Midterm 2: Introduction to Microelectronic Circuits - Spring 2008, Exams of Microelectronic Circuits

Download Midterm 2: Introduction to Microelectronic Circuits - Spring 2008 and more Exams Microelectronic Circuits in PDF only on Docsity! EE 40: Introduction to Microelectronic Circuits Spring 2008: Midterm 2 Venkat Anantharam 03/19/08 Total Time Allotted : 50 min Total Points: 50 1. This is a closed book exam. However, you are allowed to bring two pages (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-g) 10 1 (h-k) 10 2 6 3 (a) 4 3 (b) 4 4 10 5 6 Total 50 1 1. (20 pts total) In the circuit shown above the current source is a DC source. The switch is open prior to time 0, and the current has reached steady state prior to time 0. At time 0 the switch closes. The reference configuration to use for each circuit element in solving the problem is shown in the figure. Please provide all answers in terms of Is, R, L, and C. (a) What is iL(0−)? (2 pts) iL(0−) = Is (because cap is open circuit in DC steady state) (b) What is vC(0−)? (2 pts) vC(0−) = 0 (because inductor is short circuit in DC steady state) (c) What is iL(0+)? (1 pt) iL(0+) = iL(0−) = Is (d) What is vC(0+)? (1 pt) vC(0+) = vC(0−) = 0 2 4R = √ L C 5 2. (6 pts) In the circuit above the two sources are at different frequencies ω1 = 200rad/sec and ω2 = 100rad/sec respectively. Use superposition and phasor analysis to determine iC(t). arctan 2 = 63.435◦ arctan 4 = 75.96◦ First consider just S1. We replace the other current source by an OPEN circuit. Observe that ωC = 200 ∗ 10 ∗ 10−3 = 2, so ZC = 1 jωC = 1 j2 = −j/2. Using current divider, we find that Ic1 = ZR ZR + ZC Is1 = 2Ω 2Ω− j2Ω Is1 = 4 4− j Is1 = 4√ 17 6 arctan −14 56 − 90◦ = 20√ 17 6 (−90◦ + arctan 1 4 ) Alternatively, using the other form of current divider (or just multiplying top and bottom of the above expression by j), we find that Ic1 = 1 Zc 1 ZR + 1ZC Is1 = j2 1 2 + j2 Is1 = 4j 1 + 4j Is1 = 46 90◦√ 176 arctan 4 56 − 90◦ = 20√ 17 6 (− arctan 4) = 20√ 17 6 −75.96◦ Now consider just S2. We replace the other current source by an OPEN circuit. Observe that ωC = 100 ∗ 10 ∗ 10−3 = 1, so ωC = 1. This time, we find that Ic1 = 1 Zc 1 ZR + 1ZC Is1 = j 1 2 + j Is1 = 2j 1 + 2j Is1 = 26 90◦√ 56 arctan 2 106 45◦ = 20√ 5 6 (135◦ − arctan 2) = 20√ 5 6 (135◦ − 63.435◦) To find the total solution, we add the previous solutions: iC(t) = iC1(t) + iC2(t) = 20√ 17 cos(200t− 75.96◦) + 20√ 5 cos(100t + 135◦ − 63.435◦) Rubric: 1 pt for correct IC1 = 2Ω2Ω− j2Ω Is1. 1 pt for iC1(t) = 20√17 cos(200t− 75.96◦). Likewise 2 pts for iC2. 2 more points for adding them. 6 3. (8 pts total) Consider the circuit above operating in AC steady state at frequency ω = 1000rads/sec. (a) What is the average power dissipated in the resistance? (4 pts) There are several ways to do this. The easiest is to work directly with the resistor. Using voltage divider, we obtain the voltage across the resistor as: VR = VS R jωL + R . Note that VS = 10V , jωL = 1Ω, and R = 1Ω, so that plugging these values in, we have VR = 10 1 j + 1 = 10√ 26 45◦ . Hence, VR(t) = 10√2 cos(1000t− 45◦). To find the average power dissipated, we note that across a resistor, Pave = |VR|2 2R = (10/2) 2 = 25. It was a bit more algebraically messy to do this problem by instead finding the average power supplied across the current source (and using the fact that inductors do not dissipate average power to equate this to the power dissipated by the resistor). One could do this as follows: Observe that we