Large Kool Mu Solutions
for PFC Boost Inductors
Power Forum
September 23, 2010
William Glass
Overview
• Large Choke material options
• Advantages of Kool Mµ
• New Kool Mµ options
• Design notes
• Design Example
Large Choke Material Alternatives
• Sendust and Kool Mµ
• Iron Powder
• Laminated Silicon Iron
• Gapped Ferrite
• MPP and High Flux
• Laminated Amorphous
Kool Mµ Advantages
• Moderate Cost
• Superior to MPP/HF, laminated alloys
• Moderate Losses
• Superior to iron powder
• High Saturation
• Superior to gapped ferrite
• Near-zero magnetostriction
• Superior to iron powder
• No Thermal Aging
• Superior to iron powder
• Soft Saturation
• Superior to gapped ferrite & laminated alloys
Kool Mµ 26 Permeability
• Most commonly used material grade for large power inductors.
• Good retained inductance at high Amp-turns (good DC bias, in
other words.)
• Other permeabilities available:
– Large E/U cores, blocks: 40µ, 60µ
– All E/U cores, blocks: 40µ, 60µ, 90µ
– Toroids: 40µ, 60µ, 75µ, 90µ, 125µ
• 14 permeability Kool Mµ now in development; available for
prototyping.
Energy Storage Equation
LI2
Energy Storage:
Henries x Amps2 = Joules
=
Ae le
0.4 π 108
µ H2
Core Size
Saturation Characteristic:
Permeability rolloff with MMF
Design Process
• Calculate LI2
– Peak current (A) & inductance needed at
that current (mH).
• Use chart to select core with LI2 in
range
L * 106
N2 =
• Determine first cut turns
H=
• Determine mmf
0.4 π N I
le
AL
Rolloff Curve
Determine inductance roll-off for Kool Mµ using calculated mmf
Design Iterations
• The method is to repeat calculations
• For the second calculation, increase turns by
increasing the target (0DC) inductance by
the same proportion that the first cut rolled
off, plus a little more since the second
calculation of mmf will be a little greater.
• Re-assess roll-off.
• The end result is the number of turns
necessary to achieve target inductance under
the peak DC bias condition.
Check Window Fill
• Determine area of wire (AW)
– 0.1824 cm2 (example)
• (Area of wire) ∙ (# of Turns (N))
• Use window area (WA) to
determine percentage of fill (KW)
KW =
N AW
WA
Material Loss Calculations
• Only AC ripple causes cores loss. DC bias
generates zero losses in the core material.
• Convert AC current to mmf swing
(Oersteds).
• Convert mmf swing to flux swing.
• Use frequency and ½ of the flux swing to
find the loss density on the core loss curve.
Inductance Swing
Determine inductance roll-off for Kool Mµ using calculated mmf
AC Flux Calculation
AC flux density computation rough cut
• Mmf = 125 ± 25 Oersteds
• µe = 75% ∙ 26
• B=µ∙H
• B = 0.75 ∙ 26 ∙ (125 ± 25)
• B = 2450 ± 500 G
• ½ Delta-B = 500G.
• However: permeability at the low mmf is
actually higher than 0.75; perm at the peak
mmf is actually lower than 0.75.
AC Flux Calculation
AC flux density computation example
• Mmf = 100 to 150 Oersteds
• µe = 80% ∙ 26 to 70% ∙ 26
• B = µ ∙H
• B = (0. 80 ∙ 26 ∙ 100) to (0. 70 ∙ 26 ∙ 150)
• B = 2080 to 2730 G
• ½ Delta-B = 325 G.
• The reduced, swinging inductance
results in lower core loss than otherwise.
Core Losses for 26µ Kool Mµ
Estimate Core Loss (mW/cm3) for AC flux density and frequency
Effect of Winding Geometry on Losses
• Leakage effects result in favorable loss
performance for open window
constructions.
• 26µ nominal loss density at 100kHz,
500G is 380 mW/cm3.
• Measured values are significantly lower
in high leakage assemblies.
• Practical consequence: Prototyping is
essential for determining actual loss
performance.
U-cores
LE Series Structure
26µ Kool Mµ measured losses
Core
Winding
Core loss
density at
100kHz, 500G
Nominal (Toroid)
Even distribution,
high turns
380 mW/cm3
65 mm U-core
One side only N=40
180 mW/cm3
72 mm U-core
One side only N=30
100 mW/cm3
80 mm U-core
One side only N=28
70 mW/cm3
80 mm U-core
Even distribution,
two sides N=28
150 mW/cm3
(same pieces as row above)
26µ Kool Mµ measured losses
Core
Winding
WL at 100kHz,
500G
Nominal (Toroid)
