Designing Information Devices and Systems I Spring 2018 Homework 8

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1 EECS 16A Designing Informtion Devices nd Systems I Spring 2018 Homework 8 This homework is due Mrch 19, 2018, t 23:59. Self-grdes re due Mrch 22, 2018, t 23:59. Sumission Formt Your homework sumission should consist of one file. hw8.pdf: A single PDF file tht contins ll of your nswers (ny hndwritten nswers should e scnned). Sumit the file to the pproprite ssignment on Grdescope. 1. (PACTICE) Thévenin nd Norton Equivlent Circuits () Find the Thévenin nd Norton equivlent circuits seen from outside of the ox. 1 10V 3 4.5Ω Ω To find the Thévenin nd Norton equivlent circuits, we re going to find the open circuit voltge etween the output ports nd the current flowing through the output ports when the ports re shorted. For finding the open circuit voltge etween the output ports, let us lel the nodes s shown in the figure elow. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 1

2 1 A 10V 3 4.5Ω 2 B 4 1.5Ω GND First, let us egin y clculting the effective resistnce etween nodes A nd GND. We hve the resistor in prllel to the 4.5Ω 1.5Ω resistnce. This gives n equivlent resistnce of Ω1.5Ω = 2Ω. Then we see tht we hve voltge divider from the positive terminl of the 10V supply. Voltge divider is mde up of two resistnces in series, where the resistnces re nd 2Ω. This gives the voltge t node A equl to V A = 10V 2Ω 2Ω = 4V To find the voltge t node B, note tht we hve nother voltge divider etween nodes A nd GND. Hence, we cn find the voltge t node B s V A 1.5Ω 4.5Ω 1.5Ω = 4V 1 4 = 1V Hence the open circuit voltge seen etween the output ports is equl to V open = V A V B = 4V 1V = 3V Now we need to choose V test to find Th. For convenience, let us choose V test = V open. Then I test = I sc = I No. This Now let us find the short circuit current flowing through the output ports. When doing this, we get the following circuit. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 2

3 1 10V C 3 4.5Ω I test Ω GND Note tht when we short the output terminls, the voltges t the nodes chnge, this is why we chnged the lel of the node elow the resistor 1. Since there is short circuit prllel to the resistor 3, there will e no current flowing through it, hence we hve I test = I 4 To find this current, let us find the equivlent resistnce due to 2 eing connected prllel to 4 when we short the output ports. We hve prellel to 1.5Ω, which gives n equivlent resistnce Ω = 1Ω We gin hve voltge divider etween the positive side of the 10 V supply nd the ground. Using this voltge divider, we clculte the voltge t node C s V C = 10V 1Ω 1Ω = 2.5V Hence, we see tht the voltge cross the resistor 4 is equl to 2.5V. Using Ohm s lw, we get Since we hve I test = I 4, we hve I 4 = 2.5V 1.5Ω = 5 3 A I test = I 4 Summrizing the results, we hve This gives Hence the Thévenin equivlent circuit is given y V test = 3V I test = 5 3 A Th = V test I test = 9 5 Ω EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 3

4 V Th where = Th nd V Th = V open, nd the Norton equivlent circuit is given y I No where = Th nd I No = I test. () Find the Thévenin nd Norton equivlent circuits seen from outside of the ox Ω 3 4.5Ω Ω 4 1.5Ω 4A As with the previous prt of this question, to find the Thévenin nd Norton equivlent circuits we re going to find the open circuit voltge etween the output ports nd the current flowing through the output ports when the ports re shorted. Let us first find the open circuit voltge etween the output ports. We cn use the symmetry in the circuit to see there will e no current flowing through 5. Then, we hve current divider where ech rnch hs the sme resistnce, hence the current 4A divides eqully etween the left nd right rnches. Hence we hve I 1 = I 2 = I 3 = I 4 = 2A This gives us the voltge cross 2, equivlently the open circuit voltge etween the output terminls, equl to V open = 2A 1.5Ω = 3V Similr to prt (), we use the Thévenin voltge s V test for convenience, since tht sets I test to the short circuit current I No. Now let us find the short circuit current cross the output terminls. Let us find this EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 4

