Power Electronics
|
|
- Roland Cobb
- 6 years ago
- Views:
Transcription
1 Prof. Dr. Ing. Joachim Böcker Power Electronics Last Name: Student Number: First Name: Study Program: Professional Examination Performance Proof Task: (Credits) (0) (0) 3 (0) 4 (0) Total (80) Mark Duration: 0 minutes Permitted auxiliaries: Nonprogrammable calculator without graphic display Drawing material (circle, triangle, ruler, pens ) Please note the following remarks: You may only attend the exam if you have registered for it in the system PAUL. If you participate without registration your generated results will not be evaluated and accepted. Please hold your student identity card with photograph in readiness! Please label every blank sheet of paper with your name and your student number. For every task please use a new blank sheet of paper. Please do not use pencils and red pens. Every numerical calculation has to be furnished with units. Non-compliance will lead to point deduction. All approaches have to be documented in a comprehensible way! Specifying a single value as final result without a recognizable approach will not be evaluated. Good Luck! Power Electronics Page of 9
2 Task : Buck Converter (0 Credits) For an automotive application a buck converter is used. The output capacitor C is very large so that the output voltage ripple can be neglected. All components are assumed to be ideal. Please assume steady state for the complete task. Figure.: Circuit diagram of buck converter with resistive load The converter operates in discontinuous conduction mode (DCM). The chosen and adjusted values are given as follows: Input voltage: U = V Load resistor: R = 0 Ω Duty cycle: D DCM = 0. Switching frequency f S = 50 khz Furthermore the plot of the input current i over two switching periods is shown in Figure.: Figure.: Input current i over two switching periods. Calculate the average value i of the input current i. Determine the power P = P = P which has to be provided by the DC voltage source.. Calculate the values of the output voltage u = U and the output current i = I..3 Determine the value of the inductance L..4 Determine the current conducting duration T off of the diode. Draw the inductor current i L in the diagram in Figure. and calculate its average value i L Power Electronics Page of 9
3 .5 Now the buck converter should operate in boundary conduction mode (BCM). Determine the minimum possible duty cycle D BCM that has to be adjusted for that operation status. How large are the current ripple i L and the average current i L?.6 Insert the known points of DCM and BCM in the diagram in Figure.3 and sketch roughly the characteristic of the average inductor current i L over the duty cycle D. Figure.3: Average inductor current i L over duty cycle D Power Electronics Page 3 of 9
4 Task : Boost Converter (0 Credits) A boost converter is used in a PC for voltage adjustment. The circuit diagram of the converter is shown in Figure.. Figure.: Topology of boost converter The following values of the converter are known: Input voltage: U = 5 V Output voltage: u = V Inductance: L = 00 µh Switching frequency: f S = 00 khz The boost converter can be considered as ideal and operates in steady state. The input DC voltage U is assumed constant. The output voltage ripple u can be neglected.. Derive the ideal control characteristic U /U = f(d) of the boost converter. Set up the equations for i L (t = DT S ) and i L (t = T S ) as functions of U, U, D and T S in order to solve for the asked voltage ratio U /U. Assume that the boost converter operates in continuous conduction mode (CCM). Calculate the duty cycle D.. Determine the current ripple i L of the inductor current i L (t)..3 The boost converter operates in CCM. The load resistor R should now be adjusted so that the converter operates in boundary conduction mode (BCM). Determine the required resistance value R BCM. Now the load resistor is further increased (R > R BCM ) to force the converter to operate in discontinuous conduction mode (DCM). In order to keep the output voltage constant the duty cycle is changed to D = Draw the transistor voltage u T (t), the inductor current i L (t) and the diode current i D (t) over two periods in the diagrams in Figure.. Determine the current conducting time span t D of the diode (after the transistor has been switched off) Power Electronics Page 4 of 9
5 Figure.: Transistor voltage, inductor and diode current.5 Calculate the average values i L and i D of the inductor current i L (t) and the diode current i D (t). Determine the value of the load resistance R Power Electronics Page 5 of 9
6 Task 3: Four-Quadrant Converter (0 Credits) A suburban train is equipped with a four-quadrant converter (4QC) to rectify the AC line voltage and to provide a nearly constant DC voltage for the traction drives and the on-board electrical systems. The system is shown in Figure 3.: Figure 3.: Transformer, four-quadrant converter and DC-link between grid and load The following physical quantities are known: Grid voltage: U AC = 5 kv Grid frequency: f = Hz Secondary-side leakage inductance: L σs = 6.5 mh DC-link voltage: U DC =.5 kv The four-quadrant converter can be considered as ideal (no losses, no interlocking times). The primary leakage inductance and the winding resistances of the transformer can be neglected. In the above diagram L σs is the transformer s leakage inductance. So, the transformer circuit symbol is considered as ideal transformer. 3. Draw the structure of the PWM unit for interleaved switching mode in Figure 3.. Input is the normalized reference voltage, output are the two switching signals for the two legs of the 4QC. Figure 3.: PWM unit for interleaved switching mode 3. What value for the switching frequency f S has to be selected so that the maximum occurring inductor current ripple i max does not exceed 00 A for interleaved switching mode? Determine the frequency of the current ripple that results from interleaved switching mode. For which values of s ref (t) does the maximum current ripple appear? Power Electronics Page 6 of 9
