EE362L_Spring06_Tests - EE362L, Test 1, 2-2z50pm, March 22,...

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Unformatted text preview: EE362L, Test 1, 2-2z50pm, March 22, 2006. Name: g BY , Page 1 of 50 minute test. Seven multiple choice questions. One 8.5x1 1” note sheet permitted (both sides). No laptop computers. Unless otherwise indicated, you must show sufficient reasoning, equations, and steps to justify your answer. Circle the correct answer. I intended for the correct answer to be one of the choices, but if not, then choose “other” (if provided) and write in your answer. Do not unstaple. Seating assignments. No partial credit. If you finish more than 10 minutes early, quietly turn in your test. Else, wait until the end of class to avoid disturbing others. Problem 1. one c. clepof aperiodic voltage waveform is shown below. What value of D yields an rms value of (0.2,; . ' ,bsgo‘aaw z T i , .... 9 2. 7,27,”, : #[WZDT + (DZD '2 " 0 Vm v(t) DT DT I L = (1)117 MAYBE $29 .a T a . 0-Way Mam—70:7, 9‘ (991245 10.5‘ D=0olZ§ :03) Problem 2. A DBR serves a constant (663507 , computer load. The DBR’s 470uF capacitor voltage is practically constant at 160Vf““Suddenly, there is a power blackout. The capacitor voltage starts to drop, and when it reaches 100V, the computer trips Off. Assuming constant load power, how long does it take for the computer to trip off? Pea/6:. MIA/vac, JZCQ/PSK - Flam-f, WW'NW m www.mawm «Muwmrn'Kmt‘flm-Yk‘mWI/mfimWWWu-MMM“,h “WM Z, : CW) (4'70x/o"’9)(lwz~’002>¢ 0,0611% 70 ) :- Diogltsec "30 :7 56c, EE362L, Test 1, 2-2250pm, March 22, 2006. Name: ¥ ' , Page 2 of Problem 3. The circuit shown below is operating in periodic steady-state. One cycle of the current i(t) is shown. The load draws constant current. Determine the peak-to-peak ripple voltage on the IOOOuF capacitor. 1(t) 9 Avg. VA/ue 0f 1‘ m =- 2,74- constant —> r C TE”’T?EWTZ5;V -_ onthei(t) scale A V p C “ '{g'ZE '“ g jog; ;: @tS—V 0 wees 3A ) 0. 375 V W“ ‘ ”“" ”" W" " “Wm-«wwme 7pm 0,373v » L , S'A‘ I O! (03.5 V - Problem 4. The conventional firing circuit of a triac light dimmer is replaced with the one shown below. If Van = 120x/5 sin(1207zt) , and the diac has a 32V breakover voltage, what is the firing angle a? ( \ R2 . ._ V3 * Light F 6m X kll‘ez RK bulb W m. mmmmmmmmmm ~W.wm..m_m.mmw w a b 6/ij : V355 err/{2,3 : 3.2 VP Aw noJ‘z em o< = 0,4713“:— 254 “W WW.W gnu-mlul—mmpwmiwmflww O+Lerts Q2; 7KSL , 04 2' £273: (0 KSL J 1'3 got K47, ‘ 3‘ 343$” Useg‘8 for the lower resistor \m/ Mr“ M N, EE362L, Test 1, 2-2z50pm, March 22, 2006. Name: K ‘1 , Page 3 of 4 Problem 5. A portion of Monday’s solar radiation data is shown below. The afternoon sky was so clear that diffuse horizontal radiation was practically negligible. Use the global horizontal m, CDMflM’Ul” together with and panel orientation, to estimate the incident solar power on our panels at ((1‘530,‘?'152100,1g§0,0,Z/:(399 hours, in W/mZ. GIG lord (05 WNW} 5M) N D "R! c-+ NDRMAi '— Global Cosine of Incidence Incidence anle zeth anle Horizontal; W/m2 Nokmnl ow pawl : (Pf/Met Nflm‘ll» (C09 lg> K 45.5 30.5 48.2 33.8 GED 51.0‘ 37.3 :3" " / z ' 1545 53.8 . 40.7 X W M 1600 560 56.7 44.3 mm: (400-9 ‘7 Brita/MY“, /900—~> 874W/Mz/ [430 ——“-> flow/m 7* b-sz/u __..._ .1 A 5 Problem 6. A buck converter operates at the following conditions: 0,3, Oj'kHz, 100W, 40V input, 20V output. L = IOOuH. Determine the rasvalue of the inductor/current. 1, -—=>§A Peak ' 0 : tgo LING" \. 