2. Table of Content
B.CMC
B1. CMC Principle
B2. CMC Design
B3. CMC Failure Example
C.Power Choke and Transformer
C1. Power Choke
C2. Coupled Inductor (Flyback Transformer)
C3. SMPS Transformer
2
3. B. Common mode choke CMC)
3
Magnetic components play key role in power conversion system
4. 4
B. Common mode choke (CMC)
Magnetic components which ferrites can make
5. 5
Input side CMC generally adopts closed-loop core (toroid) or
ET core. On the high frequency output side, HI configuration is also
popular (usually mirror lapping required, easy for auto winding)
Hi-Lo side isolation
is required by the
spacers of bobbin
B. Common mode choke (CMC)
6. 6
B. Common mode choke (CMC)
SMD Pulse Transformer SMD Common Mode Choke
Conventional Newer
磁環
7. 7
B1. CMC Principle
CMC is used to filter out the “common mode” noise with the
minimal effect on the main electrical quantity (no matter signal or power)
(EN55022 requires EMI conduction reduction in 150kHz~500kHz)
Two independent choke
(Two cores)
CMC
(One core)
8. 8
B1. CMC Principle
Ampere’s right-
hand rule
The main (load) current
flows opposite direction in
the two windings, so its flux
canceled each (differential
mode), only leakage can
cause the core to saturate.
CM noise generates flux in
the same direction, so it
sees high impedance
𝒁 = 𝟐𝝅𝒇 ∙ 𝑳
𝑓: noise frequency，
L: CM inductance
12. 12
B2. CMC Design
The impedance of CMC (resistance and reluctance) are from
the complex permeability of the material (u‘ and u“), and
the frequency of u’ is the key to CMC design
13. 13
Ignoring the wiring resistance, (leakage) inductance, and parasitic
capacitance, the impedance of CMC 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
Using Q as the quality
index of CMC or power
choke must consider its
load situation ( and
frequency, voltage)
Q in power and signal
level are not equivalent
to each other！
Core loss factor
tan 𝛿 𝑚 =
𝑅
𝜔𝐿
=
𝜇"
𝜇′ ≈
1
𝑄
B2. CMC Design
14. 14
Critical in common mode
choke design selection
𝜇𝑖 × 𝑓𝑟𝑒𝑞𝑢𝑛𝑒𝑐𝑦 ≈ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
B2. CMC Design
15. 15
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
A2. Specifications – Power/High Perm Materials
For higher perm material to have wider frequency response is against
the law of physics.
16. 16 Characteristics of Mn-Zn and Ni-Zn Ferrite in the sense of ui vs.
frequency
All governed by Snoek limit.
A2. Specifications – Power/High Perm Materials
Limit is here
roughly by the
advances of
ferrite research
17. 17 CMC is in transformer configuration (coupled
inductor) yet is not used in transformer way
Usually by adding a pair of small capacitor (Y capacitors) on the AC
input side to form an L-C low pass filter can suppress the noise better
the L2-C configuration will have -3dB corner frequency 𝑓𝑐
𝑓𝑐 =
1
2𝜋 𝐿3 𝐶
low pass corner frequency.
The parasitic capacitor affects the performance (by changing
|Zcmc| in a more complex way
𝑓𝑐 ≪ 𝑓𝑝𝑎𝑟𝑎 =
1
2𝜋 L3 𝐶 𝑝𝑎𝑟𝑎
B2. CMC Design
18. 18 Case 1: A10EP7L failed in 150k~200kHz EMI conduction
Z_peak @ 300kHz?!
Turns’ number: 80
B2. CMC Design
19. 19 Case 1: A10EP7L failed in 150k~200kHz EMI conduction
By material u’ and u”, the frequency of Z_peak > 1MHz
B2. CMC Design
20. 20
The parasitic
capacitance cannot be
ignored in this case
𝑍 𝑐𝑜𝑟𝑒 = (𝑁2 𝜔𝜇0
𝐴 𝑒
𝑙 𝑒
) 𝑢" + 𝑗𝑢′
𝑍 𝑐𝑎𝑝 = 1/𝑗𝜔𝐶 𝑝
𝑍 𝑐𝑜𝑟𝑒 =
1
𝑍 𝑐𝑜𝑟𝑒
+
1
𝑍 𝑐𝑎𝑝
−1
Case 1: A10EP7L failed in 150k~200kHz EMI conduction
B2. CMC Design
21. 21 Case 1: A10EP7L failed in 150k~200kHz EMI conduction
Just by 4pF, the amplitude and peak of the impedance of CMC totally changed
B2. CMC Design
22. 22 Case 1: A10EP7L failed in 150k~200kHz EMI conduction
In the end, it is A05-90T that works!
