Electromagnetic and Electromechanical Engineering Principles Notes 01 Basics
Marc T. Thompson, Ph.D. Thompson Consulting, Inc. 9 Jacob Gates Road Harvard, MA 01451 Phone: (978) 456-7722 Email: [email protected] Website: http://www.thompsonrd.com
© Marc Thompson, 2006-2008 Portions of these notes reprinted from Fitzgerald, Electric Motors, 6th edition
Electromechanics Basics 1-1 Background of Instructor • BSEE (’85), MS (’92), and PhD (’97) from MIT • Doctoral work done in the Laboratory for Electromagnetic and Electronic Systems (L.E.E.S.) at MIT in electrodynamic Maglev • Consultant in analog, magnetics, power electronics, electromagnetics, magnetic braking and failure analysis • Adjunct Professor at Worcester Polytechnic Institute, Worcester MA teaching graduate courses in analog design, power electronics, electromechanics and power distribution • Have also taught for University of Wisconsin since 2006
Electromechanics Basics 1-2 Lens Actuator • For high speed laser printing z Iron r Coil Back iron
Permanent Nd-Fe-Bo Lens Magnet
Air gap
Electromechanics Basics 1-3 High Power Laser Diode Driver Based on Power Converter Technology
See: 1. B. Santarelli and M. Thompson, U.S. Patent #5,123,023, "Laser Driver with Plural Feedback Loops," issued June 16, 1992 2. M. Thompson, U.S. Patent #5,444,728, "Laser Driver Circuit," issued August 22, 1995 3. W. T. Plummer, M. Thompson, D. S. Goodman and P. P. Clark, U.S. Patent #6,061,372, “Two-Level Semiconductor Laser Driver,” issued May 9, 2000 4. Marc T. Thompson and Martin F. Schlecht, “Laser Diode Driver Based on Power Converter Technology,” IEEE Transactions on Power Electronics, vol. 12, no. 1, Jan. 1997, pp. 46-52 Electromechanics Basics 1-4 Magnetically-Levitated Flywheel Energy Storage System • For NASA; P = 100W, energy storage = 100 W-hrs
Guidance and Suspension
S N N S N S Flywheel (Rotating) S N
Stator Winding
N S S N
N S S N
z
N S S N S N N S
r
Electromechanics Basics 1-5 Transrapid Maglev • Currently in operation from Pudong (Shanghai), connecting airport to subway line • Operational speed is 431 km/hr
Electromechanics Basics 1-6 Japanese EDS Maglev • Slated to run parallel to the high-speed Shinkansen line • On December 2, 2003, this three-car train set attained a maximum speed of 581 km/h in a manned vehicle run
Reference: http://www.rtri.or.jp Electromechanics Basics 1-7 Japanese EDS Guideway • This view shows levitation, guidance and propulsion coils
Electromechanics Basics 1-8 MIT Maglev Suspension Magnet
M. T. Thompson, R. D. Thornton and A. Kondoleon, “Flux-canceling electrodynamic maglev suspension: Part I. Test fixture design and modeling,” IEEE Transactions on Magnetics, vol. 35, no. 3, May 1999 pp. 1956-1963 Electromechanics Basics 1-9 MIT Maglev Test Fixture
M. T. Thompson, R. D. Thornton and A. Kondoleon, “Flux- canceling electrodynamic maglev suspension: Part I. Test fixture design and modeling,” IEEE Transactions on Magnetics, vol. 35, no. 3, May 1999 pp. 1956-1963
Electromechanics Basics 1-10 MIT EDS Maglev Test Facility • 2 meter diameter test wheel • Max. speed 1000 RPM (84 m/s) • For testing “flux canceling” HTSC Maglev • Sidewall levitation
References: 1. Marc T. Thompson, Richard D. Thornton and Anthony Kondoleon, “Scale Model Flux-Canceling EDS Maglev Suspension --- Part I: Design and Modeling,” IEEE Transactions on Magnetics, vol. 35, no. 3, May 1999, pp. 1956-1963 2. Marc T. Thompson and Richard D. Thornton, “Scale Model Flux-Canceling EDS Maglev Suspension --- Part II: Test Results and Scaling Laws,” IEEE Transactions on Magnetics, vol. 35, no. 3, May 1999, pp. 1964-1975
Electromechanics Basics 1-11 Permanent Magnet Brakes • For roller coasters • Braking force > 10,000 Newtons per meter of brake
Reference: http://www.magnetarcorp.com
Electromechanics Basics 1-12 Halbach Permanent Magnet Array • Special PM arrangement allows strong side (bottom) and weak side (top) fields • Applicable to magnetic suspensions (Maglev), linear motors, and induction brakes
Electromechanics Basics 1-13 Halbach Permanent Magnet Array • 2D FEA modeling
Electromechanics Basics 1-14 Photovoltaics
S. Druyea, S. Islam and W. Lawrance, “A battery management system for stand-alone photovoltaic energy systems,” IEEE Industry Applications Magazine, vol. 7, no. 3, May-June 2001, pp. 67-72
Electromechanics Basics 1-15 Offline Flyback Power Supply
P. Maige, “A universal power supply integrated circuit for TV and monitor applications,” IEEE Transactions on Consumer Electronics, vol. 36, no. 1, Feb. 1990, pp. 10-17 Electromechanics Basics 1-16 Transcutaneous Energy Transmission
Reference: H. Matsuki, Y. Yamakata, N. Chubachi, S.-I. Nitta and H. Hashimoto, “Transcutaneous DC-DC converter for totally implantable artificial heart using synchronous rectifier,” IEEE Transactions on Magnetics, vol. 32 , no. 5, Sept. 1996, pp. 5118 - 5120
Electromechanics Basics 1-17 50 KW Inverter Switch
Electromechanics Basics 1-18 Transformer Failure Analysis
Electromechanics Basics 1-19 Non-Contact Battery Charger • Modeled using PSIM
Electromechanics Basics 1-20 High Voltage RF Supply
Electromechanics Basics 1-21 60 Hz Transformer Shielding Study • Modeled using Infolytica 2D and 3D
Electromechanics Basics 1-22 Course Overview • Day 1 --- Basics of power, 3 phase power, harmonics, etc. • Day 2 --- Magnetics, transformers and energy conversion • Day 3 --- Basic machines
• Numerous examples are done throughout the 3 days, some from Fitzgerald, Kingsley and Umans
Electromechanics Basics 1-23 Course Overview --- Day 1
Electromechanics Basics 1-24 Course Overview --- Day 2
Electromechanics Basics 1-25 Course Overview --- Day 3
Electromechanics Basics 1-26 Basic Circuit Analysis Concepts • Some background in circuits to get us all on the same page • First-order and second-order systems • Resonant circuits, damping ratio and Q • Reference for this material: M. T. Thompson, Intuitive Analog Circuit Design, Elsevier, 2006 and Kusko/Thompson Power Quality in Electrical Systems, McGraw-Hill, 2007
Electromechanics Basics 1-27 First-Order Systems • A system with a single energy-storage element is a “first- order” system • Step voltage driven RC lowpass filter
−t τ o −= eVtv )1()( V −t ti )( = e τ r R 1 τ = RC ω = h τ 35.0 τ R = ω f τ R = 2.2 τ f = h h h 2π
Electromechanics Basics 1-28 First-Order Systems --- Some Details
• Frequency response:
• Phase response:
• -3 dB bandwidth:
Electromechanics Basics 1-29 Step Response 10 - 90% Risetime • Often-measured figure of merit for systems • Defined as the time it takes a step response to transition from 10% of final value to 90% of final value • This plot is for a first-order system with no overshoot or ringing
Electromechanics Basics 1-30 Relationship Between Risetime and Bandwidth • Exact for a first-order system (with one pole):
• Approximate for higher-order systems
Electromechanics Basics 1-31 First-Order Step and Frequency Response • Single pole at -1 rad/sec.
