Download 3 Problems with Solution of Components and Circuits Lab - Midterm Exam | ECE 65 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE65 (Winter 2008), Midterm Notes: 1. Write your answers on these three sheets. 2. For each problem, 20% of points are alloacted for the correct final answer. 3. Use the following information in designing circuits: OpAmps are powered by ±15 V power supplies (power supplies not shown), have a unity-gain bandwidth of 106 Hz, a short-circuit current limit (maximum output current limit) of 50 mA, and a slew rate of 1 V/µs. In circuit design, use 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 (×10n where n is an integer). You can also use 5 mH inductors. Problem 1. Computer iL and show that it is independent of RL (Assume OpAmp is ideal). (10pts) L L s on p − + R i v vv v − + R R R R Ideal OpAmp: Ip ≃ In ≃ 0. We have negative feedback: Vn ≈ Vp. Using node-voltage method: Node Vn Vn − Vs R + Vn − Vo R = 0 → Vo − 2Vp = Vs Node Vp Vp − 0 R + Vp − Vo R + Vp − 0 RL = 0 → Vo − ( 2 + R RL ) Vp = 0 where we have substituted for Vn ≈ Vp in the first equation. Since we are interested in iL = Vp/RL, we solve the above two equations in two unknowns (Vp and Vo) to find Vp. The simplest way is to subtract the two equations from each other to get: − R RL Vp = Vs → Vp = − RL R Vs iL = Vp RL = − Vs R Therefore, iL is independent of RL (i.e., this circuit is a current source) Problem 2. Design a circuit with a transfer function of H(jω) = −4 1 + j ω/105 and a minimum input impedance of 10 kΩ. (10pts) The transfer function is in the general form for first-order low-pass filters: H(jω) = K 1 + jω/ωc with K = −4 and ωc = 10 5 rad/s. As |K| > 1, we need to use an active filter. The prototype of the circuit is shown below with i o 2 21 V V C RR + − H(jω) = − R2/R1 1 + jω/ωc ωc = R2C2 The input impedance of this filter is Zi|min = R1. Comparing the transfer function of this prototype circuit with the desired one, we get: R2 R1 = 4 ωc = 1 R2C2 = 105 Zi|min = R1 ≥ 10 kΩ Choosing commercial value of R1 = 10 kΩ, we get: R2 = 4R1 = 40 kΩ → 39 kΩ (Commercial) C2 = 1 105R2 = 1 4 × 109 = 0.25 × 10−9 → 0.24 nF (Commercial) We need to consider the impact of the bandwidth of OpAmp chip. (1+A)×fc = fu = 10 6 leads to fc = 10 6/(1 + 4) = 200 ≫ fu = 100/(2π) = 16 kHz. So the circuit should work properly.