B2. CMC Design
23. 23 Case 1: A10EP7L failed in 150k~200kHz EMI conduction
In the end, it is A05 that works!
Measurement matched the
calculation from the material
table
B2. CMC Design
24. 24 Case 2：Geometric Fantasy of CMC
Due to the size and material limitation, someone was thinking if bigger Ae (or
higher Ae/le) available by the following topology change.
To replace the simple toroid
B2. CMC Design
25. 25
As long as the windings are still on two sides of the core, any of the above ways
cannot increase Ae or yield more flux passing surface. Only increase the leakage
of the CMC which is unwanted.
B2. CMC Design
Case 2：Geometric Fantasy of CMC
26. The same for this one
B2. CMC Design
Case 2：Geometric Fantasy of CMC
27. 27
If winding are on the center part, than the right (EE) and left cores are
equivalent.
Q: Why did not see CMC doing this way?
A: next page
B2. CMC Design
Case 2：Geometric Fantasy of CMC
28. 28
28
High-low side insulation is the basic safety regulation
requirement.
B2. CMC Design
Case 2：Geometric Fantasy of CMC
30. 30
u
Z
This point is by LC parallel
resonance, where L is
pure winding, C is layer
parasitic. Baiscally it
becomes space EM issue,
nothing to do with ferrite.
Solution: (1) increase
layer space (2) change
wire diameter (3) change
turns number or core size
B2. CMC Design
31. 31
Winding is a kind of art, the unique
technic for winding house to survive
B2. CMC Design
Not really there
34. 34
B3. CMC Failure Example
Failed replacement attempt: - A12URT19C
Trying to use the custom made A12URT19C to replace A10T20x11x10C, but the
customer found when the load current > 3A, CMC failed like not there.
1. To replace, check if Ae/le is better first, as least should be equivalent.
For T20x11x10C, Ae/le=0.951 only 21 turns needed for A10 to have 5mH low
frequency inductance
Outer D Inner D Height
d1 d2 h
mm 20 11 10 Nominal
Nominal
C1 1.051E+00 mm^-1 uo 1.25664E-06
C2 2.406E-02 mm^-3 ui 10000
Le 45.911 mm AL 1.19567E-05
Ae 43.683 mm^2 N 21
Ve 2,005.527 mm^3 L 5.273E-03
36. 36
2. Check if new material yields a better frequency response
B3. CMC Failure Example
37. 37
2. . Check if new material has better frequency response
EN-55022 EMI Conduction frequency range is 150kHz~500kHz，
A12URT19C fails!
B3. CMC Failure Example
38. 38
3. For current > 3A, CMC lost its function, the conduction noise like the case of no CMC
Only leakage will cause CMC to saturate then fail
B3. CMC Failure Example
39. 𝑉
𝜔
= 𝐿 ∙ 𝐼 = 𝑁𝐵𝑐 𝐴 𝑒 𝐵𝑐 =
𝐿∙𝐼
𝑁𝐴 𝑒
R12k (27Ts) 𝐵𝑐 =
𝐿∙𝐼
𝑁𝐴 𝑒
=
38∙10−6∙3 2
27∙20∙10−6 = 299𝑚𝑇
A07(A121 36Ts) 𝐵𝑐 =
𝐿∙𝐼
𝑁𝐴 𝑒
=
65∙10−6∙3 2
36∙20∙10−6 = 383𝑚𝑇
A10 (31Ts) 𝐵𝑐 =
𝐿∙𝐼
𝑁𝐴 𝑒
=
48∙10−6∙3 2
31∙20∙10−6 = 328𝑚𝑇
34
B3. CMC Failure Example
3. For current > 3A, CMC lost its function, the conduction noise like the case of no CMC
Only leakage will cause CMC to saturate then fail
40. 35
B3. CMC Failure Example
3. For current > 3A, CMC lost its function, the conduction noise like the case of no CMC
Only leakage will cause CMC to saturate then fail
41. 4. ui and bandwidth of A12 is inferior than R12K (in this case), by the same core A12 of
36 turns equals to 27 turns of R12. Surely A12 has more leakage.