Even distribution,
high turns
380 mW/cm3
130 mm E-core
Center leg N=10
170 mW/cm3
145 mm E-core
Center leg N=14
150 mW/cm3
160 mm E-core
Center leg N=14
150 mW/cm3
160 mm E-core
One side only N=14
150 mW/cm3
160 mm E-core
Divided on two
sides N=14
230 mW/cm3
160 mm E-core
Even distribution
all 4 sides N=14
280 mW/cm3
Flux Model – 80mm U-cores, even
winding distribution
Flux Model – 80mm U-cores,
winding one side only
Current and New Large Geometries
65mm E-core
130mm E-core
72mm E-core
145mm E-core
80mm E-core
160mm E-core
65mm U-core
55mm Block
72mm U-core
47mm Block
80mm U-core
60mm Block
Available in 26,40, and 60 perm
Multiple Design Configurations
•
•
•
•
•
U-U Combination
E-E Combination
Large E-E Combination (130 mm+)
Block/Custom Configuration
Large Toroids (102 mm+)
Main Advantages:
– Much larger structures; much higher power
handling
– Assembly flexibility
– Pre-wound coils; foil winding
60mm & 55mm Block
Configuration
47mm Block Configuration
47mm Stack Block Configuration
Large Toroid Options
Assembly Considerations
• Uncoated Kool Mµ is conductive. Dielectric
isolation from windings must be considered.
• Conductive fixturing (copper, steel, aluminum,
etc.) in very close proximity to core material
surface may result in eddy current heating in
the conductive material.
• Bare Kool Mµ is not dense like ferrite. The
surface texture is rougher. Cosmetics and
potting must be considered differently.
Calculated vs. Actual
Inductance
• In powder cores with low perms the AL can be
calculated, provided the window area (WA) is
relatively small in comparison with the core area
(Ae) .
• Theoretical AL value can be calculated using
– AL = 4πAeµ/ Le
• As the ratio of window area (WA) to core area (Ae)
increases, the estimated and actual values for AL
become significantly different, due to leakage
effects.
Ratio of Measured L to Theroetical
L
3.00
2.80
2.60
2.40
2.20
2.00
1.80
1.60
1.40
1.20
1.00
0.00
5.00
10.00
15.00
20.00
Window to Core Area Ratio
Squares = LE Core Series
Triangles = Blocks
Circles = U Core Series
Diamonds = E Core Series
Effect of Winding Design on Inductance
• Leakage effects result in increases in
measured L for low turns.
• The measurement increase due to core
geometry is compensated in the
published AL, but that is done with high
turns (several hundred) to remove most
of the coil influence.
• Measured values may shift higher than
calculated in low turns inductors.
• Practical consequence: Prototyping is
essential for determining actual
measured inductance.
Design Example Parameters
– 100 Amps
– 100 µH
– 50 kHz
Impractical Designs
Core Size
65mm E-core
72mm E-core
80mm E-core
160mm E-core
55mm Block
47mm Block
60mm Block
% Window Fill
136%
155%
85%
16%
11%
14%
10%
Practical Designs
Core Size
% Window Fill
65mm U-core
53%
72mm U-core
62%
80mm U-core
35%
130mm E-core
23%
145mm E-core
31%
Detail of two parts
Part Size
Ae
Le
WA
Ve
H
N
AL
65mm U
2.70 21.90 19.60 59.10 195 57
89
72mm U
1.84 21.00 18.70 38.60 221 64
74
Detail of two parts
Part Size
Aw
% Fill
No Load
L (µH)
H
Full Load
L (µH)
65mm U
10.40
53%
289
327
110
72mm U
11.67
62%
303
383
107
Approximate LI2 Design Table
Part #
LI2 (mH • A2)
K-6527-E026
K-7228-E026
K-8020-E026
K-6527-U026
K-8020-U026
K-130L-E026
K-145L-E026
K-160L-E026
300 - 900
300 - 800
400 - 1200
1000 - 2700
1500 - 2800
3100 - 6300
2100 - 4300
4600 - 7700
Approximate LI2 for Specific Block
Configurations
Part #
LI2 (mH • A2)
K-5528-B026
K-4741-B026
K-6030-B026
8000 - 11900
7000 - 13800
6900 - 9200
Design Example
PFC Boost
Power: 650W
Input: 85-260 Volts DC input
Output: 370 Volts DC output
Frequency: 65 kHz
General Boost Circuit Schematic
Vd = 1V
I avg
Vin = 85VDC Min
Vin = 260VDC Max
Io
Vo = 370VDC
Design inputs
Power = 650 Watts
Frequency = 65 kHz
1
T = = 15.4 µ sec .
f
I out
650Watts
=
= 1.76 Amps
370Volts
Dmax
Vin min
=1−
= 0.77
Vout
Dmin
Vin max
= 1−
= 0.30
Vout
Inductor Current
Inductor Current
At Low Line Voltage
At High Line Voltage
I avg
 1 
= I out 

