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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 1
METU
Magnetic Circuits
Magnetic Circuits
byProf. Dr. Osman SEVAİOĞLU
Electrical and Electronics Engineering Department
Shanghai Pudong Airport
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 2
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Magnetic Circuits
Principles of Magnetism
Permanent Magnet
A permanent magnet is a piece of metal withcharacteristics of attracting certain metals
N
S
Horseshoe Magnets
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 3
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Magnetic Circuits
Principles of Magnetism
Permanent Magnet
NS
Magnetic flux lines form closed paths thatare close together where the field isstrong and farther apart where the field isweak.
Flux lines leave the north-seeking end of amagnet and enter the south-seeking end.
When placed in a magnetic field, acompass indicates north in the directionof the flux lines.
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Magnetic Circuits
Principles of ElectromagnetismMagnetic Field around a wire
A current in a wire creates a magneticfield around the wire as shown in thefollowing figure
Right Hand Rule
Wrap your right hand fingers aroudthe wire while your thumb fingerpoints the direction of current flow
Other fingerswill point thedirection offield lines
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Magnetic Circuits
Principles of ElectromagnetismMagnetic Field around a wire
A current in a wire creates a magneticfield around the wire as shown in thefollowing figure
If this wire is wound around an ironcore, the field lines are superposedas shown in the following figure
Electromagnet
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Magnetic Circuits
Application of Right Hand Rule to ElectromagnetsThe coil is grasped with the fingerspointing the direction of current flow,the thumb finger points the direction ofthe magnetic field in the coil
Wire is grasped with the thumb fingerpointing the direction of current flow,the fingers encircling the wire pointthe direction of the magnetic field
Field lines
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Magnetic Circuits
Electromagnet
Consider again the coil woundaround the iron core as shown
Electromagnet Horseshoe Electromagnet
If the iron core is shaped as an “U” shape , weobtain a “horseshoe” electromagnet
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Magnetic Circuits
Electromagnet
Iron Core may be closed to form a magnetic “Loop”
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Magnetic Circuits
Magnetic Ohm’s Law
Electrical Ohm’s Law Magnetic Ohm’s Law
Voltage(Volt)
Current I (Ampere)
Resistance(R)
+
V = R x I (Volt) (Ohm) (Ampere)
F = R x Ф(Ampere x Turn) ( ) (Weber)
Flux Ф (Weber)
Reluctance( R )
F = N x I
Electro Motive Force (emf) Magneto Motive Force (mmf)
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 10
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Magnetic Circuits
Magneto Motive Force (mmf)
Definition F = N x I (Ampere x Turn) (Turn) (Amper)
Flux Ф (Weber)
Reluctance( R )
F = N x I
Please note that flux is proportional toboth;• Current I,• Number of Turns, N,i.e. flux depends on the product: N x I,
called Magneto Motive Force (mmf)
Flux Ф (Weber)
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 11
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Magnetic Circuits
Electrical Resistance
Electrical Resistance
R = l / A
where, R is the resistance of conductor, is the resistivity coefficient,
= 1 / 56 Ohm-mm2 /m (Copper)
1 / 32 Ohm-mm2
/m (Alumin.)l (m) is the length of the conductor A (mm2 ) is the cross sectional area
of conductor
Resistance of a cable is proportional tothe length and inversely proportional tothe cross sectional area of the cable
A
(Area- m 2 ) (meter)
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 12
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Magnetic Circuits
Magnetic Resistance (Reluctance)
Reluctance
R = (1/ µ) l / A
where, R is the resistance of conductor,
µ is the magnetic permeabilitycoefficient,
µ 0 = 4 10 -7 (Air)
l (m) is the length of the material, A (mm2 ) is the cross sectional area
of the material
The reluctance of a magnetic material isproportional to the mean length andinversely proportional to the crosssectional area, and the permeability ofthe magnetic material
l (meter) (Area- m 2 )
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 13
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Magnetic Circuits
Magnetic Resistance (Reluctance)
Reluctance
The reluctance of a magnetic material isproportional to the mean length andinversely proportional to the product ofthe cross sectional area and thepermeability of the magnetic material
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 14
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Magnetic Circuits
SummaryReluctance
R = (1/ µ)l
/ A
A (Area- m 2 ) l (meter)
Resistance
R =l
/ A
A (Area- m 2 ) l (meter)
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 15
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Magnetic Circuits
Current Density
Current Density
Current density in a cable is the currentflowing through per unit area in a planeperpendicular to the direction of currentflow
J = I / A (Amper / m2 )
A (Area- m 2 ) I (Amper)
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 16
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Magnetic Circuits
Flux Density
Flux Density
Flux density in a magnetic material isthe flux flowing through per unit area ina plane perpendicular to the direction offlux flow
B = Ф / A ( Weber / m2 )
A (Area- m 2 ) Ф (Weber)
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 18
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Magnetic Circuits
Magnetizing ForceDefinition
B = µ x
H
Please note that the µ coefficient in
the following expression is notconstant, but function of H.
