W08 314 HW02 solutions

W08 314 HW02 solutions - Homework 2 Solutions Problem 1...

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Homework 2 Solutions EECS 314 Winter 2008 Problem 1 Part1 : 1) The problem statement gives us the following information: ± K R R P X + = 2 b a Vout + = In addition, we know because of voltage division that ± ± ² ³ ´ ´ µ = P X S R R V Vout . If you substitute the given information for X R , you get: ± ² ³ ´ µ + ± ± ² ³ ´ ´ µ = · K R R V Vout P P S 2 1 ± ± ² ³ ´ ´ µ + = P S R K V Vout 2 1 P S S R KV V Vout + = 2 1 By matching coefficients, we can see that: V V a S 6 2 1 = = deg 06 . 0 deg 50 3 V V R KV b P S = = =
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Homework 2 Solutions EECS 314 Winter 2008 Part 2 -80 -60 -40 -20 0 20 40 60 80 1 2 3 4 5 6 7 8 9 10 11 alpha (degrees) V o u t ( V ) Vout vs Alpha alpha=linspace(-75,75,10000); vout=6+ 0.06*alpha; plot(alpha,vout); xlabel( 'alpha (degrees)' ); ylabel( 'Vout (V)' ); grid; title( 'Vout vs Alpha' ); Part 3 If we rearrange ± 50 3 6 + = Vout , we get: () 6 3 50 ± = Vout ² . ° ± ² ± = ± = ° 33 . 33 3 100 6 4 3 50 4 ³ ° ² = ± = ° 67 . 16 3 50 6 7 3 50 7 Vout ± = ± 3 50 ° = = ± 1667 . 010 . 3 50
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Homework 2 Solutions EECS 314 Winter 2008 We can rewrite the error above as: b Vout ± = ± ² . We also know that P S R KV b = . When you substitute, you get: S P P S V K R Vout R KV Vout ± ± ² = ² = ² ³ . By inspection, you can see that ± will decrease when P R is decreased, when K is increased, or when S V is increased. Part 4: No , you cannot use the device above. When 0 = ± , S V Vout 2 1 = . Since we know that we want V Vout 5 . 2 = when 0 = , we have: ) 0 ( 2 1 P S S R KV V Vout + = S V 2 1 5 . 2 = V V S 5 = Next, when 90 = , we would like to have . 5 V Vout = P S S R KV V Vout + = 2 1 () 90 5 5 2 1 5 P R K + = 180 1 = P R K Therefore, the necessary changes to the circuit are as follows: V V S 5 = 180 1 = P R K Any answer that has the above ratio is correct.
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EECS 314 Winter 2008 Homework set 2 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 3 Problem 2 Wheatstone bridge, Comparator, and MOSFET 10 points for each circuit analysis done correctly Consider the circuits built with photoresistors whose resistance R PR decreases when the incident light intensity increases, comparators and MOSFETs: the diagrams are shown on the next page. Such circuits will be used to maintain the height H of liquid in the cylindrical tank between the preset levels H1 and H2 (see the side view below). The big idea is that the liquid attenuates light emitted by the LED. In other words, if the light beam goes through the liquid, its intensity drops. As shown on the side view, two light- emitting diodes LED1 and LED2 are installed at two heights H1 and H2; both are turned on all the time. Light from each LED is detected with a a photoresistor (PR). The mechanical design prevents crosstalk, in other words, light from LED1 reaches only PR1 but it does not reach PR2; similarly, light from LED2 reaches only PR2 but not PR1.
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This homework help was uploaded on 04/10/2008 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.

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W08 314 HW02 solutions - Homework 2 Solutions Problem 1...

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