7 In Figure 3.3 the root mean square value of the secondary current I and the power P drawn from the grid were recorded during the ride of the train from one station to the next. The ride required a total amount of energy of W = 3.5 kwh. Figure 3.3: Diagrams showing secondary side current (top), active power (centre) and reactive power (bottom) 3.3 Calculate the transformer turns ratio ü with help of the given diagrams above. Determine the duration t total of the train ride. 3.4 Draw the grid-side phasor diagrams for the times t = t (end of acceleration phase), t < t < t (constant power phase) and t = t (start of the decelerating phase). Calculate the phase angles φ (angle between grid voltage and transformed converter voltage) for the three cases. It can be assumed that the controller operates in power factor correction mode (cos(φ ui ) = ) leading solely to active power drawn from or fed into the grid. Use the following scales: 0 kv 5 cm; 00 A 5 cm. 3.5 Calculate the reactive power Q which is required by the transformer and delivered by the 4QC for the three cases of subtask 3.4), insert the calculated values in the power diagram in Figure 3.3 and sketch roughly its course. Task 4: Line-Commutated Rectifier (0 Credits) Power Electronics Page 7 of 9
8 A 3-pulse thyristor-controlled converter is connected to the 3-phase grid via a Dy-transformer configuration and supplies a resistive load. The complete configuration is shown in Figure 4.: Figure 4.: Line commutated rectifier in M3 configuration The transformer and converter can be considered as ideal (no losses, no magnetic leakage). 4. Sketch the output voltage u d (t) in the diagrams in Figure 4. for the control angles α = 0, 45 and 35. What is the control angle limit value α L that represents the boundary control angle between CCM and DCM operation? Figure 4.: Normalized output voltage u d /u for different control angles 4. Calculate and draw in Figure 4.3 the control characteristic U d /U dmax = f(α) of the converter for the continuous conduction mode (α < α L ) and discontinuous conduction mode (α > α L ) range Power Electronics Page 8 of 9
9 Figure 4.3: Control characteristic of converter for CCM and DCM 4.3 Determine the active power P R which is consumed by the load resistance R as a function of the control angle α for the continuous conduction mode (CCM). 4.4 The converter supplies a resistive load. Describe in full sentences why reactive power appears on the mains side as well? Assign the different types of losses to its causes. The following mathematical relations might be needed for some of the subtasks: sin(x ± y) = sin (x) cos (y) ± cos (x) sin (y) cos(x ± y) = cos (x) cos (y) sin (x) sin (y) cos (ax)dx = x + sin (ax) 4a sin (ax)dx = x 4a sin(ax) Power Electronics Page 9 of 9
10 Solution Task ) Buck Converter [0 Credits].) Average value of input current and power drawn from the DC source: i = D DCM T S i max = D T S DCM i max = 0, A = A P = U i = V A =.05 W [3 Credits].) Output voltage and output current: U = P R =.05 W 0 Ω = 3.4 V I = U R [ Credits] = 3.4 V 0 Ω = 0.34 A.3) Value of inductance: + i L = U U L [ Credits] D DCM T S L = U U D i DCM T S = L V 3.4 V A 0, s = 40 µh.4) Off-time, inductor current and average value of inductor current: + i Lmin = i Lmax U T L off = 0 T off = Li Lmax H A = = 0.8 µs U 3.4 V i L = (D DCM T S +T off ) i max T S = (0. 0 µs+0.8 µs) 0,876 A = 0.34 A = I 0 µs [4 Credits].5) Duty cycle for BCM, inductor current ripple and average inductor current: Power Electronics Page 0 of 9
11 In BCM, value of average inductor current is half of the inductor current ripple: i L = i L = D( D)U T S L i L = U = DU R R Equating the two equations gives: D( D)U T S L = DU R ( D)T S L = R D = L T S R D = L 40 µh = = 0.4 = 0.6 T S R 0 µs 0 Ω i L = D( D)U T S L = i L = i L =.44 A = 0.7 A [5 Credits] 0.6( 0.6) V 0 µs 40 µh =.44 A.6) Average inductor current over duty cycle: + [4 Credits] Task ) Boost Converter [0 Credits] Power Electronics Page of 9
12 .) Control characteristic and duty cycle: U U = f(d) =? Setting up the two formulas for the inductor current (steady state is assumed): i L (t = DT S ) = U L DT S + i L (t = 0) i L (T S ) = i L (t = DT S ) + U U ( D)T L S = i L (t = 0) Solving for the searched voltage ratio: U DT L S + i L (t = 0) + U U L U DT L S + U U L U D + U U L L ( D)T S = 0 ( D) = 0 U D + (U U )( D) = 0 U D + U U U D + U D = 0 U U + U D = 0 U + U D = U U U D = U U ( D) = U ( D)T S = i L (t = 0) U = U D Adjusted duty cycle: D = U U D = U = U U V 5 V = = U U V [6 Credits] Power Electronics Page of 9
13 .) Inductor current ripple: i L = U L DT S = 5 V 00 µh µs = 0.95 A [ Credit].3) Resistance value for boundary conduction mode: i D = i R = U = i R L ( D) = i L ( D) max U = i L ( D) R R BCM BCM = U = V = Ω i L ( D) 0.95 A ( 0.583) [3 Credits].4) Current conduction time span of diode and diagrams: ++ i Lmax = U L DT S = 5 V 00 µh µs = 0.75 A i L (t = t Lück ) = U U t L D + i Lmax = 0 A t D = (0 A i Lmax )L U U [7 Credits] = L i Lmax 00 µh 0.75 A = =.5 µs U +U 5 V+ V Power Electronics Page 3 of 9