5 using nodl nlysis on the resulting circuit when we short the output ports. To help do the nlysis, let us lel the nodes s shown in the figure elow. A I test 1 4.5Ω 3 4.5Ω 5 C B 2 1.5Ω 4 1.5Ω 4A GND Now wht re the unkown node voltges? We do not know the voltge t node A nd B. On the other hnd, ecuse node C is connected y short circuit to GND, we know its voltge is equl to the GND which we set s 0; hence voltge t node C is not n unkown. Next, ecuse there is short circuit cross resistor 2, there will e no current flowing through it. Let us write KCL t the nodes 4A = I 1 I 3 (Node A) I 3 = I 4 I 5 (Node B) I test = I 1 I 5 (Node C) Now let us relte the currents I 1, I 2, I 3, I 4 nd I 5 to node voltges using Ohm s lw. We hve I 1 = V A V C 1 = V A 4.5Ω since V C = 0V ecuse it is connected to the ground y short circuit. Furthermore, we hve I 2 = 0A, I 3 = V A V B = V A V B 3 4.5Ω I 4 = V B = V B 4 1.5Ω, I 5 = V B V C 5 Plugging these into the first two KCL equtions, we get = V B 4A = V A 4.5Ω V A V B 4.5Ω V A V B 4.5Ω = V B 1.5Ω V B EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 5

6 These equtions re solved y Using the KCL t node C, we get V A = 9.9V, V B = 1.8V I test = I 1 I 5 = V A 4.5Ω V B = Ω 1.8 = 2.8A Summrizing the results, we hve This gives V open = 3V I test = 2.8A Th = V test I test = Ω Hence the Thévenin equivlent circuit is given y V Th where = Th nd V Th = V open, nd the Norton equivlent circuit is given y I No where = Th nd I No = I test. 2. Whetstone Bridge Thévenin equivlence is powerful technique we cn use to solve the Whetstone ridge circuit shown elow. This circuit is used in mny sensor ppliction where sensing element is the "ridge" resistor, 3. It is often useful to find the current through the ridge resistor or the voltge cross the ridge resistor. Intuitively, knowing I 3 or V 3 llows us to solve the rest of the circuit. In this prolem, we wnt to find the current I 3 flowing through the ridge resistor 3. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 6

7 V () First, let s remove the ridge resistor 3. Clculte the Thévenin equivlent voltge V th etween the two terminls nd, for the circuit shown elow, where the ridge resistor hs een removed V Notice in the ove circuit tht there re two voltge dividers, so we cn clculte v nd v quickly. v = V v = V Thus, the Thévenin voltge is simply the difference etween the two voltges: V th = v v = ( ) 5V. () Is the Thévenin voltge V th you found in prt () equl to the ctul voltge V 3 cross the ridge resistor? Why or why not? EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 7

8 No, the Thévenin voltge we found in prt () is the open-circuit voltge. If we dd 3 ck into the originl circuit, 3 would lod the other resistors (or, equivlently, the Thévenin resistnce), so the Thévenin voltge is not equl to the ctul voltge cross the ridge resistor. (c) Find the Thévenin resistnce th etween the two terminls nd for the ove circuit. Drw the Thévenin equivlent etween the terminls nd for the circuit ove. We find the Thévenin resistnce y replcing the voltge source with short nd clculting the resistnce etween the two terminls nd. th = ( 1 4 ) ( 2 5 ), where denotes the prllel opertor. = Using V th = ( ) 5V from prt (), we cn construct the Thévenin equivlent circuit ( ) 5V (d) With this equivlent circuit, clculte the current I 3 cross the ridge resistor. Using the equivlent circuit, we now dd 3 ck in. through the ridge resistor nd the voltge V ( ) 5V 3 ( ) V I 3 = ( ) V 3 V 3 = I 3 3 = EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 8

9 3. Mximum Horsepower You re n engineer working on n electric cr. Your jo is to design motor to e used the cr. Specificlly you re designing the resistnce of this motor. The ttery used y this cr hs some series resistnce, nd it is modeled y the circuit shown elow. s V s You ttch your motor to the ttery s shown elow. s V s motor () Clculte the power P s delivered y the voltge source in terms of V s, s, nd motor. V s I s = s motor V 2 s P s = I s V s = s motor Using pssive sign convention, we otin the power dissipted y the source. This is negtive quntity, implying tht the source is delivering power. The power delivered y the source is then () Now clculte the power P motor dissipted y the lod resistor in terms of V s, s, nd motor. V motor = motor s motor V s P motor = V motor 2 motor = motor ( s motor ) 2 V s 2 V 2 s s motor. (c) Suppose we wnted to mximize the power dissipted cross the lod. Find the optiml vlue for motor in terms of s. Hint: Use clculus. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 9

