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Unformatted text preview: Solution of ECE65 Final (Winter 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. OpAmps have a unitygain bandwidth of 10 6 Hz, a maximum output current limit of 100 mA, and a slew rate of 1 V/ s. OpAmps are powered by 15 V power supplies, 4) Use the following information in designing circuits: BJT Si transistors have = 400, min = 200, r = 3 k, and r o = 100 k. 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 Thevenin Resistance of the following circuit (Assume OpAmp is ideal). (10 pts) R 2 R 1 + V T v p v n v o I T R + To find the Thevenin resistance of the circuit we attach a voltage source, V T , to the circuit and compute I T (with current flowing into the sub circuit). Then, R T = V T /I T . Since OpAmp is ideal i n = i p = 0. Because of the negative feedback v n v p . Since v p = V T , we will have v p = v n = V T . Using Node voltage method (choosing the ground as shown): Node v n v n R 1 + v n v o R 2 = 0 V T R 1 + V T v o R 2 = 0 Since i p = 0, i T will flow in resistor R . Then, by Ohms Law: i T = v p v o R = V T v o R Substituting for V T v o = Ri T in the node equation, we get: v T R 1 + i T R R 2 = 0 v T = RR 1 R 2 i T R T = V T i T = RR 1 R 2 Note that this circuit act as a negative resistor. Solution of ECE65 Final (Winter 2007) 1 Problem 2. Find i in terms of v in the circuit below (The Si diode has a Zener voltage of 4 V). (10 pts) v D i D i 1 1k + v i + 1k Equations governing this circuit are: v = v D 10 3 i D v = 10 3 i 1 i 1 = 10 3 v i = i 1 i D = 10 3 v i D Case 1: Diode is ON ( v D = v = 0 . 7 V, i D > 0): v = . 7 10 3 i D i D = 10 3 ( v + 0 . 7) i = i 1 i D = 10 3 v + 10 3 ( v . 7) = 10 3 (2 v + 0 . 7) For diode being ON, i D > 0: i D = 10 3 ( v + 0 . 7) > ( v + 0 . 7) < v < . 7 V So, for v < . 7 V, diode will be ON and i = 10 3 (2 v + 0 . 7). Case 2: Diode is OFF ( i D = 0 , V Z < v D < v ): i = i 1 i D = i 1 = 10 3 v For diode being OFF, V Z < v D < v . v = v D 0 = v D V Z < v D < v  V Z < v < v + V Z > v > v So, for . 7 < v < 4 V, diode will be OFF and i = 10 3 v . Case 3: Diode is Zener region ( v D = V Z = 4 V, i D < 0): v = 4 10 3 i D i D = 10 3 ( v 4) i = i 1 i D = 10 3 v + 10 3 ( v 4) = 10 3 (2 v 4) For diode being in Zener Region, i D < 0: i D = 10 3 ( v 4) < ( v 4) > v > 4 V So, for v > 4 V, diode will be in the Zener region and i = 10 3 (2 v 4)....
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 Winter '08
 najamanverde

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