Electromechanics Basics 1-32 Second-Order Mechanical System • 2 energy storage modes in mass and spring
Spring force:
Newton’s law for moving mass:
Differential equation for mass motion:
Guess a solution of the form:
Electromechanics Basics 1-33 Second-Order Mechanical System
Put this proposed solution into the differential equation:
This solution works if:
Electromechanics Basics 1-34 Second-Order Mechanical System • Electromechanical modeling; useful for modeling speakers, motors, acoustics, etc.
Reference: Leo Beranek, Acoustics, Acoustical Society of America, 1954 Electromechanics Basics 1-35 Second-Order Electrical System
1 sv )( 1 1 ω 2 sH )( o == Cs = = = n sv )( 1 2 RCsLCs ++ 1 2 2ζss 2 2 ss ++ ωζω2 i LsR ++ nn 2 ++ 1 Cs ω ωnn 1 Natural frequency ωn = LC ω RC 1 R 1 R ς n === Damping ratio 2 2 L 2 Zo C
Electromechanics Basics 1-36 Second-Order System Frequency Response 1 jH ω)( = 2 jςω ⎛ ω 2 ⎞ ⎜1 −+ ⎟ ⎜ 2 ⎟ ωn ⎝ ωn ⎠ 1 jH ω)( = 2 2 ⎛ 2ςω ⎞ ⎛ ω 2 ⎞ ⎜ ⎟ ⎜1 −+ ⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎝ ωn ⎠ ⎝ ωn ⎠ ⎛ 2ςω ⎞ ⎜ ⎟ ⎜ ω ⎟ ⎛ 2ςωω ⎞ jH ω −=∠ tan)( −1 ⎝ n ⎠ −= tan −1 ⎜ n ⎟ ⎛ ω 2 ⎞ ⎜ − ωω22 ⎟ ⎜1 − ⎟ ⎝ n ⎠ ⎜ 2 ⎟ ⎝ ωn ⎠
Electromechanics Basics 1-37 Second-Order System Frequency Response
Electromechanics Basics 1-38 Quality Factor, or “Q”
Quality factor is defined as: E ω stored Pdiss where Estored is the peak stored energy in the system and Pdiss is the average power dissipation.
Electromechanics Basics 1-39 Q of Series Resonant RLC
1 ω = LC 1 = LIE 2 stored 2 pk 1 = 2 RIP diss 2 pk ⎛ 1 ⎞ L ⎜ LI 2 ⎟ ⎛ 1 ⎞ pk C Z Q = ⎜ ⎟⎜ 2 ⎟ == o LC ⎜ 1 2 ⎟ R R ⎝ ⎠⎜ RI ⎟ ⎝ 2 pk ⎠
Electromechanics Basics 1-40 Q of Parallel Resonant RLC
1 ω = LC 1 E = CV 2 stored 2 pk 1 V 2 P = pk diss 2 R ⎛ 1 ⎞ ⎜ CV 2 ⎟ ⎛ 1 ⎞ pk R R Q = ⎜ ⎟⎜ 2 ⎟ == ⎜ 2 ⎟ ⎝ LC ⎠ 1 Vpk L Zo ⎜ ⎟ ⎝ 2 R ⎠ C
Electromechanics Basics 1-41 Relationship Between Damping Ratio and “Quality Factor” Q • A second order system can also be characterized by it’s “Quality Factor” or Q. 1 sH )( == Q =ωωn 2ς
• Use Q in transfer function of series resonant circuit: 1 1 sH )( = = 2 2ζ 2 ssss 2 ++ 1 2 ++ 1 ω nn ωωωnn Q
Electromechanics Basics 1-42 Second-Order System Step Response • Shown for varying values of damping ratio.
Electromechanics Basics 1-43 Undamped Resonant Circuit
di v dv − i L = c c = L dt L dt C
2vd 1 di v c L −=−= c 2 Cdt dt LC
We find the resonant frequency by guessing that the voltage v(t) is sinusoidal with v(t) = Vosinωt. Putting this into the equation for capacitor voltage results in:
1 − 2 ωωt)sin( −= ωt)sin( LC
This means that the resonant frequency is the standard (as expected) resonance:
1 ω 2 = r LC
Electromechanics Basics 1-44 Energy Methods
Storage Mode Relationship Comments Capacitor/electric field 1 = CVE 2 storage elec 2 Inductor/magnetic field 1 B 2 LIE 2 == dV storage mag ∫ 22 μo Kinetic energy 1 = MvE 2 k 2
Rotary energy 1 2 I ≡ mass moment of inertia = IE ω 2 r 2 (kg- m ) Spring 1 k ≡ spring constant (N/m) E = kx 2 spring 2
Potential energy p Δ=Δ hMgE Δh ≡ height change Thermal energy C ≡ thermal capacitance Δ=Δ TCE TH T TH (J/K)
Electromechanics Basics 1-45 Energy Methods
By using energy methods we can find the ratio of maximum capacitor voltage to maximum inductor current. Assuming that the capacitor is initially charged to Vo 2 volts, and remembering that capacitor stored energy Ec = ½CV and inductor 2 stored energy is EL = ½LI , we can write the following:
1 1 2 = LICV 2 2 o 2 o
Electromechanics Basics 1-46 Energy Methods
What is the inductor current? We can solve for the ratio of Vo/Io resulting in:
Vo L ≡= Zo Io C
The term “Zo” is defined as the characteristic impedance of a resonant circuit. E.g with C = 1 microFarad and L = 1 microHenry, the resonant frequency is 10 6 radians/second (or 166.7 kHz) and that the characteristic impedance is 1 Ohm.