239%: 29415“- ;WHCQ c/osec/j "' VIN ' Wt .YQJR Vz‘z/Qfljf’c g]; 3 VI ,.Vm+: 60/ Cuaflew flt‘fies filmy/0 m)? 24 dde' f by v /OQ;:H when Swill-LL C/Ofietfl,dx’flzefif €025 +0 Q ’figgeJD/t’ilfig M (allay/W5” “Thus IPQAL :: rm, .l— AIL; :3 We“, BOW-2r, T: 33.9599 7723/érézasec/ AI»: 3,331+, 1,1“: +3?” = 0.674} 40MB, 7’: aE/xsecz, ‘F/zr-r firs/w, 43;: 2.55 , rpm“ mgr 5.2% 7OKH?’ [‘f/ 3,1,0ch ‘77; : 7,14%;“7 41: 3 1,42%”, a gwélgasflmr EE362L, Test 1, 2-2c50pm, March 22, 2006. Name: E Fig l , Page 4 of 4 Problem 7. Considfie act V a1 fl cue for a solar panel. A boost converter is connected between the solar panel and i 0 load resistor. For the sun and panel orientation conditions represented by the P- \ curve: what duty cycle D will draw maximum power from the solar panel? Panel Power versus Voltage _ w \ ,,,_.l,,, WNW. ........... . ._ m ,me i t...“ 110 - l V 100 T r 90 e 80 1 7o — 60 e ‘ 50 ~ 1 i 1 4oe ; ; ' 7 = ' 3o — 3 i I l i I ‘ I 3 20 — 10 ~ Power - W . 01 1‘ l 'l\ 1 l r i Voltage - V Z3 V V2 *2 LOAD F b I Rump MM >VLMD”8/mx LDAD' l 2 ~L—‘2 2 AM Vw/w " (HQVPM; $0 VLmo 14>) VFW I a *2. 4 S0/ KM) (l4) ZVPAWL ) (“DB2 :%%0 __ : Nu- V 2% ~ . .4) .— S W —— — ; EL w —— :2 (l 3 ) a DJ W D “1 metro) 3‘“); v a 7L Croap=0qzr (VD :QBJVP :Wbusw (9% W Melvwlgxr W “M” '1’ (on) [094696 H EE362L, Test 2, 2—2250pm, May 3, 2006. Name: \ K E , Page 1 of 4 Work any SIX of the SEVEN problengi‘aQVDICATE HERE: DO NOT GRADE PROBLEM One 8.5x11” note sheet permitted (bot 1 es). No laptop computers. You must show sufficient reasoning, equations, and steps to justify your answer. Circle the correct answer, or answer as otherwise indicated. I intended for the correct answer to be one of the choices, but if not, then choose “other” (if provided) and write in your answer. Do not unstaple. Seating assignments. No partial credit. If you finish more than 10 minutes early, quietly turn in your test. Else, wait until the end of class to avoid disturbing others. x. Problem 1 (3 points). The input side of an optoisolator is turned on and off b “a 12Vdc”, ‘ower supply. For proper tum-on, the opto requires 20mA @ 1.5V. To achieve this, w “series resistance must be inserted between the opto and the power supply? IK: *IJ'V; [or-SNV __ /0/; w a 4 R .. (9,020 m 585's?” a. 8259 b. 3759 c. 6759 d. 5259 - v Problem 2 (3 points). A MOSFET’s gate is a capacitor. COnsider a case where a 12Vdc driver chip turns on a MOSFET through a 1000 series resistor. When the driver output goes “high,” VGS increases to 4Vdc ir@y.2use§é Compute the capacitance of the MOSFET’s gate. ‘ Hint - use charging eiipression v(t) = 12(1_ e—t / RC )_ EE362L, Test 2, 2-2:50pm, May 3, 2006. Name: L K E I , Page 2 of 4 Problem 3 (3 points). The MOSFET firing circuit shown below is not exactly like the one you used in your inverter - I have changed the output stage and created a problem. What is the problem, and what will be its impact? Explain in two or three sentences. Answer: 77m diode is flew/356°, ' 7T] {9 mew/vb flirt-r the. II' Mower MN“ Jim» 6242 Micki: +Mo Mm 110.: I“ Eesfe-fge. guide will HA6, (220012.» flfiél'fi'iwfl. Th {5 (‘5 QWI.M WI! ‘ We ww. Problem 4 (3 points). Every capacitor has a current rating. Consider the lOuF, 50V bipolar high- frequency capacitor that you used in your inverter’s output filter. This capacitor is used along with a 100 uH inductor. Find the rms current in the capacitor for the (following situation: 0 No load attached to output Vac. a