1 − D 
I avg
 1 
= 1.76
 = 7.65 Amps
 1 − 0.77 
I avg
 1 
= 1.76
 = 2.51Amps
 1 − 0.30 
Ripple
• Max Current Ripple = 40%
This is arbitrary. The inductance and loss
calculations depend on this value. Actual result
will undershoot because the worst case
inductance and ripple do not occur together.
Design can be iterated to improve ripple or
improve cost/space.
Looking closer at the inductor Current Iin
Ipeak
ΔI
Iavg
Imin
ton
ton + toff = 15.4 µ seconds
toff
ton
Duty Cycle( D) =
15.4 µ sec
Equivalent Circuits
ton
•Iin ramps from Imin to Ipk
•Iin charges L
•Cap holds up Io + Vo
toff
•Iin = Io + Ic
•Ic charges cap
•Iin drops from Ipk to Imin
Iin is greater than Io by the ratio
1
(1 − D )
Worst case ripple is at high line voltage
Ipk =3.51A
Iavg =2.51A
40%
Imin =1.51A
∆I = 2.51(40% )(2 )
∆I = 2 A
I pk = 3.51 A
V across inductor
(Dmin )(t )
L=
∆I
260 − 1
(0.30)(15.4)
L=
2.00
L = 598µH
Worst case Ipk is at low line voltage
Ipk 8.48A
Iavg 7.65A
11%
Imin 6.82A
85 − 1
(0.77 )(15.4)
∆I =
598
∆I = 1.66 A
I pk
1.66
= 7.65 +
= 8.48 A
2
I pk = 8.48 A
L = 598µH
Core Selection Process
(
)
LI = (0.598) 8.48 = 43
2
2
From the catalog core selector chart: Next slide
Kool Mµ Part Number 77439-A7
µ = 60
Ve = 21.3 cm
AL = 135
Aw = 4.27 cm
le = 10.74 cm
MLT = 8.66 cm ( full )
3
2
Determine # of Turns
 1 
Assume 50% roll off, wind for: 
598 µH = 1.2 mH
 0.5 
( ) ⇒ N = 94
1.2mH = N (135) 10
2
−6
.4π (94 )(8.48)
H=
= 93 Oer ⇒ 54% rolloff from cata log curve
10.74
( )
(
)
L = (0.46 ) 94 2 (135) 10 −3 = 548µH
Turns could be added to achieve the 598µH target,
but 548 is not an unreasonable result
N = 94
L at no load = 1190µH
µ eff = 27.4 = 46% of initial perm
Recalculate Inductor Current
High Line Voltage
.4π (94)(3.51)
= 38.6 Oer ⇒ 23% rolloff
Initial I pk = 3.51A ⇒ H =
10.74
( )
(
L = 0.77 94 (135) 10
2
−3
) = 919µH
260 − 1
(0.30)(15.4) = 1.30 A
∆I =
919
Recalculated peak current
3.16
26%
2.51
1.86
1.30
I = 2.51 ±
A
2
I pk = 3.16 A ⇒ 34.8Oer ⇒ 21%rolloff
Recalculate Inductor Current
Low Line Voltage
Initial I pk = 8.48 A ⇒ 93.3 Oer ⇒ 54% rolloff
L = 548µH
85 − 1
(0.77 )(15.4) = 1.82 A
∆I =
548
I pk = 8.56 A ⇒ H = 94.1 Oer ⇒ 55% rolloff