Electromechanics Basics 1-47 Simulation
Initial conditions at t = 0: capacitor voltage = 1; inductor current = 0.
Electromechanics Basics 1-48 Typical Resonant Circuit • Model of a MOSFET gate drive circuit
1 sv )( 1 1 sH )( o == Cs = = sv )( 1 2 RCsLCs ++ 1 2 2ζss i LsR ++ 2 ++ 1 Cs ω ωnn 1 ω = n LC ω RC 1 R 1 R ς n === 2 2 L 2 Zo C Electromechanics Basics 1-49 Resonant Circuit --- Underdamped • With “small” resistor, circuit is underdamped
1 ω n == Mrad sec/200 LC
n 8.31 MHzf L Z 5Ω== o C ω RC 1 R 1 R ς n ==== 001.0 2 2 L 2 Z o C
Electromechanics Basics 1-50 Resonant Circuit --- Underdamped Results • This circuit is very underdamped, so we expect step response to oscillate at around 31.8 MHz • Expect peaky frequency response with peak near 31.8 MHz 1 ω n == Mrad sec/200 LC
n 8.31 MHzf L Z 5Ω== o C ω RC 1 R 1 R ς n ==== 001.0 2 2 L 2 Z o C
Electromechanics Basics 1-51 Underdamped Resonant Circuit, Step Response
• PSPICE result: rings at around 31.8 MHz
Electromechanics Basics 1-52 Underdamped Resonant Circuit, Frequency Response
• Frequency response peaks at 31.8 MHz
Electromechanics Basics 1-53 Resonant Circuit --- Critical Damping • Now, let’s employ “critical damping” by increasing value of
resistor to R = 2Zo = 10 Ohms • This is also a typical MOSFET gate drive damping resistor value
Electromechanics Basics 1-54 Critical Damping, Step Response • Note that response is still relatively fast (< 100 ns response time) but with no overshoot • “Critical damping” results in fastest step response without overshoot
Electromechanics Basics 1-55 Critical Damping, Frequency Response • No overshoot in the transient response corresponds to no peaking in the frequency response
Electromechanics Basics 1-56 Impedance of a Series-Resonant Circuit
• The impedance is capacitive below the resonant frequency • Impedance is inductive above resonance
Reference: N. Mohan, Power Electronics, 3d edition
Electromechanics Basics 1-57 Impedance of a Parallel-Resonant Circuit • The impedance is inductive below the resonant frequency
Reference: N. Mohan, Power Electronics, 3d edition
Electromechanics Basics 1-58 Power, Complex Numbers and RMS • Basic single-phase circuits • AC voltage, current and power • Complex numbers • 3-phase power • Root-mean square (RMS)
Electromechanics Basics 1-59 Sinewaves
120V RMS 60 Hz sinewave • A pure sinewave can 200 be expressed as v(t) = 150 Vpksin(ωt) 100
50 • Vpk = peak voltage • ω = radian frequency 0 • In Hz, ω = 2πf where f -50 is in Hz -100
-150 • VRMS = Vpk/sqrt(2) =
-200 120V for sinewave with 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 Time [sec.] peaks at ±170V • More on RMS later
Electromechanics Basics 1-60 Sinewave Voltage Source with Resistive Load • v(t) and i(t) are in phase and have the same shape; i.e. no harmonics in current - Time representation -Phasor representation -In this case, V and I have the same phase
Electromechanics Basics 1-61 Sinewave with Inductive Load • For an inductor, remember that v = Ldi/dt • So, i(t) lags v(t) by 90o in an inductor
Electromechanics Basics 1-62 Sinewave with L/R Load • Phase shift (also called angle) between v and i is somewhere between 0o and -90o
−1⎛ωL ⎞ −=∠ tan ⎜ ⎟ ⎝ R ⎠
Electromechanics Basics 1-63 Sinewave with Capacitive Load • Remember that i = Cdv/dt for a capacitor • Current leads voltage by +90o
- Phasor representation
Electromechanics Basics 1-64 Phasor Representation of L and C
In inductor, current lags In capacitor, voltage lags voltage by 90 degrees current by 90 degrees Reference: N. Mohan, Power Electronics, 3d edition
Electromechanics Basics 1-65 Response of L and C to pulses
Reference: N. Mohan, Power Electronics, 3d edition
Electromechanics Basics 1-66 Review of Complex Numbers • In “rectangular” form, a complex number is written in terms of real and imaginary components • A = Re(A) + j×Im(A)
- Angle
−1 ⎛ A)Im( ⎞ θ = tan ⎜ ⎟ ⎝ A)Re( ⎠
- Magnitude of A
= ()()AA 2 + A)Im()Re( 2
Electromechanics Basics 1-67 Find Polar Form • Assume that current I = -3.5 + j(4.2)
I 2 2 =+−= 5.5)2.4()5.3( A 180o −= γθ ⎛ 2.4 ⎞ γ = tan −1⎜ ⎟ = 2.50 o ⎝ 5.3 ⎠ θ o o =−= 8.1292.50180 o ∴ AI ∠= 8.1295.5 o Electromechanics Basics 1-68 Converting from Polar to Rectangular Form
Re{ }= AA θ )cos( Im{}= AA θ )sin(
Electromechanics Basics 1-69 National Electrical Code (NEC) • Put out by National Fire Protection Agency (NFPA) as NFPA 70 • Initially developed in 1897, and is updated every 3 years • Sets out requirements for wire sizing (phase conductors, neutrals, grounds) and other rules for installation of power in residential single phase and industrial 3 phase systems
Electromechanics Basics 1-70 120V, 2-Wire System • Older system, largely (hopefully) replaced in residences • Neutral = “grounded conductor” • Voltage between hot and neutral is 120 VAC (or 120 VRMS) • Voltage stepped down at pole-mount or pad-mount transformer; enters house service entrance at 120V between HOT and NEUTRAL