10 motor P motor = ( s motor ) 2 V s 2 dp motor = ( s motor ) 2 motor ( s motor ) d motor ( s motor ) 4 Vs 2 = 0 2 s 2 motor = 0 2 motor = 2 s motor = s (d) Now you ve swiched tems to designing the ttery. Your jo is now to pick the optiml s for mximizing the power delivered to the motor. Wht vlue of s should you pick? Hint: Don t use clculus. To mximize the power delivered to the motor, we need to mximize the voltge cross the motor. This is done when there is no s. Thus, the optiml vlue of s is 0Ω. 4. Digitl to Anlog Converter (DAC) For some outputs, such s udio pplictions, we need to produce n nlog output, or continuous voltge from 0 to V s. These nlog voltges must e produced from digitl voltges, tht is sources, tht cn only e V s or 0. A circuit tht does this is known s Digitl to Anlog Converter. It tkes inry representtion of numer nd turns it into n nlog volge. The output of DAC cn e represented with the eqution shown elow: where ech inry digit n is multiplied y 1 2 n. N 1 V out = V s n=0 2 n n () We know how to tke n input voltge nd divide it y 2: V s V out To divide y lrger powers of two, we might hope to just cscde the ove voltge divider. For exmple, consider: V s V out EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 10

11 Clculte V out in the ove circut. Is V out = 1 4 V s? We first find the potentil V x. V x V s V out V x = V out = 2 V 3 s = 2 3 V s = 2 5 V s V x = V s = 1 5 V s 1 4 V s () The - ldder, shown elow, hs very nice property. For ech of the circuits shown elow, find the equivlent resistnce looking in from points nd. Do you see pttern? i. ii. iii. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 11

12 i. eq = = ii. We find the equivlent resistnce for the resistors from left to right. eq = = iii. Agin, we find the equivlent resistnce for the resistors from left to right. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 12

13 The equivlent resistnce is lwys eq =. eq = = (c) The following circuit is n - DAC. To understnd its functionlity, use superposition to find V out in terms of ech V k in the circuit. V out V 1 V 2 V 3 V 1 : EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 13

14 V x V y V out1 V 1 We first find the potentil V x. To do this, we cn simplify the circuit. V x eq V 1 eq = ( ( ( ))) = We cn then find V x using the voltge divider formul. V x = V 1 = V 1 Similrly, we use the voltge divider formul to find V y in terms of V x. V y = 6 ( ) ( ) V 5 x = 6 5 V x = V 1 = 3 16 V 1 Applying the voltge divider formul gin gives us V out1. V 2 : V out1 = V y = V 1 = 1 8 V 1 EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 14

15 V z V out2 V 2 We first find the potentil V z. To do this, we cn simplify the circuit. V z eq1 eq2 V 2 eq1 = ( ) = = eq2 = = 3 We cn then find V z using the voltge divider formul. V z = 6 3 ( 3) V 5 2 = 6 5 V 2 = 3 8 V 2 Applying the voltge divider formul gin gives us V out2. V 3 : V out2 = V z = V 2 = 1 4 V 2 EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 15

16 V out3 V 3 We cn simplify this circuit. V out3 V 3 V out3 = V 3 = 1 2 V 3 V out = V out1 V out2 V out3 = 1 8 V V V 3 (d) We ve now designed 3-it - DAC. Wht is the output voltge V out if V 2 = 1V nd V 1 = V 3 = 0V? V out = 1 8 0V 1 4 1V 1 2 0V = 1 4 V (e) Drw the Thévenin equivlent of the ove circuit, looking in from the terminls nd with V 2 = 1V nd V 1 = V 3 = 0V. V th = 1 4 V th = EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 16

