Preview

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Cookbook
for Discontinuous Conduction Mode
Flyback Converters
Content
Cookbook
2
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Product overview
04
Introduction 05
Smart Transformer Selector
06
Transformer design 08
Step-by-step
to flyback converter design 09
1st step:
Compile specifications
12
2nd step:
Define primary
inductance and peak current 13
3rd step:
Definition of turns ratio 14
4th step:
Selection of the core 15
5th step:
Define the number of turns
and calculate core losses 16
6th step:
Define wire thickness
and calculate copper losses 18
Transformer construction 21
General example of step-by-step
construction of a transformer 22
Glossary
23
3
Product overview
Power Magnetics
Transformers for AC/DC Converters
Transformers for DC/DC Converters
Power Inductors
Sense Transformers
Signal and Communications
Analog Modem Transformers
xDSL Transformers
ISDN Transformers
Metering Signal & Communications
Transformers
EMC Components
Power Common Mode Chokes
Signal Common Mode Chokes
Filter Chokes
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Introduction
This cookbook shows you an example of
how to design and wind a transformer.
We hope it helps you to understand the
foundation of transformer design. When
you’re ready to move forward, we’ll design
it for you and offer free samples.
CUSTOM
SOLUTIONS
Possible
Refer to our Custom Capabilities Catalog to find
out what packages we offer for manufacturing.
Ask for a copy of the Custom Capabilities
Catalog or browse the electronic version at
www.we-online.com/customcapabilities.
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expect
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pabilities
Custom Ca
ers
Transform
Toroids
PLEASE NOTE:
Although great care has been taken
to provide accurate and current
information, neither the authors nor
the publisher, nor anyone else associated
with this publication, shall be liable for
any loss, damage, or liability directly or
indirectly caused by this book.
All appropriate material is only valid for
low power applications. For applications
with 60VDC /48VAC or more, please refer to
relating safety regulations.
5
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Benefits
Smart search for all flyback transformers on the website
- Discontinuous Mode
- Boundary Mode
n Simple search needs only power supply parameters
- Input voltage and switching frequency
- Output voltages and current
n Finds all parts for your application that will:
- Not saturate
- Provide acceptable output voltages
n Samples available
n
Features
n
6
RA
Losses & temperature rise
Compare multiple parts at once
n Schematic showing how to connect
n
n
R
T
Analyzes transformer in defined application
- Voltage levels
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Enter power supply input and output
criteria and click on “Search Now”.
STEP 3
STEP 2
STEP 1
Easy as 1 n 2 n 3!
From the resulting list of parts, click
on the transformer of choice.
Review transformer performance in
either summary tab or any of the other
three tabs, which provide more detail.
7
Transformer design
The following example gives you an idea how to design
a transformer for a flyback converter.
Compile specifications
Fig. 1 shows a flowchart for the approach in designing
a DCM flyback transformer. As you can notice, designing
a transformer is a highly iterative process.
Define primary inductance
and peak current
Further transformer designs for forward converters
and push-pull converters are integrated in Würth
Elektronik’s design guide, Trilogy of Magnetics.
Define turns ratio
Select the core
Define the number of turns
and calculate core losses
Core loss ok?
no
yes
Define wire thickness and
calculate copper losses
Order Code: English version 744 006
Copper loss ok?
yes
Fig. 1: Flow chart for the approach to
design a DCM flyback transformer
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Step-by-step to flyback
converter design
Fig. 2 shows the basic schematics of a flyback
converter. The switch S1 is a controlled switch,
e.g. a MOSFET. To understand the basic function
of the flyback converter, the switching processes
are described as follows:
Fig. 2: Circuit diagram of a flyback converter
1. Switch closed
The closed switch applies the input voltage on
the transformer´s primary winding. As a result
of the inductance, a current rises linearly on
the primary side. The polarity of the transformer is that the diode blocks the current on the
secondary side. The energy fed is stored in the
gap of the transformer.
2. Switch open
With the switch open, the current is interrupted
on the primary side. The inductance of the transformer tries to maintain the flow of energy, so
that the polarity of the secondary side changes.
The diode becomes conducting, and a linear
declining current flows on the secondary side.