36
B3. CMC Failure Example
42. 4. By comparing the datasheet, A12 is as good as R12K (though not fully sintered),
but sometimes it is just not the case in real product.
37
B3. CMC Failure Example
43. 43
Big part of MnZn CMC complaints, beside Z and bandwidth, are
related to the coating damage caused by winding process
(NiZn has no this issue)
Flux linkage from load current
𝑖𝑙𝑜𝑎𝑑 cancels each other, so
the differential voltage across
CMC 𝑉𝑑𝑚 =0
B3. CMC Failure Example
44. 44
When short happens, due to high
𝐿 𝑚 𝑎𝑛𝑑 𝑖 𝐿𝑜𝑎𝑑, 𝑖𝑛𝑠𝑡𝑎𝑛𝑡 𝑉𝑑𝑚 = 𝐿 𝑚
𝑑𝑖 𝐿𝑜𝑎𝑑
𝑑𝑡
is high thus it will cause sparks or even
crack the core when the coating
insulation cannot sustain.
B3. CMC Failure Example
45. 45
When windings are partially shorted, if the coating layer
sustains, the imbalance on the MMF will heat up the core till it
breaks down.
B3. CMC Failure Example
46. 46
B3. CMC Failure Example
When windings are partially shorted and the coating layer
broken, the instantaneous high current may cause the core to
explode.
47. 47
EN55022 specifies the EMI/EMC regulation, in HF
band, NiZn is used and it’s the winding skill that matters most,
beside the ferrite property.
B. Common mode choke (CMC)
48. 48
C. Power Choke and Transformer
Frequent asked questions:
1. When measuring core loss of a gapped part, it is much higher than
calculating it using the core curves in the catalog. WHY?
(explained)
2. How to derive the core height from a fixed core footprint for a given
power level in power transformer? Or for a fixed core geometry, what is
the most fit height for maximal power density?
This is a difficult to answer question and might be no answer. The only
certain thing is that for a fixed core geometry, there is a maximal power
density it can go for two-winding model (and it is SMPS topology
dependent), there is no universal formula for any winding number and any
switching topologies.
Switching transformer design is still a case-by-case issue to make
the engineer’s career interesting.
49. 49
2. How to derive the core height from a fixed core footprint for a
given power level in power transformer? Or for a fixed core
footprint, what is the most fit height for maximal power density?
Even better winding arrangement can increase the power handling
capability of a transformer in the sense of temperature rise.
C. Power Choke and Transformer
51. 51
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
+ -
For the below half-bridge converter, its equivalent is a buck
converter, with Vin=10V, Vout=5V, Pout=100W, and switching frequency=
100kHz. Determine the inductance and design the inductor.
Half-Bridge Converter is a Buck Topology
Note: In practical design, there are many factors to consider. There is some degree
of freedom in choosing the core type, but not with no limitation at all.
C1. DC-DC Power Choke
53. 53
Vin
12
Rload
CD
LSW
2
1
3
+ +
-
-
Vo
+ -
Assuming CCM mode,
Then the V-I relationships are
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
∆𝑉𝑜/𝑉𝑜 is called voltage regulation and is determined by L
and C.