14 .5) Average value of inductor and diode current and value of resistor: i L = i D = i Lmax 0.6T S = i T S 4 L max = A i Lmax 0.5T S = i T S 8 L max = A R = U = U = V = Ω i R i D A [3 Credits] Power Electronics Page 4 of 9
15 Task 3) Four-Quadrant Converter [0 Credits] 3.) PWM unit structure for interleaved operation mode of 4QC: [3 Credits] 3.) Leakage inductance for requested maximum current ripple: i _INTERLEAVED = s ( s ) U DC f S L σs Maximum current ripple for s ref (t) = ±0.5: i _max_interleaved = U DC 8f S L σs f S = Pulse frequency: f p = f S = 500 Hz = 000 Hz [4 Credits] U DC i _max _IL 8 L σs =.5 kv 00 A mh = 500 Hz 3.3) Transformer turns ratio and duration of train ride: ü = I = I 000 A = = 0 I AC P/U AC 750 kw/5 kv W = P(t = t ) t + P(t < t < t ) (t t ) + P(t = t ) (t total t ) W = P(t = t ) t + P(t < t < t ) 3t + P(t = t ) t t = W = [ P(t = t ) + P(t < t < t ) 3 + P(t = t )] t W P(t=t )+P(t <t<t ) 3+ = 3.5 kwh P(t=t ) 750 kw+50 kw kw t = 3.5 kwh 3600 s h 55 kw = 4 s Power Electronics Page 5 of 9
16 t total = 5t = 0 s [4 Credits] 3.4) Grid-side phasor diagrams for three cases: Case : End of acceleration phase (t = t ) P = U AC I AC cos(φ ui ) I AC = 750 kw 5 kv cos (0 ) = 50 A I AC U AC j L s ü² I AC üu φ = arctan ( ω L σs ü I AC U AC ) = arctan ( π Hz H 0 50 A ) = 4. 5 kv Case : Constant power phase (t < t < t ) 0.5 P = U AC I AC cos(φ ui ) I AC = 50 kw 5 kv cos (0 ) = 0 A I AC U AC üu j L s ü² I AC φ = arctan ( ω L σs ü I AC U AC ) = arctan ( π Hz H 0 0 A ) = kv Case 3: Start of deceleration phase (t = t ) Power Electronics Page 6 of 9
17 0.5 P = U AC I AC cos(φ ui ) I AC = 600 kw 5 kv cos ( 80 ) = 40 A üu j L s ü² I AC 0.5 I AC U AC 0.5 φ = arctan ( ω L σs ü I AC U AC [6 Credits] ) = arctan ( π Hz H 0 40 A ) = kv 3.5) Reactive Power (required by transformer): Case : Q = π f L σs (I ) = π Hz H (000 A) Q = kva Case : Q = π f L σs (I ) = π Hz H (00 A) 0.5 Q = 6.8 kva Case 3: Q = π f L σs (I ) = π Hz H (800 A) 0.5 Q = kva Reactive power has to be provided by the 4QC. The reactive power oscillates between transformer and 4QC Power Electronics Page 7 of 9
18 [3 Credits] Power Electronics Page 8 of 9
19 Task 4) Line-Commutated Rectifier [0 Credits] 4.) Output voltage waveforms for three different control angles: [4 Credits] 4.) Control characteristic of line-commutated rectifier: CCM (0 α < 30 ): 50 +α U d = p 3 u sin(ωt) dωt = π 30 +α U d = 3 u [ cos(50 + α) + cos(30 + α)] π π 50 +α u [ cos(ωt)] 30 +α U d = 3 u [ cos(50 )cos (α) + sin(50 ) sin(α) + cos(30 ) cos(α) sin(30 ) sin (α)] π U d = 3 u [ cos(50 )cos (α) + cos(30 ) cos(α)] π U d = 3 π u [ 3 cos(α) cos(α)] = u cos (α) π U dmax α=0 = 3 3 π u f CCM (α) = U d = U d max 3 3 u cos (α) π = cos (α) 3 3 π u DCM (30 α < 50 ): Power Electronics Page 9 of 9
20 80 U d = p 3 u sin(ωt) dωt = π 30 +α π U d = 3 u [ cos(80 ) + cos(30 + α)] π 80 u [ cos(ωt)] 30 +α U d = 3 u [ cos (80 ) + cos(30 ) cos(α) sin(30 ) sin (α)] π U d = 3 3 u [ + cos(α) sin (α)] π U dmax α=0 = 3 3 π u (Universal reference value for CCM and DCM) 3 f DCM (α) = U d π = u [+ 3 cos(α) sin (α)] = + cos(α) sin (α) = [ + cos (α + 30 )] U 3 3 d max π u Control characteristic of converter for R-load: ++ [0 Credits] Power Electronics Page 0 of 9
21 4.3) Active power in dependency of control angle: CCM (0 α < 30 ): P R = U d RMS R = P R = 3 u R π [ωt p π 50 +α u sin (ωt)dωt 30 +α = 3 u R R π 30 +α sin(ωt) 4 P R = 3 u 5π [ 6 +α R π P R = 3 u R π [5π 50 +α 3 u ] = 30 +α sin (300 +α) 4 π 6 +α R π [50 +α sin (60 +α) + ] 4 sin(300 ) cos(α)+cos(300 ) sin (α) π α sin (ωt) dωt sin( 50 +α) 30 +α 4 + sin( 30 +α) ] 4 sin(60 ) cos(α)+cos(60 ) sin (α) + ] 4 P R = 3 u R π [4π 3 cos(α) 3 sin(α) + cos(α) + sin (α)] P R = 3 u R π [π cos (α)] DCM (30 α < 50 ): P R = U d RMS R = P R = 3 u R π [ωt P R = 3 u R π [π p π 80 u sin (ωt) dωt 30 +α = 3 u 80 R R π 30 +α sin(ωt) α 3 u ] = 30 +α sin (360 ) π α 4 R π [80 sin( 80 ) 4 P R = 3 u R π [5π α cos(α) + 8 sin(α)] [4 Credits] sin (ωt)dωt 30 +α sin(60 ) cos(α)+cos (60 ) sin(α) + ] 4 + sin ( 30 +α) 4 ] 4.4) Reasons for reactive power occurring on the mains side: For each control angle α the output voltage results in a periodic but non-sinusoidal voltage that contains harmonics. These voltage harmonics lead to current harmonics (independent of the load type) which also appear in the phase currents. Distortion reactive power D In addition to that the control angle α also determines the zero crossing of the fundamental phase currents and in consequence influences the phase difference φ between grid-side voltage and current of the same phase. Fundamental reactive power Q [ Credits] Power Electronics Page of 9
ET4119 Electronic Power Conversion 2011/2012 Solutions 27 January 2012
ET4119 Electronic Power Conversion 2011/2012 Solutions 27 January 2012 1. In the single-phase rectifier shown below in Fig 1a., s = 1mH and I d = 10A. The input voltage v s has the pulse waveform shown
More informationFachgebiet Leistungselektronik und Elektrische Antriebstechnik. Test Examination: Mechatronics and Electrical Drives
Prof. Dr. Ing. Joachim Böcker Test Examination: Mechatronics and Electrical Drives 8.1.214 First Name: Student number: Last Name: Course of Study: Exercise: 1 2 3 Total (Points) (2) (2) (2) (6) Duration:
More informationTSTE25 Power Electronics. Lecture 3 Tomas Jonsson ICS/ISY
TSTE25 Power Electronics Lecture 3 Tomas Jonsson ICS/ISY 2016-11-09 2 Outline Rectifiers Current commutation Rectifiers, cont. Three phase Inrush and short circuit current Exercises 5-5, 5-8, 3-100, 3-101,