Fig. 3 shows the current and voltage profile on the primary and secondary sides of the transformer.
Fig. 3: Current and voltage profiles of a DCM flyback converter
9
Step-by-step to flyback converter design
Two flyback converter operating modes are distinguished depending on the current profile.
1. Continuous mode:
In continuous mode (trapezoid operation or
continuous conduction mode, CCM) energy is
still stored at the end of the switching cycle. The
linear decline in current does not return to zero.
2. Discontinuous mode:
In discontinuous mode (triangular operation
or discontinuous conduction mode, DCM) the
current on the secondary side will be zero at the
end of the cycle. There are current gaps in which
no current flows, neither on the primary nor on
the secondary side.
Prior to design, the following parameters must be known.
n
Input voltage range
n
Output voltage
n
Output power or output current
n
Switching frequency
n
Operating mode
n
Maximum duty cycle of the IC
n
Safety requirements
n
Ambient temperature
n
Size requirements
Safety requirements such as dielectric withstand voltage, creepage distance and clearance distance
should be especially considered in the design phase, as a transformer requires a larger package if
these requirements are considered. Special care should be taken for offline applications.
An idea about the creepage distance and clearance distance and the dielectric withstand voltages are
given in Tab. 1 and Tab 2. The values, therein, are based on IEC 60950.
10
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Creepage distances for working voltages according to IEC 60950 for pollution degree 2
RMS
working
voltage
50
63
80
100
125
160
200
250
320
400
500
630
800
1000
Creepage distance pollution degree 2 (mm)
Basic insulation
Reinforced insulation
CTI>600
400<CTI<600
CTI<400
CTI>600
400<CTI<600
CTI<400
0.60
0.63
0.67
0.71
0.75
0.80
1.00
1.25
1.60
2.00
2.50
3.20
4.00
5.00
0.85
0.90
0.90
1.00
1.05
1.10
1.40
1.80
2.20
2.80
3.60
4.50
5.60
7.10
1.20
1.25
1.30
1.40
1.50
1.60
2.00
2.50
3.20
4.00
5.00
6.30
8.00
10.00
1.20
1.26
1.34
1.42
1.50
1.60
2.00
2.50
3.20
4.00
5.00
6.40
8.00
10.00
1.7
1.8
1.8
2.0
2.1
2.2
2.8
3.6
4.4
5.6
7.2
9.0
11.2
14.2
2.4
2.5
2.6
2.8
3.0
3.2
4.0
5.0
6.4
8.0
10.0
12.6
16.0
20.0
*CTI (Comparative Tracking Index)
Tab. 1
Dielectric withstand voltages according to IEC 60950
Dielectric withstand voltage (V)
Operating voltage
peak value or DC
Basic insulation
Reinforced insulation
50
100
125
150
200
250
300
400
600
800
1000
1000
1000
1000
1000
1000
1500
1500
1500
1893
2164
2399
2000
2000
2000
2000
2000
3000
3000
3000
3000
3000
3000
Tab. 2
11
1
1st Step
Compile specifications
We will now show the step-by-step design process for a DCM flyback converter. The following example
will help to illustrate the design steps.
Input voltage range (VIN MIN – VIN MAX): 36 - 57V
Output voltage VOUT: 5V
Output current IOUT: 2A
Operating mode: DCM
Maximum duty cycle DMAX: 48% (use 45%)
Switching frequency: 100kHz
Safety requirements: Functional insulation
Output diode drop voltage: VD = 0.5V
D
Assume transformer efficiency, η = 0.9. The on time will be TON = MAX
f
Output transformer power:
POUT = (VOUT+Vd) · IOUT = (5+0.5) · 2 = 11W (including output diode loss Vd)
Input power:
PIN = POUT/ η = 11/0.9 = 12.22W
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2
2nd Step
Define primary
inductance and
peak current
In a discontinuous mode, all of the energy stored in the transformer is delivered to the output during
each cycle.
