introduction to power electronics, Summaries of Electrical and Electronics Engineering

Download introduction to power electronics and more Summaries Electrical and Electronics Engineering in PDF only on Docsity! MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.334 Power Electronics Practice Exam 1 Problem 1 qt Ty Eo _—— + Fig. 1 The buck/boost converter of Pig. 1 contains a very large inductor, L, with 2 winding resistance, R, = 0.19. The circuit is operating under the following conditions: D=0% =10A vy, =40V Determine 13. Problem 2 is Vp é a gts + xy y reek - uC N at al Fig. 2 The network N of Fig. 2 is supplied by a sinusoidal voltage soarce, », = Voinwt, and draws a current i, as shown in the figure. What is the power factor of the network N at terminal pair « - a? MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Compater Science 6.334 Power Electronics Practice Exam 1 Problem 3 Figure 20.2 shows an inductor made of a three-legged core with a winding of N tums on the center leg. The two outer legs have cross-sectional areas different from that of the center leg. What is the inductance of the winding? a) Figure Magn . The etic structure with one winding of N turns and three legs. m2 a outer legs have cross-sectional area A). and the inner leg has area A,. The core material has permeability 4,  MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.334 Power Electronics Practice Exam 4 Problem | A ferrite core, shown in Fig. 1, has a mean core length /. of 10 em and a cross- sectional area of 2 em’. Its sz turation flux density is 0.3 T, and the permeability of the core is 1250, (where x, = 4 *107 Him), a. 1f10 turns of wire are placed on the core, how much current can the saturating the core? without b. What is the minimum number of tums on the core necessary so that the core will not saturate when a 120 VW (rms !}, 6 Hz sinusoidal voltage is applied across the coil? Depth: 2 cm w= 1250H9 Figure 1 : Beat = 0.3T faire Problem 2 (a) What is the minimum value of V, needed to permit the circuit of Fig. 2 to operate at D = 0.27 (b) Sketch and dimension i,,%,, and vs for D= 0.2. Fig. 2  a G234 Buser Elechromics |, Racha Bee L Solo [Posi] a DI,- (->) <M, eur: (2-t ae (+(-dy\,- Gee 150. SE DY FOOD, TAR EO Lk ~via se —G-D yy, = (we )2g - DY, - -[v,=-2 = -2. * D(i-0) TR _. — = == 80 a a! jo. a Re SE ee es a - < Peoe Zh N's. ems Fe _ Ass ssa x si IY reer pore or oO _ ee oe <P R= =\ Nise © A@ = YE 95 9¢ ein ! 9 —— Sv r — = a a re en cs Noa B oe 6334 Pouee Electronics Tobie Exam 1 Sol’as Peso 3) Note! TOS \6 Cxanpk ZO. 10 ksy 4 ke Zhi ky g, PAY Raa ky 7 Rat "Kt zee eat CER Ae i= - : eA Ar a Nb ot is fa ao A- = fick 8 oad NeweAAa | | AiO (CE Rse ey | — + =: eee a | 7 = md | : | — _— a  64 Zrser Cleckomes edu Exam 3 Sal as D a Z| ra | 5 , Be Se. off: Ngo = 2412 = ZNV Su. Sa 3 Ass = WSR \O= YOR _ i a eed, suo tel hes Ms he, = en Noth Ha es “ost ek 4c" _ - (34 berteonre cha) _ : Sok Soff “Sot Sion Hof S, on 4, ok. of / 1D <P2= AR > Asa £ o oe Sects | ce eee ee ee es a seo eee | ee ore eee ae f @ GCA3H Rover Elockeantcs Prackiie Exan % Solas [Reb 2) Vea. {EQEND) ee . ae oo mined oe le eile NPD = vou = 2 VT os@, + 2V,T6s(@,) | <es EMEP HINTS O50, _ wae a ae oe = e / = i a ea x i Pais Vems Fens z a (v2 ave itteatary oo — K Ecos @ + YaTs Cos Oy a PS Se ee — \ See iran yp  CQ BAA Fewer Electronics Pacha Exand Solas \Rootla A= Ne = NBA, _ Nt 2 NYeAG so b> eo Fa — ge tere =LA. > NoA,= PBFA » a A iin = Bsaz Re a 37 Coun) ea Nw Gorsoamest) a § Va j . << e > » Coe Ceayrce B = Bcnt as N= BAe @ Az Svat = NSA } i @ 65142 , Wwe 233 col/se , mex Cvae «Soe t 3 Cycle TT 1202 TSO = -\-- ot ss oreo = aa, WH ile soPde = = Ga) Te saps VD = = (Ve “VO G- ey So trans Seamer Coce oll reset dann, 1-0 hee gered. _ ee an oe _ - D oe - oo a oo se Ne = Nistepnt Vie eee — ee JL jg magi oO _ z s— ee \ Y > =k ss = =e, ioe nec ect | ee = ----f- -p---- of TTT + + -— LNoe etd es hides clema yatta se \s fae glad ot his SP —_ 4 eet