More information6.3. Transformer isolation
6.3. Transformer isolation Objectives: Isolation of input and output ground connections, to meet safety requirements eduction of transformer size by incorporating high frequency isolation transformer inside
More information2005 AP PHYSICS C: ELECTRICITY AND MAGNETISM FREE-RESPONSE QUESTIONS
2005 AP PHYSICS C: ELECTRICITY AND MAGNETISM In the circuit shown above, resistors 1 and 2 of resistance R 1 and R 2, respectively, and an inductor of inductance L are connected to a battery of emf e and
More informationChapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode
Chapter 11 AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode Introduction 11.1. DCM Averaged Switch Model 11.2. Small-Signal AC Modeling of the DCM Switch Network 11.3. High-Frequency
More informationREACTANCE. By: Enzo Paterno Date: 03/2013
REACTANCE REACTANCE By: Enzo Paterno Date: 03/2013 5/2007 Enzo Paterno 1 RESISTANCE - R i R (t R A resistor for all practical purposes is unaffected by the frequency of the applied sinusoidal voltage or
More informationEN Power Electronics and Machines
EN 206 - Power Electronics and Machines Phase Controlled Rectifiers Suryanarayana Doolla Department of Energy Science and Engineering Indian Institute of Technology, Bombay suryad@iitb.ac.in Prof. Doolla
More informationEE Branch GATE Paper 2010
Q.1 Q.25 carry one mark each 1. The value of the quantity P, where, is equal to 0 1 e 1/e 2. Divergence of the three-dimensional radial vector field is 3 1/r 3. The period of the signal x(t) = 8 is 0.4
More informationECE 2210 Final given: Spring 15 p1
ECE 2 Final given: Spring 15 Closed Book, Closed notes except preprinted yellow sheet, Calculators OK. Show all work to receive credit. Circle answers, show units, and round off reasonably 1. (15 pts)
More informationEE 3120 Electric Energy Systems Study Guide for Prerequisite Test Wednesday, Jan 18, pm, Room TBA
EE 3120 Electric Energy Systems Study Guide for Prerequisite Test Wednesday, Jan 18, 2006 6-7 pm, Room TBA First retrieve your EE2110 final and other course papers and notes! The test will be closed book
More informationScheme I SAMPLE QUESTION PAPER I
SAMPLE QUESTION PAPER I Marks : 70 Time: 3 Hours Q.1) A) Attempt any FIVE of the following. a) Define active components. b) List different types of resistors. c) Describe method to test following passive
More informationECE Circuit Theory. Final Examination. December 5, 2008
ECE 212 H1F Pg 1 of 12 ECE 212 - Circuit Theory Final Examination December 5, 2008 1. Policy: closed book, calculators allowed. Show all work. 2. Work in the provided space. 3. The exam has 3 problems
More informationAC Circuits Homework Set
Problem 1. In an oscillating LC circuit in which C=4.0 μf, the maximum potential difference across the capacitor during the oscillations is 1.50 V and the maximum current through the inductor is 50.0 ma.
More informationGATE 2010 Electrical Engineering
GATE 2010 Electrical Engineering Q.1 Q.25 carry one mark each 1. The value of the quantity P, where P = xe dx, is equal to (A) 0 (B) 1 (C) e (D) 1/e 2. Divergence of the three-dimensional radial vector
More informationFALL 2004 Midterm Exam #2, Part A
Physics 152 FALL 2004 Midterm Exam #2, Part A Roster No.: Score: 17 points possible Exam time limit: 50 minutes. You may use a calculator and both sides of ONE sheet of notes, handwritten only. Closed
More informationChapter 7 DC-DC Switch-Mode Converters
Chapter 7 DC-DC Switch-Mode Converters dc-dc converters for switch-mode dc power supplies and dc-motor drives 7-1 Block Diagram of DC-DC Converters Functional block diagram 7-2 Stepping Down a DC Voltage
More informationECE 201 Fall 2009 Final Exam
ECE 01 Fall 009 Final Exam December 16, 009 Division 0101: Tan (11:30am) Division 001: Clark (7:30 am) Division 0301: Elliott (1:30 pm) Instructions 1. DO NOT START UNTIL TOLD TO DO SO.. Write your Name,
More informationChapter 3. Steady-State Equivalent Circuit Modeling, Losses, and Efficiency
Chapter 3. Steady-State Equivalent Circuit Modeling, Losses, and Efficiency 3.1. The dc transformer model 3.2. Inclusion of inductor copper loss 3.3. Construction of equivalent circuit model 3.4. How to
More informationBasics of Electric Circuits
António Dente Célia de Jesus February 2014 1 Alternating Current Circuits 1.1 Using Phasors There are practical and economic reasons justifying that electrical generators produce emf with alternating and
More informationmywbut.com Lesson 16 Solution of Current in AC Parallel and Seriesparallel
esson 6 Solution of urrent in Parallel and Seriesparallel ircuits n the last lesson, the following points were described:. How to compute the total impedance/admittance in series/parallel circuits?. How
More informationCircuit Analysis-III. Circuit Analysis-II Lecture # 3 Friday 06 th April, 18
Circuit Analysis-III Sinusoids Example #1 ü Find the amplitude, phase, period and frequency of the sinusoid: v (t ) =12cos(50t +10 ) Signal Conversion ü From sine to cosine and vice versa. ü sin (A ± B)
More informationELECTROMAGNETIC OSCILLATIONS AND ALTERNATING CURRENT
Chapter 31: ELECTROMAGNETIC OSCILLATIONS AND ALTERNATING CURRENT 1 A charged capacitor and an inductor are connected in series At time t = 0 the current is zero, but the capacitor is charged If T is the
More informationElectromagnetic Oscillations and Alternating Current. 1. Electromagnetic oscillations and LC circuit 2. Alternating Current 3.
