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# exam2solution - EE321 Exam 2 Spring 2010 Notes You must...

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Unformatted text preview: EE321 Exam 2 Spring 2010 Notes: You must show work for credit. The last page of exam is blank for extra paper if needed. Good luck, and have a safe and relaxing spring break! 1) 34 pts. Measuring the a—phase impedance looking into a stepper motor, it is found that the inductance varies between 4 mH and 20 mH, and has 8 maxima as it is rotated over one revolution, and that the resistive part of the impedance is constant at 4 Q. If the machine is fed from a drive with a l2 V dc source, and neglecting forward semiconductor voltage drops, what is the maximum load torque that can be applied if the machine is to step forward correctly using single step excitation? What is the minimum load torque (i.e. the most negative value of load torque) which can be applied if the machine is to step forward correctly using single step excitation? R7”? LA +Le’ 3 20m” Quip, = 4MP) LA: l2mH “52919014 (4190917) == 122+ % («9&997 MH 9’ 7? Te " “'1 La RT [034 An RTGrM [a a‘ ,. 2 9 ldc. e 2 -Jixﬁxex? ,5“ \$97.,“ to“ NM Ta = «43-29% \$50897,“ Mm 7"MW 2 i Temx 2 1.. 0-299: 0'lAl1NM ?~ 2. 4dvammmznt Two 2) 22 pts. A separately excited dc machine has a armature resistance of 0.2 Ohms, a ﬁeld resistance of 10 Ohms, L,,/ of 150 mil. We wish to determine the currents (armature and field) in such a way as to achieve a torque of 2 Nm, while minimizing the loss. What should the armature and ﬁeld current be ? What will the machine loss be at this point ? Note: you may assume you are not against any voltage limits. I Resz‘m'i/e £09m; Othr an arm/vane amt bldd mum/1093 9> 3,0452 /a22ra+/Z“r; £55“ éqnua/(Z'Ly qufféﬂ DC machine 73 = Lap/alt: a) /a/,c = 72> or 1;: 723 M L0; Lap/a (QM/375: A055 5“ 57”“ 05% (Cam alto 60 ark/whim?) (firm; 0/) 1;) 12095 2 3211:34- (f; )2? Lag/a anz‘wge £055 5/ @055 e 0 £674 19 \$2720“ : ZWa/a ~2 Efﬁgy“ ., W Ia’ ”0 “‘> (52* = 22 Le? 7a 64/. a: agar/u (0.4.. )W 0”— (Emma Firm = 377/ n! 3.) 22 pts. A permanent magnet dc machine is being fed from a single quadrant chopper, as we discussed in class. The armature resistance is 100 m9, and the back emf constant is 0.2 Vs, and the armature inductance is 0.25 mH. The chopper is supplied from a 25 V dc source, has a 1 V forward transistor drop, and a 0.75 V forward diode drop. At a speed of 100 rad/s, the chopper is in discontinuous mode, and has a peak current of 5 A. How long is the transistor on? How long does the diode conduct (i.e. what is the duration of non-zero current in the diode within a cycle) ? WW Wéi9h>zr 1'; on m DC mam/time. Voltage WWW do ’A VOW “ lam + Vrsw + La Ma ”(D 072" m *‘V-VLUY lq: L“!!! '41 Oilq 2 lMK 2 “‘1 ~—-——~ L 'TDA Fran/JD 3’5 lmx A ”+5”; VW Lam 7’2)- . * L .. 4a “M 2 o-Qéxs moa’ , 0-553MA Var V95m*'l‘a'?a ~l<vzm 3-75 St‘milaxbj “9m Wanei‘slec (‘5 01H) NEE; : ﬁvd" Erarkvwa Low. :> Ed 2 lmx Laa __. 53y cameo” 2 0435‘? m4 var + Tam-i mm m 4.) 22 pts. A permanent magnet dc machine is supplied by a four-quadrant converter with a hysteresis control. The machine has no armature resistance, and the converter has no forward semiconductor drops (wouldn’t it be nice). The desired armature current is i: and the hysteresis level is denoted h. When the armature current falls below i: —h , the converter is switched such that the armature voltage is vdc. When the armature current goes above i: +h , the converter is switched such that the armature voltage is —vc,c. Derive an expression for the switching frequency in terms of h, vdc , Ln” , the voltage constant k‘, , and the rotor speed (or. W list < (it: ”“14. Twang/'5 {w an V'olc = Kvwz *Laa CL“ 0“? at? = an. at? 73/0 ’5 LQQ 1?}; :: V4L~I<Vw7 07 7:9“); (QALaq ’7' ON Vdc‘/<VWY Lia. > 6a M Tanner‘s {77’ a“ - Vote 2 may +Laa dict M \$ch ~ ~2 a m To“ >> La“ ‘2“ I; Vdc+kvwv 03‘ be'f :: 2A£aﬂ ”’7“ lawn/m kw: 729M {-TDH; 2 QALCM 4W0 2. Va! A “(Ki/wrlz a / 2- 0H; ( V Y3 m (Var. c kvwﬂ (Vac «Ho/cur) WW 4hLacc Vdc 414 Laa Vac \q E ...
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