Lecture Set 1 Introduction to Magnetic Circuits S.D. Sudhoff Spring 2017 1
Goals Review physical laws pertaining to analysis of magnetic systems Introduce concept of magnetic circuit as means to obtain an electric circuit Learn how to determine the flux-linkage versus current characteristic because it will be key to understanding electromechanical energy conversion 2
Review of Symbols Flux Density in Tesla: B (T) Flux in Webers: Φ (Wb) Flux Linkage in Volt-seconds: λ (Vs,Wb-t) Current in Amperes: i (A) Field Intensity in Ampere/meters: H (A/m) Permeability in Henries/meter: µ (Η/m) Magneto-Motive Force: F (A,A-t) Permeance in Henries: P (H) Reluctance in inverse Henries: R (1/H) Inductance in Henries: L (H) Current (fields) into page: Current (fields) out of page: 3
Ampere s Law, MMF, and Kirchoff s MMF Law for Magnetic Circuits 4
Consider a Closed Path 5
Enclosed Current and MMF Sources Enclosed Current i Γ = N i + N i N i enc, a a b b c c MMF Sources F F a b = N i = N i b b a a F c = N i c c Conclusion i enc, Γ = Fs SΓ = {' a ',' b',' c '} s S Γ 6
MMF Drops Going Around the Loop MMF Drop F y Thus Or H d l = H d l + H d l + H d l + H d l l l l lγ l Γ = H d l l y H d l = α β δ H d l = F + F + F + F l Τ F d DΓ = {' α ',' β ',' γ ',' δ '} l d DΓ Γ α β γ δ 7
Ampere s Law Ampere s Law H d l = i l Γ Recall i enc, Γ Thus = s S Γ enc, Γ F F = s s S d D Γ s Γ F d H d l = l Γ d D Γ F d 8
Kirchoff s MMF Law Kirchoff s MagnetoMotive Force Law: The Sum of the MMF Drops Around a Closed Loop is Equal to the Sum of the MMF Sources for That Loop 9
Example: MMF Sources 10 conductors, 3 A = 5 conductors, 2 A = 10
Example: MMF Drops A quick example on MMF drop: 11
Another Example: MMF Drops How about F cb? 12
Magnetic Flux, Gauss s Law, and Kirchoff s Flux Law for Magnetic Circuits 13
Magnetic Flux Φ m = B S m ds 14
Example: Magnetic Flux Suppose there exist a uniform flux density field of B=1.5a y T where a y is a unit vector in the direction of the y-axis. We wish to calculate the flux through open surface S m with an area of 4 m 2, whose periphery is a coil of wire wound in the indicated direction. 15
Example: Magnetic Flux 16
Example: Magnetic Flux 17
Kirchoff s Flux Law Gauss s Law Thus Or B d S = S Γ o O Γ S o Φ o = o O Γ 0 B ds O Γ = {' α ',' β ', ' γ ', } 0 18
Kirchoff s Flux Law Kirchoff s Flux Law: The Sum of the Flux Leaving A Node of a Magnetic Circuit is Zero 19
Kirchoff s Flux Law A quick example on Kirchoff s Flux Law 20
Magnetically Conductive Materials and Ohm s Law for Magnetic Circuits 21
Ohm s Law On Ohm s Law Property for Electric Circuits On Ohm s Law Property for Magnetic Circuits 22
Types of Magnetic Materials Interaction of Magnetic Materials with Magnetic Fields Magnetic Moment of Electron Spin Magnetic Moment of Electron in Shell These Effects Lead to Six Material Types Diamagnetic Paramagnetic Ferromagnetic Antiferromagnetic Ferrimagnetic Superparamagnetic 23