Electromagnetic Oscillations and Alternating Current 1. Electromagnetic oscillations and LC circuit 2. Alternating Current 3. RLC circuit in AC 1 RL and RC circuits RL RC Charging Discharging I = emf R
More informationR. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 2.4 Cuk converter example L 1 C 1 L 2 Cuk converter, with ideal switch i 1 i v 1 2 1 2 C 2 v 2 Cuk
More informationSinusoidal Response of RLC Circuits
Sinusoidal Response of RLC Circuits Series RL circuit Series RC circuit Series RLC circuit Parallel RL circuit Parallel RC circuit R-L Series Circuit R-L Series Circuit R-L Series Circuit Instantaneous
More informationPower Factor Improvement
Salman bin AbdulazizUniversity College of Engineering Electrical Engineering Department EE 2050Electrical Circuit Laboratory Power Factor Improvement Experiment # 4 Objectives: 1. To introduce the concept
More informationModule 4. Single-phase AC Circuits
Module 4 Single-phase AC Circuits Lesson 14 Solution of Current in R-L-C Series Circuits In the last lesson, two points were described: 1. How to represent a sinusoidal (ac) quantity, i.e. voltage/current
More informationAnalysis of AC Power RMS and Phasors Power Factor. Power Factor. Eduardo Campero Littlewood
Power Factor Eduardo Campero Littlewood Universidad Autónoma Metropolitana Azcapotzalco Campus Energy Department Content 1 Analysis of AC Power 2 RMS and Phasors 3 Power Factor Recommended Bibliography
More informationChapter 33. Alternating Current Circuits
Chapter 33 Alternating Current Circuits 1 Capacitor Resistor + Q = C V = I R R I + + Inductance d I Vab = L dt AC power source The AC power source provides an alternative voltage, Notation - Lower case
More informationReview of DC Electric Circuit. DC Electric Circuits Examples (source:
Review of DC Electric Circuit DC Electric Circuits Examples (source: http://hyperphysics.phyastr.gsu.edu/hbase/electric/dcex.html) 1 Review - DC Electric Circuit Multisim Circuit Simulation DC Circuit
More informationEE313 Fall 2013 Exam #1 (100 pts) Thursday, September 26, 2013 Name. 1) [6 pts] Convert the following time-domain circuit to the RMS Phasor Domain.
Name If you have any questions ask them. Remember to include all units on your answers (V, A, etc). Clearly indicate your answers. All angles must be in the range 0 to +180 or 0 to 180 degrees. 1) [6 pts]
More information+ 2. v an. v bn T 3. L s. i c. v cn n. T 1 i L. i a. v ab i b. v abi R L. v o T 2
The University of New South Wales School of Electrical Engineering & Telecommunications Lecture 11. Effect of source inductance in three-phase converters 11.1 Overlap in a three-phase, C-T, fully-controlled
More informationRLC Circuit (3) We can then write the differential equation for charge on the capacitor. The solution of this differential equation is
RLC Circuit (3) We can then write the differential equation for charge on the capacitor The solution of this differential equation is (damped harmonic oscillation!), where 25 RLC Circuit (4) If we charge
More informationLECTURE 8 Fundamental Models of Pulse-Width Modulated DC-DC Converters: f(d)
1 ECTURE 8 Fundamental Models of Pulse-Width Modulated DC-DC Converters: f(d) I. Quasi-Static Approximation A. inear Models/ Small Signals/ Quasistatic I V C dt Amp-Sec/Farad V I dt Volt-Sec/Henry 1. Switched
More informationCross Regulation Mechanisms in Multiple-Output Forward and Flyback Converters
Cross Regulation Mechanisms in Multiple-Output Forward and Flyback Converters Bob Erickson and Dragan Maksimovic Colorado Power Electronics Center (CoPEC) University of Colorado, Boulder 80309-0425 http://ece-www.colorado.edu/~pwrelect
More informationReview of Basic Electrical and Magnetic Circuit Concepts EE
Review of Basic Electrical and Magnetic Circuit Concepts EE 442-642 Sinusoidal Linear Circuits: Instantaneous voltage, current and power, rms values Average (real) power, reactive power, apparent power,
More information11. AC Circuit Power Analysis
. AC Circuit Power Analysis Often an integral part of circuit analysis is the determination of either power delivered or power absorbed (or both). In this chapter First, we begin by considering instantaneous
More informationThe output voltage is given by,
71 The output voltage is given by, = (3.1) The inductor and capacitor values of the Boost converter are derived by having the same assumption as that of the Buck converter. Now the critical value of the
More informationConverter System Modeling via MATLAB/Simulink
Converter System Modeling via MATLAB/Simulink A powerful environment for system modeling and simulation MATLAB: programming and scripting environment Simulink: block diagram modeling environment that runs
More informationEXP. NO. 3 Power on (resistive inductive & capacitive) load Series connection
OBJECT: To examine the power distribution on (R, L, C) series circuit. APPARATUS 1-signal function generator 2- Oscilloscope, A.V.O meter 3- Resisters & inductor &capacitor THEORY the following form for
More informationChapter 15 Power And Harmonics in Nonsinusoidal Systems
Chapter 15 Power And Harmonics in Nonsinusoidal Systems 15.1. Average power in terms of Fourier series 15.2. RMS value of a waveform 15.3. Power factor THD Distortion and Displacement factors 15.4. Power
More informationMidterm Exam 2. Prof. Miloš Popović
Midterm Exam 2 Prof. Miloš Popović 100 min timed, closed book test. Write your name at top of every page (or initials on later pages) Aids: single page (single side) of notes, handheld calculator Work
More informationSECTION - A. A current impulse, 5δ(t), is forced through a capacitor C. The voltage, Vc(t), across the capacitor is given by. 5 t C.