Electromechanics Basics 1-71 120/240, 3-Wire System • Most common residential service; 3 wires are derived by center-tapping the distribution transformer secondary • Voltage between one phase and neutral is 120V • Voltage between phases is 240V: used for hot water heaters, dryers, air conditioners, etc. • Neutral is grounded at the service entrance
Electromechanics Basics 1-72 Typical Single-Phase Residential Service • Typical 7.2 kV distribution voltage • Output voltage is 120V with 2 single phases • You’ll run low-power loads (i.e. lights, electronics) from a single phase • High power loads (i.e. clothes dryer, hot water heater) run phase-to-phase
Electromechanics Basics 1-73 Single-Phase Service --- Simulation Results • Transformer step-down
Electromechanics Basics 1-74 200A Breaker Panel • Main breaker at top (200A) • Multiple branch circuits below: black = hot wire, white = neutral
Electromechanics Basics 1-75 Single-Phase 60 Hz Power
Reference: A. Kusko and M. Thompson, Power Quality in Electrical Systems, McGraw-Hill, 2007
Electromechanics Basics 1-76 Three-Phase Power • Three-phase is used in higher power applications • You can carry more power with less copper using 3 phase • Large generators and motors are more efficient and smaller using 3-phase compared to single-phase • 3-phase voltages are created by 3-phase generators
Electromechanics Basics 1-77 Three-Phase Generators • More on 3-phase motor/generators later
Y-connection 2-pole 4-pole of windings
Electromechanics Basics 1-78 Three-Phase, 480V 60 Hz Power • Line-line voltage is 480V RMS • Line-neutral voltage is 277V (or 480/sqrt(3)), which has peaks at ± 392V
Reference: A. Kusko and M. Thompson, Power Quality in Electrical Systems, McGraw-Hill, 2007
Electromechanics Basics 1-79 208Y/120, 3-Phase, 4 Wire • Can drive small 3 phase loads or single phase loads • Voltage phase-phase = 208V (i.e. from A to B)
• Voltage phase-neutral = = 120V
Electromechanics Basics 1-80 480/277, 3-Phase, 4 Wire • Typical industrial voltage for larger 3-phase loads • Large 3-phase loads can be connected to the three phases • Smaller 277V single-phase loads can be connected between phase and ground • Phase-phase = 480V; phase-neutral = 277V
Electromechanics Basics 1-81 480V 4-Wire System
• Line-line = 480V; line-neutral = 277V
Reference: Ralph Fehr, Industrial Power Distribution, Prentice Hall, 2002
Electromechanics Basics 1-82 Some Comments on Delivery Voltage • A typical factory utilization voltage is 480Y/277V, meaning that a 4-wire Y connected service is provided with 480V line-line and a line-neutral voltage of 277V • Service for industrial facilities can be supplied by a utility at distribution voltage (2.4kV to 34.5 kV) or at subtransmission or transmission voltages (46 kV to 230 kV) • 1888: Polyphase AC system invented by Tesla • First three-phase system was installed between Lauffen and Frankfurt am Main in Germany in 1891
Electromechanics Basics 1-83 Root Mean Square (RMS) • Used for description of periodic, often multi-harmonic, waveforms • Calculation of RMS is done by taking the square root of the average over a cycle (mean) of the square of a waveform 1 T I = 2 )( dtti RMS ∫ T 0 • RMS current of any waveshape will dissipate the same amount of heat in a resistor as a DC current of the same value
– DC waveform: Vrms = VDC – Symmetrical square wave: IRMS = Ipk – Pure sine wave: IRMS=0.707Ipk • Example: 120 VRMS line voltage has peaks of ±169.7 V
Electromechanics Basics 1-84 Intuitive Description of RMS • The RMS value of a sinusoidal or other periodic waveform dissipates the same amount of power in a resistive load as does a battery of the same RMS value • So, 120VRMS into a resistive load dissipates as much power in the load as does a 120V battery
Electromechanics Basics 1-85 RMS Value of Various Waveforms • Following are waveforms typically found in power electronics, power systems, and motors, and their corresponding RMS values • Reference: R. W. Erickson and D. Maksimovic, Fundamentals of Power Electronics, 2nd edition, Kluwer, 2001
Electromechanics Basics 1-86 DC Voltage
• Battery
Electromechanics Basics 1-87 Pure Sinewave
• AC line
Electromechanics Basics 1-88 Square Wave
• This type of waveform can be put out by a square wave converter or full-bridge converter
Electromechanics Basics 1-89 DC with Ripple
• Buck converter inductor current (DC value + ripple)
Electromechanics Basics 1-90 Triangular Ripple, No DC Value • Capacitor ripple current in some converters (no DC value)
Electromechanics Basics 1-91 Pulsating Waveform • Buck converter input switch current (assuming small ripple)
Electromechanics Basics 1-92 Pulsating with Ripple
• i.e. buck converter switch current • We can use this result to get RMS value of buck diode current
Electromechanics Basics 1-93 Triangular
Electromechanics Basics 1-94 Piecewise Calculation
• For I1, I2, etc. at different frequencies
Electromechanics Basics 1-95 Piecewise Calculation --- Example • What is RMS value of DC + ripple (shown before)?