So, for each cycle, to make sure that the stored energy is delivered, we need to calculate the
maximum inductance as:
V2
D 2Ƞ
36 2 · 0.45 2 · 0.9
LMAX = IN, MIN MAX
LMAX =
= 107.36µH
2fPOUT
2 · 100,000 · 11
Consider 10% inductor tolerance + 5% safety margin = 15%. The new value of L is: LMAX = 91.25µH
(Choose primary inductance L = 91µH )
The Primary Peak Current is determined from: IPK,PRI =
2 · POUT
L· f ·
IPK,PRI =
2 · 11
= 1.64A
91x10 -6 · 100000 · 0.9
13
3
3rd Step
Definition of turns ratio
Turns ratio and duty cycle determine each other; i.e. if one of the parameters is defined, so is the
other.
The maximum duty cycle and the highest currents are occurring at the minimum input voltage. This is
the worst case. In fast transient response, the duty cycle can be higher for a short time.
Design Tip 1: Keep a little safety margin to the maximum allowed duty cycle of the IC.
The relationship between maximum duty cycle and turns ratio is given by the following formula:
NS
(V + VD)(1 - DMAX)
= OUT
VIN,MINDMAX
NP
For our example, we calculate a turns ratio of:
Care should be taken on the breakdown
voltage of the MOSFET. The voltage
between drain and source of this MOSFET
during the off time is:
DMAX
VIN,MIN
TON, OFF
NP, NS
VOUT
VD
Maximum duty cycle: DMAX = TON /(TON+TOFF)
Minimum input voltage
MOSFET ON time, OFF time
Primary and secondary number of turns
Output voltage
Output diode drop off voltage
NS
(5 + 0.5)(1 - 0.45)
=
= 0.187
36 · 0.45
NP
VDS = VIN +
NP
*
· VOUT
+VLEAKAGE
NS
VLEAKAGE is the voltage spike resulting from leakage inductance
(Typically add 20-30% of VIN depending if a snubber is used or not)
V*OUT = VOUT + VD output voltage including diode voltage drop
Design Tip 2: Use a MOSFET with a sufficient safety margin in breakdown voltage, as the
voltage spike from discharge of leakage inductance can destroy the MOSFET.
14
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4
4th Step
Selection of the core
For frequencies between 25 and
500kHz, the best choices for core
material are so called power ferrites,
MnZn ferrites with a permeability of
2400. The saturation flux density Bs of
this material is 390mT at 100°C.
Fig. 4 shows the specific losses for
given frequencies and flux densities.
Pcv - Bm
5588.4492
1E+4
200kHz
14.465505
1035.2894
100˚C
Core loss Pcv (kW/m3)
100kHz
4.677008
1E+3
1056.6629
64kHz
2.306808
594.96247
The package type depends on the
power to be transformed. A starting
point is transformer indexes. Tab. 3
shows a power table in the Custom
Capabilities Catalog.
500kHz
200kHz
32kHz
1E+2
1.053128
222.66027
100kHz
25kHz
1.030651
154.6565
1E+1
25kHz
64kHz
32kHz
1E+0
1E+1
1E+2
1E+3
Flux Density (mT)
Fig. 4: Specific losses against the change in flux density
Power
DC/DC Flyback
Power Level
(W) at 100kHz
1
2
2
3
3
3
4
5
6
6
8
8
8
8
8
10
10
10
10
10
14
14
14
Offline Power
Level (W) at
100kHz
5
7
Package Size
Mount
Pins
EP5
ER9.5
ER11.5
EP7
EP7
EP7
EPX7
ER14.5
EPX9
RM4
EP10
EP10
EP10
EPC13
EPC13
EE13/7/4 (EF12.6)
EE13/7/4 (EF12.6)
EE13/7/4 (EF12.6)
EE13/7/4 (EF12.6)
RM5
EFD15
EE13/7/6
EFD15
SMD-H
SMD-V
SMD-V
SMD-H
TH-H
SMD-H
SMD-H
SMD-V
SMD-H
TH-V
SMD-H
TH-H
SMD-H
TH-H
SMD-H
TH-H
SMD-H
TH-H
SMD-H
TH-V
SMD-H
TH-H
TH-H
6
8
12
6
6
8
8