SETION - A. This question consists of TWENTY-FIVE sub-questions (..5) of ONE mark each. For each of these sub-questions, four possible alternatives (A,B, and D) are given, out of which ONLY ONE is correct.
More informationNZQA registered unit standard version 2 Page 1 of 6
Page 1 of 6 Title Demonstrate and apply knowledge of capacitance, inductance, power factor, and power factor correction Level 3 Credits 7 Purpose This unit standard covers an introduction to alternating
More informationSinusoidal Steady-State Analysis
Sinusoidal Steady-State Analysis Almost all electrical systems, whether signal or power, operate with alternating currents and voltages. We have seen that when any circuit is disturbed (switched on or
More information12. Introduction and Chapter Objectives
Real Analog - Circuits 1 Chapter 1: Steady-State Sinusoidal Power 1. Introduction and Chapter Objectives In this chapter we will address the issue of power transmission via sinusoidal or AC) signals. This
More informationNovel DC-AC Converter Topology for Multilevel Battery Energy Storage Systems. Mario Gommeringer, Felix Kammerer, Johannes Kolb, Michael Braun
Elektrotechnisches Institut (ETI) Prof. Dr.-Ing. Michael Braun Prof. Dr.-Ing. Martin Doppelbauer Prof. Dr.-Ing. Marc Hiller Kaiserstr.12. 76131 Karlsruhe 13. Sept. 216 Title: Novel DC-C Converter Topology
More informationSingle Phase Parallel AC Circuits
Single Phase Parallel AC Circuits 1 Single Phase Parallel A.C. Circuits (Much of this material has come from Electrical & Electronic Principles & Technology by John Bird) n parallel a.c. circuits similar
More informationEN Power Electronics and Machines
1/19 - Power Electronics and Machines Transformers Suryanarayana Doolla Department of Energy Science and Engineering Indian Institute of Technology, Bombay suryad@iitb.ac.in Lecture Organization - Modules
More informationR. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder Part II" Converter Dynamics and Control! 7.!AC equivalent circuit modeling! 8.!Converter transfer
More informationModule 4. Single-phase AC circuits. Version 2 EE IIT, Kharagpur
Module 4 Single-phase circuits ersion EE T, Kharagpur esson 6 Solution of urrent in Parallel and Seriesparallel ircuits ersion EE T, Kharagpur n the last lesson, the following points were described:. How
More informationECE 585 Power System Stability
Homework 1, Due on January 29 ECE 585 Power System Stability Consider the power system below. The network frequency is 60 Hz. At the pre-fault steady state (a) the power generated by the machine is 400
More informationGATE 2008 Electrical Engineering
GATE 2008 Electrical Engineering Q.1 Q. 20 carry one mark each. 1. The number of chords in the graph of the given circuit will be + _ (A) 3 (B) 4 (C) 5 (D) 6 2. The Thevenin'a equivalent of a circuit operating
More informationLecture 05 Power in AC circuit
CA2627 Building Science Lecture 05 Power in AC circuit Instructor: Jiayu Chen Ph.D. Announcement 1. Makeup Midterm 2. Midterm grade Grade 25 20 15 10 5 0 10 15 20 25 30 35 40 Grade Jiayu Chen, Ph.D. 2
More informationSinusoidal Steady State Power Calculations
10 Sinusoidal Steady State Power Calculations Assessment Problems AP 10.1 [a] V = 100/ 45 V, Therefore I = 20/15 A P = 1 (100)(20)cos[ 45 (15)] = 500W, 2 A B Q = 1000sin 60 = 866.03 VAR, B A [b] V = 100/
More informationElectronic Power Conversion
Electronic Power Conversion Review of Basic Electrical and Magnetic Circuit Concepts Challenge the future 3. Review of Basic Electrical and Magnetic Circuit Concepts Notation Electric circuits Steady state
More informationCircuit Analysis-II. Circuit Analysis-II Lecture # 5 Monday 23 rd April, 18
Circuit Analysis-II Capacitors in AC Circuits Introduction ü The instantaneous capacitor current is equal to the capacitance times the instantaneous rate of change of the voltage across the capacitor.
More informationALTERNATING CURRENT. with X C = 0.34 A. SET UP: The specified value is the root-mean-square current; I. EXECUTE: (a) V = (0.34 A) = 0.12 A.