∆𝐼𝐿=
𝑉𝑜
𝐿
(1 − 𝐷)𝑇𝑠 = 𝐿=
𝑉𝑜
∆𝐼 𝐿
(1 − 𝐷)𝑇𝑠
C1. DC-DC power choke
54. 54
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
By volt – second balance
C1. DC-DC power choke
55. 55
𝑉𝑖𝑛 = 10𝑉 𝑉𝑜𝑢𝑡 = 5𝑉, so the duty ratio D=0.5. The switching period
𝑇𝑠=
1
2𝑓𝑠
=
1
2
1
100𝑘
= 5𝜇𝑠 (two switches in half-bridge configuration)
Determine the Inductance L: 𝐼 𝑎𝑣𝑒 =
𝑃 𝑜
𝑉𝑜
=
100
5
= 20𝐴
If ∆𝐼𝐿≅ 25% 𝐼 𝑎𝑣𝑒 = 5𝐴 (This is a design spec that must be given)
𝐿=
𝑉𝑜
∆𝐼 𝐿
1 − 𝐷 𝑇𝑠 =
5
5
1 − 0.5 ∙ 5𝜇 = 2.5μH
Why the switching frequency gets 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 Size and windings of the
inductor will substantially increase
C1. DC-DC power choke
56. 56 How to realize the inductance?
Tackle it from the “sustained power/energy” concept
“Area product” 𝐴 𝑃 this parameter links the electro-mechanical nature of a
core.
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 current density on the wire, safety related
𝐵 𝑚𝑎𝑥: NOT 𝐵𝑠𝑎𝑡, the maximal allowable flux density under this opwerational
condition, it directly relates to the volt-second balance.
𝐴 𝑃 = 𝐴 𝑤 ∙ 𝐴 𝑒
𝐴 𝑤: winding space means the current sustainability
𝐴 𝑒: core space means the voltage sustainability
C1. DC-DC power choke
60. 60 Core type and winding are determined, now need to select the
core material to work normally under the frequency. Bmax should be as high
as possible (1st), Pv minimal (2nd), efficiency high (by the trade-off with $$)
The material can affect all the above design choices. In DC-DC choke
design, the inductance is set by 𝜇 𝑑𝑖𝑓𝑓 or 𝜇 𝑟𝑒𝑣, i.e., with DC content
HH
rev
AC
H
B
µ
µ
0
1
∆B (B 𝑚𝑎𝑥) ∝ B 𝑠𝑎𝑡
The upper limit of 𝐻 𝑑𝑐 also
is material related.
Note：most of Hdc will drop
across the gap, but the H
across the core still follows
the original BH curve
C1. DC-DC power choke
62. 62
ITEM FINISHED PRODUCT
+ -
A 10.40 0.15 0.20
B 5.30 max
C 8.85 0.10 0.15
D 4.40 0.10 0.10
E 0.80 0.10 0.10
F 2.95 0.10 0.05
H 8.05 0.15 0.15
K 8.35 0.15 0.15
T 2.15 0.10 0.05
A1 10.40 0.15 0.20
C1 8.85 0.10 0.15
I 2.00 0.05 0.05
Real complaint:
C1. DC-DC power choke –failure example
64. 64 “Soft Ferrite” is not soft at all.
Skipping the tedious calculation, it has proven that by only adding
extra 4um gap, the Bmax on the core dropped from 407mT (H=76A/m)
to 389mT(H=69A/m), this tiny 20mT means the difference between
pass and fail！
Note that 56Adc generates H=3800A/m, but most of Hdc is
dropped across the gap, only 76 or 69A/m is on the core.
C1. DC-DC power choke –failure example
65. 65
B-H curve of no
gap
B-H curve of with
gap
By 𝑁 ∙ 𝑖 = 𝐻𝑐 𝑙 𝑐 + 𝐻𝑔 𝑙 𝑔, the H drop on the core is still the same as the material B-H,
excess H is shared by the air gap.
C1. DC-DC power choke –failure example
66. 66
Illustration of the trade
off between
inductance and
current handling
capability of ferrite due
to the limit of Hmax it
allows.
C1. DC-DC power choke –failure example
“Soft Ferrite” is not soft at all.
67. 67 Uneven magnetic path in the loop will not work.
A_center 39.04
A_1 side 26.83
A_plate 19.93
This kind of geometry is no good for dc-dc
choke，Ae_min section will saturate first,
waste the rest core area and space.
A(mm): 15±0.4
B(mm): 9.8
C(mm): 7.05±0.2
D(mm): 11.7±0.3
E(mm): 1.9±0.1
F(mm): 3.7±0.1
𝐴 𝑠𝑖𝑑𝑒
𝐴 𝑐𝑒𝑛𝑡𝑒𝑟
𝐴 𝑝𝑙𝑎𝑡𝑒
C1. DC-DC power choke –failure example
68. 68
Flyback topology and 與 waveforms of
DCM
The “transformer” (coupled inductor) stores
and release energy through its magnetizing
inductance Lm (by Ip & Is), and decide the
converter behaviors.