Electromechanics Basics 1-96 Apparent, Real and Reactive Power • “Power” has many shapes and forms – Real power (P, in Watts). This is the power that does work – Apparent power (S, in VA). This is the total “apparent”
power seen by the source. It’s the product of VRMS and IRMS – Reactive power (Q, expressed in VAr) • Reactive power does not do real work
– Instantaneous power = titvtp )()()( 1 T – Average power tp )( = ∫ )()( dttitv T 0
Electromechanics Basics 1-97 Vector Relationship Between P, Q and S • P = real power (Watts) • Q = reactive power (VAr) • S = apparent power (VA) • PF = power factor (unitless)
+= QPS 22 P PF == θ )cos( S Electromechanics Basics 1-98 Apparent Power from Voltage and Current • We’ll use the “Poynting vector” ×= IVS * • V = complex phase voltage, I* = complex conjugate of phase current • If A = α + jβ, then A* = α -jβ • A×A*= α2 + β2
Electromechanics Basics 1-99 Apparent Power from Voltage and Current • Result: = IVP θ )cos( = IVQ θ )sin(
Electromechanics Basics 1-100 Example: Finding P, Q and S • Let’s find P, Q, S and Z given V and I
- Find apparent power S using Poynting vector IVS * o o ∠=−∠∠=×= 20840)107)(30120( o += )287(789 VAjS
Electromechanics Basics 1-101 Example: Finding P, Q and S
- Find real power P = VIP θ = o = 798)20cos()7)(120()cos( W - Find imaginary power Q = VIQ θ = o = 287)20sin()7)(120()sin( VAr
- Find load impedance Z V ∠30120 o Z == ∠Ω= 2014.17 o I ∠107 o
Electromechanics Basics 1-102 Finding Load Current in Single Phase Systems • We must know the current magnitude to properly size conductors. The NEC (National Electrical Code) gives guidance on sizing of conductors once we know the RMS conductor current • Current magnitude results in I2R loss and temperature rise in conductors P VAVI == PF P I = × PFV
Electromechanics Basics 1-103 Examples: Finding Load Current • P=10 kVA, PF=1, 120V single phase 10000 I == 3.83 A 120 • P=50 kVA, PF=0.9 lagging, 240V single phase 50000 I = = 5.231 A )9.0)(240(
Electromechanics Basics 1-104 Sizing of Power Cables • Large power cables are available in copper and aluminum • Wire is sized by AWG (American Wire Gage) or thousands of circular mils (kcmils) • He smallest wire used in power distribution is #14 AWG, typically a solid conductor with an outside diameter of 0.0641”, or 64.1 mils • The largest AWG is #4/0, with a diameter of approximately 0.522” for a 7-strand conductor
Electromechanics Basics 1-105 AWG • Wire size is denoted by AWG (American Wire Gage) • Wire diameter varies by a factor of 2 every 6 AWG • #36 AWG is defined to be 0.005” (5 mil) diameter • “Circular mils” is the diameter in mils, squared. (1 mil = 0.001”) To find wire diameter d based on AWG number:
36−AWG In inches: d ×= 92)"005.0( 39
36−AWG In millimeters: d ××= 92)"005.0()4.25( 39
Electromechanics Basics 1-106 Circular Mils • The area of larger conductors is expressed in circular mils. To find the area in cmils, square the diameter in mils by itself. For example, a #10 solid conductor with a diameter of 162 mils has an area of (162 mils)2 = 26244 cmils, or 26.2 kcmil
Electromechanics Basics 1-107 DC Resistance of Power Cables • Resistance is a function of length, cross-sectional area, and electrical conductivity l R = σA
• For copper, σ≈5.8×107 Ohm-1m-1 at room temperature. For aluminum, σ≈3.5×107 Ohm-1m-1 at room temperature
Electromechanics Basics 1-108 Copper Temperature Coefficient of Resistivity • Copper resistivity goes up as temperature goes up • So, the total resistance of a piece of wire goes up as it heats up • Temperature coefficient of resistivity is about +0.004/oC, or about 0.4% per degree C • Resistance goes up as temperature goes up. = [ +α TRR − )298(1 ] T2 25 2
• R25 = resistance at 25C (298K), α = temperature coefficient ≈ 0.00385 for copper and ≈ 0.00395 for aluminum; rule of thumb for copper is +0.4% per degree C
Electromechanics Basics 1-109 Issues With Aluminum Wire • Aluminum has higher resistivity than copper (about 70% higher), with resistivity ρ = 2.9×10-8 Ω-m • A 15-amp branch circuit made with #14 copper would require #12 aluminum, by the NEC • Similar TC to copper • Aluminum is much lighter than copper (about 33% of the weight per unit volume) • Often used in power transmission (good ratio of resistivity to weight) • Corrosion when joining aluminum to copper lugs; differing thermal expansion
Electromechanics Basics 1-110 Power Cables and Cable Impedance • Power cables have a finite resistance (due to the resistivity of the conductor) and a finite inductance (due to the geometry of the cable) • Cable impedance results in voltage drops, power loss, and heating in the cable
Electromechanics Basics 1-111 AC Cable Resistance • DC resistance of a piece of wire: l R = σA • A = cross sectional area of wire, l = length, and σ = electrical conductivity • However, AC resistance is higher than DC resistance, due to skin effect and proximity effect • Wire resistance can be found in a “wire chart” for an isolated wire (i.e. in air, no metallic conduit, no other current-carrying wires nearby)
Electromechanics Basics 1-112 Copper Wire Data
Electromechanics Basics 1-113 Copper Wire Data
Reference: P. Krein, Elements of Power Electronics, 3d edition
Electromechanics Basics 1-114 Causes of Skin Effect • Self-field of wire causes current to crowd on the surface of the wire, raising the AC resistance • Method: – Current i(t) ⇒ changing magnetic flux density B(t) ⇒ reaction currents
Reference: http://ece-www.colorado.edu/~pwrelect/book/slides/Ch12slide.pdf
Electromechanics Basics 1-115 Effects of Skin Effect • For high frequencies, current is concentrated in a layer approximately one skin depth δ Skin depth in copper at 300K cond 5.90E+07 thick mu 1.26E-06 • Skin depth varies with frequency f skin depth (meter) 12 δ == 1 6.55E-02 fμσπωμσ 10 2.07E-02 100 6.55E-03 • σ = electrical conductivity = 1000 2.07E-03 5.9×107 Ω-1 m-1 for copper at 1.00E+04 6.55E-04 1.00E+05 2.07E-04 300K 1.00E+06 6.55E-05 • μ = magnetic permeability of material = 4π×10-7 H/m in free space
Electromechanics Basics 1-116 Skin Effect --- Increase in Wire Resistance • For high frequencies, resistance of wire increases • DC resistance of wire: l RDC = 2 ()πσrw • For frequencies above critical frequency where skin depth R(f) equals wire radius rw: l ⎛ 2δ ⎞ R = = R ⎜ ⎟ AC ()2 r δπσDC ⎜ r ⎟ w ⎝ w ⎠ RDC 1 fcrit = 2 f f rw μσπ crit • Result: for high frequency operation, don’t bother making wire radius > δ – Skin depth in copper at 60 Hz is approximately 8 mm
Electromechanics Basics 1-117 Proximity Effect • In multiple-layer windings in inductors and transformers, the proximity effect can also greatly increase the winding resistance • Field from one wire affects the current profile in another
Currents in opposite Currents in same direction direction
Electromechanics Basics 1-118 Cable Inductance • Inductance is defined by the geometry of the wire loop • Included in the calculation is wire radius, and loop length and shape
Circular loop of wire Parallel-wire line R a 2R
d
μ l d 1 d 8a o ⎡ ⎤ ⎡ ⎛ ⎞ ⎤ L =+−ln⎢ ⎥ La= μ o ⎢ln⎜ ⎟ − 175 . ⎥ π ⎣ R 4 l ⎦ ⎣ ⎝ R ⎠ ⎦
Electromechanics Basics 1-119 AC Service with Real-World Impedance • All electrical services have finite impedance; shown
here the impedance is Rs + jωLs • These types of power distribution drawings rarely show inductance explicitly; rather the inductive
reactance jXs is used. • Ls may be the sum of wiring inductance, transformer leakage, etc.
