Solution of ECE65 Final(Winter 2006 Notes 1...

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