0,0000
0,0005
0,0010
0,0015
0,0020
0,0025
0,0030
0,0035
0,0040
0,0045
0,0050
0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000
Mild Steel Sheet
µ(H)
Magnetizing Force H (AT / m)
µ-H Characteristics
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 19
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Magnetic Circuits
Magnetizing ForceDefinition
Magnetizing Force is the mmf perunit length of the magnetic material
NI /l
= H
B = µ x H
or
H = F / l (AT / meter)
Relation between B and H;
A
(Area- m 2 ) l
(meter)
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 20
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Magnetic Circuits
B-H CharacteristicsDefinition
The importance of B-H
characteristics is that, it isindependent of the crosssectional area A and length l, i.e.it is independent of the shape
and volume of the material,In other words, B-Hcharacteristics exhibits themagnetic property of the
material for per unit length andand cross sectional area,
Hence, it is provided by themanufacturer of the magnetic
material
B-H Characteristics
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 400 1200 2000 2800 3600 4400 5200 6000
Magnetizing Force H (At/m)
M a g n e t i c
f l u x d e n s i t y B ( T = T e s l a )
Mild steel sheetCast steelCast iron
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 21
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Magnetic Circuits
B-H Characteristics
B-H Characteristics
Please note that µ is a function of H
µ-H Characteristics
Please note that B-H characteristicssaturates at high values of H
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 400 1200 2000 2800 3600 4400 5200
0,0000
0,0005
0,0010
0,0015
0,0020
0,0025
0,0030
0,0035
0,0040
0,0045
0,0050
0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000
Mild Steel Sheet
µ(H)
Magnetizing Force H (AT / m)6000Magnetizing Force H (At/m)
M a
g n e t i c f l u x d e n s i t y B
( T = T e s l a )
Mild steel sheet
Cast steel
Cast iron
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 22
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Magnetic Circuits
B-H Characteristics of Free Air
B-H Characteristics
Please note that µ 0 is constant
µ0-H Characteristics
B-H characteristics of free air is linearexhibiting no saturation effect
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 400 1200 2000 2800 3600 4400 5200
µ0(H)
6000Magnetizing Force H (At/m)
M a
g n e t i c f l u x d e n s i t y B
( T = T e s l a )
0,0000011
0,0000012
0,0000013
0,0000014
0,0000015
0,0000016
0,0000017
0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000Magnetizing Force H (AT / m)
µ 0 = 4 10 -7 µ 0 = 4 10
-7
The ratio µ/ µ 0
is called “Relative Permeability”
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 23
METU
Magnetic Circuits
Ф- F Characteristics
Ф - F Characteristics
Vertical and horizontal axes in theB-H Characteristics may bemultiplied by A and l, respectivelyyielding the Ф - F Characteristics
B x A = Ф (Weber)
H x l = F (AT)
F (mmf) (AT)
Ф
( F l u x ) ( W e b e r )
Please note that the shape of the B-Hcharacteristics is unchanged, while onlythe figures on the axes are changed
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
400 1200 2000 2800 3600 4400 5200 6000
Mild steel sheetCast steel
Cast iron
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 24
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Magnetic Circuits
Knee Point
