Power Electronics Exam: Final Exam for EE Students, Philadelphia University, 2010/2011, Study notes of Power Electronics

Download Power Electronics Exam: Final Exam for EE Students, Philadelphia University, 2010/2011 and more Study notes Power Electronics in PDF only on Docsity! 1 Philadelphia University Student Name: Faculty of Engineering Student Number: Dept. of Electrical Engineering Final Exam, First Semester: 2010/2011 Course Title: Power Electronics /Drives Date: 24 /1/2011 Courses No: (610581) + (640324) Time Allowed: 2 Hours Lecturer: Dr. Mohammed Tawfeeq No. of Pages: 3 Question 1: (10 Marks) Objectives: Understanding the principles of power electronics circuits and devices. Choose the correct answer for the following: 1. A single-phase full-wave fully-controlled rectifier has a resistive- inductive load with high L/R ratio and a supply of 110V at 50 Hz, The average load voltage at firing angle of 75° is: (a) 25.6 V (b) 30.6 V (c) 40.7 (d) 70.14V, . 2. A single-phase full-wave half-controlled rectifier (semiconverter) has a resistive- load with high R = 10 Ω and a supply of 220V at 50 Hz , The average load current at firing angle of 90° is : (a) 4.90 A (b) 9.90A (c) 7.86 A (d) 10.86 A. 3. A 3-phase full-wave controlled bridge rectifier, if the phase input voltage is 240 V, at 50 Hz , then the average output voltage at firing angle α = 30° is : (a) 233.91 V (b) 336.95 V (c) 486.0 V (d) 501.28 V. 4. In a dc chopper, if ton is the on-period and f is the chopping frequency, then output voltage in terms of input voltage Vs is given by (a) V s . ton / f (b) Vs .f/ ton (c) Vs / f. .ton (d) Vs . f . ton 5. The output voltage waveform of a single-phase DC-to AC converter of parallel inverter type with resistive is a non-sinusoidal wave contains: (a) Even and odd harmonics (b) Odd harmonics only (c) Even harmonics only 2 Question 2: (10 Mark) Objectives: Understanding the principles of AC-DC converter. A three-phase half- wave controlled thyristor converter shown in Fig.1 has a resistive load of 10Ω and a supply phase voltage of 240V (rms) at 50Hz. (a) Determine the values of average load voltage and current, and load power for firing angle α = 30°. (b) If the load is replaced by a dc separately excited motor with Ra = 0.3 Ω and La = 15 mH. The load torque of the motor is constant and requires an average current of 25 A. The motor constant Ke Φ = 0.1 V/rpm. Calculate the speed of the motor for firing angle α = 60°. Fig.1 Question 3: (10 Marks) Objectives: Understanding the principles of DC-DC converters and DC drives. In the chopper circuit shown in Fig.2, the input voltage Vi = 220V, the load is resistive- inductive with L = 15 mH and R = 5.5 Ω. It is operating with T = 3 ms and ton = 1.5 ms. (a) Determine the duty cycle and the average values of load voltage. (b) Determine the minimum, maximum and average values of the load current. (c) Determine the rms values of load voltage and average value of the input current. (d) Calculate the ripple factor and the chopper frequency. (e) Sketch approximately to scale the output voltage and current wave forms. Fig.2 TH Dfw Load io vo Question Z Cia Marks) @) Vinca = V2 X24e ~ q bo DE Vde - sve Vmax ¢ “= OS XK e bar = B43 KVR X 240 cas3e = 2 = 243V. ” Vde . 243 _ / . Tae = a7 = 2-24 3A Po= Tae R (24-3) x0 = sqooW . com (b) For ytofur lead P , __ Eb= keg on be Ve Ebr TLaRa o—>f-—_— V =Kedn + Taka ' ! | = o.l 14 29X03 le o wo ne NEES ol but VeVde = BE Vina Cos, — WERER240 coe 66 zt 27 = 14¢-3V - vo nM - 1403-5 _ i328 rpm: er [so] Quespon 3: Cla Marks) Solanion ; For the Chopper cirewit Showa Load Voltage Waveform in Fig : Vv Ve 4 Ca) Vo =Voy = Ve pil ae _ tan LSms Os gms 7 oS te of Vay = e-S4220 = NOV. fom] b +, , + 25-58 | Taagsbs Load (©) Lonag = Mie 4 BE vy seen To Ganon =dto , 45i® _[- Tay 720A 55 2xisne® ings hays “ 18 ' = 204 5.5 = 255A ~~ ST . u Sims sas Lin = oe Eel Vay 2b fe) @m) = 20-56 =14.5,A. ' Lay — Lm +l 25.5414. an = amen Soma = ASUS - 20, Cam Ce) The rms Value of the okt pur voltage: Vorms = VE Vi = Yak X220 = 1555V The cwerangee, vekue. of the tn pur Curren’: sin gay =¥loy = 0.5x20= 10A A) ripple Pacts Fe yf/ind 2 (A) ripple factor RF ye = HSS. Che reguency Pao . Lo 4 _ = - peer Tay Te = = Sayer 73S ne - [2m] _ 3- Qvestion 4 C10 Marks) Solu Ve =< loo Sintet o Va = ioo and Vs = 92 we) Veams = Vey de (Carmex) + Sinan) = Foor L(w-Byebsmnae wap cat win ce) onpur vag 2 7) 0.ce6 + Ot waveferm [23m ) = $3.42vV Thame = Med feapedome = BLY ayegneneg = Bai s127- A. Lamu] (c) The Ourput Power fe = Vive Liem = 68-4EXE2 F = G04 [am If the Fail to seperate , Hie yout wy Qh behave as half-wave , Single ~ phase rechrferr, the cout pur Yolrage writ be de with qweraga V vakue of : Vale = Vm (t+Cosax) = 1O8 (i+ cos 63) wv = FY amy ath,