Electromechanics Basics 1-120 Example: Finding Load Voltage and Current • Find load voltage and load current • Note that line resistance is 0.1Ω and line reactance is 0.2Ω
277 I = −= 54.04.27 j L j ++ 101.02.0
LOAD RIV LL −== 4.5274 j
• Note voltage drop across line impedance of a few Volts
Electromechanics Basics 1-121 National Elec. Code Impedance Estimates • Note conduit type affects resistance and reactance
Electromechanics Basics 1-122 Power Cable Types
Electromechanics Basics 1-123 Power Cable Types
Electromechanics Basics 1-124 National Elec. Code Ampacity Estimates
Electromechanics Basics 1-125 National Elec. Code Ampacity Estimates
Electromechanics Basics 1-126 “Single-Line Drawings” • Also called “one-line” drawing • Shorthand to describe power distribution systems • Symbols include power busses, transmission lines, disconnect switches, air and oil-filled circuit breakers, transformers, motors, generators, etc. • Shown here is a “radial” power distribution system
Electromechanics Basics 1-127 Single-Line Drawing Symbols • There are lots more, but these are some of the important ones
Electromechanics Basics 1-128 Typical Single-Line Drawing
Electromechanics Basics 1-129 Typical Single-Line Drawing
Reference: J. C. Das, “Passive Filters --- Potentialities and Limitations,” IEEE Transactions on Industry Applications, vol. 40, no. 1, January/February, 2004, pp. 232-241 Electromechanics Basics 1-130 Power Factor and Power Factor Correction • Power factor is an important concept in power circuits, power electronics, and electric motors • Power factor is a measure of how easily you deliver real power to a load
Electromechanics Basics 1-131 Power Factor • Ratio of delivered power to the product of RMS voltage and RMS current < P > PF = IV RMSRMS • Power factor always <= 1 • With pure sine wave and resistive load, PF = 1 • With pure sine wave and purely reactive load, PF = 0 • Whenever PF < 1 the circuit carries currents or voltages that do not perform useful work • The more “spikey” a waveform is the worse is its PF – Diode rectifiers have poor power factor • Power factor can be helped by “power factor correction”
Electromechanics Basics 1-132 Example: Power Factor Calculation • A 3-phase load consumes 100 kW and 50 kVAr • Find P, Q, S and power factor PF
100 kWP == 105 Q = 50 kVAr
2 S 25 ()4 =×+= 8.111105)10( kVA P 100 PF === 894.0 S 8.111 Electromechanics Basics 1-133 Example: Another PF Calculation • A load consumes 500 kW at 0.85 PF lagging. Find apparent power, reactive power, and power factor angle
kWP ×== 105500 5 P P ×105 5 PF S ==⇒= = 588 kVA S PF 85.0 222 PSQSQP 22 2 2 =−=−=⇒=+ 310500588 kVAr θ = −1 PF = −1 −= 8.31)85.0(cos)(cos o
Electromechanics Basics 1-134 Causes of Low Power Factor --- Nonlinear Load • Nonlinear loads include: • Variable-speed drives • Frequency converters • Uninterruptible power supplies (UPS) • Saturated magnetic circuits • Dimmer switches • Televisions • Fluorescent lamps • Welding sets • Arc furnaces • Semiconductors • Battery chargers
Electromechanics Basics 1-135 Further Comment on Power Factor
• Power factor (kp) can be broken up into 2 pieces: st • Displacement power factor kθ, due to phase shift between 1 harmonic of current and input voltage
• Distortion power factor (kd) due to harmonics • Total power factor is product of these 2 = kkk p d θ
Electromechanics Basics 1-136 HWR with Resistive Load • Note that power is delivered in periodic blips; line current has harmonics so PF < 1
Electromechanics Basics 1-137 HWR with Resistive Load --- Power Factor • We can calculate P.F. by inspection: V v = pk ,RMSin 2
⎛Vpk ⎞⎛ ⎞⎛ 11 ⎞ Vpk i ,RMSin = ⎜ ⎟⎜ ⎟⎜ ⎟ = ⎝ 2 ⎠⎝ RL ⎠⎝ 2 ⎠ 2RL 2 ⎛ V ⎞ V 2 2 pk ⎟ pk >=< , Rip LRMSinL = ⎜ ⎟ RL = ⎝ 2RL ⎠ 4RL ⎛ V 2 ⎞ ⎜ pk ⎟ ⎜ ⎟ p >< 4RL 2 FP .. = L = ⎝ ⎠ == 707.0 iv ,, RMSinRMSin ⎛ pk ⎞⎛ VV pk ⎞ 2 ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2RL ⎠
Electromechanics Basics 1-138 FWR with Resistive Load --- PSPICE Simulation • We need dummy resistor to keep PSPICE happy
Electromechanics Basics 1-139 FWR with Resistive Load
Electromechanics Basics 1-140 FWR with Resistive Load --- Power Factor V v = pk ,RMSin 2 ⎛V ⎞⎛ 1 ⎞ V pk ⎜ ⎟ pk i ,RMSin = ⎜ ⎟⎜ ⎟ = ⎝ 2 ⎠⎝ RL ⎠ 2RL 2 ⎛ V ⎞ V 2 >=< 2 Rip = ⎜ pk ⎟ R = pk , LRMSinL ⎜ ⎟ L ⎝ 2RL ⎠ 2RL ⎛ V 2 ⎞ ⎜ pk ⎟ ⎜ ⎟ p >< 2RL 2 FP .. = L = ⎝ ⎠ == 0.1 iv ⎛ ⎞⎛ VV ⎞ 2 ,, RMSinRMSin pk ⎜ pk ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ 22 RL ⎠
Electromechanics Basics 1-141 Half Wave Rectifier with RC Load • If RC >> 1/f then this operates like a peak detector and the output voltage
Electromechanics Basics 1-142 Half Wave Rectifier with RC Load • Note poor power factor due to peaky input line current
Electromechanics Basics 1-143 Unity Power Factor --- Resistive Load • Example: purely resistive load – Voltage and currents in phase = sin)( ωtVtv V ti = sin)( ωt R V 2 titvtp == sin)()()( 2 ωt R V 2 tp )( >=< 2R V V = RMS 2 V I = RMS R 2 V 2 tp )( >< PF = = 2R = 1 IV ⎛ ⎞⎛ VV ⎞ RMSRMS ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ R 22 ⎠
Electromechanics Basics 1-144 Causes of Low Power Factor --- Reactive Load
• Example: purely inductive load – Voltage and currents 90o out of phase = sin)( ωtVtv V ti = cos)( ωt ωL V 2 titvtp == cossin)()()( ωωtt ωL tp >=< 0)( • For purely reactive load, PF=0
Electromechanics Basics 1-145 Why is Power Factor Important? • Consider peak-detector full-wave rectifier
• Typical power factor kp = 0.6 • What is maximum power you can deliver to load ?
