Ferrite Specifications and ACME Ferrites (2)

ACME Electronics Corporation 1
Ferrite Specification
&
ACME Ferrites
Technical Aspects
By Ray Lai, FAE
June 2015
With Supports of
RD & Marketing Teams

ACME Electronics Corporation
Table of Content
1. Specifications of Ferrites – Materials & Products
2. ACME ferrite road map and development trend
3. Technical Application Example: CMC
4. Technical Application Example: DC-DC choke
5. Technical Application Example: SMPS transformer
6. Appendix A: Further on ferrite specifications
7. Appendix B: (a) Fringing effect of gapped core (b)
Manipulating magnetizing curve
8. Appendix C: An analogy and differentiation on R, C, and L
and why magnetic components are so UNIQUE
Q & A
2

3
Common mode choke usually employs closed-loop cores: ring or ET
But for output side, H-I core is getting ground.
High-low side
insulation, by
bobbin or spacer
partition

4
Purpose of CMC – filtering our the “common mode” noise without
distorting/decay the desired main quantity (either signal or power level)

5
Ampere’s Right Hand Rule
Common mode noise has
the flux in the same direction,
the core-winding
combination will act like an
inductor (Choke)
Main circuit input/output in
different direction so the flux
canceled each other, ideally
there will be no “saturation”
possible, but, there is
leakage inductance.

6
Measuring CM impedance Measuring DM
impedance
(leakage)

1
L1 20
1 2
0
3
K
COUPLING=
K2
1
TX22_14_13_3E27
L1 = L2
K
COUPLING=
K1
1
TX22_14_13_3E27
L1 = L1
R1
750
2
V1
TD = 0
TF = 1u
PW = 49u
PER = 100u
V1 = -1
TR = 1u
V2 = 1
L2 20
1 2
R2
0.1
R3
0.1
7
Q: Why
A: 𝑍 𝐿 = 𝑗𝜔𝐿
 Low pass
filtering

8
1
L1 20
1 2
0
K
COUPLING=
K1
1
TX22_14_13_3E27
L1 = L1
L2 = L2
R1
750
2
V1
TD = 0
TF = 1u
PW = 49u
PER = 100u
V1 = -1
TR = 1u
V2 = 1
L2 20
1 2
R2
0.1
R3
0.1

9
Complex permeability u’ and u” plays important roles in composing the
CMC impedance Z

10
Ignoring the winding resistance (and stray capacitance), the impedance Z by
the core is
𝑍 𝐿_𝐶𝑀𝐶 = 𝑗𝜔𝐿 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 = 𝑗𝜔𝑁2 𝑢 𝑜
𝐴 𝑒
𝑙 𝑒
(𝜇′ − 𝑗𝜇")
𝑍 𝐿_𝐶𝑀𝐶 = 𝜔𝑁2
𝑢 𝑜 𝜇"
𝐴 𝑒
𝑙 𝑒
+ 𝑗𝜔𝑁2
𝑢 𝑜
𝐴 𝑒
𝑙 𝑒
𝜇′
= 𝑅 𝐶𝑀𝐶 + 𝑗𝜔𝐿 𝐶𝑀𝐶
𝑍 𝐿_𝐶𝑀𝐶 = 𝜔𝑁2 𝑢 𝑜
𝐴 𝑒
𝑙 𝑒
𝜇′2
+ 𝜇"2

CMC is a transformer configuration yet not applied in the transformer
way.
If a single CMC cannot achieve the EMI conduction suppression
effect, usually two identical capacitors will be connected to create a
Y network (two symmetrical LC low pass filter) for better
suppression effect
The corner frequency for the L2-C combination will be
𝑓𝑐 =
1
2𝜋 𝐿2 𝐶
to filter out the noise further.

Critical in common mode
choke design selection

Material goal
 higher µi with improved frequency stability
 basically against physical principles
where:
fg – gyromagnetic critical frequency
γ ~0.22 ΜΗz m/A is the gyromagnetic ratio for an electron
i.e. the ratio of magnetic moment and torque
Bs – saturation flux density
μi,0– initial permeability * J. L. Snoek, Physica 14, 207, 1948
sig Bf 
3
4
)1( 0,  Snoek Limit

Characteristics of Mn-Zn and Ni-Zn Ferrite in the sense of ui vs. frequency
 All governed by Snoek limit.