Ferromagnetic Materials Iron Nickel Cobalt 24
Ferrimagnetic Materials Iron-oxide ferrite (Fe 3 O 4 ) Nickel-zinc ferrite (Ni 1/2 Zn 12 Fe 2 O 4 ) Nickel ferrite (NiFe 2 O 4 ) 25
Behavior of Ferromagnetic and Ferrimagnetic Materials 26
Behavior of Ferromagnetic and Ferrimagnetic Materials M19 MN80C 27
Modeling Magnetic Materials Free Space B = µ H 0 Magnetic Materials B = µ 0( H + M ) M = χ H B = µ H + M Either Way B Or B 0 M = µ χ H = µ 0 (1 + χ) H B = µ 0µ r H µ r = 1+ χ = µ H µ = µ 0µ r 0 28
Modeling Magnetic Materials B = µ H ( H ) H B = µ B ( B) H 29
Back to Ohm s Law 30
Relationship Between MMF Drop and Flux F = R ( ξ ) Φ a a a R a ( ξ ) la Form 1: ξ = Fa Aa µ H ( Fa / la ) = la Form 2 : ξ = Φ Aa µ B( Φa / Aa ) a Φ = P ( ξ ) F a a a P a ( ξ ) = A µ ( F / l ) a H a a l a A µ ( Φ / A ) a B a a l a Form 1: ξ = F Form 2 : ξ = Φ a a 31
Derivation 32
Derivation 33
Construction of the Magnetic Equivalent Circuit 34
Consider a UI Core Inductor 35
Inductor Applications 36
UI Core Inductor MEC 37
Reluctances 38
Reluctances 39
Reduced Equivalent Circuit 40
Example Let us consider a UI core inductor with the following parameters: w i = 25.1 mm, w b = w e = 25.3 mm, w s = 51.2 mm, d s = 31.7 mm, l c = 101.2 mm, g =1.00 mm, w w = 38.1 mm, and d w = 31.7 mm. The winding is composed of 35 turns. Suppose the magnetic core material has a constant relative permeability of 7700, and that a current of 25.0 A is flowing. Compute the flux through the magnetic circuit and the flux density in the I-core. 41
Example 42
Example 43
Solving the Nonlinear Case R ( Φ ) = R ( Φ ) + R ( Φ ) + 2 R ( Φ ) + 2R eq ic buc luc g Φ = R eq Ni ( Φ) 44
Translation of Magnetic Circuits to Electric Circuits: Flux Linkage and Inductance 45
Flux Linkage λ = N Φ x x x 46
Inductance L x, abs = λ i x x i, λ x1 x1 L inc = λx i x i x1 47
Example Suppose λ = 0.05(1 e 0.2i ) + 0.03i Find the absolute and incremental inductance at 5 A 48
Computing Inductance Φ x N x i x + - R x 49
Computing Inductance 50
Nonlinear Flux Linkage Vs. Current 51
Nonlinear Flux Linkage Vs. Current 52
Nonlinear Flux Linkage Vs. Current % Computation of lambda-i curve % for UI core inductor % using a simple but nonlinear MEC % % S.D. Sudhoff for ECE321 % January 13, 2012 % clear work space clear all close all g=1*mm; % air gap (m) N=100; % number of turns mu0=4e-7*pi; % perm of free space (H/m) Bsat=1.3; % sat flux density (T) mur=1500; % init rel permeability Am=w*d; % mag cross section (m^2) % start with flux linkage lambda=linspace(0,0.08,1000); % assign parameters cm = 1e-2; % a centimeter mm = 1e-3; % a millimiter w=1*cm; % core width (m) ws=5*cm; % slot width (m) ds=2*cm; % slot depth (m) d=5*cm; % depth (m) % find flux and B and mu phi=lambda/n; B=phi/(w*d); mu=mu0*(mur-1)./(exp(10*(b-bsat)).^2+1)+mu0; 53
