Midterm 1: Introduction to Microelectronic Circuits - Spring 2008, Exams of Electrical Engineering

Download Midterm 1: Introduction to Microelectronic Circuits - Spring 2008 and more Exams Electrical Engineering in PDF only on Docsity! EE 40: Introduction to Microelectronic Circuits Spring 2008: Midterm 1 Venkat Anantharam February 22, 2008 Total Time Allotted : 50 min Total Points: 50 1. This is a closed book exam. However, you are allowed to bring one page (8.5 x 11) notes, with writing on both sides. 2. No electronic devices, i.e. calculators, cell phones, computers, etc. 3. Show the steps used to arrive at your answer, where necessary. Partial credit will be given if you have the proper steps but an incorrect answer. A correct answer for a problem involving multiple steps where it is not clear how you arrived at the answer may not be given full credit. 4. Write your final answers into the boxes. 5. Remember to put down units. Last (Family) Name: First Name: Student ID: Lab Section: Signature: Max Score 1 (a) 4 1 (b) 4 2 (a) 3 2 (b) 3 2 (c) 3 2 (d) 3 2 (e) 3 2 (f) 3 3 10 4 6 5 (a) 3 5 (b) 5 Total 50 1 unknown circuit 1A 2Ω + v − 2Ω 2Ω 2Ω − + Vs 2Ω i2 2Ω i1 2Ω Figure 1: Circuit 1 1. In the circuit in Figure 1, each of the seven visible resistors has value 2Ω. Unfortunately, the designers forgot to specify what is in the part of the circuit in the box on the left. The voltage Vs is also unknown. You are given that v equals 5V . (a) Find i1. Firstly, we recognize that the current through the 2Ω resistor in the middle is im = v 2Ω = 5 2 A Next, we consider a supernode around the unknown circuit and the two vertical aligned resistors next to it and write down the corresponding KCL equation: i1 = 1A + im = 7 2 A i1 = 7 2A (b) Find i2 in terms of Vs. Setting up the loop equation (KCL) on right side yields Vs = i22Ω + (i2 + 1A)2Ω + (i2 + 1A + im)2Ω We solve for i2 Vs − 2V − 7V = i26Ω i2 = Vs − 9V 6Ω i2 = Vs−9V 6Ω 2 Is R R i Is 2R −+ Vs Figure 3: Circuit 3 3. In the circuit in Figure 3, use superposition to find the current i in terms of R, Vs and Is. (Note: You MUST use superposition to answer this question. Solutions using other techniques will not be accepted. Show the details of your work.) Zeroing all sources with the exception of the current source on the right yields the circuit in Figure 4. All current from the current source is flowing through the center resistor. Hence, R R i Is 2R Figure 4: Circuit 7 R R i 2R −+ Vs Figure 5: Circuit 8 i|1 = Is (3) Secondly, the two current sources are zeroed, calculating the impact of the voltage source on the current through the center resistor. The Circuit is depicted in Figure 5. We see that no current is flowing through the center resistor. i|2 = 0 (4) Lastly, we consider the impact of the left current source and zero the right current source and the voltage source. We get the circuit in Figure 6. We directly see that all the current from the current source is flowing through the center resistor i|3 = Is (5) Now we apply the superposition principle and calculate the total current through the center resistor from equations (3), (4), and (5). i = i|1 + i|2 + i|3 = 2Is i = 2Is 5 Is R R i 2R Figure 6: Circuit 9 R Rs R R a 2R 2R b R Figure 7: Circuit 4 4. Consider the resistive circuit in Figure 7. Find the equivalent resistance Rab across the terminals a and b in terms of R and Rs. By symmetry, we know that there is no current through resistor Rs. Hence, we can analyze the circuit as if this resistor was not there. Rab = (2R + 2R)||(R + R + R + R) = 2R Rab = 2R 6 − + Vs R R b a Figure 8: Circuit 5 5. (a) Considering the circuit in Figure 8, find the Thévenin equivalent circuit across the terminals a and b. We first zero the independent voltage source in order to find the Thévenin resistance. We get a circuit with two resistors with value R in parallel. Hence, we have Rth = R||R = R 2 Next, we calculate the open circuit voltage. Here we know that the current through the terminals a and b is both 0A and we can use the voltage divider formula to get vth = Vs R R + R = Vs 2 Thévenin Equivalent Circuit: − + Vs 2 R 2 a b 7