Download UC Berkeley EECS 40, Spring 2007 Midterm 1 - Circuit Analysis and Differential Equations and more Exams Microelectronic Circuits in PDF only on Docsity!
EECS 40, Prof. Chang Midterm #1 Spring 2007-Hasnain
Total Time Allotted: 8^ February 21, 2007 Total Points: 1000 minutes
- This is a closed book exam. However, you are allowed to bring one page (8.5โ x 11โ),No electronic devices, i.e.SHOW all the steps on the exam. Answers without steps will be given only a small single-sided notes. calculators, cell phones, computers, etc.
- percentage of credits. Partial credits will be given if you have proper steps but no^ final answers.Draw BOXES around your final an Remember to put down units. Points will be taken off for answers without units.swers. Last (Family) Name:___ First Name: ___Peter_____________________________________________________Perfect_____________________________________________ Student ID: __ Signature: _____________________________________ 00000000 _________________ Discussion Session: ________________________________________
Score: Problem 1 (26 pts) Problem 2 (28 Problem 3 (32 pts):pts) Problem 4 (14 Total pts)
- (26 pts) Circuit Analysis
V 1 =^ (a) 2 i(4 pts) 1 Express i^1 in terms of V^1 and constants given in this problem. i 2 = 3^ (b) โ^ V(4 pts) 1 / 2 Express i^2 in terms of V^1 AND^ V^2 and/or constants given in the problem.
10+^ (c)V 1 โ (10 pts)^ KCL or KVL.) V 2 = 0 Write two equations in V.^1 and V^2 that can be^ used to solve the circuit^ (Hint: Use 2 x(3 โ V 1 /2)+8x(V 1 /2) โ V 2 = (d) V 1 = (8 pts) 2 V Solve for V 1 , V 2 , i 1 and i 2. V i i 121 == 2 A = 12 1 A V
c) (12 pts) Provide the steps or explanation for your answers, e.g. using KCL/KVL, etc. Find the current and voltages after a very, very long time. i i 12 1.11A0A i i 34 - 0.11A0A v v 12 11.11V25.56V
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- (32 pts) Second-Order Circuit. Remember to put down units. 3 V
1 _ 1 / 6 F V^ +( - t) 2 / 3 H
t= 0
The switch is closed at t=0. The goal is to find the voltage across the capacitor, a.) (2 pts) For t<0, assume Find V(0-). that the switch was open and remained open for a very long time. V(t). V b.)(0 - (4 pts)) = 3 V What is V(0+)? Explain. V Because voltage across capacitor can not change abruptly. c.)(0+) (10 pts)=3V. Derive the second-order differential equation for the circuit. (Hint: Use KVL/KCL) Using KVL: !
" 3 + R ( iC ( t ) + iL ( t )) + V ( t ) = 0 !
" 3 + R^ $ % & C dV dt^ ( t )+ (^1) L (^) 0 #^ t + V ( t ') dt '+ iL ( 0 +)' ( ) + V ( t ) = 0 ! or using KCL:^ RC^ d^2^ dtV^^2 (^ t )+^ R^ L^ V^ ( t )^ +^ dV^ dt^ (^ t )=^0 !
V ( t R ) (^) " (^3) + iC ( t ) + iL ( t ) = 0
!
V ( t R ) " (^3) + C dV dt ( t )+ (^1) L (^) 0 #^ t + V ( t ') dt '+ iL ( 0 +)= 0
! In standard form:^ C^ d^2^ dtV^^2 (^ t )+^^ L^1^ V^ ( t )^ +^1^ R^ dV^ dt^ (^ t )=^0 !
d^2 dtV ( 2 t )+ (^) RC 1 dV dt ( t )+ (^) LC (^1) V ( t ) = 0
!
d^2 dtV ( 2 t )+ 6 dV dt ( t )+ 9 V ( t ) = 0
- (14 pts) Equivalent Circuit. Remember to put down units.
a.) 1500x1000/(1500+1000)=600 Vin (^) = 5 x(5 pts) What is 600/(600+400)= Vin? 3 V ฮฉ
b.) Vab (= (^4) = pts) What is-- ggmmx3Vin x 5000 V x 5000 V Vab? = - gmx15000 V
c.) V RThTh (5 pts) What is Answer in terms of g =V= VabTh =/I - sc g =m x15000 V- gm (^) x15000 /(the Them.venin- gm (^) x3) =5000equivalent circuit for terminals a ฮฉ -b? (Find RTh,VTh).
5 !V^ + โ^1.^5 k" VIN^ +!^ +gmVIN^5 k"^ a b
5 !V^ + โ^1.^5 k"^1 k" VIN^ +!^ +gmVIN^5 k"^ a b
1 k"
