Lecture Set 1:Introduction to Magnetic Circuits

Lecture 1

S.D. SudhoffSpring 2021

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Lecture Set 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

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Review of Symbols• Flux Density in Tesla: B (T)• Flux in Webers: (Wb) • Flux Linkage in Volt-seconds: (Vs,Wb-t)• Current in Amperes: i (A)• Field Intensity in Ampere/meters: H (A/m)• Permeability in Henries/meter: 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’s MMF 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 i

a a aF N i

b b bF N i

,enc ss S

i F

{' ', ' ', ' '}S a b c

c c cF N i

MMF Drops

• Going Around the Loop

• MMF Drop

• Thus

• Or

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y

yl

F d H l

l l l l l

d d d d d

H l H l H l H l H l�

dd Dl

d F

H l�

{' ', ' ', ' ', ' '}D

l

d F F F F

H l�

Ampere’s Law

• Ampere’s Law

• Recall

• Thus

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,encl

d i

H l�

dd Dl

d F

H l�,enc ss S

i F

s ds S d D

F F

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 =

Example: MMF Drops

• A quick example on MMF drop:

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Another Example: MMF Drops

• How about Fcb ?

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Lecture 2

Magnetic Flux, Gauss’s Law, andKirchoff’s Flux Law for Magnetic Circuits

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Magnetic Flux

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m

m d S

B S

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

0d

S

B S�

ood

O S

B S {' ', ' ', ' ', } O

0oo

O

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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Lecture 3

Magnetically Conductive Materials

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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

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Ferromagnetic Materials

• Iron• Nickel• Cobalt

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Ferrimagnetic Materials

• Iron-oxide ferrite (Fe3O4)• Nickel-zinc ferrite (Ni1/2Zn12Fe2O4) • Nickel ferrite (NiFe2O4)

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Behavior of Ferro/Ferrimagnetic Materials

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Behavior of Ferro/Ferrimagnetic Materials

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M19                                                       MN80C

Modeling Magnetic Materials

• Free Space

• Magnetic Materials

• Either Way

• Or

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0B H

0 ( )B H M

0B H M

M H

0M H

0 (1 )B H 0 rB H 1r

B H 0 r

Modeling Magnetic Materials

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( )HB H H ( )BB B H

Lecture 4

Ohm’s Law for Magnetic Circuits

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Ohm’s Law

• On Ohm’s Law Property for Electric Circuits

• On Ohm’s Law Property for Magnetic Circuits

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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

B

Form 1:( / )

( ) Form 2 :

( / )

aa

a H a aa

aa

a a a

l FA F l

Rl

A A

( )a a aP F B

( / ) Form 1:( )

( / ) Form 2 :

a H a aa

aa

a a aa

a

A F l Fl

PA A

l

Derivation

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Derivation

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Lecture 5

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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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

NiR

( ) ( ) ( ) 2 ( ) 2eq ic buc luc gR R R R R

Lecture 6

Flux Linkage and Inductance

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Flux Linkage

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x x xN

Inductance

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1 1

,,x x

xx abs

x i

Li

1x

xinc

x i

Li

Example

ie i 03.0)1(05.0 2.0

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• Suppose

• Find the absolute and incremental inductance at 5 A

Computing Inductance

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+

-xxiN

x

xR

Mutual Inductance

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Lecture 7

Nonlinear Flux-Linkage Versus Current

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Approach

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Matlab Code

% 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 spaceclear allclose all

% assign parameterscm = 1e-2; % a centimetermm = 1e-3; % a millimiterw=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 turnsmu0=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 linkagelambda=linspace(0,0.08,1000);

% find flux and B and muphi=lambda/N;B=phi/(w*d);mu=mu0*(mur-1)./(exp(10*(B-Bsat)).^2+1)+mu0;

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Matlab Code

% Assign reluctances;Ric=(ws+w)./(Am*mu);Rbuc=(ws+w)./(Am*mu);Rg=g/(Am*mu0);Rluc=(ds+w/2)./(Am*mu);

% compute effective reluctancesReff=Ric+Rbuc+2*Rluc+2*Rg;

% compute MMF and currentF=Reff.*phi;i=F/N;

% find lambda-i characteristicfigure(1)plot(i,lambda);xlabel('i, A');ylabel('\lambda, Vs');grid on;

% show B-H characteristicH=B./mu;figure(2)plot(H,B)xlabel('H, A/m');ylabel('B, T');grid on;

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Results

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Lecture 8

Accuracy of the MEC

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Hardware Example – Our MEC

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

0.08Log10 Energy Density J/m3

x,m

y,m

Our MEC – With Leakage and Fringing

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Lecture 9

Mapping Magnetic Circuits to Electrical Circuits

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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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dtdirv x

xxx

xxx iL

dtdiLirv x

xxxx

Energy in a Magnetically Linear Inductor

• We can show

• Proof

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221

xxx iLE

Faraday’s Law with Varying Inductance

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Lecture 10

Case Study: UI Core Inductor – Variables and Constraints

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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

Design Requirements

• Current Level: 40 A• Inductance: 5 mH• Packing Factor: 0.7• Slot Shape dw: 1• Maximum Resistance: 0.1

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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 m

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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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Lecture 11

Case Study: UI Core Inductor – Metrics and Results

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Step 3: 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

Step 3: Design Metrics

• Wire Volume

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Step 3: 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

Step 3: Design Metrics

• Material Cost

• Mass

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Lecture 12

Case Study: UI Core Inductor – Approach

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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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Lecture 13

Case Study: UI Core Inductor – Results and Reflections

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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

Pareto-Optimal Front

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