12
8
6
8
8
8
10
10
8
9
9
10
6
12
8
8
NEW
Safety






Length
(mm)
Width
(mm)
Height
(mm)
Bobbin
Page
Number
6.6
10.0
13.0
10.2
10.2
9.8
10.2
16.0
10.2
11.4
13.3
13.3
13.3
14.6
14.6
14.7
13.8
13.7
13.7
14.0
17.2
15.0
16.8
8.3
12.1
12.7
13.4
8.3
9.1
9.1
16.8
10.2
11.4
15.2
11.7
15.2
14.7
20.9
16.8
23.7
20.5
19.7
14.0
22.2
15.8
16.8
5.6
6.0
6.4
8.6
9.8
10.5
12.3
7.6
12.7
11.2
11.4
12.6
11.6
8.5
8.3
12.7
11.0
10.2
10.5
11.2
8.9
18.5
8.9
070-4426
070-6051
070-6058
070-5801
070-2150
070-4436
070-4434
070-4477
070-5103
070-5754
070-6052
070-2365
070-4410
070-5483
070-4887
070-4849
070-6258
070-5662
070-4820
070-2250
070-4265
070-5449
070-2745
51
73
74
52
53
54
64
75
65
88
55
57
56
66
67
22
24
23
25
89
41
26
39
Tab. 3: Core geometries and typical transformable power at 100kHz DC/DC
For the total needed power of 12.22W, in our example we choose EFD15 core size, with SMD-H,
10-pin bobbin, as the estimated power level for this package (100KHz DC/DC Flyback) is 14W.
15
5
5th Step
Define the number of
turns and calculate core losses
The minimum number of turns is defined by the saturation flux density for a given core. The ferrite
material 1P2400 has a saturation flux density of 312mT (BMAX derated to 80%). Thus, the minimum
number of turns is:
NP >
LPRI · IPK,PRI
91µH · 1.64A
=
= 31.88, choose 32
BSAT · AE
0.312T · 15mm2
NS = NP · n = 32 · 0.187 = 5.98
As we need a complete number of turns, and to have a little safety margin, we choose NS = 6:
NP =
NS
∕n = 6∕0.187 = 32.08
NP = 33
Calculate the core loss due to change of the flux density as following:
The flux variation is:
∆B =
LPRI · IPK,PRI
91µH · 1.64A
=
= 301mT
NP · AE
33 · 15mm2
Divide ΔB by 2 for a unipolar waveform to calculate Bpk from which we will determine the core loss.
Out of Fig. 4 we can determine the specific loss, and together with effective volume of Tab. 4, we can
calculate the core losses. Please use only half of ΔB to calculate the specific core loss.
Bpk = ∆B / 2
= 301mT / 2
= 150.5mT
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Core geometries and parameters
Core geometry
AE (mm2)
LE (mm)
VE (mm3)
RTH (K/W)
ER11/5
ER14.5
EFD15
EFD20
EE13/7/4 (EF12.6)
EE16/8/5 (EF16)
EE20/10/6 (EF20)
EE25/13/7 (EF25)
11.00
17.30
15.00
31.00
12.40
20.10
32.00
51.40
14.70
19.00
34.00
47.00
29.60
37.60
46.00
57.80
161.70
328.70
510.00
1457.00
367.04
755.76
1472.00
2970.92
134
99
75
45
94
76
46
40
winding window
height (mm)
1.60
2.74
1.80
2.25
1.80
2.51
3.15
4.01
Tab. 4
Calculating core loss (using Fig. 4 and Tab. 4):
From Fig. 4, PV = 120kW/m3 (at 100kHz and 100˚C)
Core loss = VE x PV = 510mm3 x 120kW/m3 = 62mW
Thermal resistance of an EFD15 SMD-H 10 pin, Rth = 75K/W
If we set a temperature rise limit of 40°C then the maximum power loss (PMAX) of the transformer is:
PMAX = ˚CRISE/RTH = 40/75 = 533mW
Dividing half of the losses to the core and half to the coil that is 533mW/2 = 266mW.
The calculated core loss of 62mW is well below 266mW which suggests that our core is too large for
this application but we’re not done yet.
17
6
6th Step
Define wire thickness
and calculate copper losses
Select the wire cross section that the total power loss and the resulting temperature rise remain within
reasonable bounds.
