Ferrite Specifications and ACME Ferrites (2)

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

Transcript of Ferrite Specifications and ACME Ferrites (2)

Page 1: Ferrite Specifications and ACME Ferrites (2)

ACME Electronics Corporation 1

Ferrite Specification&

ACME Ferrites

Technical Aspects

By Ray Lai, FAEJune 2015

With Supports of RD & Marketing Teams

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Table of Content

1. Specifications of Ferrites – Materials & Products2. 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 & A2

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3. Technical Application Example: CMC

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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

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3. Technical Application Example: CMC

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

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3. Technical Application Example: CMC

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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.

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3. Technical Application Example: CMC

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Measuring CM impedance Measuring DM impedance (leakage)

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1

L 1 2 0

1 2

0

3

K

COUPLING=

K 2

1TX2 2 _ 1 4 _ 1 3 _ 3 E 2 7

L 1 = L 2

K

COUPLING=

K 1

1TX2 2 _ 1 4 _ 1 3 _ 3 E 2 7

L 1 = L 1R 17 5 0

2

V 1

TD = 0

TF = 1 uP W = 4 9 uP E R = 1 0 0 u

V 1 = -1

TR = 1 u

V 2 = 1

L 2 2 01 2

R 2

0 . 1

R 3

0 . 1

3. Technical Application Example: CMC

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Q: WhyA:

Low pass filtering

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3. Technical Application Example: CMC

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1

L 1 2 0

1 2

0

K

COUPLING=

K 1

1TX2 2 _ 1 4 _ 1 3 _ 3 E 2 7

L 1 = L 1

L 2 = L 2R 17 5 0

2

V 1

TD = 0

TF = 1 uP W = 4 9 uP E R = 1 0 0 u

V 1 = -1

TR = 1 u

V 2 = 1

L 2 2 01 2

R 2

0 . 1

R 3

0 . 1

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3. Technical Application Example: CMC

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

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3. Technical Application Example: CMC

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Ignoring the winding resistance (and stray capacitance), the impedance Z by the core is

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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

to filter out the noise further.

3. Technical Application Example: CMC

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3. Technical Application Example: CMC

Critical in common mode choke design selection

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3. Technical Application Example: CMCMaterial goal higher µi with improved frequency stability

basically against physical principles

where:fg – gyromagnetic critical frequencyγ ~0.22 ΜΗz m/A is the gyromagnetic ratio for an electron

i.e. the ratio of magnetic moment and torqueBs – saturation flux densityμi,0– initial permeability * J. L. Snoek, Physica 14, 207, 1948

sig Bf 34)1( 0, Snoek Limit

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3. Technical Application Example: CMCCharacteristics of Mn-Zn and Ni-Zn Ferrite in the sense of ui vs. frequency

All governed by Snoek limit.

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3. Technical Application Example: CMC

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 A151100k 1.648E+03 3.343E+03150k 2.568E+03 4.109E+03200k 3.517E+03 4.609E+03500k 9.086E+03 5.938E+031000k 1.486E+04 5.938E+03

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3. Technical Application Example: CMC

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High Permeability Material List

Symbol UnitMeasuring 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 factortan δ/μi 10-6

10KHz< 0.25mT 25°C < 4 < 8 < 8 < 10 < 10 < 10 < 10

100KHz < 15 < 30 < 30 < 60 < 60 < 60 <110Saturation Flux Density

Bms mT 10KHz H=1200A/m

25°C 440 400 440 410 380 380 400100°C 300 200 280 210 180 180 170

RemanenceBrms mT 10KHz H=1200A/

m25°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 -1~120~70°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 110Resistivity 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

AutomotiveRecommended

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3. Technical Application Example: CMC

Commonly seen CMC defects: Mostly epoxy coating & winding issues

The fluxes of cancel each other in CMC configuration, so ideally the voltage across the CMC is zero,

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3. Technical Application Example: CMC

Commonly seen CMC defects: Mostly epoxy coating & winding issues

is huge, as both and are high, it will exceed what the core can sustain and crack will happen.

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3. Technical Application Example: CMC

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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

is huge, as both and are high, it will exceed what the core can sustain and crack will happen.

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3. Technical Application Example: CMC

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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

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4. Technical Application Example: DC-DC choke

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Ci

CoN2

N1

D1

Ci

D2

LR

S1

2

1

3

S2

2

1

3

N2

Vo

>iL

-

+

-

-

++

Vi Vin

12

RloadCD

LSW2

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.

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4. Technical Application Example: DC-DC choke

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4. Technical Application Example: DC-DC choke

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Vin

12

RloadCD

LSW2

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) :

where and

As called voltage variation, it’s an important power supply specification and specifies L and C.

=

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4. Technical Application Example: DC-DC choke

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time

S1 on S1 off

Ts 2Ts

DTs (1-D)Ts

3Ts

time

ON

OFF

0

Vin-VoV

-Vo

time

 

0

A

i1

i2

1

2∆ 𝐼 𝐿∝𝐿

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4. Technical Application Example: DC-DC choke

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As and , so duty ratio D=0.5. The switching frequency (why? the half bridge employs two switches)

Determine the inductance L:

Assuming (this is a design specification that should be given)

=

This explains why switching frequency goes higher and higher. If , then

= A much bigger inductor is needed!

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4. Technical Application Example: DC-DC choke

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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” is the index for this capability and links electrical with mechanical factors.

: peak current : maximal allowed current density on wire : This is not , it is the flux density expected under the operated load condition : the area product : winding area of the core : the core’s effective cross-sectional area

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4. Technical Application Example: DC-DC choke

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: peak current

: winding area of the core

Windings are with insulation and unusable space

is the fraction of the core window area that is filled by copper

Typical values of :0.5 for simple low-voltage inductor0.25 to 0.3 for off-line transformer0.05 to 0.2 for high-voltage transformer (multiple kV)0.65 for low-voltage foil-winding inductor

: fill factor of wires in the winding window

By

to select core

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4. Technical Application Example: DC-DC choke

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By choosing EE22 (chosen because is the easiest to get),

CORESDIMENSIONS (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

With turns number known, now determine the gap of the core.

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4. Technical Application Example: DC-DC choke

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Exact doesn’t matter for power choke actually.

Check copper area needed

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 OK!

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4. Technical Application Example: DC-DC choke

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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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Open Discussions

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