o The inverter’s voltage (on the lgftside of the ac filter) has 25Vrms of 60Hz, a El lOth’lS of b _:'— Eutpm 40kHz. ' ww'i * o No other frequencies are present. Hint - the total squared rrns current is the sum of squared individual rrns harmonic currents. tumwfl/ Fan. «90 Hazy M} g , ‘— FEM 40Ksz 325’“; + l ' i ,7; 0.4 05’ A a. 0.58-0.62A b. 0.38-0.42A c. 0.98-1.02A d.0.78—0.82A Véesr'iw . M4: W Wiesfwé. Vqu - : io P. V1535 “210,415 OlgIOA', :5 hawk-20.6 mmwwumwmmvwmww~mqutwm* EE362L, Test 2, 2-2z50pm, May 3, 2006. Name: K i; , Page 3 of 4 Problem 5 (3 points). An unfiltered PWM inverter has a sinusoidal 60Hz Vcont input. The inverter’s triangle wave frequency is 20kHz. The FFT of the inverter out wish?) 5 a strong 40kHz cluster component that is 8.1dB down from the 60Hz component. If Vdé‘jOV, nd ma = 0.8, what is the rms value of the 40kHz cluster voltage in volts? (Note — do not use t ewifi: table from the lab document). v0 , Véa :1 WA é: Rm: :2 22w V60Hz,rms 487.! V _....—« linear overmodulation saturation g: w :5 0l Variation of RMS no-load 60Hz inverter output voltage with ma Ammummmmmmumu Vemim) V...=-4:v V4131: WSW . “wW EE362L, Test 2, 2-2z50pm, May 3, 2006. Name: t , Page 4 of 4 Problem 6 (3 points). When using your inverter as an audio amplifier, you tested it with a pure 1kHz sinusoidal input. Assume that the FFT of the amplifier output showed 0 a 3kHz component whose magnitude is of the lkHz component. M (end 3, IN V o a 5kHz component whose magnitude the lkHz component. Fem “N H» 0‘? o no other significant frequency component‘s”. . , WV Jnmgflfig I Compute the total harmomc dlstortlon (THD) of the amplifier s output voltage. THU? \5 (0,04%? (0,0202 r: 0,0471 fHD’tM: “mm. fimwmmawvmwhm W Problem 7 (7a, 7b, and 70 form one problem). No justification is needed for your answer. Problem 7a (1 point). You pay about $0.12 per kWH for retail electric energy. About how much does it cost electric utilities to produce wind energy (in $ per kWH)? .w a. 0.11-0.14 .0i02-o.05 c. 0.08-0.11 7d.0.05-0.08 e.0.14-0.l7 Problem 7b (1 point). Roy Blackshear’s wind farm in west Texas has more than 100 wind turbines. What is the power rating (in MW) of each of his wind turbines? r my ' i I Problem 7c (1 point). Regarding our tour of the Fine Arts Building utility room, what is the approximate cost range of motor drives like the one we saw (in $ per horsepower)? /‘ a. 500 .250 i . 50 ‘ e. b 7 d 750 , ' \ r . . /‘t EE362L, Spring 06, Final Exam. Name: r ' ' Four equally weighted problems. To receive credit, you must show all work. Do not unstaple. Problem 1. Determine the rms value of the voltage waveform shown. B is a variable. Hint — divide the waveform into the triangle section, and the constant section, and then ‘ combine the squared rms values of the two sections. 0 T/2 T 3T/2 2T vflfgs E Ave/mgr; value; mt" «the. figufimfl wwvaJEiM/l. FDR MM. 7?le "Tl; Sada/Ut’5) the, fluent/.396, gin/m.ch value, he slow» w w (vatrwzuvr, Ma 7cm, We, tkfwgk shown, HM; W2 (Jg +143 (sz1 iii!”- 32 "Ti. : ma as» = 2% w a): erg-«i G ' PW: £11: .