Design Tip 3: For small parts the temperature rise should be less than 40°C.
Design Tip 4: A good starting point is to select a current density of 4A/mm².
The copper losses are calculated by Ohm´s law. For the thin wires, it is reasonable to disregard eddy
current losses in the first step.
Check if the selected wire fits into the winding build of the bobbin. By using Fig. 5 you can determine
the number of layers you need. Note that this figure is only valid if you don´t need creepage and
clearance distances.
Fig. 5: Number of turns per layer for different packages and wires
By multiplying the number of layers with the outer wire diameter (Tab. 5) we calculate the winding
build. Calculate the total winding build by adding the winding heights of all windings. Check if the total
winding build is less than the build of the winding window (Tab. 4)
18
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Winding wires and parameters
Wire
diameter
(mm)
AWG
0.1
0.15
0.2
0.28
0.3
0.35
0.4
0.5
38
34
32
29
28
27
26
24
Outer
diameter
ER11.5 ER14.5
(mm)
0.125
0.177
0.239
0.329
0.337
0.387
0.459
0.566
57.18
24.00
13.10
6.55
5.68
4.13
3.14
1.97
71.47
30.00
16.38
8.19
7.10
5.16
3.92
2.47
DCR/Turn (mΩ/Turn)
EFD15
EFD20 EE13/7/4 EE16/8/5
69.62
29.22
15.96
7.98
6.91
5.03
3.82
2.40
90.26
37.89
20.69
10.34
8.96
6.52
4.95
3.12
63.53
26.66
14.56
7.28
6.31
4.59
3.49
2.19
92.65
38.89
21.23
10.62
9.20
6.69
5.09
3.20
EE20/10/6 EE25/13/7
103.23
43.33
23.66
11.83
10.25
7.46
5.67
3.57
139.76
58.66
32.03
16.01
13.88
10.10
7.67
4.83
Tab. 5
As with storage chokes, first the currents have to be calculated. The effective current, the average
current and the peak current can be distinguished by examining the current curves.
The effective or RMS current is that with which the copper losses are calculated. It is the current
averaged over the period. For the secondary side we calculate:
IRMS, SEC = IPK, SEC 1-DMAX
3
( NN )
we know, LSEC =
IPK, SEC =
s
p
2
IRMS,SEC effective current on
secondary winding
where IPK, SEC =
2 · POUT
LSEC · ƒ
· LPRI = (0.187)2 · 91µH = 3.182µH
2 · 11
= 8.31A
3.182 x 10-6 x 100000
Hence, IRMS, SEC = 8.31 · 1-0.45 = 3.56A
3
19
For the effective current on the primary winding we calculate:
IRMS, PRI = IPK, PRI
DMAX
= 1.64
3
0.45 = 0.635
3
In our example, we have an RMS current of 0.635A on primary and 3.56A on secondary
side. At 4A/mm², we need cross sections of 0.158mm² and 0.89mm² on primary and
secondary, the corresponding diameters of which are 0.44mm and 1.06mm respectively.
We choose a wire diameter of 2 strands of 0.28mm on primary side and 2 strands of
0.5mm wire on secondary side. These results in a resistance of 132mΩ for primary
winding and about 7mΩ maximum for the secondary side (see Tab 5). According to
Ohm´s law, we calculate winding losses of 53mW and 88mW on primary of secondary
windings respectively.
Now we have fixed the design and can start with the winding of the transformer:
1. Core and bobbin: EFD15
2. Primary 2 * 33 turns ø 0.28mm wire
3. Insulation tape between primary and secondary
4. Secondary: 2 * 6 turns ø 0.5mm wire
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Transformer
construction
Now that the steps are completed, you can begin the construction of the transformer.
Review the following questions 1-9 to see if anything was missed in the steps leading up
to the construction process.
Q1: Is the transformer required to
meet safety agency standards
that are intended to reduce risks of fire, electric shock or injury to personnel?
What Material Group/CTI rating is required for the materials?
What are the creepage/
clearance distances?
Q2: Is the transformer required to meet an insulation system?
Q3: In what environment will the
transformer operate?
Q4: What power supply and trans
former topology will be used?