Definition
Knee Point on the Ф - F Characteristics is the point belowwhich the characteristics may beassumed to be linear
Ф - F Characteristics
0,00
0,20
0,40
0,60
0,80
1,00
1,20
1,40
1,60
1,80
2,00
0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000
Knee Point
F (mmf) (AT)
( F l u x ) ( W e
b e r )
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Magnetic Circuits
An Alternative Definition of Reluctance
Definition
Reluctance is the inverse of theslope of the chord drawn by joining the origin and a point onthe Ф - F Characteristics
Ф - F Characteristics
0,00
0,20
0,40
0,60
0,80
1,00
1,20
1,40
1,60
1,80
2,00
0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000
Knee Point
F (mmf) (AT)
( F l u x ) ( W e b e r )
R = (1/ µ) l / A
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 26
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Magnetic Circuits
An Alternative Definition of Reluctance
Definition
Please note that reluctance isconstant in the region below theknee point
Ф - F Characteristics
0,00
0,20
0,40
0,60
0,80
1,00
1,20
1,40
1,60
1,80
2,00
0 500 1000 1500 2000 2500 3000 3500
Knee Point
F (mmf) (AT)
( F l u x ) ( W e b e r )
R = (1/ µ) l / A
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 27
METU
Magnetic Circuits
Magnetic Kirchoff’s Laws
Kirchoff’s First Law
Summation of fluxes entering in a junction is equal to that of leaving
1 2
0
V0 = 0
R13
R20
+
3
I 1 I 2
I 3V 1 V 2
Ф 1 = Ф 2 + Ф 3 (Weber)
Electrical Analog
I 1 = I 2 + I 3 (Amper)
R12
R30
+
R10
+ ++
1 2
0
3Ø1 Ø2
Ø3
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 28
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Magnetic Circuits
Magnetic Kirchoff’s Laws
Kirchoff’s Second Law
Summation of magneto motive forcesin a closed magnetic circuit is zero
R0
I
F = NI = R 1 Ф + R 2 Ф + R 0 Ф + R 3 Ф
Electrical Analog
V = R 1 I + R 2 I + R 0 I +R 3 I
R1
R2
R3
R1
R2
R0
R3
V 1
+
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 29
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Magnetic Circuits
R1
R2
R0
R3
Magnetic Kirchoff’s Laws: Application
Kirchoff’s Second Law
Reluctance of the parts of the magneticmaterial
R 1 = l1 / µ A
R 2 = l2 / µ A
R 3 =l
3 / µ A
Reluctance of the air gap
R 0
= l0 / µ
0 A
µ 0 = 4 10 -7
Total reluctance
R T = R 1 + R 2 + R 3 + R 0
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 30
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Magnetic Circuits
Magnetic Kirchoff’s Laws: Application
Kirchoff’s Second Law
Then flux may be calculated as;
Ф = F / R T = NI / ( R 1 + R 2 + R 3 + R 0 )
or Then flux may be calculated as;
F = Ф ( R 1 + R 2 + R 3 + R 0 )
= Ф R 1 + Ф R 2 + Ф R 3 + Ф R 0 = F 1 + F 2 + F 3 + F 0 KVL for magnetic circuits
R1
R2
R0
R3
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 31
METU
Magnetic Circuits
Magnetic Kirchoff’s Laws: Application
Kirchoff’s Second Law
Then flux density may be written as;
B = Ф / A
= ( F / ( R 1 + R 2 + R 3 + R 0 )) / A
= (NI / A)/( l1 /Aµ + l2 /Aµ + l3 /Aµ + l0 /Aµ 0 )
= NI / ( l 1 / µ + l 2 µ + l 3 / µ + l 0 µ 0 )
Ampere’s Law: ∑∫ =⋅ Nid lHMagnetomotive force may then be written as;F = NI = F 1 + F 2 + F 3 + F 0
= H 1 l 1 + H 2 l 2 + H 3 l 3 + H 0 l 0
R1
R2
R0
R3
M ti Ci it
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 32
METU
Magnetic Circuits
Example 1
Question
The magnetic circuit shown on theRHS is operated at saturated fluxvalues, hence linear techniques cannot be applied.
