09W-final-sol

# 09W-final-sol - Solution of ECE65 Final(Winter 2009 Notes 1...

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Unformatted text preview: Solution of ECE65 Final (Winter 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: OpAmps have a unity-gain bandwidth of 10 7 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. A) Show that the circuit below is a current source (assume OpAmp is is deal for this part only), B) For v s = 2 V and R = 10 kΩ, find the maximum R L that the current source can drive. (10pt) L L s L L- + +- i R v i v +- R Part A: Because OpAmp is ideal, I p ≈ 0 and I n ≈ 0. Then, by KCL, current i L flows through resistor R . Negative feedback: v n ≈ v p = v s Ohm’s law across R : i L = ( v n − 0) /R = v s /R = constant Since the output current is a constant and independent of v L (or R L ), this is a current source. Part B: For v s = 2 V and R = 10 kΩ, i L = 0 . 2 mA. Since a DC voltage v S is applied to the circuit, we only need to consider saturation and maximum current limits: Maximum output current: i L = 0 . 2 < i sc = 100 mA, does not apply. Saturation: v o ≤ v sat = 15. v o = ( R L + R ) i L ≤ 15 R L ≤ 15 i L − R = 75 − 10 = 65 kΩ Therefore, the maximum R L that the current source can drive is 65 kΩ. Solution of ECE65 Final (Winter 2009) 1 Problem 2. In the circuit below with Si diodes, find i . (10 pt) D 2 D 1 i 1 i 2 v 1 v 2 1k 1k 1k i 15 V-15 V Note: v 1 = v D 1 and v 2 = − v D 2 and by KCL i 1 = i + i D 1 i 2 = i + i D 2 i 1 i 2 v 1 v 2 1k 1k 1k i 15 V-15 V Case 1: Assume both diodes are off ( i D 1 = i D 2 = 0, v D 1 < v γ , and v D 2 < v γ ) . We can replace both diodes with open circuit to arrive at the circuit on the right. Then: i 1 = i and i 2 = i and i = 15 − ( − 15) 10 3 + 10...
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09W-final-sol - Solution of ECE65 Final(Winter 2009 Notes 1...

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