Lecture Set 1 Introduction to Magnetic Circuits
Transcript of Lecture Set 1 Introduction to Magnetic Circuits
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Lecture Set 1
Introduction to Magnetic Circuits
S.D. Sudhoff
Spring 2017
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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
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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:
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Ampere’s Law, MMF, and Kirchoff’sMMF Law for Magnetic Circuits
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Consider a Closed Path …
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Enclosed Current and MMF Sources
• Enclosed Current
• MMF Sources
• Conclusion
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,enc a a b b c ci N i N i N iΓ = − + −
a a aF N i= −
b b bF N i=
,enc ss S
i FΓ
Γ∈
= ∑ ' ', ' ', ' 'S a b cΓ =
c c cF N i= −
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MMF Drops
• Going Around the Loop
• MMF Drop
• Thus
• Or7
y
y
l
F d= ⋅∫H l
l l l l l
d d d d dα β δγΓ
⋅ = ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫H l H l H l H l H l
dd Dl
d FΓΓ ∈
⋅ = ∑∫ H l ' ', ' ', ' ', ' 'D α β γ δΓ =
l
d F F F Fα β γ δΤ
⋅ = + + +∫ H l
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Ampere’s Law
• Ampere’s Law
• Recall
• Thus
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,enc
l
d iΓ
Γ⋅ =∫ H l
dd Dl
d FΓΓ ∈
⋅ = ∑∫ H l,enc ss S
i FΓ
Γ∈
= ∑
s ds S d D
F FΓ Γ∈ ∈
=∑ ∑
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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
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Example: MMF Sources
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10 conductors, 3 A =
5 conductors, 2 A =
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Example: MMF Drops
• A quick example on MMF drop:
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Another Example: MMF Drops
• How about Fcb ?
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Magnetic Flux, Gauss’s Law, andKirchoff’s Flux Law for Magnetic Circuits
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Magnetic Flux
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m
m dΦ = ⋅∫S
B S
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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.
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Example: Magnetic Flux
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Example: Magnetic Flux
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• Gauss’s Law
• Thus
• Or
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Kirchoff’s Flux Law
0dΓ
⋅ =∫S
B S
oo
dΓ∈
⋅∑ ∫O S
B S ' ', ' ', ' ', α β γΓ = ⋅ ⋅ ⋅O
0oo Γ∈
Φ =∑O
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Kirchoff’s Flux Law
• Kirchoff’s Flux Law:
The Sum of the Flux Leaving A Node of a Magnetic Circuit is Zero
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Kirchoff’s Flux Law
• A quick example on Kirchoff’s Flux Law
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Magnetically Conductive Materials and Ohm’s Law for Magnetic Circuits
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Ohm’s Law
• On Ohm’s LawProperty for Electric Circuits
• On Ohm’s LawProperty for Magnetic Circuits
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Types of Magnetic Materials
• Interaction of Magnetic Materials with Magnetic FieldsMagnetic Moment of Electron SpinMagnetic Moment of Electron in Shell
• These Effects Lead to Six Material TypesDiamagneticParamagnetic FerromagneticAntiferromagneticFerrimagneticSuperparamagnetic
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Ferromagnetic Materials
• Iron
• Nickel
• Cobalt
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FerrimagneticMaterials
• Iron-oxide ferrite (Fe3O4)
• Nickel-zinc ferrite (Ni1/2Zn12Fe2O4)
• Nickel ferrite (NiFe2O4)
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Behavior of Ferromagnetic and FerrimagneticMaterials
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Behavior of Ferromagnetic and FerrimagneticMaterials
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M19 MN80C
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Modeling Magnetic Materials
• Free Space
• Magnetic Materials
• Either Way
• Or28
0B Hµ=
0( )B H Mµ= +
0B H Mµ= +
M Hχ=
0M Hµ χ=
0(1 )B Hµ χ= + 0 rB Hµ µ= 1rµ χ= +
B Hµ= 0 rµ µ µ=
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Modeling Magnetic Materials
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( )HB H Hµ= ( )BB B Hµ=
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Back to Ohm’s Law
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Relationship Between MMF Drop and Flux
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( )a a aF R ξ= Φ
B
Form 1:( / )
( )
Form 2 :( / )
aa
a H a aa
aa
a a a
lF
A F lR
l
A A
ξµ
ξξ
µ
== = Φ Φ
( )a a aP FξΦ =B
( / ) Form 1:
( )( / )
Form 2 :
a H a aa
aa
a a aa
a
A F lF
lP
A A
l
µ ξξ
µ ξ
== Φ = Φ
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Derivation
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Derivation
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Construction of the Magnetic Equivalent Circuit
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Consider a UI Core Inductor
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Inductor Applications
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UI Core Inductor MEC
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Reluctances
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Reluctances
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Reduced Equivalent Circuit
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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.