For a flux density of 1.6 Weber / m2
,determine the current I and flux Ф
Flux Ф (Weber)
Reluctance
( R )
F = N x I
Parameters
Cross sectional area = A = 10 cm2
Mean length of flux path =l
m = 30 cmNumber of turns = N = 500
M ti Ci it
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 33
METU
Magnetic Circuits
Example 1
Solution
Flux Ф (Weber)
Reluctance ( R )
F = N x I
Parameters
Cross sectional area = A = 10 cm2
Mean length of flux path = l m = 30 cm
Number of turns = N = 500
The first thing to do, is to use B-H
curve for finding the value of Hcoreesponding to flux density of 1.6Weber / m2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 400 800 1200 1600 2000 2400 2800 3200 3600 4000 4400 4800 5200 5600Magnetizing Force H (At/m)
M a g n e t i c f
l u x d e n s i t y B ( T = T e
s l a )
B = 1.6 Wb/m2→ H = 2400 AT / m
6000
M ti Ci it
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 34
METU
Magnetic Circuits
Example 1
Solution
Flux Ф (Weber)
Reluctance R
F = N x I
Parameters
Cross sectional area = A = 10 cm2
Mean length of flux path = l m = 30 cm
Number of turns = N = 500
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 400 800 1200 1600 2000 2400 2800 3200 3600 4000 4400 4800 5200 5600
Magnetizing Force H (At/m)
M a g n e t i c f l u x d e n s i t y B ( T = T e s l a )
B = 1.6 Wb/m2 → H = 2400 AT / mF = NI = H l m → I = H x l m / N
= 2400 x 0.3 / 500= 1.44 Amper
Ф = B x A = 1.6 x 10 x 10 -4 = 0.0016 Wb
6000
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 35
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Magnetic Circuits
Example 2
Flux density = B = 0.15 Wb / m2
µ = 1000 x µ0 (B-H curve is linear)Question
Determine the flux and mmf required to
produce the flux in the toroid shown on theRHS wound around a certain ferromegneticmaterial called “Ferrite”
Outer diamater = 0.10 mInner diameter = 0.08 mDept = 0.02 m
The ratio µ/ µ 0 is called “RelativePermeability”
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Magnetic Circuits
Example 2
Flux density = B = 0.15 Wb / m2
µ = 1000 x µ0(B-H curve is linear)
Solution
Outer diamater = 0.10 mInner diameter = 0.08 mDept = 0.02 m
Thickness of the core = (0.10 – 0.08) / 2 = 0.01 mDiameter mean = (0.10 + 0.08 ) / 2 = 0.09 m
Cross sectional area = 0.02 x 0.01 = 0.0002 m2
Then, flux = Ф = B x A = 0.15 x 0.0002 = 0.00003 Wb
Cross sectional area A
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 38
METU
Magnetic Circuits
The need for the Load Line Technique
When the material is operated within the
saturated region of the B-Hcharacteristics, solution of the circuitwhen the mmf is given, but the flux isunknown is rather difficult, due to the
fact that µ of the material is a funcitonof flux Ф, but flux is unknown, i.e. theequation;
Ф = F / ( R total + R 0 )= F / ( l total / ( µ( Ф ) x A)) + R 0 )
Load Line Technique
In other words, both sides of the above
equation involve the unknown variable Ф
R1
R2
R0
R3
F = NI
Implicit nonlinar equation
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 39
METU
Magnetic Circuits
Question
B-H curve of the magnetic
circuit shown on the RHS isgiven below. Fluxes in thebranches are given as;
Ø a = 0.003 Weber,
Ø b = 0.003 Weber,Ø c = 0.003 Weber
find the direction andmagnitudes of the currents
0,00
0,20
0,40
0,60
0,80
1,00
1,20
1,40
1,60
1,80
0 500 1000 1500 2000 2500 3000H (AT/m)
B ( W b / m 2 )
Øa Øc
Øb
