EECS314_W08_HW11sol - EECS 314 Winter 2008 HW 11 Solutions...

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Part 1 Mesh analysis makes for a quick solution due to the presence of the current source. The left mesh current name is taken from the I B controlling current, and the right mesh name is taken from the ± ·I B current source. Apply KVL to both loops to generate two simultaneous equations, taking care to incorporate V out into the right-side loop: V in I B R B I B β I B R E 0 V cc β I B R C V out 0 Solve to f nd the output as a function of the input: I B V in R B β 1 R E V cc β I B R C V out 0 V out V cc β R C R B β 1 R E V in EECS 314 Winter 2008 HW 11 Solutions Problem 1 PDF processed with CutePDF evaluation edition www.CutePDF.com
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Part 2 Using the given input signal, substitute into the result from Part 1 to f nd the output: V out V cc β R C R B β 1 R E V in , dco β R C R B β 1 R E V in , max sin ω t where: V out , dco V cc β R C R B β 1 R E V in , dco AC Gain β R C R B β 1 R E Because all quantities ± , R E , R C , and R B are non-negative, the sign of AC Gain is negative and hence the ampli f er is inverting. Part 3 Given the circuit parameters, AC Gain ( ± = 100) = ± 47.17. Changing ± gives AC Gain ( ± = 50) = ± 44.64 and AC Gain ( ± = 200) = ± 48.54. The percent change is 8.036%. Part 4 Solving algebraically, increasing R C to 15.9 k ² gives the requisite gain magnitude of 50. Decreasing to 7.95 k ² reduces the gain to 25. EECS 314 Winter 2008 HW 11 Solutions Problem 1
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EECS 314 Winter 2008 Homework set 11 Student's name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2008 Alexander Ganago Page 1 of 4 Problem 2 Electrostatic Discharge (ESD) and its prevention: 1 The Big Picture This problem requires basic knowledge of Electrostatic Discharge (ESD). To answer the questions below, read 3 documents on the web: http://www.esda.org/documents/esdfunds1print.pdf http://www.esda.org/documents/esdfunds2print.pdf http://www.esda.org/documents/esdfunds3print.pdf Grading policy: As usual, maximal score is 40 points per problem. Deduct 1 point for each wrong answer, and deduct 1.5 points for each skipped question. Questions: 1. Static electricity was discovered by Ben Franklin in 18 th century True / False (circle one) 2. Static electricity did not present any practical problems until the electronics industry was developed in 20 th century True / False (circle one) 3. The sensitivity of electronic devices to static electricity reduces as the size of devices is reduced and the speed of operation is increased True / False (circle one) 4. ESD presents a great problem for product reliability, manufacturing costs, and profitability of electronics industry True / False (circle one) 5. Typical cost of devices damaged by ESD does not exceed a few cents True / False (circle one) 6. Static electricity is defined as the transfer of charges between bodies at different
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This note was uploaded on 09/04/2008 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.

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EECS314_W08_HW11sol - EECS 314 Winter 2008 HW 11 Solutions...

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