Why this is “inductor” instead of
transformer??
One current in one winding at any instant
C2. Coupled Inductor (Flyback Transformer)
69. 69
Popular Flyback Converter Design (3-winding)
Low cost adapter
CMC
DC
choke
50/60Hz
AC
Vin
C2. Coupled Inductor (Flyback Transformer)
70. 70
The design formulae are irrelevant to he magnetic component knowhow,
thus are skipped here.
Comparing the coupled inductor and the DC-DC choke, their similarities
and differences are
Similarity:
1. Bsat(Bmax) as high as possible
2. Gap is needed for different reason. Coupled inductor uses gap for energy
management, DC-DC choke needs gap for consuming Hdc
Difference:
1. The permeability of coupled inductor is amplitude permeability 𝜇 𝑎, DC-DC choke
one is differential permeability 𝜇 𝑑𝑖𝑓𝑓
2. As ∆B ≪ B 𝑠𝑎𝑡，usually Pv in general DC-DC choke is not an issue (Except the
way of Vicor P61 operation), coupled inductor is operated under the first quadrant
of V-I (B-H) chart , but its ∆B ≈ B 𝑠𝑎𝑡 to maximally drive the core, so Pvis an
important factor like the case in transformer.
C2. Coupled Inductor (Flyback Transformer)
71. 71
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 ∆𝐼𝐿∝ 𝐿
Ip Is’
DC-DC choke
Coupled inductor
time
A
∆𝐼𝐿∝ 𝐿
C2. Coupled Inductor (Flyback Transformer)
All follow V-T
balance principle
73. 73
Real Design Example：
a.. By below EP0714 core drawing, get Ae/Ve/Le (P4/P47 mat’l) and compare with EP7
b. What is the power rating(W) ??
c.. If power rating < 5W, feel free to redefine the core just keeping the core’s L/W/H
dimensions
Frequency 100kHz, inductance 840uH (not given in the first place, means the engineer
has no enough experience)
C2. Coupled Inductor (Flyback Transformer)
74. 74
a. IEC60205 for core’s geometry factor is embedded on the right
EP0714 EP7
ui 2400 2400
uo 1.2566E-06 1.2566E-06
Ae 1.49E-05 1.03E-05
le 2.79E-02 1.57E-02
AL 1.604E-06 1.979E-06 0 gap
N 106 106 Given
L 8.40E-04 8.40E-04
AL(g) 7.476E-08 7.476E-08
lg 2.38E-04 1.67E-04 m
0.238 0.167 mm
C2. Coupled Inductor (Flyback Transformer)
75. 75
b. Power Rating
By original condition, none of two designs reach 5W power
Hdc 60 60 設計選擇
Idc 3.39E-01 2.35E-01
Energy 4.8314E-05 2.3234E-05 Joule
Power 4.83142813 2.3233873 Watt
C2. Coupled Inductor (Flyback Transformer)
Inductor is an “energy
storing” device. The power
rating actually is just the
energy stored multiplied by
the frequency
76. 76
b. Power Rating
By modifying the turns’ number to 120, EP0714 can sustain 6W. This is to set
Hdc=60A/m (higher). Under full load, Bmax may close to saturation with lower
efficiency. But it is no problem to work under 5W (with lower Hdc_max)
c. No need to do so, solved.
EP0714 EP7
N 120 120 Suggested
L 8.40E-04 8.40E-04
AL(g) 5.83333E-08 5.83333E-08
lg 3.08E-04 2.15E-04 m
0.308 0.215 mm
Hdc 60 60
Idc 3.84E-01 2.66E-01
Energy 6.19193E-05 2.97764E-05 Joule
Power 6.19 2.978 Watt
WHY?
