ECE 110 Exam Two SP2005 Solutions

ECE 110 Exam Two SP2005 Solutions - ECE 110 Professors...

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Unformatted text preview: ECE 110 Professors Brunet and Trick March 14, 2005 HOUR EXAMINATION #2 Last Name (use capital letters): First Name (use capital letters): _ ,“ t Signature: / at i ‘ Circle your section: AL1(lpm)-Trick BL1(noon)-Brunet A. Write or print clearly. Answer each problem on the exam itself. If you need extra paper, there is an extra sheet at the end of this exam. Clearly identify the problem number on any additional pages. The Boolean Algebra identities are also given at the end of the exam. ' — B. In order to receive partial or full credit, you must show all your work, e.g., your solution process, the equation(s) that you use, the values of the variables used in the equation(s), etc. You must also include the unit of measurement in each answer. Students caught cheating on this exam will earn a grade of F for the penalties may include suspension and/or dismissal from the university. Problem 1 (20 points) Alex is using a, 15 V voltage source and a RIOOMA voltage regulator with a 5 V, 2 W Zener diode to power the XPSOO circuit which is rated at 5 V and 200 mA. Unfortunately, Alex finds that under maximum load conditions VL = 3 V. Advise Alex how he can modify the RIOOMA circuit so that it will supply a constant 5 V to the XPSOO circuit over the current range 0 5 IL < 200 Problem 2 (20 points) 3 V Assume: 2 k9 Show 3L1 of + B = 100 your work VIN VOUT VON = VBEON = 0.7 V for all parts. VCESAT = 0.2 V VIN: 1.5 +0.5 sin (ot- a) (2 pts.) Give the range of values for VIN: S VIN .<_ b) (pm) For the given VIN explore all possible states of the BJT in the circuit. If a state is possible, give the range of VIN values for that state (if not, check “not possible”). IE- I:- E- Idlwx : "gyiOKZl/z: 1‘6 3 MO ll 3R. :2 :15 c) (9pm.) Compute the values of i3, ic, and VOUT when the BJT is active and for the extreme values of VIN (i.e., for VMIN and VMAX) found in b). c ' ‘~ g 37%?) / (0' B Problem 3 (20 points) a) Convert the Boolean function F = KEY 0 fi to a SOP used in the conversion. ‘ -t’ tfi"“ :Da/b/TKMS [~— :_ form. State the identity or method you b) Complete the truth table for the Boolean function in part 3a). We) . Wm: @i" 1:: I 35%?“ (fa/L73; xwdfiy ‘0 c) For the given truth table, write a _’ Boolean function for it i anomca ’ M Z 0 0 O 1 SOP form. x 7 fl.‘ ' ' O O 1 __. 0 [’;29—+XV*+%R'010 0- . - a a _ \-~~~—/ o 1 1 0 gig) if ' i100 1X5. 1 1 0 O 1 1 1 0 d) Find the ptimal S P Boolean function for the truth table in SC). , HINT The imal corresponding circuit has 3 gates and F: [7’th +X'7g ix??? ,_ ._ =QZ+><M7§ +~><'\7(:+&> = 7’32 “(Y Problem 4 (20 points) a) (10 pts.) Complete the truth table for the CMOS circuit below; A low voltage is a logic “0” and a high voltage is a logic “1.” b) (10 pts.) For the given truth table, draw the CMOS circuit that implements the logic function specified by the truth table below. Problem 5 (20 points) a) (12 pts.) Fill in the truth tables for the circuit below. Show your work. a X , b Y c (CA) d e f a l l g > ’ 3:13;“ ay 3 display 4 «fine :e:$ 211:6: ' Q~ 2 V - D aw what is displayed: :3 :I Q ‘3: a . b ‘_‘ C d 6 f g 7-segment display ...
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This note was uploaded on 09/16/2009 for the course ECE 110 taught by Professor Haken during the Spring '08 term at University of Illinois at Urbana–Champaign.

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ECE 110 Exam Two SP2005 Solutions - ECE 110 Professors...

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