l BB’ = 0.1 ml BAA’B’ = l BCC’B’ = 0.4 m
ABB’ = 5 cm2
ABAA’B’ = ABCC’B’ = 20 cm2
Example 4
Magnetic Circuits
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 40
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Magnetic Circuits
Solution
A’ B’
V1
RBCC’B’
+ +
C’
I 1 I 2
I 3V 1 V 2
RAA’BB’
A B C
Solution
Writing down magnetic KVL for the loops
of the magnetic circuitF 1 = N 1 I 1 = H BB’ l BB’ + H BAA’B’ l BAA’B’ (1)
F 2 = N 2 I 2 = H BCC’B’ l BCC’B’ + H BB’ l BB’ (2)
Magnetic Circuits
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 41
METU
Magnetic Circuits
Solution
Solution
Now, flux densities
BBB’ = Ø b / ABB’ = 0.0008 / ( 5 x 10 -4 )
= 1.6 Wb/m2
BBAA’B’ = Ø a / ABAA’B’ = 0.0030 / (20 x 10 -4 )
= 1.5 Wb/m2
BBCC’B’ = Ø c / ABCC’B’ = 0.0022 /(20 x 10 -4 )
= 1.1 Wb/m2
Now, find magnetizing forces by using thegiven B-H curve
BBB’ = 1.6 Wb/m2 HBB’ = 3000 AT/m
BBAA’B’ = 1.5 Wb/m2 HBAA’B’ = 2000 AT/m
BBCC’B’ = 1.1 Wb/m2 HBCC’B’ = 400 AT/m
0,00
0,20
0,40
0,60
0,80
1,00
1,20
1,40
1,60
1,80
0 500 1000 1500 2000 2500 3000H (AT/m)
B ( W b / m 2 )
Magnetic Circuits
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 42
METU
Magnetic Circuits
Solution
Solution
Now, substitute the above magnetizing
forces into equatins (1) and (2)
100 I 1 = 3000 x 0.01 + 2000 x 0.4 = 1100
Thus, I 1 = 1100 / 100 = 11.0 Ampers
300 I 2 = 400 x 0.4 – 3000 x 0.1 = -140
Thus I 2 = -140.0 / 300 = - 0.467 Ampers
A’ B’
V1
RBCC’B’
+ +
C’
I 1 I
2 I 3V 1 V 2
RAA’BB’
A B C
Please note that I 1 is in the same direcitionas shown in the figure, but I 2 is in the
opposite direction
Magnetic Circuits
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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİ O Ğ LU, Page 43
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Magnetic Circuits
Description
F = H c l c + H 0 l 0 (1)
where, H c and H 0 are the magnetizing
forces in the material (core) andair gap,l c and l 0 are the lengths of the
core and air gap, respectively
Graphical Solution (Load Line Technique)
Writing Amper’s equation for the
magnetic circuit
Now, noting that;
H 0 = B0 / µ 0 = Bc / µ 0
And substituting this into (1)
l 0
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or writing the above equation in
terms of Hc
Graphical Solution (Load Line Technique)
l 0Derivation of the Load Line Equation
F = NI = H C l c + ( Bc / µ 0 ) l 0 (2)
H C = - l 0 / (µ 0 l c ) Bc + NI / l c (3)
Load Line Equation:
y = - a x + b
The Load Line is then drawn on the B-Hcharacteristics and the intersection point
of the two curves is found
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Graphical Solution
Solve the magnetic circuit shown on the
RHS with the B-H characteristics
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 400 1200 2000 2800 3600 4400 5200
Magnetizing Force H (At/m)
M a g n e t i c f l u x d e n s i t y B ( T = T e s l a )
Load Line
B-H Characterstics
l 0
Graphical Solution (Load Line Technique)
5600
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l 0 = 10-3 mQuestion
Example 3
Solve the magnetic circuit shown on the
RHS with the B-H characteristics givenbelow by using the Load Line Technique
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 400 800 1200 1600 2000 2400 2800 3200 3600 4000 4400 4800 5200 5600Magnetizing Force H (At/m)
M a g n e t
i c f l u x d e n s i t y B ( T =
T e s l a )
6000
