Lecture Set 1
Introduction to
Magnetic Circuits
S.D. Sudhoff
Spring 2016
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
ienc ,Γ = − N a ia + Nb ib − Nc ic
• MMF Sources
Fa = − N a ia
Fb = N b ib
Fc = − N c ic
• Conclusion
ienc , Γ =
∑F
s
s∈S Γ
SΓ = {' a ', ' b ', ' c '}
6
MMF Drops
• Going Around the Loop
∫ H ⋅ dl = ∫ H ⋅ dl + ∫ H ⋅ dl + ∫ H ⋅ dl + ∫ H ⋅ dl
lΓ
lα
lβ
lγ
lδ
• MMF Drop
Fy = ∫ Hidl
ly
• Thus
∫ Hidl = Fα + Fβ + Fγ + Fδ
lΤ
• Or
∫ H ⋅ dl = ∑ F
d
lΓ
d ∈DΓ
DΓ = {'α ', ' β ', ' γ ', ' δ '}
7
Ampere’s Law
• Ampere’s Law
∫ H ⋅ dl = i
enc , Γ
lΓ
• Recall
ienc , Γ =
∑ F ∫ H ⋅ dl = ∑ F
s
d
s∈S Γ
lΓ
• Thus
∑F
s
s∈SΓ
=
d ∈DΓ
∑F
d
d ∈DΓ
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 Fcb ?
12
Magnetic Flux, Gauss’s Law, and
Kirchoff’s Flux Law for Magnetic Circuits
13
Magnetic Flux
Φm =
∫ B⋅dS
Sm
14
Example: Magnetic Flux
• Suppose there exist a uniform flux density field of
B=1.5ay T where ay is a unit vector in the direction
of the y-axis. We wish to calculate the flux through
open surface Sm with an area of 4 m2, 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
∫ B⋅dS = 0
SΓ
• Thus
∑ ∫ B ⋅ dS = 0
O Γ = {' α ', ' β ', ' γ ',⋯}
o∈ΟΓ So
• Or
∑Φ
o
=0
o∈Ο Γ
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 (Fe3O4)
• Nickel-zinc ferrite (Ni1/2Zn12Fe2O4)
• Nickel ferrite (NiFe2O4)
25
Behavior of Ferromagnetic and
Ferrimagnetic Materials
26
Behavior of Ferromagnetic and
Ferrimagnetic Materials
M19
MN80C
27
Modeling Magnetic Materials
• Free Space
B = µ0 H
• Magnetic Materials
B = µ0 ( H + M )
M = χH
B = µ0 H + M
M = µ0 χ H
• Either Way
B = µ0 (1 + χ ) H
B = µ0 µr H
µr = 1 + χ
• Or
B = µH
µ = µ0 µr
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
Fa = Ra (ξ )Φ a
Φ a = Pa (ξ ) Fa
la

 A µ (F / l )
 a H a a
Ra (ξ ) = 
la

 Aa µB (Φ a / Aa )
 Aa µ H ( Fa / la )

la

Pa (ξ ) = 
 Aa µ B (Φ a / Aa )