ATENATING UENT 3 3 IDENTIFY: i Icosωt and I I/ SET UP: The specified value is the root-mean-square current; I 34 A EXEUTE: (a) I 34 A (b) I I (34 A) 48 A (c) Since the current is positive half of the time
More informationInduction_P1. 1. [1 mark]
Induction_P1 1. [1 mark] Two identical circular coils are placed one below the other so that their planes are both horizontal. The top coil is connected to a cell and a switch. The switch is closed and
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) ELECTRICAL & ELECTRONICS ENGINEERING EXAMINATION SEMESTER /2018
ENG009 SCHOOL OF ENGINEERING BENG (HONS) ELECTRICAL & ELECTRONICS ENGINEERING INTERMEDIATE ELECTRICAL PRINCIPLES & ENABLING POWER ELECTRONICS MODULE NO: EEE5003 Date: 15 January 2018 Time: 10.00 12.00
More informationECE2210 Final given: Fall 13
ECE22 Final given: Fall 3. (23 pts) a) Draw the asymptotic Bode plot (the straight-line approximation) of the transfer function below. Accurately draw it on the graph provided. You must show the steps
More informationENGG Fundamentals of Electrical Circuits and Machines Final Examination
Name: Lecture Section: ID#: ENGG 225 - Fundamentals of Electrical Circuits and Machines Final Examination Tuesday, April 28, 2015 Time: 12:00-3:00 PM Red Gymnasium (L01-L03) Auxiliary Gymnasium (L04) L01
More informationPhysics 142 AC Circuits Page 1. AC Circuits. I ve had a perfectly lovely evening but this wasn t it. Groucho Marx
Physics 142 A ircuits Page 1 A ircuits I ve had a perfectly lovely evening but this wasn t it. Groucho Marx Alternating current: generators and values It is relatively easy to devise a source (a generator
More informationLecture 17 Push-Pull and Bridge DC-DC converters Push-Pull Converter (Buck Derived) Centre-tapped primary and secondary windings
ecture 17 Push-Pull and Bridge DC-DC converters Push-Pull Converter (Buck Derived) Centre-tapped primary and secondary windings 1 2 D 1 i v 1 v 1s + v C o R v 2 v 2s d 1 2 T 1 T 2 D 2 Figure 17.1 v c (
More informationChapter 21: RLC Circuits. PHY2054: Chapter 21 1
Chapter 21: RC Circuits PHY2054: Chapter 21 1 Voltage and Current in RC Circuits AC emf source: driving frequency f ε = ε sinωt ω = 2π f m If circuit contains only R + emf source, current is simple ε ε
More informationRefresher course on Electrical fundamentals (Basics of A.C. Circuits) by B.M.Vyas
Refresher course on Electrical fundamentals (Basics of A.C. Circuits) by B.M.Vyas A specifically designed programme for Da Afghanistan Breshna Sherkat (DABS) Afghanistan 1 Areas Covered Under this Module
More informationECE2210 Final given: Spring 08
ECE Final given: Spring 0. Note: feel free to show answers & work right on the schematic 1. (1 pts) The ammeter, A, reads 30 ma. a) The power dissipated by R is 0.7 W, what is the value of R. Assume that
More informationSECOND ENGINEER REG III/2 MARINE ELECTRO-TECHNOLOGY. 1. Understands the physical construction and characteristics of basic components.
SECOND ENGINEER REG III/ MARINE ELECTRO-TECHNOLOGY LIST OF TOPICS A B C D Electric and Electronic Components Electric Circuit Principles Electromagnetism Electrical Machines The expected learning outcome
More informationPhysics 240 Fall 2005: Exam #3. Please print your name: Please list your discussion section number: Please list your discussion instructor:
Physics 240 Fall 2005: Exam #3 Please print your name: Please list your discussion section number: Please list your discussion instructor: Form #1 Instructions 1. Fill in your name above 2. This will be
More informationC R. Consider from point of view of energy! Consider the RC and LC series circuits shown:
ircuits onsider the R and series circuits shown: ++++ ---- R ++++ ---- Suppose that the circuits are formed at t with the capacitor charged to value. There is a qualitative difference in the time development
More information12 Chapter Driven RLC Circuits
hapter Driven ircuits. A Sources... -. A ircuits with a Source and One ircuit Element... -3.. Purely esistive oad... -3.. Purely Inductive oad... -6..3 Purely apacitive oad... -8.3 The Series ircuit...
More informationSolution to Tutorial 5 Isolated DC-DC Converters & Inverters
ELEC464 University of New South Wales School of Electrical Engineering & Telecommunications Solution to Tutorial 5 Isolated DC-DC Converters & Inverters. i D O d T The maximum value of L m which will ensure
More informationEngineering Unit 1: Engineering Principles
Write your name here Surname Other names Pearson BTEC Level 3 Extended Certificate, Foundation Diploma, Diploma, Extended Diploma Centre Number Learner Registration Number Engineering Unit 1: Engineering
More informationUNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level
UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level *3242847993* PHYSICS 9702/43 Paper 4 A2 Structured Questions October/November 2012 2 hours Candidates
More information1 Phasors and Alternating Currents
Physics 4 Chapter : Alternating Current 0/5 Phasors and Alternating Currents alternating current: current that varies sinusoidally with time ac source: any device that supplies a sinusoidally varying potential
More informationELEC 2501 AB. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work.
It is most beneficial to you to write this mock midterm UNDER EXAM CONDITIONS. This means: Complete the midterm in 3 hour(s). Work on your own. Keep your notes and textbook closed. Attempt every question.
More informationEECE 2150 Circuits and Signals Final Exam Fall 2016 Dec 16
EECE 2150 Circuits and Signals Final Exam Fall 2016 Dec 16 Instructions: Write your name and section number on all pages Closed book, closed notes; Computers and cell phones are not allowed You can use
More informationSINUSOIDAL STEADY STATE CIRCUIT ANALYSIS
SINUSOIDAL STEADY STATE CIRCUIT ANALYSIS 1. Introduction A sinusoidal current has the following form: where I m is the amplitude value; ω=2 πf is the angular frequency; φ is the phase shift. i (t )=I m.sin
More informationModule 4. Single-phase AC Circuits. Version 2 EE IIT, Kharagpur 1
Module 4 Single-phase A ircuits ersion EE IIT, Kharagpur esson 4 Solution of urrent in -- Series ircuits ersion EE IIT, Kharagpur In the last lesson, two points were described:. How to represent a sinusoidal
More informationPhysics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe. Useful Information. Your name sticker. with exam code
Your name sticker with exam code Physics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe SIGNATURE: 1. The exam will last from 4:00 p.m. to 7:00 p.m. Use a #2 pencil to make entries on the
More informationENGG Fundamentals of Electrical Circuits and Machines Final Examination
Name: Lecture Section: ID#: ENGG 225 - Fundamentals of Electrical Circuits and Machines Final Examination Thursday, April 17, 2014 Time: 8:00-11:00 AM Red Gymnasium (L01-L03) Auxiliary Gymnasium (L04)
More informationHighpowerfactorforwardAC-DC converter with low storage capacitor voltage stress
HAIT Journal of Science and Engineering B, Volume 2, Issues 3-4, pp. 305-326 Copyright C 2005 Holon Academic Institute of Technology HighpowerfactorforwardAC-DC converter with low storage capacitor voltage
More informationSinusoidal Steady State Analysis (AC Analysis) Part II
Sinusoidal Steady State Analysis (AC Analysis) Part II Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/
More informationQ. 1 Q. 5 carry one mark each.