Nonlinear Flux Linkage Vs. Current % Assign reluctances; Ric=(ws+w)./(Am*mu); Rbuc=(ws+w)./(Am*mu); Rg=g/(Am*mu0); Rluc=(ds+w/2)./(Am*mu); % compute effective reluctances Reff=Ric+Rbuc+2*Rluc+2*Rg; % compute MMF and current F=Reff.*phi; i=f/n; % find lambda-i characteristic figure(1) plot(i,lambda); xlabel('i, A'); ylabel('\lambda, Vs'); grid on; % show B-H characteristic H=B./mu; figure(2) plot(h,b) xlabel('h, A/m'); ylabel('b, T'); grid on; 54
Nonlinear Flux Linkage Vs. Current 55
Hardware Example Our MEC
Non-Idealities: Fringing and Leakage Flux Log10 Energy Density J/m3 0.08 0.06 0.04 y,m 0.02 0-0.02-0.04-0.06 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 x,m 57
Our MEC With Leakage and Fringing 58
Faraday s Law Faraday s Law, voltage equation for winding vx = rxix dλ + x dt If inductance is constant λ x = L x i x Combining above yields vx = rxix + di L x x dt 59
Energy Stored in a Magnetically Linear Inductor We can show Proof E = x 1 2 L x i 2 x 60
Mutual Inductance 61
Mutual Inductance 62
Case Study: Lets Design a UI Core Inductor 63
Architecture Depth into page = d Node 8 Node 7 Node 1 Node 2 w Node 3 g Node 4 N Turns d s w Node 6 Node 5 w ws w 64
Design Requirements Current Level: 40 A Inductance: 5 mh Packing Factor: 0.7 Slot Shape η dw : 1 Maximum Resistance: 0.1Ω 65
Information Cost of Ferrite: 230 k$/m 3 Density of Ferrite: 4740 Kg/m 3 Saturation Flux Density: 0.5 T Cost of Copper: 556 $/m 3 Density of Copper: 8960 Kg/m 3 Resistivity of Copper: 1.7e-8 Ωm 66
Step 1: Design Variables Depth into page = d Node 8 Node 7 Node 1 Node 2 w Node 3 g Node 4 N Turns d s w Node 6 Node 5 w ws w 67
Step 2: Design Constraints Constraint 1: Flux Density 68
Step 2: Design Constraints Constraint 2: Inductance 69
Step 2: Design Constraints Constraint 3: Slot Shape 70
Step 2: Design Constraints Constraint 4: Packing Factor 71
Step 2: Design Constraints Constraint 5: Resistance 72
Step 2: Design Constraints 73
Approach (1/6) 74
Approach (2/6) 75
Approach (3/6) 76
Approach (4/6) 77
Approach (5/6) 78
Approach (6/6) 79
Design Metrics Magnetic Material Volume Depth into page = d Node 8 Node 7 Node 1 Node 2 w Node 3 g Node 4 N Turns d s w Node 6 Node 5 w ws w 80
Design Metrics Wire Volume 81
Design Metrics Circumscribing Box Depth into page = d Node 8 Node 7 Node 1 Node 2 w Node 3 g Node 4 N Turns d s w Node 6 Node 5 w ws w 82
Design Metrics Material Cost Mass 83
Circumscribing Box Volume 84
Material Cost 85
Mass 86
Least Cost Design N = 260 Turns d = 8.4857 cm g = 13.069 mm w = 1.813 cm a w = 21.5181 (mm)^2 d s = 8.94 cm w s = 8.94 cm V mm = 0.00066173 m^3 V cu = 0.0027237 m^3 V bx = 0.0075582 m^3 Cost = 153.7112 $ Mass = 27.5408 Kg 87
Loss and Mass 88
Design Criticisms 89
Manual Design Approach Flux Density B/m 2 0.08 Analysis 0.06 0.04 0.02 y,m 0 Design Equations Numerical Analysis Design Revisions -0.02-0.04-0.06-0.08-0.1-0.05 0 0.05 0.1 x,m Final Design 90
Formal Design Optimization Evolutionary Environment Detailed Analysis Fitness Function 91
Another Design Trade-Off 92
Lunar Rover Propulsion Drive Pareto-Optimal Front 93