Recalculate Inductor Current (con’t)
8.56
7.65
12%
6.74
Iterate:
85 − 1
(0.77 )(15.4) = 1.86 A
∆I =
536
I pk = 8.58
L = 536 µH
8.58
7.65
6.72
12%
RMS Current
I pk = 8.58 A
7.65 A
2.51A
I min = 1.86 A
I rms = 2.51 +
1
2
(7.65 − 2.51) = 6.14 A
Wire
For 6.1 A current use AWG #17 Wire
R = 16.57 mΩ/m
Wa = 0.0122 cm2
NWa 94(0.0122)
=
= 27%
Fill Factor is
4.27
Aw
A larger wire size
could be used to have a
more nominal window
area fill
For AWG #16 Wire
R = 13.19 mΩ/m
Wa=0.0152 cm2
Fill = 33%
NOTE: AC Ripple at 65 kHz will result in skin effect losses. Multistrand wire equivalent to the #16 gauge would actually be used.
Flux Density Calculations
At Low Line Voltage
I pk = 8.58 ⇒ H pk = 94.4Oer
I min = 6.72 ⇒ H min = 73.9Oer
From AC magnetization curve (next slide)
B pk = 4810 Gauss
1
2
Bmin = 4040 Gauss
∆B = 385 Gauss
NOTE: Curve fit formula is available in catalog
Flux Density Calculations (con’t)
At High Line Voltage
H pk = 34.8Oer
H min = 20.5Oer
B pk = 2170 Gauss
1
2
Bmin = 1340 Gauss
∆B = 415 Gauss
Core Losses
P=B f
2
for 60 µ Kool Mµ
1.46
P = (0.385) (65)
1.46
= 66 mW cm3
High Line
1.46
P = (0.415) (65)
= 76 mW cm 3
Low Line
2
2
Ve = 21.3 cm
3
Power Loss = (
mW
)(cm )
3
cm 3
Core losses are 1400 − 1620 mW
Copper Losses
For #16 Wire
Rcoil = MLT (N )(R length )
Rcoil = (8.66 cm turn )(94T )(0.1319 mΩ cm )
Rcoil = 107 mΩ
Power LossCopper = (I ) (R )
2
Pcu = (6.14) (0.107 ) = 4030mW
2
NOTE: This neglects AC losses. Litz or multistrand wire should be used.
Total Losses and Estimated
Temperature Rise
Total losses 5.4 - 5.7 Watts
Temperature rise with no active air flow
Wound inductor surface area S
OD = 6.3cm, Hgt = 3.8cm
  6.3  2 
2

S = π (6.3cm )(3.8cm ) + 2 π 
=
138
cm
  2  


 mW 
∆T ≈ 

S


0.833
 5700 
=

138


With airflow, ΔT would improve
0.833
= 22  C
Summary
77439 Kool Mµ Toroid
N=94 turns of multistrand equivalent to AWG#16,
giving a fill factor of 33%
L=1190µH at no load
L=536µH at peak (8.58A)
Inductor Max Ripple = 26%
Core losses = 1.4-1.6 W
Copper losses = 4.0 W
ΔT estimate ≈ 22°C
Questions?
William Glass
Sales Manager
Magnetics, A Division of Spang & Co.
wglass@spang.com