Power Magnetic Devices: A Multi-Objective Design Approach ...sudhoff/ECE61014/Lecture...p kdt T fB dt B αβα− = 2 1 0 2cossin k kh d π πθθααβα−−θ = S.D. Sudhoff, Power

Power Magnetic Devices: A Multi-Objective Design Approach

Chapter 6: Magnetic Core Loss

S.D. Sudhoff, Power Magnetic Devices: A Multi-Objective Design Approach

6.1 Eddy Current Losses

• Eddy current lost is a resistive power loss associated with induced currents

• Let’s consider a rectangular conductor

2S.D. Sudhoff, Power Magnetic Devices: A Multi-Objective Design Approach


• Defining

• We can show

1 min( , )k w d=

2k w d= −

214 3 2 2 3 4

1 2 1 2 1 2 1 22 0

2 114 8 2 2 ln128

Tkl dBP k k k k k k k k dtk T dt

σ += + + − +



• Derivation



• Example 6.1A. Eddy current loss in a toroid

Table 6.1A-1 Aluminum alloy test samples.

Material

σ(MS/m)

rμ ir

(mm) or

(mm)

w (mm)

d (mm)

6061T6 25.3 1.23 140 150 10 12.7 6013 23.3 1.20 140 150 10 27.2 9S.D. Sudhoff, Power Magnetic Devices: A Multi-Objective Design Approach


• Example 6.1A. Normalized power loss

• Thus, with our expression for eddy current loss, we have

2

0

1n TPpdBV dt

T dt

=

4 3 2 2 3 4 11 2 1 2 1 2 1 2

2

214 8 2 2 ln

128n

kk k k k k k k kk

pwd

σ ++ + − + =



• Example 6.1A. Measured power loss. We can show

0

1 Tps p

s

NP v i dt

N T=



• Example 6.1A. Derivation of average power.



• Example 6.1A. Now we measured power, we can find the measured normalized power loss density

• Clearly – we need to know B. There are two approaches.

2

0

1n TPpdBV dt

T dt

=



• Example 6.1A. Approach 1 to finding B (predicted)• We use

0

2r piBr

μ μπ

=



• Example 6.1A. Approach 1 to finding B (measured)• We use

s

s

dB vdt wdN

=



• Example 6.1A. Results:



• Sinusoidal excitation

• Periodic excitation

( )cospk eB B tω=2

2 2

0

1 12

T

e pkdB dt B

T dtω =

, ,1

cos( ) sin( )K

a k e b k ek

B B k t B k tω ω=

= +

( )2

2 2 2 2, ,

10

1 12

T K

e a k b kk

dB dt k B BT dt

ω=

= +



• Thin laminations. Suppose k2 >> k1• Then we can show

• Can be approximated as

4 3 2 2 3 4 11 2 1 2 1 2 1 2

2

214 8 2 2 ln

128n

kk k k k k k k kk

pwd

σ ++ + − + =

23

2 10

112

Tl dBP k k dtT dt

σ =



• To show this



• It follows that

• To show this, we start with

22

0

112

Tw dBp dtT dt

σ =

23

2 10

112

Tl dBP k k dtT dt

σ =



• Continuing …


6.2 Hysteresis Loss and the B-H Loop

• Domain wall motion



• Sample B-H characteristics



• The area of a B-H trajectory represent energy loss.

• The first step to do this is to show

• To do end, consider the toroid0

TB

B

e HdB=



• Proceeding …



• Now let us consider a closed path

1 2

0 1

3 0

2 3

Term 2Term 1

Term 3 Term 4

B B

B B

B B

B B

e HdB HdB

HdB HdB

= +

+ +



• Minor loop loss

01

0 1

Term 2Term 1

BB

B B

e HdB HdB= +



• Now that we have established the energy loss per cycle, the average power loss may be expressed

h BHp fe=


6.3 Empirical Modeling of Core Loss

• The Steinmetz Equation1

pkhBH

b bb

Bfkef Bf

βα − =

pkh h

b b

Bfp k

f B

βα =



• The Modified Steinmetz Equation (MSE)– Motivated by observation that localized eddy current due