– VAC x current x kp x rectifier efficiency – (120)(15)(0.6)(0.98) = 1058 Watts • Assume you replace this simple rectifier by power electronics module with 99% power factor and 93% efficiency: – (120)(15)(0.99)(0.93) = 1657 Watts Electromechanics Basics 1-146 Power Factor Correction • A toaster can draw 1500W from a 120V/15A line • Typical offline switching power converter can draw <1000W from the line since it has poor power factor • High power factor results in: lower utility bills, increased system capacity, better voltage quality, reduced heating losses • Methods of power factor correction – Passive: add capacitors or inductors – Active
Electromechanics Basics 1-147 Power Factor Correction --- Passive • Switch capacitors in and out as needed as load changes
Electromechanics Basics 1-148 Power Factor Correction --- Active • Fluorescent lamp ballast application
Electromechanics Basics 1-149 Power Factor Corrected Power Supplies
Without PFC
With PFC
Reference: On Semiconductor, “Power Factor Correction Handbook,” www.onsemi.com
Electromechanics Basics 1-150 Power Supply with Power Factor Correction
Reference: On Semiconductor, “Power Factor Correction Handbook,” www.onsemi.com
Electromechanics Basics 1-151 Harmonics
• Joseph Fourier (1768-1830) worked out that any periodic waveform can be broken up into a fundamental frequency plus harmonics • The harmonics need to have the proper amplitude and phase relationship to the fundamental • The result is a “Fourier series” for a periodic waveform
Electromechanics Basics 1-152 Fourier Series • Here’s the recipe:
• Finding the Fourier coefficients:
Reference: http://www.efunda.com/math/fourier_series/fourier_series.cfm
Electromechanics Basics 1-153 Harmonics of Square Wave • A 50% duty cycle square wave:
⎛ 4 ⎞ ⎛ 4 ⎞ ⎛ 4 ⎞ tv )( = ⎜ ⎟ ωt)sin( + ⎜ ⎟ ωt)3sin( + ⎜ ⎟ ωt)5sin( ⋅⋅⋅+ ⎝π ⎠ ⎝ 3π ⎠ ⎝ 5π ⎠
Electromechanics Basics 1-154 Building Up a Square Wave • Building up a 50% duty cycle square wave:
Reference: http://mathworld.wolfram.com/FourierSeries.html
Electromechanics Basics 1-155 Triangle Wave
⎛ 8 ⎞ ⎛ 8 ⎞ ⎛ 8 ⎞ tv )( = ⎜ ⎟ ωt)sin( − ⎜ ⎟ ωt)3sin( + ⎜ ⎟ ωt)5sin( ⋅⋅⋅+ ⎝ π 2 ⎠ ⎝ 3 π 22 ⎠ ⎝ 5 π 22 ⎠
Electromechanics Basics 1-156 Harmonics in the Power Line • Harmonics are created by nonlinear circuits – Rectifiers • Half-wave rectifier has first harmonic at 60 Hz • Full-wave has first harmonic at 120 Hz – Switching DC/DC converters • DC/DC operating at 100 kHz generally creates harmonics at DC, 100 kHz, 200 kHz, 300 kHz, etc. • Line harmonics can be treated by line filters – Passive – Active
Electromechanics Basics 1-157 Harmonics in the Power Line • The issue of harmonics has become much more important in recent years in power systems • Harmonic sources include – Switching power supplies – Variable speed drives (VSDs) – Arc furnaces – Welders – Fluorescent lamp ballasts
Electromechanics Basics 1-158 Half-Wave Rectifier, Resistive Load • Simplest, cheapest rectifier • Line current has DC component; this current appears in neutral • High harmonic content, Power factor = 0.7
P FP .. = avg IV RMSRMS
Reference: Mohan, Undeland and Robbins, Power Electronics, Converters, Applications and Design, John Wiley, 2003, pp. 80 Electromechanics Basics 1-159 Half Wave Rectifier with Resistive Load --- Power Factor and Average Output Voltage Average output voltage:
1 π Vpk t=πω Vpk vd >=< pk ωωtdtV )()sin( []−= ωt)cos( ωt=0 = 2π ∫0 2π π Power factor calculation:
2 2 1 ⎛ I pk ⎞ I pk P =>< ⎜ ⎟ R = R 2 ⎝ 2 ⎠ 4 IV V pk == pk RMS R 22 I 1 I = pk RMS 2 2 I 2 pk R P >< PF = = 4 = 707.0 IV RMSRMS ⎛ I pk ⎞⎛ I pk 1 ⎞ ⎜ ⎟⎜ ⎟ ⎝ R ⎠⎝ 22 2 ⎠ Electromechanics Basics 1-160 Half-Wave Rectifier, Resistive Load --- Spectrum of Load Voltage
Electromechanics Basics 1-161 Half Wave Rectifier with RC Load • More practical rectifier • For large RC, this behaves like a peak detector
Electromechanics Basics 1-162 Half Wave Rectifier with RC Load • Note poor power factor due to peaky line current • Note DC component of line current
Electromechanics Basics 1-163 Half Wave Rectifier with RC Load --- Spectrum of Line Current
Electromechanics Basics 1-164 Total Harmonic Distortion • Total harmonic distortion (THD) – Ratio of the RMS value of all the nonfundamental frequency terms to the RMS value of the fundamental
2 ∑ I n,RMS 2 2 n≠1 RMS − II ,1 RMS THD = 2 = 2 ()I ,1 RMS I ,1 RMS
• Symmetrical square wave: THD = 48.3% • Symmetrical triangle wave: THD=12.1%
Electromechanics Basics 1-165 Crest Factor • Another term sometimes used in power engineering • Ratio of peak value to RMS value • For a sinewave, crest factor = 1.4 – Peak = 1; RMS = 0.707 • For a square wave, crest factor = 1 – Peak = 1; RMS = 1
Electromechanics Basics 1-166 Harmonics and THD --- Pure Sinewave • THD = 0%
Number of harmonics N = 1 THD = 0 % 1.5
1
0.5
0
-0.5
-1
-1.5 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07
Electromechanics Basics 1-167 Harmonics and THD - Sinewave + 3rd Harmonic • THD = 33.3%
Electromechanics Basics 1-168 Harmonics and THD --- Sinewave + 3rd + 5th Harmonic • THD = 38.9%
Electromechanics Basics 1-169 Harmonics --- Up to N = 103 • THD = 48%
Electromechanics Basics 1-170 Single Phase, Full-Wave Rectifier • Draws significant harmonics from the line
Electromechanics Basics 1-171 Full-Wave Rectifier with RC Filter --- PSPICE
Electromechanics Basics 1-172 Three-Phase Circuits and Three-Phase Power • Three phase is used extensively in high power applications • For high power, it has multiple advantages over single phase
