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Unformatted text preview: EE362L, Test 1, 22z50pm, 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 .
0Way 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  Flamf, WW'NW m www.mawm «Muwmrn'Kmt‘ﬂmYk‘mWI/mﬁmWWWuMMM“,h “WM Z, : CW) (4'70x/o"’9)(lwz~’002>¢ 0,0611% 70 ) : Diogltsec
"30 :7 56c, EE362L, Test 1, 22250pm, March 22, 2006. Name: ¥ ' , Page 2 of Problem 3. The circuit shown below is operating in periodic steadystate. One cycle of the current
i(t) is shown. The load draws constant current. Determine the peaktopeak 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 gnumlul—mmpwmiwmﬂww O+Lerts Q2; 7KSL , 04 2' £273:
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\m/ Mr“ M N, EE362L, Test 1, 22z50pm, 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: (4009 ‘7 Brita/MY“, /900—~> 874W/Mz/ [430 ——“> ﬂow/m 7* bsz/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" \.
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when SwillLL C/Oﬁetﬂ,dx’ﬂzeﬁf €025 +0 Q ’ﬁggeJD/t’ilﬁg M (allay/W5”
“Thus IPQAL :: rm, .l— AIL; :3 We“, BOW2r, T: 33.9599 7723/érézasec/ AI»: 3,331+, 1,1“: +3?” = 0.674}
40MB, 7’: aE/xsecz, ‘F/zrr 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, 22c50pm, 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
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(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 tumon, 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; [orSNV __ /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, 22:50pm, 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 ﬂirtr the. II' Mower MN“ Jim» 6242
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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. 0.580.62A b. 0.380.42A c. 0.981.02A d.0.78—0.82A Véesr'iw . M4: W Wiesfwé. Vqu  : io P. V1535 “210,415 OlgIOA', :5 hawk20.6 mmwwumwmmvwmww~mqutwm* EE362L, Test 2, 22z50pm, 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 noload 60Hz inverter output voltage with ma Ammummmmmmumu Vemim)
V...=4:v
V4131: WSW . “wW EE362L, Test 2, 22z50pm, 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. 0.110.14 .0i02o.05 c. 0.080.11 7d.0.050.08 e.0.140.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
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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
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p0%}b/e 9 ~+ww 26 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
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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 peaktopeak 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..
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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 +
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. . . . . . . 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
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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.
 Fall '11
 GRADY

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