to domain movement being tied rate of change• MSE equivalent frequency

mx mnB B BΔ = −2

0

1 T dBB dtB dt

= Δ

2

2 20

2 Teq

dBf dtB dtπ

= Δ 32S.D. Sudhoff, Power Magnetic Devices: A Multi-Objective Design Approach


• MSE loss equation1

2eqh

BHbbb

fkeBff

α β− ΔΒ =

1

2eq

h hbb b

f fp kBf f

α β− ΔΒ =



• Note



• MSE with sinusoidal flux density



• Example 6.3A. MSE with triangular excitation



• Generalized Steinmetz Equation (GSE)– Assumes instantaneous loss is a function of rate of

change and flux density value

– A suggested choice is

,idBp f Bdt

=

1i

b b b

dB Bp k

f B dt B

α β α−

=



• The GSE then becomes

where we choose

0

11 Th

b b b

dB Bp k dt

f B dt BT

α β α−

=

( )2

1

02 cos sin

hkkd

πα α β α θπ θ θ− −

=



• Example 6.3B. Consider MN60LL ferrite with α=1.034, β=2.312, kh = 40.8 W/m3

• Let us compare MSE and GSE models for

• Cases– Case 1: B1=0.5 T, B3=-0.05T– Case 2: B1= 0.45T, B3=0 T– Case 3: B1= 0.409 T, B3=0.0409 T

1 3cos(2 ) cos(6 )B B ft B ftπ π= +



• Example 6.3 flux density waveforms



• Results– MSE model: 87.4, 86.5, 86.2 kW/m3

– GSE model: 86.9, 86.5, 86.8 kW/m3

• Comments on MSE and GSE models



• Combined loss modeling– Eddy current loss

– Hysteresis loss (MSE)

– Combined loss model

2

0

1 Te e

dBp k dtT dt

=

1

2eq

h hbb b

f fp kBf f

α β− ΔΒ =

1 2

0

12

Teq

t h ebb b

f f dBp k k dtBf f T dt

− ΔΒ = + α β



• Example 6.3C. Lets look at losses in M19 steel with kh=50.7 W/m3, α=1.34, β=1.82, and ke=27.5·10-3Am/V. We will assume

• Since the flux density is sinusoidal

cos(2 )pkB B ftπ=

pkh h

b b

Bfp k

f B

βα =

2 2 22e e pkp k B fπ=



• Example 6.3C results


6.4 Time Domain Modeling of Core Loss

• These models can capture effects such as – waveforms with dc offset– aperiodic excitation– waveforms with minor loops

• Two approaches– Jiles-Atherton– Praisach



• Jiles-Atherton model0 ( )B H Mμ= +

irr revM M M= +

M ( )(M ( ) )an irrirr

an irr

H MdM dHk H Mdt dtδ α

− = − − 1 0

0 0

1 0

dHdt

dHdtdHdt

δ

>= =−


• Preisach model is based on behavior of a hysteron

• Total magnetization based on

Q( , ) P( , )U

M U V U V dVdU∞

−∞ −∞

= 49S.D. Sudhoff, Power Magnetic Devices: A Multi-Objective Design Approach


• Saturated magnetization

• Normalized magnetization

• Normalized hysteron density

P( , )U

sM U V dVdU∞

−∞ −∞

=

s

MmM

=

1p( , ) P( , )s

U V U VM

=


6.4 Time Domain Modeling of Core Loss• Lets consider behavior of model• Start in State 1 – all hysterons neg• State 1 to State 2

• State 2 to State 3

1

2 1 2 p( , )H U

s sm m U V dVdU−∞ −∞

= +

1 1

2

3 2 2 p( , )H H

s sH V

m m U V dUdV= −


6.4 Time Domain Modeling of Core Loss• State 3 to State 4

3

2 2

4 3 2 p( , )H U

s sH H

m m U V dVdU= +


6.4 Time Domain Modeling of Core Loss• Reconsider State 3 to State 4

• Suppose field decreases after State 4

2 2

3 2 p( , )H U

sH H

m m U V dVdU= +

2

2 p( , )H

H

dm H V dVdH

=

3 3

4 2 p( , )H H

sH V

m m U V dUdV= − 3

2 p( , )H

H

dm U H dUdH

= 54S.D. Sudhoff, Power Magnetic Devices: A Multi-Objective Design Approach


• Completing the modeldm dm dHdt dH dt

=

sM M m=

0B H Mμ= +



• Relative advantages and disadvantages of approaches


Power Magnetic Devices: A Multi-Objective Design Approach ...sudhoff/ECE61014/Lecture...p kdt T fB dt B αβα− = 2 1 0 2cossin k kh d π πθθααβα−−θ = S.D. Sudhoff, Power

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