émfl +1“; Ave/Mara. atria/31ml mam; [g 82, ““ l ' (Om bf“ (£15 I VKM:: : Big; 4"" 7] ,. ’7 3 _ A ‘ ‘m 2 Wins: 82PM: la 826%} B [3%. IIZ‘WB (OK amt émw‘q Chat) Naggm Ivo+ Mug/r clam/1% for: fiNswe/es 0a l—s/‘e/e +££ p0%}b/e 9 ~+ww 2-6 fl/lmyte, I 94' {2% fiwswma Hurt hum ‘7'" ,‘M +59% EE362L, Spring 06, Final Exam. Name: E E z Four equally weighted problems. To receive credit, you ust show all work. Do not unstaple. ‘ Problem 2. Last Thursday was a brilliant solar day in Austin. A screen shot of our rotating shadowband pyranometer data is shown below. Graphically estimate the kWH that would have been produced by a horizontal solar panel having 1m2 of surface area and 14% efficiency. " IE'EEQIIIII Aria ‘12 512‘ “a DN " i ‘ x g I naaaaam / Al 2L 1’ 2r 2a 27 k \ nun HAVEN HEM w M65111y¥2008 ‘ i ' ClockiNmn(Standetdlirfie1_ L .41” l 7 ne‘htvurgridspecing, A A hOI‘iiEWJ‘Ml pawl sates global homzmmi( (é»H3 INaLegZ/mle, 44% MM UNJe:Q Hm GH Came) Wham. 6min WK 1'5 (ZOOW/ML)°(LMJ: @wa (3; Whale“ boxes) pill!» Lyn”: box?» Plug, WWQWM My?) =, az—l 95+ 3% :2 32+5fi+é = 3? tom 33 Wm " 0.200 =‘--» 7. @ {cwH/mz Muwg % 41:, 044) was , Nome—v N0+ matrix ween/:4 750,4 ‘uswd/ +Le “Jaw; Wflve. All/960825 A» {M41434 \, W” +0 KW'H, like, KWH) 5AM” have baa/u Wig/70H «(flag/7+ La 57‘~uc/m/s wz‘w [mud 'e/(W/é’lé’w’de, / EE3 62L, Spring 06, Final Exam. Name: 3/ Four equally weighted problems. To receive credit, you must show all work. Do not unstaple. " ' Problem 3. Help me design a boost converter for a solar array that will be moved from ETC to ENS this year. The maximum power condition for the array is lkW @ 64V. The boost converter will drive light bulbs at 120V, using a switching frequency of SOkHz. a. Determine the smallest input inductor (in pH) that limits the array’s peak-to-peak ripple current to 1A at the maximum power condition. a b. Neatly sketcgfithe input inductor current, and detenéu its rms value. 9009+” CON WNW—l, Vow. :3: LIN » JFW jaw-$1.. IWD WW (2.0V When $50”ch duff, V - I I : VJ“ “if; IN AI: w. T: VIN D g d t / '6 L— ) L D L Amiga"; V Awake" ‘ ~ IN USINQ “MD” ) \ —.D : VIN D a” { w m ‘ ‘ a; i m Vomw ) “W VOW -— ’ 2C) "(J1 l ' ‘, :V.‘D “I'VE/VD, 7; £4) LMW) Rewmle. aI .3; LMW," WIAI mmmwwmwmmvmpmwx mmuww. amwwmhvwmimnmwmmmmwumx :WZZflm +63) m mwmmmlum.mnu4~m vw.m.._..mmw /000w __ V 64V - 15.634 EE362L, Spring 06, Final Exam. Name: Four equally weighted problems. To receive credit, you must show all work. Do not unstaple. Problem 4. The output of your PWM inverter is connected to the same LC filter that you used in the lab. No load is attached to the filter output. 100” H + + _(YYL:L . . . . . . . Filter ConSIder a s1tuation where the in ut to our 1nverter 1s a erfect 3kHz sme Inverter I . . P. y P . ~ output lOuF output wave. DUe to nonlmear1t1es, your inverter output has the des1red V3 kHz _ ————iL — ~ _ . ‘ VJ component, but also a V9 kHz (1.e., 3rd harmonic) component. No other 4. imm—T ' + frequencies are present in the inverter’s output. (No 44 k H E I e +0 “We When you measure the THDv at the filter’s output, you get a value of 0.01. What is the THDv at your inverter’s output? Hint — you will need to use the magnitude of the output filter’s transfer function at 3kHz and 9kHz. Also, remember that THD E (rms value of harmonics) divided by (rms value of fundamental). Hem. we lmtwe a, ‘FMNciflmemlfl/e (429“) V3.4”; WWQ ON 8. hfiflMONfC (qun>v‘ WMA Opel WANDA/"C pRQsQ/o-t I fiAQN THD :: WWW)- .. ‘ i V3 we] Elmer [1/ WWA. 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This note was uploaded on 04/03/2012 for the course EE 462 taught by Professor Grady during the Fall '11 term at University of Texas at Austin.

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EE362L_Spring06_Tests - EE362L, Test 1, 2-2z50pm, March 22,...

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