C2. Coupled Inductor (Flyback Transformer)
77. 77
C3. SMPS transformer
Basic Prototype
Ideal Model
Flux linkage and leakage
Applicable Model (1st order, winding
capacitance ignored)
78. 78
Wire intrinsic inductance and leakage
Core nature defined by B-H curve
𝐵 𝑚 = 1/𝐿 𝑚
𝐺𝑐 = 1/𝑅 𝑐
C3. SMPS transformer
79. 79
From ideal model, it is seen that 𝐿 𝑚 = ∞ so the efficiency can
be100%. But this is impossible as current needs passing 𝐿 𝑚 to build up the
induced voltage to create the “transformer” action
When all electrical parameters determined, like DC-DC choke, the
transformer core is determined by area product Ap, to have the suitable
geometry for electro-magnetic conversion.
𝑺 𝒎𝒂𝒙 = 𝑿 𝟏 𝑓𝑟𝑒𝑞 𝑘 𝑐𝑢 𝐽 𝑚𝑎𝑥 𝐵 𝑚𝑎𝑥 𝐴 𝑒A 𝑤 𝑆 𝑚𝑎𝑥: maximal power,
𝑋1: power coefficient for different
excitation and core type
C3. SMPS transformer
80. 80
EE core for example, the Ae is
uneven along the flux path. The
center post has Ae_min to yield a
larger winding space.
To optimize 𝐴 𝑒A 𝑤, cores for transformer
usually has sectionalized Ae
C3. SMPS transformer
81. 81
If even Ae is required, the EE
core will look like the left
figure
The Ae obtained by IEC60205
tends to overestimate the
transformer’s power handling
capability.
In transformer design,
the Ac used must be
Ae_min for power
calculation
In this way, Ap=Ac x Aw cannot be maximize.
C3. SMPS transformer
82. •Turns:
EE5: 22 : 22
EP5: 28 : 28
•I have different cores:
But with some I could reach the inductance with some not and also I could
not reach the current with all of them!!
82
Real example: Engineers forgot to look at the basics.
EE5 or EP5 core
Inductance: 350 µH (between PIN 2 and PIN 3), 100kHz, 100mV,
8mA DCBias
Saturation current: approximately 2 A DC (drop of 20%)
Rated current of course at least the same
Operating frequency: 1 – 125 kHz
Application: Transformer for LAN application
A121 EP5 N42 EP5 A10 EE5 A07 EE5
Of Course NOT！
C3. SMPS transformer
83. 83 Geometric facts of EE5 and EP5
This is a telecom transformer with POE (Power Over Ethernet) requirement.To
have 350uH with 11 turns on EE5 and 14 turns on EP, the permeability ui must be
the value shown below
i.e., no gap allowed EE5 EP5
Ae 2.66E-06 3.00E-06
le 1.25E-02 9.70E-03
uo 1.25664E-06 1.25664E-06
N 11 14
L 3.50E-04 3.50E-04
AL 2.89256E-06 2.89256E-06
ur 10817 7443
C3. SMPS transformer
84. 84 Under no gap situation, when 1Adc flows through one winding,
(center-tapped for split the 2A)
𝐻 𝑑𝑐 =
𝑁∙𝐼 𝑑𝑐
𝑙 𝑒
=
11∙1
12.50∙10−3 = 880A/m for EE5
𝐻 𝑑𝑐 =
𝑁∙𝐼 𝑑𝑐
𝑙 𝑒
=
14∙1
9.70∙10−3 = 1443A/m for EP5 Core is saturated like air!!
P4 B-H Curve
C3. SMPS transformer
Only when Hmax
(<150A/m)，there
is 𝜇 or permeability，
over it the core goes
to saturation and no
usable inductance
for SMPS
Bsat= 520mT obtained
at H =1200A/m, just for
brag!
For ferrites, only this short portion is good for SMPS/filtering application
85. 85 Winding technics and way of copper’s layout is the key to the
success of inductor and transformer. This is also one of the factors that
why magnetic component design is still a case by case issue. Good wiring
can evenly distribute H can reduce leakage and noise, also can reduce iron
and copper losses.
C. Power Choke and Transformer
86. 86
C. Power Choke and Transformer
Winding technics and way of copper’s layout is the key to the
success of inductor and transformer. This is also one of the factors that
why magnetic component design is still a case by case issue. Good wiring
can evenly distribute H can reduce leakage and noise, also can reduce iron
and copper losses.
Minimal copper losses can be obtained through careful layout and
wire type selection
