Differential Equation - Introduction to Microelectronic Circuits - Solved Exam, Exams of Microelectronic Circuits

Download Differential Equation - Introduction to Microelectronic Circuits - Solved Exam and more Exams Microelectronic Circuits in PDF only on Docsity! UC BERKELEY EECS 40, Fall 2006 Page 1 of 8 EECS 40, Fall 2006 Prof. Chang-Hasnain Midterm #1 September 27, 2006 Total Time Allotted: 50 minutes Total Points: 100 1. This is a closed book exam. However, you are allowed to bring one page (8.5” x 11”), single-sided notes 2. No electronic devices, i.e. calculators, cell phones, computers, etc. 3. SHOW all the steps on the exam. Answers without steps will be given only a small percentage of credits. Partial credits will be given if you have proper steps but no final answers. 4. Draw BOXES around your final answers. 5. Remember to put down units. Points will be taken off for answers without units. Last (Family) Name:_____Perfect___________________________________________ First Name: ___________Peter_____________________________________________ Student ID: __________00000001__________Discussion Session: _______000______ Signature: _____________________________________________________________ Score: Problem 1 (50 pts) Problem 2 (50 pts): Total UC BERKELEY EECS 40, Fall 2006 Page 2 of 8 1. (50 pts) Equivalent circuit. (a) (5 pts) What is the current i1 through the 5 Ohm resistor? Ai 51 = (b) (5 pts) Use KVL, write down the equation for Vx in terms of V1 and/or V2 21 −=VVx (c) (5 pts) Use KCL, write down the equation for V1 and solve for V1 VV V VV VV VVV V VV x xx x x 3 0824 0)2(710 0710 0610 0 2 3 2 5 1 1 11 1 1 1 = =⋅+− =−⋅++− =⋅++− =+⋅++− =+⋅++− (d) (5 pts) Use KCL, write down the equation for V2 and solve for V2 VV VV VV VV 28 25 025 0 5 5 2 12 12 12 = += =−+− = − +− UC BERKELEY EECS 40, Fall 2006 Page 5 of 8 2. For t<0, the switch was open and Vout=0. At t = 0s, S1 closes. NOTE: μ=10-6 ; k=103; e-1=0.37; e-2=0.14 Remember to put down units. (a) (12 pts) Construct the differential equation of Vout in terms of all the given quantities. Hint:you may solve this use Mesh or Nodal analysis, or, even simpler, Thevnin equivalent circuit. Write all your steps. Thevenin Equivalence: Rewrite the 10V source and R1 into a Nodal Equivalent Circuit: 10V source becomes 1A source R1 is now in parallel with the 1A source. Combine R1 and R3 together to create a 5k ohm resister. Rewrite the 1A source and 5k ohm resister into Thevenin Equivalent Circuit. 1A source becomes 5V source 5k ohm resister is in series with the 5V source. Combine R1||R3 with R2 to yield 20k ohm resister. We now have a 5V source in series with a 20k ohm resister in series with a 1uF capacitor. Using the predetermined equations, we can fill in the variables and obtain the equation show below. Nodal Analysis: 015 4 1 8 3 0) 4 1 8 3( 15 1 15 8 23 0238 022333 30 0 15 0 151010 2 2 222 2 222 =⋅+−− =++− + = =−− =−++− =+ − = − ++ − dt dV CkVVV dt dV CVV kk V VVV VVV VVVVV kbysidesbothmultiply dt dV C k VV k VV k V k VV out outinout out outin out outin outin outin outout outin 10V Vout R2=15k C=1uF S1 R3=10k V2 R1=10k ic i3 UC BERKELEY EECS 40, Fall 2006 Page 6 of 8 V dt dV msV dt dV uFkV V dt dV CkV V dt dV CkV out out out out in out out in out out 520 5115 2 120 8 315 4 3 =+ =⋅+ =⋅+ =⋅+ (b) (5 pts) Write a closed-form expression for Vout(t) for t>0 )1(5 20/ msteVout −−= (c) (8 pts) Plot Vout as a function of time t = 0 to t = 100ms. Label the y-axis and all key points: starting value, 1 time constant value, value at infinity. 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 X: 0.02 Y: 3.161 Vout vs Time time (s) V ou t (V ) X: 0.12 Y: 4.988 (Note at 20ms, Vout = 3.15 using the above approximation for e-1) (Note at infinity, Vout should approach 5V) UC BERKELEY EECS 40, Fall 2006 Page 7 of 8 (d) (5 pts) As t approaches infinity, what value will i3 approach? Because at infinity, the capacitor becomes an open, mA kRRR VI 2 1 20 10 21 10 == + == (e) (5 pts) Now, suppose someone disturbed the circuit and S1 is re-opened at 40 ms again! Construct the new differential equation. If switch S1 is open, R1 becomes irrelevant because it is connected to an open circuit. Therefore we combine R2 and R3 to yield a 25k ohm resister. Again we have a predetermined form and therefore the equation is 025 0125 0 =+ =⋅+ =+ dt dV msV dt dV uFkV dt dV RCV out out out out out out (f) (6 pts) What is the new time constant? What is the new expression for Vout(t) for t>40 ms. msmst mst msmsms mst eVout timeshiftmsawith KeVout K KeemstVout KeVout msRC 25/)40( 25/ 25/020/40 25/ 3.4 40 3.4 3.4 3.4)1(5)40( 25 −− − −− − = = = ==−== = ==τ