Electromechanics Basics 1-173 A Three-Phase, Four-Wire System • A common neutral wire is assumed
• If there are 3 equal linear loads, in = 0 • However, if there are nonlinear loads, there can be a neutral current
Electromechanics Basics 1-174 Balanced, Three-Phase, Four-Wire System
• Result shows that for balanced load, in = 0
Electromechanics Basics 1-175 Three-Phase, Four-Wire System, Unbalanced Load
Electromechanics Basics 1-176 Three-Phase, Full-Bridge Rectifier • Commonly used in high power applications • Also called “6-pulse rectifier” • This circuit generates line harmonics
Electromechanics Basics 1-177 6-Pulse Rectifier with Current Source Load • 3-phase source; draws 5th, 7th, 11th, 13th, … harmonics
Electromechanics Basics 1-178 6-Pulse Rectifier: Redrawn • Two groups with three diodes each
Electromechanics Basics 1-179 6-Pulse Rectifier Waveforms • Shown for output DC current source load
Electromechanics Basics 1-180 6-Pulse Rectifier: Line Current • Assuming output current to be purely dc and zero ac-side inductance
Electromechanics Basics 1-181 6-Pulse Rectifier with Current Source Load Simulation
• Simplified with Ls = 0 (no line inductance); and current source load
Electromechanics Basics 1-182 6-Pulse Rectifier with Current Source Load
Electromechanics Basics 1-183 6-Pulse Rectifier with Resistive Load Simulation
• Simplified with Ls = 0 (no line inductance)
Electromechanics Basics 1-184 6-Pulse Rectifier with Resistive Load --- Output • Note that fundamental of ripple frequency = 360 Hz • Note that average output voltage is higher than single line voltage. Peak value is sqrt(3) x peak of line = 294V
Electromechanics Basics 1-185 6-Pulse Rectifier with Resistive Load and Capacitor Filter • Let’s add filter capacitor • Note that a smaller capacitor can be used for the 3 phase rectifier compared to single phase rectifier, because (1) Ripple is smaller and (2) Ripple frequency is higher
Electromechanics Basics 1-186 6-Pulse Rectifier with Resistive Load and Capacitor Filter
Electromechanics Basics 1-187 12-Pulse Rectifier • Add 2 phase-shifted 6-pulse rectifiers; 5th and 7th are eliminated • Only harmonics are the 11th, 13th, 23rd, 25th …
Reference: R. W. Erickson and D. Maksimovic, Fundamentals of Power Electronics, 2d edition Electromechanics Basics 1-188 Three-Phase, Four-Wire System with Nonlinear Load • With single-phase nonlinear load, there can be a neutral current
Reference: Mohan, Undeland and Robbins, Power Electronics, Converters, Applications and Design, John Wiley, 2003, pp. 101 Electromechanics Basics 1-189 Neutral Current in a 3-Phase, Four-Wire System • The neutral current can be very high if driving nonlinear loads line to neutral which generate 3rd or other harmonics • If line currents are highly discontinuous, the neutral current can be as large as 1.73xline current 3rd harmonic •Note 3rd harmonic here
Reference: Mohan, Undeland and Robbins, Power Electronics, Converters, Applications and Design, John Wiley, 2003, pp. 102 Electromechanics Basics 1-190 Simulation of Simple Case
Electromechanics Basics 1-191 Simulation of Simple Case --- Neutral Current vs. Time
Electromechanics Basics 1-192 Simulation of Simple Case --- Spectrum of Neutral Current • Note high 3rd harmonic
Electromechanics Basics 1-193 Example: Three-Phase Line Current Calculation • Assume 60 Hz, 3-phase 4-wire, 480 V phase to phase (277V phase-neutral). Loads on phases A and B are 48 Ohms to neutral. Load on phase C is 24 Ohms to neutral. We’ll find phase currents and neutral current
Electromechanics Basics 1-194 Example: Three-Phase Line Current Calculation
• Let’s find phase currents: 480 v o ∠=∠= 02770 o an 3 480 v o −∠=−∠= 120277120 o bn 3 480 v o −∠=−∠= 240277240 o cn 3 v i = an A∠= 077.5 o a 48Ω v i = bn A −∠= 12077.5 o b 48Ω v i = cn A −∠= 24054.11 o c 24Ω Electromechanics Basics 1-195 Example: Three-Phase Line Current Calculation • Phase currents
Phase currents 20
15
10
5
0
-5
-10
-15
-20 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035
Electromechanics Basics 1-196 Example: Neutral Current Calculation • Calculation of neutral current: remember that in a balanced, 3-phase system with linear loads the neutral current sums to zero. • In this case, we have an imbalance because the amplitude of phase C current is higher than the others. • The “leftover” current equals the neutral current
++= iiii cban o o ()o −∠+−∠+−∠+∠= 24077.524077.512077.5077.5 o −∠= 24077.5 o
Electromechanics Basics 1-197 Example: Neutral Current Calculation
++= iiii cban o o ()o −∠+−∠+−∠+∠= 24077.524077.512077.5077.5 o o Neutral current −∠= 24077.5 10
8
6
4
2
0
-2
-4
-6
-8
-10 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035
Electromechanics Basics 1-198 Thyristor Converter --- Single Phase with Resistive Load
• Angle α is called “firing angle”
V v pk ()+>=< cos1 α d 2π
Electromechanics Basics 1-199 Thyristor Converter --- Single Phase with Resistive Load • Control characteristic: average output voltage vs. firing angle V pk 1 vd ()+>=< cos1 α 2π 0.9
0.8
0.7
0.6
0.5
0.4 Normalized
0.2
0.1
0 0 20 40 60 80 100 120 140 160 180 Angle, degrees Electromechanics Basics 1-200 Three-Phase Thyristor Converter • AC-side inductance is included
Electromechanics Basics 1-201 Today’s Summary • Today we’ve covered: • Signal basics, step response, resonance, etc. • Power in its various forms • Root-mean square (RMS) • Power cables and cable impedance • Power factor and power factor correction (PFC) • Fourier series and harmonics • Basic three-phase circuits
Electromechanics Basics 1-202