GATE 2016 General Aptitude - GA Set-6 Q. 1 Q. 5 carry one mark each. Q.1 The man who is now Municipal Commissioner worked as. (A) the security guard at a university (B) a security guard at the university
More informationPhysics 102 Exam 2 Fall 2012
Physics 102 Exam 2 Fall 2012 Last Name: First Name Network-ID Discussion Section: Discussion TA Name: Turn off your cell phone and put it out of sight. Keep your calculator on your own desk. Calculators
More informationToolbox: Electrical Systems Dynamics
Toolbox: Electrical Systems Dynamics Dr. John C. Wright MIT - PSFC 05 OCT 2010 Introduction Outline Outline AC and DC power transmission Basic electric circuits Electricity and the grid Image removed due
More informationProject Components. MC34063 or equivalent. Bread Board. Energy Systems Research Laboratory, FIU
Project Components MC34063 or equivalent Bread Board PSpice Software OrCAD designer Lite version http://www.cadence.com/products/orcad/pages/downloads.aspx#pspice More Details on the Introduction CONVERTER
More information2 Signal Frequency and Impedances First Order Filter Circuits Resonant and Second Order Filter Circuits... 13
Lecture Notes: 3454 Physics and Electronics Lecture ( nd Half), Year: 7 Physics Department, Faculty of Science, Chulalongkorn University //7 Contents Power in Ac Circuits Signal Frequency and Impedances
More informationECE Spring 2015 Final Exam
ECE 20100 Spring 2015 Final Exam May 7, 2015 Section (circle below) Jung (1:30) 0001 Qi (12:30) 0002 Peleato (9:30) 0004 Allen (10:30) 0005 Zhu (4:30) 0006 Name PUID Instructions 1. DO NOT START UNTIL
More informationUNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level
UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level *8055009334* PHYSICS 9702/43 Paper 4 A2 Structured Questions May/June 2012 2 hours Candidates answer on
More informationSE]y[-2017(02)-I ELECTRICAL ENGINEERING. Paper. Please read each of the following instructions carefully before attempting questions.
RoU No. 300218 Candidate should write his/her Roll No. here. Total No. of Questions : 7 No. of Printed Pages : 7 SE]y[-2017(02)-I ELECTRICAL ENGINEERING Paper Time : 3 Hours ] [ Total Marks : 300 Instructions
More informationPHA3/W PHYSICS (SPECIFICATION A) Unit 3 Current Electricity and Elastic Properties of Solids
Surname Centre Number Other Names Candidate Number Leave blank Candidate Signature General Certificate of Education January 2005 Advanced Subsidiary Examination PHYSICS (SPECIFICATION A) PHA3/W Unit 3
More informationENGG Fundamentals of Electrical Circuits and Machines Final Examination
Name: Lecture Section: ID#: ENGG 225 - Fundamentals of Electrical Circuits and Machines Final Examination Monday, April 18, 2011 Time: 3:30-6:30 PM Red and Auxiliary Gymnasium L01, L02 - Norm Bartley L03
More informationPHYA4/2. General Certificate of Education Advanced Level Examination June Unit 4 Fields and Further Mechanics Section B
Centre Number Surname Candidate Number For Examiner s Use Other Names Candidate Signature Examiner s Initials Physics A Unit 4 Fields and Further Mechanics Section B General Certificate of Education Advanced
More informationOscillations and Electromagnetic Waves. March 30, 2014 Chapter 31 1
Oscillations and Electromagnetic Waves March 30, 2014 Chapter 31 1 Three Polarizers! Consider the case of unpolarized light with intensity I 0 incident on three polarizers! The first polarizer has a polarizing
More informationEXPERIMENT 07 TO STUDY DC RC CIRCUIT AND TRANSIENT PHENOMENA
EXPERIMENT 07 TO STUDY DC RC CIRCUIT AND TRANSIENT PHENOMENA DISCUSSION The capacitor is a element which stores electric energy by charging the charge on it. Bear in mind that the charge on a capacitor
More information3 d Calculate the product of the motor constant and the pole flux KΦ in this operating point. 2 e Calculate the torque.
Exam Electrical Machines and Drives (ET4117) 11 November 011 from 14.00 to 17.00. This exam consists of 5 problems on 4 pages. Page 5 can be used to answer problem 4 question b. The number before a question
More informationAlternating Current. Symbol for A.C. source. A.C.
Alternating Current Kirchoff s rules for loops and junctions may be used to analyze complicated circuits such as the one below, powered by an alternating current (A.C.) source. But the analysis can quickly
More informationEEL 5245 POWER ELECTRONICS I Lecture #14: Chapter 4 Non-Isolated DC-DC Converters PWM Converters DCM Analysis (DCM)
EE 545 POWER EECTRONICS I ecture #4: Chapter 4 Non-Isolated DC-DC Converters PWM Converters DCM Analysis (DCM) Objectives Overview of Discontinuous Conduction Mode Buck Converter Analysis (DCM) Simulation
More information