would like to find Pave = Re{IsVs}/2. Recall that Is = Vs/Zeq, so we need to find 1/Zeq. Using the formula for parallel elements, we see that 1Zeq = 1 (jωL) + 1 (R+jωL) = −j + 11+j = −j + (1− j)/2 = (1− j3)/2. Hence Pave = Re IsVs/2 = Re I2s 2Zeq = Re(1002 1+3j 2 ) = 100 4 = 25. This agrees with our answer using the other method. (you could also find Is in polar notation, and then draw a triangle to find cos(θ), where θ is the angle between current and voltage, and then use Pave = VrmsIrms cos(θ)). Pavg = 25 7 5. (6 pts) R vs(t) C + vC(t) - Figure 1: Circuit −4T −3T −2T −T 0 T 2T 3T 4T 5T 0 V_0 t v s (t ) Figure 2: Voltage source In the circuit above (Fig. 1) the voltage source is periodic with period 2T and alternates between a value of V0 for T seconds, followed by a value of 0 for T seconds as shown (Fig. 2). Sketch the steady state response vC(t), labeling the graph completely, i.e. vC(t) should be determined exactly. You need not explain the details by which you arrived at vC(t). However, you should explain why your sketched vC(t) is correct. (Hint: Try to think about the values that the steady state vC(t) should have when t lies in an interval that is an even multiple of T and the value it should have when t lies in an interval that is an odd multiple of T . Write two equations, relating these two values, using your understanding of RC circuits.) Without writing anything down, we can observe that when the source is on, the voltage across the capacitor should increase exponentially (i.e. quickly at first, then more slowly) towards the source voltage. When the source voltage is turned to 0, then the capacitor should discharge exponentially towards 0 (more quickly at first, then more slowly). This gives rise to a saw tooth pattern (2 pts for getting this far). Note that VC(t) should be continuous unless you anticipate infinite current! Further, note that it will not quite achieve the asymptotic values of 0 and V0, because it is given finite time (and you are given no information indicating that the time constant is smaller than your period T ) (2 more points for getting this far). For full points, we can quantify things: Recall the formula v(t) = Vfinal + (Vinitial − Vfinal)e−t/τ . Applying it to our situation, we find that On odd intervals [Tn, T (n + 1)], we have Vc(t) = V0 + (Vc(0)− V0)e−(t−Tn)/RC On even intervals [T (n + 1), T (n + 2)], we have Vc(t) = 0 + Vc(T )e−[t−T (n+1)]/RC . 10 To find the precise values vc(0) and vc(T ), we enforce continuity at the boundary points: Vodd(end) = V0 + (Vc(0)− V0)e−T/RC = Vc(T )e0 = Veven(start) and Vodd(start) = Vc(0) = Vc(T )e −T/RC = Veven(end) We find V0 − V0e−T/RC = Vc(T )e0 − Vc(0)e−T/RC = Vc(T )e0 − Vc(T )e−2T/RC = Vc(T )(1− e−2T/RC) Therefore Vc(T ) = V0(1− e−T/RC)/(1− e−2T/RC), and Vc(0) = V0(e−T/RC − e−2T/RC)/(1− e−2T/RC). (note that we had better get that the amount it rises in an odd interval plus the amount it falls in an even interval sum to 0, so (Vc(0)− V0) + Vc(T ) = 0. Indeed, we can check that this holds.) The thick green lines on the sketch indicate the actual capacitor voltage, while the thin ones show how it would continue if the source did not change value. 0 T 2T 3T 4T 0 V_0 V_C(0) V_C(T) t V C (t ) Figure 3: Capacitor Voltage Rubric: 2 pts for realizing it has an exponential (with the right concavities) sawtooth pattern. 2 more for realizing it does not quite reach the extreme points 0 and V0. 2 final points for the precise values it reaches at the ends of each interval. 11