For CMC, it’s not always the
higher ui the better
Note: great chance that A151
in mass production cannot
sustain such high ui through all
frequencies
Z (Ω)
Hz A07H A151
100k 1.648E+03 3.343E+03
150k 2.568E+03 4.109E+03
200k 3.517E+03 4.609E+03
500k 9.086E+03 5.938E+03
1000k 1.486E+04 5.938E+03

16
High Permeability Material List
Symbol Unit
Measuring Conditions High Permeability Materials
Freq. Flux den. Temp. A05 A07 A07H A10 A102 A121 A151
Initial Permeability μi ≤ 10KHz 0.25mT 25°C
5000±
25%
7000±
25%
7000 ±
25%
10000±
30%
10000±
30%
12000±
30%
15000±
30%
Realative Loss factor
tan δ/μi 10-6 10KHz
< 0.25mT 25°C
＜ 4 < 8 < 8 ＜10 ＜10 ＜10 ＜10
100KHz ＜ 15 < 30 < 30 ＜ 60 ＜60 ＜60 <110
Saturation Flux Density
Bms mT 10KHz
H=1200A/
m
25°C 440 400 440 410 380 380 400
100°C 300 200 280 210 180 180 170
Remanence
Brms mT 10KHz
H=1200A/
m
25°C 80 150 80 140 95 130 220
100°C 90 110 60 110 75 110 100
Temperature Factor of
Permeability
αF 10-6
/°C 10KHz < 0.25mT
0~20°C 0~2 -1 ~ 1 -1~1 0~1.5 -1～1 0~1.5 -1~1
20~70°C 0~2 -1 ~ 1 -1~1 -0.5～1 -1～1 -0.5~1 -1~1
Hysteresis Material
Constant
ηB 10-6
/mT 10KHz 1.5-3.0mT 25°C ＜ 0.8 ＜ 1.2 < 1.2 < 0.5 ＜ 1 ＜0.5 ＜0.5
Disaccommodation
Factor
DF 10-6 10KHz < 0.25mT 25°C ＜ 3 ＜ 2 < 2 ＜ 2 ＜2 ＜2 ＜2
Curie Temperature Tc °C 140 130 ≥ 145 130 120 110 110
Resistivity ρ Ωm 0.2 0.35 0.35 0.15 0.15 0.12 0.1
Density d g/cm3
4.85 4.9 4.9 4.9 4.9 4.9 5
Automotive
Recommended

Commonly seen CMC defects: Mostly epoxy coating & winding issues
The fluxes of 𝑖𝑙𝑜𝑎𝑑 cancel each
other in CMC configuration, so
ideally the voltage across the
CMC 𝑉𝑐𝑚 is zero,

Commonly seen CMC defects: Mostly epoxy coating & winding issues
𝑉𝑐𝑚 = 𝐿 𝑚
𝑑𝑖 𝐿𝑜𝑎𝑑
𝑑𝑡
is huge, as both 𝐿 𝑚 and
𝑖 𝐿𝑜𝑎𝑑 are high, it will exceed what the core
can sustain and crack will happen.

19
Partially short in one or two windings, but the epoxy coating on the
core is intact, will cause the unbalance in different mode flux (load flux)
cancellation  Core will saturate and the core will be heated up to
break point
𝑉𝑐𝑚 = 𝐿 𝑚
𝑑𝑖 𝐿𝑜𝑎𝑑
𝑑𝑡
is huge, as both 𝐿 𝑚 and
𝑖 𝐿𝑜𝑎𝑑 are high, it will exceed what the core
can sustain and crack will happen.

20
Partially short in two windings, and the epoxy coating on the core
is damaged also, this is equal to the line-ground short  Core will be
rapidly damaged (explode) due to the instant power short of the two
wires

21
Ci
Co
N2
N1
D1
Ci
D2
L
R
S1
2
1
3
S2
2
1
3
N2
Vo
>iL
-
+
-
-
+
+
Vi Vin
12
Rload
CD
LSW
2
1
3
+ +
-
-
Vo
+ -
A half-bridge converter, the secondary equivalent Vin=10V, Vout=5V
with 100W output and the switching frequency is 100kHz. Determine the
inductance and design the inductor
Half-Bridge Converter and its equivalent circuit taking out the transformer
Real system design takes many considerations and need updated knowledge in
the development trend. Even the magnetic parts will be different.