17 1 4 V V out (f) Suppose tht we now ttch speker to the DAC with resistnce of 3. Why is the voltge cross the speker lower thn wht we computed in prt (d)? Wht is the ctul output voltge? Attching the speker will lod the DAC. The speker drws some current from the DAC, so the voltge cross the output will e lower thn expected. We cn use the Thévenin equivlent circuit to clculte the ctul output voltge. 1 4 V 3 V out 3 V out = V = V = 1 16 V 5. Mechnicl Circuits with Cpcitors nd esistors Find the voltges cross nd currents flowing through ll of the cpcitors t stedy stte. C 1 3F 10V C 3 6F C 2 4F C 4 3F For cpcitor C k, let us denote the voltge cross it y v Ck, the current flowing through it y i Ck, nd its chrge y Q Ck. In stedy stte (tht is, fter the current hs een running for very long time), direct current EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 17

18 (DC) cpcitors ct s open circuits. Hence, we see tht there is no current flowing through the cpcitors, tht is, i C1 = i C2 = i C3 = i C4 = 0A. For finding the voltges cross the cpcitors, let us lel nodes on the circuit s shown in the following figure. C 1 3F 10V A C 3 6F C 2 4F B C 4 3F GND We re going to use the following four properties to find the voltges cross the cpcitors: () Chrge is lwys conserved. () The chrge Q stored in cpcitor is given y the eqution Q = CV. (c) The chrges cross series cpcitors re equl to ech other. (d) The voltge cross prllel cpcitors is equl. As n exmple use of property (c), we hve the chrge on the cpcitor C 3 equl to the chrge on the cpcitor C 4. Let us strt y writing the eqution for conservtion of chrge t node A: Q C1 = Q C2 Q C3 By property (), tht is, Q = CV, we cn equivlently write this eqution for chrge conservtion in terms of node voltges s (10V v A )3F = v A 4F (v A v B )6F, which, fter simplifiying the eqution, gives Let us then write the chrge conservtion eqution t node B; we hve 30V = 13v A 6v B. (1) Q C3 = Q C4. As efore, we cn write this chrge conservtion eqution in terms of the node voltges s (v A v B )6F = v B 3F, EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 18

19 which, fter simplifiction, gives 2v A = 3v B. (2) Equtions 1 nd 2 give us two linerly independent equtions in two unknowns. Solving the system, we get v A = 10 3 V, v B = 20 9 V. Using the node voltges, we cn clculte the voltges cross the cpcitors s v C1 = 10V v A = 20 3 V, v C2 = v A = 10 3 V, v C3 = v A v B = 10 9 V, v C4 = v B = 20 9 V. We write the currents cross the cpcitors gin here for reder s convenience: i C1 = i C2 = i C3 = i C4 = 0A 6. Mglev Trin Height Control System One of the fstest forms of lnd trnsporttion re trins tht ctully trvel slightly elevted from ground using mgnetic levittion (or mglev for short). Ensuring tht the trin stys t reltively constnt height ove its trcks (the trcks in this cse re wht provide the force to levitte the trin nd propel it forwrd) is criticl to oth the sfety nd fuel efficiency of the trin. In this prolem, we ll explore how the mglev trins use cpcitors to keep them elevted. (Note tht rel mglev trins my use completely different nd much more sophisticted techniques to perform this function, so if you e.g. get contrct to uild such trin, you ll proly wnt to do more reserch on the suject.) () As shown elow, let s imgine tht ll long the ottom of the trin, we put two prllel strips of metl (T 1, T 2 ), nd tht on the ground elow the trin (perhps s prt of the trck), we hve one solid piece of metl (M)... T 1 T 2 W W L trin T 1 T 2 h h M Side View. M Top View. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 19