Q5: How much space is allowed for
the transformer on the printed circuit board?
Q6: What is the lowest and highest
frequency of operation?
Q7: What is the wattage rating of
the transformer?
Q8:
What are the input and output voltages and currents of the transformer, and how many windings are needed?
Q9: Are the materials suitable for a lead-free solder reflow process?
Here are some basic guidelines to follow when building a transformer. By following these
guidelines, you will minimize the manufacturing costs, while optimizing the electrical
performance. Note these guidelines are not intended to show all possible methods of
construction. The accompanying photographs show an example of a surface mount construction.
21
General example of step-by-step
construction of a transformer
22
Step 1 – Bare Bobbin
Step 2 – Shelf Tape
Step 3 – Wind 1
Step 4 – Wrapper Tape 1
Step 5 – Wind 2
Step 6 – Wrapper Tape 2
Step 7 – Wind 3
Step 8 – Finish Tape
Step 9 – Solder
Step 10 – Core
Step 11 – Core Tape
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Glossary
Margin (Shelf) Tape – Determine if a safety
isolation barrier is required and where that barrier
will be located. In the example, margin (shelf) tape
is applied to one side of the bobbin (coil former).
The number and placement of margin tape will
affect magnetic coupling and leakage inductance.
An alternative to margin tape is double or triple
insulated wire. This wire may be cost prohibitive
on high turn windings.
Wire Strands/Wire Gauge – Choose the type of
wire, number of strands, and wire gauge based on
the frequency of operation and current carrying
ability. Be aware that heavy gauge or multi-stranded
wire may solder bridge together on adjacent
terminals.
Turns Per Layer (TPL) – Pick a turns per layer of
wire that fills the winding area of the bobbin. On
low turns per layer windings, it may be necessary to
space the turns of wire evenly across the bobbin.
This also applies to high turn, multilayered windings
where the last layer does not entirely fill the bobbin.
Minimize the number of layers of wire to reduce
leakage inductance and eddy current losses.
Pinout – A number of factors will affect the bobbin
pinout, including safety agency requirements and
circuit board layout. Typically, the primary windings
are terminated on one side of the bobbin, and the
secondary windings are terminated on the other.
Ideally, the pinout for a particular winding will be
dictated by the number of layers of wire, whether
odd or even, although other factors will also affect it.
If the winding ends on the side of the bobbin that is
opposite from the intended finish terminal, bring the
wire across the coil at a 90˚ angle. Place the wire
in an area where it will be the least disruptive to
subsequent windings and the ferrite core set.
Tape can be used to hold the wire down at the
bend. It may be necessary to place a piece of tape
under this wire to insulate it from its own winding to
prevent cut-through and subsequent shorted turns.
Pulling this wire across the coil at an angle other
than 90˚ will cause the subsequent windings to not
lay uniformily and evenly.
Interlayer Insulation – Interlayer tape may be
required if there is a high voltage potential between
each layer of wire within the same winding.
Wrapper Tape/Finish Tape – Select a wrapper
tape that is slightly wider than the distance between
the bobbin flanges. This extra width allows the tape
to lap up the sides of the flanges without folding
over. This ensures isolation between the windings,
minimizing the risk of wire-to-wire contact and
potential dielectric breakdown. The higher
temperatures associated with a lead-free solder
reflow process may cause the standard polyester
tapes to shrink. Also, smaller transformer packages
will absorb more heat, causing more tape shrinkage.
This tape shrinkage will have a direct affect on
dielectric breakdown strength and the integrity of
the safety isolation barrier. High temperature
polyamide tapes are available, but their comparative
tracking index (CTI) is lower with a resulting change
in the material group. This results in a greater creepage/clearance distance requirement.
Core Set/Core Tape/Insulation Tape – Choose
the appropriate core set and AL inductance factor.
Secure the core set to the coil with 2 layers of tape.
Do not stretch the tape during the application
process. It may be necessary to apply insulation
tape to one or both sides of the core set to insulate
the core from the terminals. Additionally the core set
may be bonded to the coil (bobbin) with an adhesive
or varnish coating.
23
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INSIGHT DESIGN 9999062 • Revised MARCH 2015
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