22

23
Vin
12
Rload
CD
LSW
2
1
3
+ +
-
-
Vo
+ -
The inductor is the only concern now
and this is the easiest part.
By buck converter basic rule in CCM
(continuous conduction mode) :
𝑉𝑜 = 𝐷𝑉𝑖𝑛 =
𝑇𝑜𝑛
𝑇𝑠
∆𝑉𝑜 =
𝑇𝑠
8𝐶
(1−𝐷)𝑇𝑠
𝐿
𝑉𝑜 =
(1−𝐷)𝜋2
2
𝑓𝑜
𝑓𝑠
2
𝑉𝑜
where 𝑓𝑠 =
1
𝑇𝑠
and 𝑓𝑜 =
1
2𝜋 𝐿𝐶
As ∆𝑉𝑜/𝑉𝑜 called voltage variation, it’s an important power supply
specification and specifies L and C.
∆𝐼𝐿=
𝑉𝑜
𝐿
(1 − 𝐷)𝑇𝑠 =  𝐿=
𝑉𝑜
∆𝐼 𝐿
(1 − 𝐷)𝑇𝑠

24
time
S1 on S1 off
Ts 2Ts
DTs (1-D)Ts
3Ts
time
ON
OFF
0
Vin-Vo
V
-Vo
time
0
A
i1
i2
1
2
∆𝐼𝐿∝ 𝐿

25
As 𝑉𝑖𝑛 = 10𝑉 and 𝑉𝑜𝑢𝑡 = 5𝑉, so duty ratio D=0.5. The switching frequency
𝑇𝑠 =
1
2𝑓𝑠
=
1
2
1
100𝑘
= 5𝜇𝑠 (why? the half bridge employs two switches)
Determine the inductance L: 𝐼 𝑎𝑣𝑒 =
𝑃 𝑜
𝑉𝑜
=
100
5
= 20𝐴
Assuming ∆𝐼𝐿≅ 25% 𝐼 𝑎𝑣𝑒 = 5𝐴 (this is a design specification that should be
given)
𝐿=
𝑉𝑜
∆𝐼 𝐿
1 − 𝐷 𝑇𝑠 =
5
5
1 − 0.5 ∙ 5𝜇 = 2.5μH
This explains why switching frequency goes higher and higher. If 𝑓𝑠= 10kHz,
then 𝑇𝑠 =
1
2𝑓𝑠
=
1
2
1
10𝑘
= 50𝜇𝑠
𝐿=
𝑉𝑜
∆𝐼 𝐿
1 − 𝐷 𝑇𝑠 =
5
5
1 − 0.5 ∙ 50𝜇 = 𝟐𝟓𝛍𝐇  A much bigger inductor is
needed!

26
With the inductance determined, it’s time to realize an inductor by the
inductance, make it from an abstract number to a real device.
As the core must be able to handle the energy it stores, it’s “area product”
𝐴 𝑃 is the index for this capability and links electrical with mechanical factors.
𝐿 ∙ 𝐼 𝑚𝑎𝑥 ∙ 𝐼 𝑎𝑣𝑒 = 𝐾𝑐𝑢 ∙ 𝐽 𝑎𝑣𝑒 ∙ 𝐵 𝑚𝑎𝑥 ∙ 𝐴 𝑃 = 𝐾𝑐𝑢 ∙ 𝐽 𝑎𝑣𝑒 ∙ 𝐵 𝑚𝑎𝑥 ∙ 𝐴 𝑤 ∙ 𝐴 𝑒
𝐼 𝑚𝑎𝑥 = 𝐼 𝑎𝑣𝑒 +
∆𝐼 𝐿
2
: peak current
𝐽 𝑎𝑣𝑒 = 𝐼 𝑚𝑎𝑥/𝐴 𝑐𝑢
: maximal allowed current density on wire
𝐵 𝑚𝑎𝑥: This is not 𝐵𝑠𝑎𝑡, it is the flux density expected under the operated load
condition 𝐻 𝑚𝑎𝑥 ∝ 𝑖 𝐿𝑜𝑎𝑑
𝐴 𝑃 = 𝐴 𝑤 ∙ 𝐴 𝑒: the area product
𝐴 𝑤: winding area of the core
𝐴 𝑒: the core’s effective cross-sectional area