Ac= 8 x 10-3 m l c = 0.55 m
F = NI = 3300 AT
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The Load Line Equation
Example 3
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
0 400 800 1200 1600 2000 2400 2800 3200 3600 4000 4400 4800 5200 5600Magnetizing Force H (At/m)
M a g n e t
i c f l u x d e n s i t y B ( T =
T e s l a )
F = NI = 3300 AT 6000
H C = - l 0 / (µ 0 l c ) Bc + NI / l c = - 3327 Bc + 6000
Load Line
B-H Characterstics
Solution: B= 1.41 Wb / m2
l 0 = 10-3 mAc= 8 x 10
-3 m l c = 0.55 m
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Flux Ф t)
Reluctance R
F (t) = N x I(t)
Energy in Magnetic Circuits
Sinusoidally ExcitedMagnetic Circuits
Consider the sinusoidally excitedmagnetic circuit shown on the RHS
I (t) = - I cos wt
F (t) = - N I cos wt
Ф (t) = F (t) / R total
= - N I cos wt /R
total
= - Ф cos wt
where Ф = NI / R total
^
^
^
^
^
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Energy in Magnetic Circuits
Sinusoidally ExcitedMagnetic Circuits
I (t) = - I cos wt
F (t) = - N I cos wt
Ф (t) = F (t) / R total
= - N I cos wt / R total
= - Ф cos wt
where, Ф = NI / R total
^^
^
^
^
Waveforms
-2,5
-2,0
-1,5
-1,0
-0,5
0,0
0,5
1,0
1,5
2,0
2,5
π /2 π 3π /2 2 π
I (t)
Ф (t)
Angle (radians)
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Voltage Induced in the Coil
Sinusoidally ExcitedMagnetic Circuits
By using Lenz’s law
e (t) = N d/dt Ф(t)
= - N d/dtФ
cos wt = N Ф w sin wt
= e sin wt
where e = N Ф w
^
^
^
Waveforms
^
^
-6,00
-4,00
-2,00
0,00
2,00
4,00
6,00
π /2 π 3π /2 2 π
Angle (radians)
e (t)
Ф (t)
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Voltage Induced in the Coil
SinusoidallyExcited Magnetic
Circuits
If we plot Ф(t) and e(t) onthe same scale
-6,00
-4,00
-2,00
0,00
2,00
4,00
6,00
π /2 π 3π /2 2 π
Angle (radians)
e (t)
Ф (t)
Magnetizing Force H (AT / m)
0 1000 2000 3000
0,00
0,40
0,80
1,20
1,60
2,00
0,40
0,80
1,20
1,60
2,00
B ( W e b e r / m 2 )
- 2 , 5
- 2 , 0
- 1 , 5
- 1 , 0
- 0 , 5
0 , 0
0 , 5
1 , 0
1 , 5
2 , 0
2 , 5
π / 2
3 π / 2
2 π
I (t)
A n g l e ( r a d i a n
s )
π
-3000 -2000 -1000
Please note that the shape of this waveform isnot actually of pure sinusoidal shape, but adistorded form of sinusoidal waveform, due to
nonlinearity of the B-H curve
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Energy Stored in Magnetic Circuits
Time duration dt required for achange d Ф in flux may be foundon the Ф – t curve
Electrical EnergyStored in a Magnetic
Circuit
-2,5
-2,0
-1,5
-1,0
-0,5
0,0
0,5
1,0
1,5
2,0
2,5
π /2 π 3π /2 2 π Angle (radians)
d Ф(t)
dt
Ф ( t ) ( W e b e r )
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Energy Stored in Magnetic Circuits
Let us now calculate theinstantaneous electrical energystored in a magnetic circuit within
a time duration dt
dW elect = e(t) I(t) dt
= N d Ф(t)/dt I(t) dt
= (NI) d Ф(t) / dt x dt = F d Ф
Electrical Energy Storedin a Magnetic Circuit
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Fringing Effect
Description
Fringing effect is the deviation ofthe flux trajectory to outside in theair gap
The effects of fringing;(a) increasing the cross sectional
area of the air gap,(b) creating nonlinarity in the flux
density in the air gap
Usual practice for handling thefringing effect is to increase the crosssectional area of the air gap incalculations by a factor, such as 20 %
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