20 Assuming tht the entire trin is t uniform height ove the trck nd ignoring ny fringing fields (i.e., we cn use the simple equtions developed in lecture to model the cpcitnce), s function of L trin (the length of the trin), W (the width of T 1 /T 2 ), nd h (the height of the trin off of the trck), wht is the cpcitnce etween T 1 nd M? How out the cpcitnce etween T 2 nd M? The distnce etween the pltes (T 1 & M or T 2 & M) is h. The re of plte for the prllel plte cpcitor is A = WL trin. Using the formul for cpcitnce of prllel plte cpcitor, we get: C = εa d C 1 = εwl trin h C 2 = εwl trin h (Cpcitnce etween T 1 nd M) (Cpcitnce etween T 2 nd M) () Any circuit on the trin cn only mke direct contct t T 1 nd T 2. Thus, to detect the height of the trin, it would only e le to mesure the equivlent cpcitnce etween T 1 nd T 2. Drw circuit model showing how the cpcitors etween T 1 nd M nd etween T 2 nd M re connected to ech other. The cpcitors C 1 nd C 2 re in series. To relize this, let s consider the trin circuit tht is in contct with T 1 nd T 2. If there is current entering plte T 1, the sme current hs to exit plte T 2. Thus, the circuit cn e modeled s follows: T 1 T 2 C 1 C 2 M (c) Using the sme prmeters s in prt (), provide n expression for the equivlent cpcitnce etween T 1 nd T 2. Since the two cpcitors re in series, the equivlent cpcitnce etween T 1 nd T 2 is given y: 1 C eq = 1 C 1 1 C 2 Thus, we get 1 h = C eq εwl trin h εwl trin C eq = εwl trin 2h (d) Let s ssume tht insted of just detecting the height (y mesuring the equivlent cpcitnce etween T 1 nd T 2 ), we lso wnt to control it. Let s ssume tht the device we use to control the height tkes in only one of only two commnds: increse the height, or decrese the height. In prticulr, this device EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 20

21 is controlled y n input voltge. If tht voltge is greter thn 2.5V, it will push the trin higher ove the trck, nd if it is less thn 2.5V, it will let the trin move down closer to the trck. Assuming tht the trin is 100m long (L trin = 100m) nd tht the T 1 /T 2 metls re ech 1cm wide (W = 1cm), design circuit tht will feed voltge into the control device to mke the trin levitte 1cm ove the trck. Be sure to show how your circuit is connected to T 1 nd T 2, nd e s specific s possile in terms of the component vlues you would use. You cn use ny comintion of switches, voltge sources, current sources, resistors, nd cpcitors tht you would like to implement this circuit. The equivlent cpcitnce etween T 1 nd T 2 is inversely proportionl to the height of the trin. We re only llowed to mesure the equivlent cpcitnce etween T 1 nd T 2, we cn mesure wht the cpcitnce is to figure out wht h is. Bsed on whether h is lower or higher, we cn construct circuit using switches nd cpcitors (to figure out whether cpcitnce nd thus height is ove or elow the desired height) whose output cn then drive the height controller mechnism. First, let s figure out wht C eq should e t the desired height: C eq = 8.85 pf m 100m 102 m m = 442.5pF Let s use circuit similr to the one in the touchscreen l to mesure C eq. (Note tht there re mny other circuits tht cn mesure the cpcitnce s well. Any consistent nd functioning solution will receive full credit.) S 1 T 1 S 2 5V C pF S 3 V mes C 2 T pF ws chosen for the fixed cpcitor, so tht V mes will e round 2.5V. If the trin is higher thn it is supposed to e (which is 1cm), then C eq < 442.5pF. After chrge shring with the fixed 442.5pF cpcitor, V mes will therefore e < 2.5V, so the control device moves the trin down. Similrly, if the trin is lower thn it is supposed to e then V mes > 2.5V, so the control device moves the trin up. (e) So fr we ve ssumed tht the height of the trin off of the trck is uniform long its entire length, ut in prctice, this my not e the cse. Suggest nd sketch modifiction to the sic sensor design (i.e., the two strips of metl T 1 /T 2 long the entire ottom of the trin) tht would llow you to mesure the height t the trin t 4 different loctions. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 21

22 .. T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8. M Top View. One importnt thing to note out this circuit is tht it works only if extr cre is tken during the cpcitnce mesurement circuit. The equivlent model for this is: T 1 C 1 C 2 T 2 T 3 C 3 C 4 T 4 T 5 C 5 C 6 T 6 T 7 C 7 C 8 T 8 Therefore, the circuit needs seprte switches on ech T, so tht you cn mesure the cpcitnce etween only two terminls (like T 1 & T 2 ) nd so tht the effect of other cpcitors is nullified. 7. Homework Process nd Study Group Who else did you work with on this homework? List nmes nd student ID s. (In cse of homework prty, you cn lso just descrie the group.) How did you work on this homework? EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 22

23 I worked on this homework with... I first worked y myself for 2 hours, ut got stuck on prolem 5, so I went to office hours on... Then I went to homework prty for few hours, where I finished the homework. EECS 16A, Spring 2018, Homework 8, Lst Updted: :35 23