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Example
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Example
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Solving the Nonlinear Case
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( )eq
Ni
RΦ =
Φ
( ) ( ) ( ) 2 ( ) 2eq ic buc luc gR R R R RΦ = Φ + Φ + Φ +
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Translation of Magnetic Circuits to Electric Circuits: Flux Linkage and Inductance
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Flux Linkage
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x x xNλ = Φ
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Inductance
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1 1
,
,x x
xx abs
x i
Li λ
λ=
1x
xinc
x i
Li
λ∂=∂
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Example
ie i 03.0)1(05.0 2.0 +−= −λ
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• Suppose
• Find the absolute and incremental inductance at 5 A
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Computing Inductance
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+
-xxiN
xΦ
xR
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Computing Inductance
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Nonlinear Flux Linkage Vs. Current
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Nonlinear Flux Linkage Vs. Current
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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; % 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)
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);
% find flux and B and mu
phi=lambda/N;
B=phi/(w*d);
mu=mu0*(mur-1)./(exp(10*(B-Bsat)).^2+1)+mu0;
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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;
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Nonlinear Flux Linkage Vs. Current
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Hardware Example – Our MEC
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Non-Idealities: Fringing and Leakage Flux
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0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
Log10 Energy Density J/m3
x,m
y,m
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Our MEC – With Leakage and Fringing
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Faraday’s Law
• Faraday’s Law, voltage equation for winding
• If inductance is constant
• Combining above yields
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dt
dirv xxxx
λ+=
xxx iL=λ
dt
diLirv x
xxxx +=
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Energy Stored in a Magnetically Linear Inductor
• We can show
• Proof
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221
xxx iLE =
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Mutual Inductance
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Mutual Inductance
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Case Study: Lets Design a UI Core Inductor
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Architecture
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Node 1
Node 3
Node 5
Node 4
Node 8
Node 7
Node 2
Node 6
Depth into page = d
g
sd
sw ww
w
w
N Turns
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Design Requirements
• Current Level: 40 A
• Inductance: 5 mH
• Packing Factor: 0.7
• Slot Shape ηdw: 1
• Maximum Resistance: 0.1Ω
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Information
• Cost of Ferrite: 230 k$/m3
• Density of Ferrite: 4740 Kg/m3
• Saturation Flux Density: 0.5 T
• Cost of Copper: 556 $/m3
• Density of Copper: 8960 Kg/m3
• Resistivity of Copper: 1.7e-8 Ωm
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Step 1: Design Variables
Node 1
Node 3
Node 5
Node 4
Node 8
Node 7
Node 2
Node 6
Depth into page = d
g
sd
sw ww
w
w
N Turns
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Step 2: Design Constraints
• Constraint 1: Flux Density
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Step 2: Design Constraints
• Constraint 2: Inductance
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Step 2: Design Constraints
• Constraint 3: Slot Shape
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Step 2: Design Constraints
• Constraint 4: Packing Factor
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Step 2: Design Constraints
• Constraint 5: Resistance
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Step 2: Design Constraints
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Approach (1/6)
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Approach (2/6)
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Approach (3/6)
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Approach (4/6)
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Approach (5/6)
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Approach (6/6)
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Design Metrics
• Magnetic Material Volume
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Node 1
Node 3
Node 5
Node 4
Node 8
Node 7
Node 2
Node 6
Depth into page = d
g
sd
sw ww
w
w
N Turns
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Design Metrics
• Wire Volume
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Design Metrics
• Circumscribing Box
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Node 1
Node 3
Node 5
Node 4
Node 8
Node 7
Node 2
Node 6
Depth into page = d
g
sd
sw ww
w
w
N Turns
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Design Metrics
• Material Cost
• Mass
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Circumscribing Box Volume
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Material Cost
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Mass
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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
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Loss and Mass
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Design Criticisms
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Manual Design Approach
Analysis
Design Equations
NumericalAnalysis
DesignRevisions
FinalDesign
-0.1 -0.05 0 0.05 0.1
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
Flux Density B/m2
x,m
y,m
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Formal Design Optimization
Evolutionary Environment
Fitness Function
DetailedAnalysis
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Another Design Trade-Off
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Lunar Rover Propulsion DrivePareto-Optimal Front
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