27
𝐼 𝑚𝑎𝑥 = 𝐼 𝑎𝑣𝑒 +
∆𝐼 𝐿
2
: peak current
𝐴 𝑤:
winding
area of
the core
Windings
are with
insulation
and
unusable
space
𝐾𝑐𝑢 is the fraction of the core
window area that is filled by copper
Typical values of 𝐾𝑐𝑢 :
0.5 for simple low-voltage inductor
0.25 to 0.3 for off-line transformer
0.05 to 0.2 for high-voltage transformer (multiple kV)
0.65 for low-voltage foil-winding inductor
𝐾𝑐𝑢: fill factor of wires in the winding
window 𝐴 𝑤
By 𝐿 ∙ 𝐼 𝑚𝑎𝑥 ∙ 𝐼 𝑎𝑣𝑒 = 𝐾𝑐𝑢 ∙ 𝐽 𝑎𝑣𝑒 ∙ 𝐵 𝑚𝑎𝑥 ∙ 𝐴 𝑤 ∙ 𝐴 𝑒
𝐴 𝑤 ∙ 𝐴 𝑒 =
𝐿∙𝐼 𝑚𝑎𝑥∙𝐼 𝑎𝑣𝑒
𝐾 𝑐𝑢∙𝐽 𝑎𝑣𝑒∙𝐵 𝑚𝑎𝑥
to select core

28
𝐴 𝑤 ∙ 𝐴 𝑒 =
𝐿 ∙ 𝐼 𝑚𝑎𝑥 ∙ 𝐼 𝑎𝑣𝑒
𝐾𝑐𝑢 ∙ 𝐽 𝑎𝑣𝑒 ∙ 𝐵 𝑚𝑎𝑥
=
2.5 ∙ 10−6
∙ 22.5 ∙ 20
0.5 ∙ 300 ∙ 104 ∙ 0.25
= 3.00 ∙ 10−9
𝑚4
= 3000𝑚𝑚4
By choosing EE22 (chosen because 𝐴 𝑤 is the easiest to get),
𝐴 𝑤 ∙ 𝐴 𝑒 = 36.26 × 𝐸 − 𝐷 × 𝐹 = 36.26 × 15.5 − 5.7 × 11.2 = 3980𝑚𝑚4
CORES
DIMENSIONS (mm) EFFECTIVE PARAMETERS
A B C D E F C1(mm-1) Le(mm) Ae(mm2) Ve(mm3) Wt(g/set)
EEL22
22.25 ±
0.30
15.26 ±
0.30
5.70 ± 0.30 5.70 ± 0.30 15.50min 11.20 ± 0.30 1.77 65.00 37.00 2405.00 11.74
From Faraday’s law of electro-magnetic induction
𝐿 ∙ 𝐼 𝑚𝑎𝑥 = 𝑁 ∙ 𝐵 𝑚𝑎𝑥 ∙ 𝐴 𝑒 
𝑁 =
𝐿 ∙ 𝐼 𝑚𝑎𝑥
𝐵 𝑚𝑎𝑥 ∙ 𝐴 𝑒
=
2.5 ∙ 10−6
∙ 22.5
0.25 ∙ 37 ∙ 10−6
= 6.08 ≅ 6
With turns number known, now determine the gap of the core.

29
L =
𝑁2 𝜇 𝑟 𝜇 𝑜 𝐴 𝑒
𝑙 𝑒+𝜇 𝑟 𝑙 𝑔
𝐿 ≅
𝑁2 𝜇 𝑜 𝐴 𝑒
𝑙 𝑔

𝑙 𝑔 ≅
𝐿
=
62 ∙ 4𝜋 ∙ 10−7 ∙ 37 ∙ 10−6
2.5 ∙ 10−6
= 0.11𝑚𝑚
Exact 𝑙 𝑔 =
𝐿
−
𝑙 𝑒
𝜇 𝑟
doesn’t matter for power choke actually.
Check copper area needed 𝐴 𝑐𝑢 =
𝐼 𝑎𝑣𝑒
𝐽 𝑎𝑣𝑒
=
20
300
= 6.67 ∙ 10−2 𝑐𝑚2 = 13157𝑐𝑚𝑖𝑙
AWG Dia-mm Max OD (mm) Area (mm^2) R-ohm/cm R-ohm/mm R-ohm/mm
9 2.9063 2.98 6.634E+00 2.599E-05 2.599E-05 2.599E-06 3.49086E-06
AWG#9 is selected.
Total copper area 𝐴 𝑐𝑢_𝑡𝑜𝑡𝑎𝑙 =
𝑁∙𝐴 𝑐𝑢
𝐾 𝑐𝑢
=
6∙6.634
0.5
= 79.61𝑚𝑚2
< 109.76𝑚𝑚2
 OK!

30
Up to now, the core type/size, winding number and gap are determined.
The remaining job is to determine material by the switching & efficiency
requirement Pv and make the samples per the production request and safety
standard.
Note that the design sequence introduced above is the most simplified
version. A lot of details are ignored for simplification purpose.

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