06S-final-sol

06S-final-sol - Solution of ECE65 Final (Spring 2006)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution of ECE65 Final (Spring 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 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/ s. OpAmps are powered by 15 V power supplies (power supplies not shown), NPN Si transistors have = 200, min = 100, r = 3 k, and r o = 100 k. NMOS transistors have K = 0 . 25 mA/V 2 and V t = 2 V. 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 (Assume OpAmp is ideal). (5 pts) v o v i 20k 20k 20k 10k i i- + Since OpAmp is ideal i n = i p = 0. Since i p = 0, by Ohms law for the 10k resistor, v p = 0. Because of the negative feedback v n v p = 0. Method 1: Using Node voltage method: Node v n v n- v i 20k + v n- v o 20k = 0 v o =- v i H v = v o v i =- 1 Method 2: Since i n = 0, current i would follow in the 20k resistor attached to v i and in the 20k resistor on the top (see circuit). By Ohms Law across these resistors: i = v i- v n 20k = v n- v o 20k v i 20k =- v o 20k v o =- v i H v = v o v i =- 1 Solution of ECE65 Final (Winter 2006) 1 Problem 2. Find v o in terms of v 1 for- 2 < v i < 6 V in the circuit below with Si diodes. (15 pts) v i v o v 1 3V D1 D2 2k 2k i Equations governing this circuit are: i = i D 1 + i D 2 = 3- v 1 2 10 3 v 1 = v D 1 + v i = v D 2 + 2 10 3 i D 2 v o = 2 10 3 i D 2 Case 1: D1 is ON ( v D 1 = v = 0 . 7 V, i D 1 > 0) and D2 is ON( v D 2 = v = 0 . 7 V, i D 2 > 0): Equation for v 1 gives: v 1 = 0 . 7 + v i = 0 . 7 + 2 10 3 i D 2 v i = 2 10 3 i D 2 i D 2 = v i 2 10 3 v o = 2 10 3 i D 2 = v i Value of i D 1 can be found from the equation for i : i = i D 1 + v i 2 10 3 = 3- (0 . 7 + v i ) 2 10 3 i D 1 = 2 . 3- 2 v i 2 10 3 Both diodes are ON if i D 1 > 0 and i D 2 > 0: i D 2 = v i 2 10 3 > v i > i D 1 = 2 . 3- 2 v i 2 10 3 > v i < 1 . 15 V Therefore, for 0 < v i < 1 . 15 V, both diodes will be ON and...
View Full Document

Page1 / 8

06S-final-sol - Solution of ECE65 Final (Spring 2006)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online