la
Form1: ξ = Fa
Form 2 : ξ = Φ a
Form1: ξ = Fa
Form 2 : ξ = Φ 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: wi = 25.1 mm, wb = we = 25.3 mm, ws = 51.2
mm, ds = 31.7 mm, lc = 101.2 mm, g =1.00 mm, ww =
38.1 mm, and dw = 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
Req (Φ) = Ric (Φ) + Rbuc (Φ) + 2Rluc (Φ) + 2Rg
Ni
Φ=
Req (Φ )
44
Translation of Magnetic Circuits to Electric
Circuits: Flux Linkage and Inductance
45
Flux Linkage
λx = N xΦ x
46
Inductance
Lx , abs =
λx
ix
∂λx
Linc =
∂ix
ix1 , λx1
ix1
47
Example
• Suppose
λ = 0.05(1 − e −0.2i ) + 0.03i
• Find the absolute and incremental
inductance at 5 A
48
Computing Inductance
Φx
+
N xix -
Rx
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
% assign parameters
cm = 1e-2;
mm = 1e-3;
w=1*cm;
ws=5*cm;
ds=2*cm;
d=5*cm;
g=1*mm;
N=100;
mu0=4e-7*pi;
Bsat=1.3;
mur=1500;
Am=w*d;
%
%
%
%
%
%
air gap (m)
number of turns
perm of free space (H/m)
sat flux density (T)
init rel permeability
mag cross section (m^2)
% start with flux linkage
lambda=linspace(0,0.08,1000);
%
%
%
%
%
%
a centimeter
a millimiter
core width (m)
slot width (m)
slot depth (m)
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
v x = rx ix +
dλ x
dt
• If inductance is constant
λ x = Lx i x
• Combining above yields
v x = rx ix + Lx
di x
dt
59
Energy Stored in a Magnetically
Linear Inductor
• We can show
• Proof
1
E x = Lxi x2
2
60
Mutual Inductance
61
Mutual Inductance
62
Case Study: Lets Design a
UI Core Inductor
63
Architecture
Node 2
Node 1
Depth into
page = d
w
Node 3
g
Node 4
ds
Node 8
Node 7
N Turns
w
Node 6
w
Node 5
ws
w
64
Design Requirements
•
•
•
•
•
Current Level: 40 A
Inductance: 5 mH
Packing Factor: 0.7
Slot Shape ηdw: 1
Maximum Resistance: 0.1Ω
65
Information
•
•
•
•
•
•
Cost of Ferrite:
Density of Ferrite:
Saturation Flux Density:
Cost of Copper:
Density of Copper:
Resistivity of Copper:
230 k$/m3
4740 Kg/m3
0.5 T
556 $/m3
8960 Kg/m3
1.7e-8 Ωm
66
Step 1: Design Variables
Node 2
Node 1
Depth into
page = d
w
Node 3
g
Node 4
ds
Node 8
Node 7
N Turns
w
Node 6
w
Node 5
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
Approach (1/6)
73
Approach (2/6)
74
Approach (3/6)
75
Approach (4/6)
76
Approach (5/6)
77
Approach (6/6)
78
Design Metrics
• Magnetic Material Volume
Node 2
Node 1
Depth into
page = d
w
Node 3
g
Node 4
ds
Node 8
Node 7
N Turns
w
Node 6
w
Node 5
ws
w
79
Design Metrics
• Wire Volume
80
Design Metrics
• Circumscribing Box
Node 2
Node 1
Depth into
page = d
w
Node 3
g
Node 4
ds
Node 8
Node 7
N Turns
w
Node 6
w
Node 5
ws
w
81
Design Metrics
• Material Cost
• Mass
82
Circumscribing Box Volume
83
Material Cost
84
Mass
85
Least Cost Design
•
•
•
•
•
•
•
•
•
•
•
•
N = 260 Turns
d = 8.4857 cm
g = 13.069 mm
w = 1.813 cm
aw = 21.5181 (mm)^2
ds = 8.94 cm
ws = 8.94 cm
Vmm = 0.00066173 m^3
Vcu = 0.0027237 m^3
Vbx = 0.0075582 m^3
Cost = 153.7112 $
Mass = 27.5408 Kg
86
Loss and Mass
87
Design Criticisms
88
Manual Design Approach
Flux Density B/m 2
0.08
Analysis
0.06
0.04
y,m
0.02
0
-0.02
Design
Equations
-0.04
-0.06
Design
Revisions
-0.08
-0.1
-0.05
0
x,m
0.05
0.1
Numerical
Analysis
Final
Design
89
Formal Design Optimization
Evolutionary
Environment
Detailed
Analysis
Fitness Function
90
Another Design Trade-Off
91
Lunar Rover Propulsion Drive
Pareto-Optimal Front
92